New Senior Mathematics Extension 1 Years 11 & 12 Student Book with eBook [Third edition.] 9781488618307, 1488618305

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EXTENSION 1 FOR YEARS 11 & 12 THIRD EDITION J.B. FITZPATRICK BOB AUS

Pearson Australia (a division of Pearson Australia Group Pty Ltd) 707 Collins Street, Melbourne, Victoria 3008 PO Box 23360, Melbourne, Victoria 8012 www.pearson.com.au

Copyright © Bernard Fitzpatrick and Bob Aus 2019 First published 20 I 3 by Pearson Australia 2023 2022 2021 2020 20 I 9 JO 9 8 7 6 5 4 3 2 I

Reproduction and communication for educational purposes The Australian Copyright Act 1968 (the Act) allows a maximum of one chapter or JO% of th e pages of this work, whichever is the greater, to be reproduced and/or communicated by any educational institution for its educational p urposes provided that the educational institution (or th e body th at administers it) has given remuneration notice to Copyright Agency under th e Act. For details of the copyright licence for educational institutions contact Copyright Agency (www.copyright.com.au). Reproduction and communication for other purposes Except as perm itted under th e Act (for example a fair d ealing for the purposes of study, research, criticism or review), no part of th is book may be reproduced, stored in a retrieval system, communicated or transmitted in any form or by any means without prior written permission. All inquiries should be made to th e p ublisher at the address above.

This book is not to be treated as a blacl - 11.

No

2

1 > 11 Is (3-2) - - .

Test x= 3:

ves "

This finds \Vhere the graph of y = 1 is above the line y = -1. 2 Thus the solution is x < 1, x > 2. x -

Example 4 1

Solve X

2

1

>

-X

X

2

- 1

.

Solution Using Method 5 (from Example 3 ): x;c0,1,-1

Using Method 3 (fro1n Example 3): 1

1

- - - > - - - - -, sox;cO, 1,-1 x(x-1) (x- l)(x+l) 1

X

1 >0 (x - l )(x + 1)

x(x - 1)

1

Solve

2

-

-X

1

2

2

x - 1= x - x x2 - 1 x= l :

Critical values are 0, 1, - 1; all must be not included (i.e. open circles on nun1ber line) • I (jl (jl (jl I •

(x+l)-x >O x(x - l )(x + 1) 1 >0 x(x - l )(x + 1)

-2

_,

0

I

2

Testx=-2:

2

Multiply by x (x- 1)2(x+ 1)2:

I

1

s

x(x- l )(x+ 1) > 0

((-2) 2

Testx=

)'

1

((-2) 2

- 1).

1

1

1 s

(2-(-~)) > (2-1).

Testx =

Yes

t:

I

1 s

No

-t:

I

X

-(-2)) >

1

1

1

((~)2 -(~)) > ((1)2-1).

No

Testx=2:

The graph is above the x-axis for -1 < x < 0, x > 1, which is the required solution.

I

1

1

1

2

s ((2)2-(2)) > ((2) - 1).

Yes

The solution is - 1 < x < 0, x > 1.

EXERCISE 1.2

RATIONAL FUNCTION INEQUALITIES (x IN DENOMINATOR)

1 When asked to solve x ~ < ~, four students began their solutions as follows: 1 StudentA: 4 1. But the don1ain is restricted to - 1 < x < 1, so the solution is-1 .!._ X

6 Solve

x < 1. (Note: This looks like a standard problem, but in fact requires son1e malysis, depending on x+ 1 which method you use.)

7 Solve

ix-:l

> 1.

(Hint: l2x + 11is known to be non-negative.)

8 Solve Ix+ 11- lx- 21> 2. 9 (a) On the sa111e axes sketchy= Ix + 11md y = Ix-S I. (c) Solve lx+ll+ lx-51> 7.

1.4

(b) Hence graph y = Ix+ 11+ Ix-S I. (d) Solve Ix+ 11+ Ix -SI= 6.

CIRCULAR AND SIMULTANEOUS INEQUALITIES

A circle divides the number plane into two regions, a finite region called its interior md ai1 infinite region called its exterior, as \veil as the set of points that n1ake up the circle. A parabola, cubic, quartic or hyperbola curve divides the nu1nber plai1e into hvo infinite regions, as well as the set of points that 1nake up the curve.

Points inside and outside a circle A circle divides the number plane into three sets of points: the sets of points on the circle, inside the circle ai1d outside the circle. The set of points on a circle of centre C(h,k) md radius r is given by the equation (x-h)2 + (y-k) 2 = r2. A point P(x, y) lies on this circle if CP = r. If CP < r, the point Pis inside the circle.

If CP > r, the point Pis outside the circle. )'

)'

,,

, , ,- - - - 1'~ exterior of the circle.

r2 gives the

Regions involving simultaneous inequalities Example 8 For the circle with centre (O, 0) and radius 3 units, sketch the region of the Cartesian plme that includes all points on or inside the circle that are also: (a) to the left of the line x = 2

(b) on or above the line x + y = 3.

In each case give the inequalities that define the region.

Chapter 1

Further work with functions

9

YEAR 11

Solution (a)

y

(b)

2

3

3

-3

,

x- + y 3

)'

,2

0

-3

X

3

0

X

' -3

-3

Example 9 Sketch the region defined by y > x2 and y < 2x + 3. Describe this region in words.

Solution To find the points of intersection, solve si1nultan eously the equations y = x2 and y = 2x + 3. This gives:

2

x =2x + 3 X

Factorise: Solve:

)'

)'

2

ys 2x + 3 3

-2x-3=0

(x + l )(x-3) =0 0

x=-1, 3

X

X

2

Substitute into y = 2x + 3: x = -1, y = l ; x = 3, y = 9

)'

(3,9)

Hence the points of intersection are (-1, 1) and (3, 9). The shaded region is the points on and above the parabola y = x2 that are also on or belo,v the line y = 2x + 3.

X

Example 10 2

Describe the region of the x-y plane whose points satisfy the inequalities y < 2 + x - x and y + 2x < 2.

Solution 2

The graph of y =2 + x - x can be obtained by con1pleting a table of values and then plotting points. It can also be obtained by co1npleting the square for x and then graphing accordin g the shape and 2 2 2 properties of y = x : 2 + x - x = 2 - ( x - x)

= 2 + ! - (x 2 - X + ! ) =2¼ - (x - ½)

2

Hence you can graph y = 2 ! - (x - 1)2, which is the graph of y = x turned upside down, n1oved 0.5 units to the right an d 1noved 2.25 units up: 2

)'

, ' (0)2),

'

2

y < 2 + x - x is the region belo,v this curve.

,o

y + 2x < 2 is the region on or below the line y + 2x =2. As shown in the diagram , the required region is the region on or below 2 the lin e y + 2x = 2 that is contained between the 'anns' of y = 2 + x- x •

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10

New Senior Mathematics Extension 1 for Years 11 & 12

\

'' ' •

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X

YEAR 11 MAKING CONNECTIONS

Simultaneous inequalities Use technology to solve simultaneous inequalities graphi cally.

EXERCISE 1.4

CIRCULAR AND SIMULTANEOUS INEQUALITIES

1 Sketch the region defined by each inequality. 2 2 2 2 (a) x + y > 16 (b) x + y < 4 2

2 Which diagra1n shows the region satisfying y < 1 - x and y > lxl - 1?

r

A

r

B

C

r

D

I X

X

X

X

3 Sketch the region defined by each inequality. 2

y < x2 + 1

(a) (x + 3)2 + y < 1

(b)

(d) y> lx l

(e) y9-x

(fl y< l2x+41

4 Sketch the region defined by the given inequalities. 2

2

(a) x +y O,y>O

(b) (d) ffl (h)

2

(c) yO

~ ~ +/ x + 2 (g) y > Ix - 21, y > 3 (i) x 2 +y2 0, y >0 y< lx l,y> O, 1 1,x>O,y< l 2

5 Which graph represents the region defined by (x - 3) + (y + 4)2 < 25?

A

C

.---y

0 6

-8

r

B

Y 0 X

D

8

y 8

X

-6

-8

-6--~

Q

X

0

6

X

6 Sketch the region of the Cartesian plane whose boundary consists of: (a) the curve y = x2, the ordinates at x = 1 and x = 2 and the x-axis (b) the lines y = 1 and y = 3 and the circle with centre (O, 2) and radius 2 units (c) the circle (x - 3)2 + (y - 4)2 = 25 and the y -axis 2 (d) the arc of the parabola y = 4 - x bet\veen x = 2 and x = -2 and the x-axis

(e) the graph of y =Ix - 11, the x-axis and the y-axis 2 2 (f) the circle x + y = 36 and the lines y = 6 and x = 6 (g) the circle centred at (-2, O), radius 2 units, and the circle centred at (-2, O), radius 1 unit (h) the parabola y = x 2 - 2 and the line y = x (i) the curve y = ✓ x , ✓e y -axis and the line y = 2

0) the semicircle y = 1- x 2 and the x-axis.

Chapter 1

Further work with functions

11

YEAR 11

7 For the shaded region in the diagran1, state whether each staten1ent is correct or incorrect. 2

2

(a) The shaded region is defined by y < lx l and x + y < 4. (b) The shaded region is the part of the interior of the circle of centre (O, O) and radius 2 that is below the lines given by y = Ix 12 2 (c) The shaded region is defined by y > lx l and x + y < 4. (d) The shaded region is the part of the circle \Vith centre (O, O) and radius 2, and its interior, that is on or belo\v the lines given by y = lx l.

(- 2,2)

-2

2

0

X

-2

CHAPTER REVIEW 1 1 The diagra1n shows the graphs of y = Ix+ 21and y = ✓4 - x

2

y •

r= lx+2I

The solution of ✓4 -x 2 < Ix+ 21 is: A O 0 through 180° about the origin. For exan1ple, iff(x) = x 3,f(-x) = (-x)3 = -x3 = -f(x). 4 Recognising that a function is odd or even 1neans that you only need to draw half of the graph in detail. The other half can then be drawn using the syn1n1etry properties. 5 Odd functions will have an inverse function, although som e may require a restriction on the do1nain. 6 Even functions will not have a single inverse function, but can be split into two parts (by restricting the don1ain), so that each part has an inverse function. (Inverse functions are covered in Chapter 5.) 7 Note that m ost functions are neither even nor odd, e.g. f(x) = x2 + x , f(x) = ti'.

Graphs and graphing software It is a good idea to use graphing software to check your sketches. Software-generated graphs will also allow you to 200111 in and to easily find the coordinates of iinportant poin ts.

Example 16 Sketch the graph of the following. (a) f(x) = x3

26

(b) f(x) = (x- 2) 3

New Senior Mathematics Extension 1 for Years 11 & 12

(c) f(x)=(x-2) 3 + 1

YEAR 11

Solution

y 3

(a) f(x) = x = 0 for x = 0

For all x < O,f(x) < 0 For all x > O,f(x) > 0

0

X

J(-a) = (-a)3,J(a) = a 3 :. f(-x) =-f(x), so the function is odd. (b) J(x) = (x - 2)3

)'

J(x) = 0 at x = 2 The graph off(x) = (x - 2) 3 can be obtained fron1 the graph off(x) = x3 by a translation of 2 units to the right, parallel to the x-axis.

2

For all x < 2,f(x) < 0

X

(0, - 8)

For all x > 2,f(x) > 0 The function does not pass through the origin, so it cannot be odd. (c) J(x) = (x- 2)3 + 1

)'

J(x) = 0 at x = 1 The graph of[(x) = (x - 2)3 + 1 can be obtained fron1 the graph of f(x) = (x - 2) by a translation of 1 unit upwards, parallel to the y -axis. In general, the graph off(x) as the graph of f(x) =x".

O

=(x + b)" + c will have the sam e general shape

I ,2

(0, - 7)

X

'

'' '

MAKING CONNECTIONS

Graphing polynomial functions Use graphing software to construct graphs of polynomial functions.

Cubic functions 3

2

A general cubic function is a polynon1ial function f of the 3rd degree, defined by f(x) = ax + bx + ex+ d, ,vhere a, b, c, dare constants and a -:I: 0. Every cubic polynomial has at least one linear factor of the forn1 (x + a), where a is a real number. Factors like this can be found using the factor theoren1.

Example 17 Sketch the graph off, where f(x) = (x + l )(x - 2)(x - 3).

Solution At the points where the graph offcrosses the x-axis, f(x) 1.e.

= 0:

(x + l )(x - 2)(x - 3) = 0 X

=-1, 2, 3

The function value will change sign fron1 positive to negative or from negative to positive at these points (x = - 1, x = 2, x = 3). • For all x < - 1, each of the three factors is negative, so f(x) < 0. • For -1 < x < 2, the factors (x + 1) > 0, (x - 2) < 0, (x- 3) < 0, so f(x) > 0. • For 2 < x < 3, the factors (x + 1) > 0, (x- 2) > 0 and (x- 3) < 0, so f(x) < 0. • For all x > 3, each of the factors is positive, so f(x) > 0.

= -2: j(-2) = -20 < 0 x = 0: f(O) = 6 > 0

e.g. e.g.

X

e.g. e.g.

= 2.5: j(2.5) = -0.875 < 0 x=4: f(4)=20>0 X

Chapter 2

Polynomials

27

YEAR 11

This infonnation can be summarised in a sign diagrrun:

.

..

+ 0

X

I

At the point where the graph off crosses the y-axis, x = 0, sof(O) = 1 x (-2) x (-3) =6. Thusthegraphcrossesthey-axis at (0,6). 3 Note: this cubic equation has a positive coefficient of x , so that as x in creases, y also increases (except in the don1ain between the two turning points). For x > 3,f(x) > 0 for all values of x.

)'

(0, 6)

0

Example 18

Sketch the graph off, where f (x) = 2x3 - x 2 - 13x - 6, an d find the values of x for which 2x3 -x2 - 13x - 6 > 0.

Solution

It is useful to find the linear factors of2x3 -x2 - 13x- 6. (Re1nen1ber there will always be at least one.) f( l ) =2-1- 13-6=-18,;tO f(- 1) =-2-1 + 13-6=4;t0

f(2) = 16-4-26-6=-20-:/:. 0 f (-2) =-16-4 +26-6 = 0

so (x- 1) is not a factor so (x + 1) is not a factor so (x - 2) is not a factor so (x + 2) is a factor 3

To get the other linear factors (if any), you cru1 divide 2x

-

x

2

-

13x - 6 by (x + 2) md then factorise the quotient.

The factors of2x2 - 5x- 3 are (2x + 1) and (x - 3):

2x 2 - 5x - 3 x+2)2x 3- x 2-13x-6 2x 3+ 4x 2

·

f(x)=(x + 2)(2x + l)(x-3) f(x) =Oatx=-2, 3

-1,

-5x 2 -13x

• For all x< -2,f(x) < 0

e.g . x=-3: f(-3) = -30 < 0

-5x 2 - l Ox - 3x-6

• For-2 0

e.g . x=-1: f(-1) =4 > 0

• For

e.g . x=O:

f(O) =-6 0

- 3x-6

-1< x < 3,f(x) < 0

• For all x > 3,f(x) > 0

0

This infonnation is sun1n1arised in the following sign diagrm1.

: Hence f(x) > 0 for -2 < x
3.

At x = O,f(O) = -6, so th e graph crosses the y-axis at (O, -6). 3

Note that this cubic equation has a positive coefficient of x , so that as x increases, y also increases (except in the don1ain between the two turning points). For x > 3,f(x) > 0 for all values of x.

28

New Senior Mathematics Extension 1 for Years 11 & 12

I

2

X

ft.x)=2x3 - x 2 - 13x - 6

YEAR 11

Example 19 Without showing too 1nuch detail, sketch graphs of the polynomial functions defined by the following rules. (a) y= (x+2)(x- 1)2(x + 1)

(b) y=x2 (x-3)(x+ 1)3

Solution 4

(a) y=(x+2)(x- 1)2(x + 1). When lx lislarge,ybehaveslikex •

)'

(x - 1)2 > 0, so the sign of y is detern1ined by (x + 2)(x + 1).

Hence y > 0 for x < -2 and for x > - 1 (equality at x = l); y< 0 for-2 0, so the sign ofy is determ ined by (x - 3)(x + 1).

y > 0 for x < -1 and for x > 3 (equality at x = -1 and x= 3); y< 0 for- 1 0

X

)'

)'

Two zeros

-4

4

-2 - 1

X

I

Does not cut .6. 0 Two zeros

-3 -

- 1

-2

0 I

X

-4

---+-+-+""o+-+- x Does not cut - 2 - ,I.'f+'- 1 2 /.6. 0: y

y 2

y 2

)'

1

I

2

X

-2

-2

_,

I

2

X

-2

Cuts three tin1es Three zeros

_,

l

2

I X

-2

-2

Cuts once Triple zero

0

-2

Cuts once and touches One zero, one double zero

_,

I O

I

2

X

Cuts once One zero

a < 0: y 2

y 2

y

)'

2

I

1

1

I X

-2 -2

_, 0 - 1

2

X

-2

_, 0 1

-2

Cuts once Triple zero

2

X

-2

Cuts three tin1es Three zeros

Cuts once and touches One zero, one double zero

A cubic polynon1ial may have one, t\vo or three real zeros. A cubic polynon1ial ahvays has at least one real zero.

30

I

-2

New Senior Mathematics Extension 1 for Years 11 & 12

Cuts once One zero

YEAR 11

D egree 4 (quartic)

P(x) = ax4 + bx3 + cx2 + dx + e a > 0: y

)'

3

2

2

1

1

-3 l

-2 -1 O

2

X

_,

- 1

Touches Quadruple zero

0

I

2

1

- 1

X

-2

X

1

Touches on ce, cuts twice Two zeros, one dou ble zero

Touch es twice Two double zeros

)'

2

y

)'

2

I

-3 -

_,

- 1

l

- 1

0

2

X

X

-2

-2

_,0

-2 - 1

-3

Cuts hvice O n e zero, on e triple zero

I

2

X

Does not cut No real zeros

Cu ts four tim es Fou r zeros

The addition of an appropriate constant to each equation can create a polynomial with n o real zeros.

a < 0: A negative a inverts each of the six graphs above so that they open do\vnwards, with the properties of their zeros the san1e.

Summary of quartic polynomials 1 A quartic polynon1ial n1ay have four, th ree, two, one or no real zeros.

2 If a quartic polyn on1ial h as only one real zero, then it 1nust be a quadruple zero. 3 If a quartic polyn on1ial h as only hvo d istin ct real zeros, then they are either a triple zero and a single zero or they are both double zeros. 4 If a quartic polynonlial has only three d istinct real zeros, then they are a double zero and two single zeros.

5 If a quartic polyn on1ial h as four distinct real zeros, then it can be factorised into four real linear factors. 6 If a quartic polyn on1ial h as no real zeros, then it cannot be factorised in to any real linear factors.

EXERCISE 2.6

POLYNOMIAL FUNCTIONS

1 Sketch the graph of each function and fin d the values of x for which: (i) J(x ) =0 (ii) f(x) > 0 (a) f(x) = x(x - 2)(x + 3) (b) f(x) = 2(x - ~)(x + 1)(2x + 3) (c) f(x) = x(x - 2) 2 2 Sketch the graph offwheref(x ) is the polyno1nial given. D raw a sign diagrrun ru1d label your sketch sh owing the poin ts of intersection with the axes. Find the values of x for which f(x) > 0. 3 3 2 2 3 3 2 (a) f(x) = x - 4x (b) f(x) = x + 4x + 4x (c) f(x) = 2x - x (d) f(x) = x - x - l Ox - B 2

3

3 (a) Find the linear factors of 6 + Sx - 2x - x • 2 3 (b) Fin d the values of x for which : (i) 6 + Sx - 2x - x = 0 2 3 (c) Sketch the graph offwhere f(x) =6 + Sx - 2x - x •

(ii) 6 + Sx - 2x

4 (a) Find the linear factors of x3 - sx2 + Bx - 4. (b) Fin d the values of x for which: (ij x 3 - sx2 + Bx - 4 = 0 3 (c) Sketch the graph offwhere f (x ) = x - sx2 + Bx - 4.

(ii) x 3 -

2

-

3

x >0

sx2 + Bx - 4 > 0

Chapter 2

Polynomials

31

YEAR 11

5 (a) Find the linear factors of -x3 + 2x!- + x - 2. (b) Draw a sign diagran1 to find the values of x for which -x3 + 2x!- + x - 2 > 0. (c) Sketch the graph off where f(x) =-x3 + 2x!- + x - 2. 6 The number of zeros of the polynon1ial f(x) A

O

B

l

D

C2

7 Describe how the graph off(x) (a) f(x) = (x + 3)3

= x3 -

xis:

3

=x3 can be transfonned into the graph of the follo\ving. (b) f(x) = (x + 3)3 - 2

= x3 3 Show that the graph off, where f(x) = x -

8 Sho\v that the graph off, where f(x)

8, cuts the x-axis at one point only.

9

x 2 - Bx + 12, cuts the x-axis at one point and touches it at anoth er.

Find the values of x at these points.

10 Show that the graph off, where f(x)

= x3 -

4x2 + Bx - 8, cuts the x-axis at one point only.

11 Sketch graphs to show the general form of the following functions. , (a) y=(x + l)(x-2)(x + 3) (b) y=x(x- l)(x-2)(x-3) (c) y=x(x- 1)2 (d) y =x· (x- 1) 2 3 (e) y = x(x + 3)2(2 - x) (f) y = (x - 1)(2x - 3)(x + 4) (g) y=x(2x- l )(x-1) (h) y = x(x + 2)(x- 1)4 (i) y = (3x- 5) 3 (x- 1)2 (j) y =x(x- 1)2(2 - x) 3 12 Part of the graph of y = P(x) is shown, where P(x) is a polynon1ial of degree four: Which of the following could be the polynon1ial P(x)? A P(x)=x2 (x + 2)2 B P(x)=(x-2) 4 C P(x)=x(x-2) 3 D P(x)=(x-1) 2(x + 2)"'

1':::::---r-

1.8

1.9

2

2.1

2.2

13 Sketch graphs to show the basic features of the following functions. 4

(a) y=x(x-2/(x+ 1) (c) y = (3 -x)( l -x) 4(x + 2)

2

2

(b) y=x(x - l )(x-1) (d) y = (2 - x) 3(1 + x)2(x- 3)

CHAPTER REVIEW 2 1 Perform the following polynon1ial divisions. (a) (x3 + 2x2 -3x + 4)+(x-l)

(b) (4.x4 -6x2 + 10x-40)+(x+3)

2 Use the ren1ainder theorem to find the re1nainder of the follo\ving. (a) x3 - 4.x2 + 3x- 5 divided by (x- 2) (b) x 4 + x3 - 5x2 + 4.x- 2 divided by (x + 1) 3 Use the factor theore1n to find the linear factors (over the rational nu1nber field) of each polynon1ial. (a) x3-2x2 -5x+6 (b) x3+7x2 + 14x+8 (c) x3 + sx2-x-5 (d) x4 -K-16x2 + 4.x + 48 4 Let P(x) = (x - l)(x + 2)Q(x) + ax + b, \Vhere Q(x) is a polynomial and a and bare real nun1bers. The polynon1ial P(x) has a factor of x + 2. When P(x) is divided by x - I the re1nainder is 6. (a) Find the values of a and b. (b) Find the ren1ainder when P(x) is divided by (x - l )(x + 2). 3

2

5 Let P(x) =x + ax - x + I be a polyno1nial \Vhere a is a real nun1ber. When P(x) is divided by x - 2 the re1nainder is 15. Find the ren1ainder when P(x) is divided by x + 3. 6 The polyno1nial P(x) =x3 + ax- b has a re1nainder of 5 when divided by (x + 1) and a ren1ainder of 2 when divided by (x - 2). Find the values of a and band hence find the re1nainder when P(x) is divided by (x - 3). 3

2

7 The polyno1nial P(x) is given by P(x) = ax + 15x + ex - 72, where a and care constants. The three zeros of P(x) are -3, 2 and a. Find the value of a.

32

New Senior Mathematics Extension 1 for Years 11 & 12

•

X

YEAR 11

8 The cubic polynomial P(x) =x3 + bx2 +ex+ d (where b, c, dare real nu1nbers) has three real zeros: -1, a and-a. (a) Find the value of b. (b) Find the value of c - d. 9 The polynon1ial P(x) =x - 4.x + kx + 12 has zeros a, {3, r (a) Find the value of a+ f3 + r (b) Find the value of af3r (c) Two of the three zeros are equal in n1agnitude but opposite in sign. Find the third zero and hence find the value of k. 3

2

10 Sketch graphs of each function. For what values of xis each function positive? (a) y=(x-l)(x+2)(x-3) (b) y=(x-2)(x + 2) 2 (c) y=x(x2- l )(x + 2)

(d) y=x2(x-2)2

Chapter 2

Polynomials

33

In the Mathen1atics Advanced course, you looked at the graphs of linear, quadratic, cubic and quartic polynon1ials. In this section you are going to look at the effects of applying other function rules to linear, quadratic and cubic polynomials and drawing their graphs.

3.1

RECIPROCAL FUNCTIONS

Given the graph ofy = fix), the graph of y = f tx) is the reciprocal function of fix) . When graphing reciprocal functions, it is important to find where f(x) = 0, as these x values ,viii give vertical asymptotes for the reciprocal function. It is also important to note that ,vherefix)

➔+

00,

ftx) ➔ 0.

Example 1 In each part, use the graph of the given function to dra,v the graph of y = f tx )" (a) Given y = x, draw y= -1 . X

(b) Given y = 2x + 1, dra,v 1

(c) Given y = 1 - x, draw 1

y= 2x+r

y

y=l-x y

)'

4

3 2

4

4

3

3

2

2

I

I

-4 - 3 - 2

-4 - 3 - 2 -

I

2

3

4

y=l - x

0

X

- 4 - 3 - 2 - !.1

-2

I

-2

-3 -4

-3

-3

-4

-4

Solution y

(a) The graph of y = .!. is undefined at x = 0, so x = 0 is a vertical X

asY1nptote. The graph approaches y = 0 fro1n above as x ➔ oo. The graph approaches y = 0 fro1n below as x ➔ -oo. y = 0 is the horizontal asY111ptote. x = .!. where x = + 1, hence the curves intersect at

3 2

,

, , ,

,,

I ,0

-3 -

y=

,,

, , ,

,

, , ,

,

I

-3

I

2

,

,

,'

(-1,-1) and ( 1, 1).

New Senior Mathematics Extension 1 for Years 11 & 12

,,

I

X

34

y=x, ~

3

I X

X

3

4

X

YEAR 11

(b) The graph of y =

1 2 x+ 1

.

is undefined at x = _.!., 2

~

.

'

I

••

so x = - 1s a vertlca1asyinptote. 2 The graph approaches y = 0 fro1n above as x ➔ oo. The graph approaches y = 0 fro1n belo,v as x ➔ --oo. y = 0 is the horizontal asyinptote. 1.

' = 2x+ 1 1y

2i. '' . 1: :,: ' I

I

,o / 1:

-3 -2

1 where x = - 1, 0. Hence the curves intersect at 2x + 1 = 2x+ l (-1, -1) an d (O, 1).

I I I

'

'

1 2x +1

y=

1

.,:•• ... .

X

3

2

I I I

(c) The graph of y =

1 is undefined at x = 1. 1-x

)'

''

' 3 y= l - x' ' '' '2

The graph approaches y = 0 fro1n belo,v as x ➔ oo. The graph approaches y = 0 fro1n above as x ➔ --oo. 1 - x = 1 ,vhere x = 0, 2. The curves intersect at (O, 1)

''

1'

1-x

-3

and (2, - 1). x = 1 is the vertical asyinptote an d y = 0 is the horizontal asyinptote.

-2

''

'

.'

- 1- 10

1 ' ,2 ''

-2 -3

'

'

'

-1-1x

y=

'

'

All these reciprocal functions are rectangular hyperbolas.

Example 2 (a) Given the graph of y = x 1

draw y= -

X

2

2 ,

(b) Given the graph of y = (x 1 • dra,v y = 2 (x-2)

•

y 4

4

y =x'

3

2

2

1

1

-1

(c) Given the graph of y = (x + 2)(1 - x), 1 dra,vy= ( x + 2)( 1-x

r

)'

)

3

-4 -3 - 2 - 1 O

2)2,

l

2

3

4

X

- 4 - 3 -2 - 1 O -1

y=

4

(x - 2) 2

3

= (x + 2)( 1 - x)

.,.,...~- y 1

1

2

3

4

X

-1 0 -1

-2

-2

-2

-3

-3

-3

-4

-4

-4

.

Solution

.

1

(a) The graph of y = _l_ is undefined at x = 0.

x2

The graph approaches y = 0 fro1n above as x ➔ oo. The graph approaches y = 0 fro1n above as x ➔ --oo. The function is never negative.

x2 = -\-where x = +1. The curves intersect at (-1, 1) and (1, 1). X

y= x2r'

s

1

'

' '' ''

'

1 1 1

'

'

\

I

'1

'

I '

- 5 -4 - 3 - 2 - 1 O -1

x = 0 is the vertical asyinptote an d y = 0 is the horizontal asyinptote.

'

I

1

2

3

4

S

X

-2 -3

Chapter 3

Graphing functions

35

YEAR 11

(b) The graph ofy =

1 2

y= (x - 2) 2 4

2 I I

y= (x - 2) 2

\Vhere x = 1, 3. The curves intersect at

\

' -s

-4 - 3 - 2 - 1 0

occurs at x

~(

) is

x + 2 1-x

4 3

, ,;

-s

4 and 9

= ;-

\

.' I .: ' . 2 -1 0

-4 - 3

I

\

y = (x+2)( 1 - x)

.:•i•'

-1

'

1 1 4 x < -2, ( x + 2 )( 1- x ) < O; -2 < x < 1, ( x + 2 )(1-x) > -9;

x>

: :

-2

·

-3

'

-s

.''

2

3

'

'

'

x = -2 and x = 1 are the vertical asy1nptotes, y = 0 is the horizontal asymptote. x =- ; is an axis of syinm etry. Each of these graphs has a vertical axis of syn1n1etry.

Example 3

x3, draw y = -¼-x

y y =r'

4

(b) Given the graph of y = x (x + l )(x - 2), 1

draw the graph of y = x(x + l)(x _ 2 )" )'

3

4

2

3

I -4 - 3 - 2 -

IO

2

I

2

3

4

I X

-2

-4 - 3 - 2

I 0 -1

-3

-2

-4

-3 -4

36

New Senior Mathematics Extension 1 for Years 11 & 12

I

3

4

X

y = x(x+ l )(x - 2)

4

5

X

'\ y =(x + 2)( 1 - x) ''

-4

< 0.

(a) Given the graph of y =

X

s

'

1 1, ( x+ 2)(1-x)

s

y

4

Thus the least positive value of (

4

-3

The m axiinum value of (x + 2)( 1 - x) is 2. and occurs

=2 .

3

-2

The graph approaches y =0 from below as x ➔ oo. The graph approaches y =0 from below as x ➔ -oo.

at X

i

I

-1

Ic ) is undefined at x = -2, 1. x+ 2 1-x

1

I

3

( x -2) (1, 1) and (3, 1). x = 2 is the vertical asyn1ptote and y = 0 is the horizontal asyinptote. x =2 is an axis of syin1netry. (c) The graph ofy = (

•

s

The graph approaches y =0 from above as x ➔ oo. The graph approaches y =0 from above as x ➔ - oo. The function is never negative. (x - 2)2 =

. ... ' ... ... ' ... ... ,,' ..• ... ' ... ' .. , '

:r

1 is undefined at x = 2. 2 (x-2)

YEAR 11

(c)

Given the graph of y = (2 - x )(x + 1)2, draw the graph of y =

1

(2-x) (x + 1)2

•

(d) Given the graph of y =,2 + 2x 1

2

draw the graph of y =

X

3

+ 1, •

2

+ 2x + 1

)'

y =(2 - x)(x + 1) 2

y

4

3 2

y=x3 +2x'+ I

1

0 - 4 - 3 - 2 - !.1

3

1

4

X

-4 - 3

2 -1O

I

-1

-2

-2

-3

-3

-4

-4

2

3

Solution

y

(a) The graph of y

4

=-¼- is undefined at x =0.

3

X

The graph approaches y =0 fro1n above as x ➔ oo. The graph approaches y =0 fro1n belo,v as x ➔ -.

x

3

X

4

2

y= - 1

I

=-¼- where x =+ 1. The curves intersect at (-1, -1) and ( 1, 1).

,

10

-4 -3 -2 -

x3

I

I

1

2

3

4

X

X

x = 0 is the vertical asymptote, y = 0 is the horizontal asyinptote.

'

'

The curve does not have an axis of syinmetry, but has rotational (poin t) sy1n1netry about the origin.

'

1

(b) The graph of y = ( ( ) is undefined at x = -1, 0, 2. x x + l ) x-2 The graph approaches y = 0 fron1 above as x ➔ oo. The graph approaches y = 0 fron1 below as x ➔ -oo. x = -1, 0, 2 are the vertical asy1nptotes, y = 0 is the horizontal asy1nptote. The curve does not have an axis of syn1n1etry.

-

'

y 4

3

.. 1

: y = (x + l :(x - 2)

., ' ' I'

1

-4 - 3 -2

Q \

,

!'-- 1 . ' .~2

,..,--,i'

'

'

1

(c) The graph of y =

2

\

(2-x)(x + 1) The graph approaches y = 0 fron1 below as x ➔ oo. The graph approaches y = 0 fron1 above as x ➔ -oo. x = -1, 2 are the vertical asyn1ptotes, y = 0 is the horizontal asyinptote. 2 It looks as though y = (2 -x)(x + 1) has a local n1axim un1 value of 4 at x = 1. (This can be shown using calculus.) Hence y =

1

(2- x)(x + 1)

2

,viii have a local 1ninin1u1n value

',

:

.•;._4 • •

• •

\

...: y ..•• ...••• .; I , .: ' ·o

-4 -3 - 2 - 1

~I

y-

X

4

.. '

I

::-3

is undefined at x = -1, 2.

3

, .'

\

I

I

''

,

, ' I

I

'•

•

. '

' ' ' '

I

I

3

4

X

1 ;2 - (2 - x)(x+ 1) 2 ;.3

of-1 at x = 1. 4

Chapter 3

f .. ...~4

Graphing functions

37

YEAR 11

1

(d) The graph ofy =

)'

is undefined at x =-2.2.

+ 2x 2 + 1 The graph approaches y = 0 fro1n above as x ➔ oo. The graph approaches y = 0 fro1n belo\v as x ➔ -oo. x = -2.2 is the vertical asymptote, y = 0 is the horizontal asy1nptote. The curves touch at (O, 1), a local minim un1 of the origin al function becomes a local 1naximu1n of the reciprocal function. X

3

4

,

3

y=x3+2x-,+ l :: I ,., ' 2 :, , :

;'' '--/

-4 - 3

, I ,' y= ~ ~ ~.~ ~.~ , x-+Lr+ l

1'

·2 _ , O

:

I

-1

2

3

4

X

I:

':

-2

-3 -4

'

I

As the Exan1ples above show, a 1naxin1u1n turning point on the original function becomes a 1niniinum turnmg point on the reciprocal function (or equivalent asyn1ptote). A 1ninin1um turning point on the original function becomes a maxin1u1n turning point on the reciprocal function (or equivalent asymptote). EXPLORING FURTHER

-----===========10

Graph ing reciproca l functions Use graphing software to explore graphs of reciprocal functions.

EXERCISE 3.1

RECIPROCAL FUNCTIONS

1 The graph of y =2x - 1 is shown. )'

4

y=2x - l

3

2

I

_,

- 4 - 3 -2 - I O

I

2

3

4

X

3

-4

Which of the following represents the graph ofy ,,

A

,,'y= 2x -

y

8

-4 -3 -2

I -4 - 3 - 2

-t 1 -?

,, ,

38

~

,,

2

,

, • , 3 ,' - 4

:

,,

, /y=2x - l

3

, , ,

C

,

4

1

1 ? = 2xl

::,'

,.

,,

'

,: I , :

, ,

y

D

4

,,'y=2x - l

3

4

.

,

New Senior Mathematics Extension 1 for Years 11 & 12

X

- 4 -3 - 2 _, . -f :,

-~

, , ,

2

I

-4 -3 -2

-! 10,' -?

-1 .:j

, : ,,'

,'y=2x - l

3

, 2

,,

,:3 / -4 ,

, ,, I

2

3

4

X

YEAR 11

2 The graph of y = (2x - l ) 2 is shown. y

3

2 1

-4 - 3 -2 - 1 0

1

-1

2

3

4

X

-2 -3 -4

Which of the following represents the graph of y =

A

8

?'

7

31

3'I

2' 1

2 'I

'

\

:' ,

'

' ' ' ' ' ' ' :

1

1

-1

(2x - 1) 2

? D

C

' ' ' I

''

'',

-3 - 2 - 1

.'

1

z'

I

I

2

3

X

- 3 -2 - 1 0 -1

-2

-2

-3

-3

'

'

1

2

3

X

-3 -2

i"

I

l•'

\: '

'

I

)

' ' () 1,

' 1

2

3

'

l•' '' ' ' ' '

X

- 3 - 2 - 1 ()

2

1

- 1•

X

3

-~ - 3,'

3

3 The graph of y = (2x - l ) is shown.

y

3

2 1

- 3 -2 - 1 0

1

-1

2

3

X

-2 -3

Which of the following represents the graph of y =

A

)'

3

2

1

-

' ' ' ' ' ' '' '' ' ' '' " ' '' 1

' ' 3 I'

'o

' '

8

3

X

1 3

(2x -1)

C

y

D

{

'

~

I

3

2

1

~

'

I

2

i

i

1

1,'

1,'

I ,:

0 - 3 - 2 - 1- 1

-2

:

1

2

3

X

- 3 -2 -

() I

l, ]!

I

,

-3 '

0 I

~: ""

1

2

3

X

- 3 - 2 - 1 () - 1,

I

1

2

3

X

- 1!

0 I

•" "

- .,, I

Chapter 3 Graphing functions

39

YEAR 11

4 Given the graph of y = 3 - 2x, draw the graph

ofy=

5 Given the graph of y = (x + 1)(2 - x), draw l the graph of y = (x+ l )( 2-x )"

l 3-2x

)'

4

3 2 _.,,..,

2 1 X

-4 -3 - 2 - 1 0 -1

-4 - 3 -2

10 -1

-2

-2

-3

-3

-4

-4

123456x

-s 6 Given the graph of y =x(x - l)(x + 2), draw l

the graph of y = x (x-1 )( x+2 )"

3

7 Given the graph of y = x + 3x2, dra\v the graph

ofy=

l x 3 + 3x 2 •

)'

)'

4

4

3

2

I

-4 - 3 -

2

3

X

-4 -

_,

-2 - I O

-2

-2

-3

-3

-4

-4

8 Given the graph of y = 4 of y=

4

x2, dra\v the graph

l ,. 4-x-

I

2

3

4

X

9 Given the graph of y = x(x + 1)(2 - x), draw the l graph of y = x (x + l )( 2-x )" )'

)'

4

4

3

-4 - 3

40

_, 0 -1

I

3

4

X

-4 - 3 - 2

I

-2

-2

-3

-3

-4

-4

New Senior Mathematics Extension 1 for Years 11 & 12

3

4

X

YEAR 11

10 Given the graph of y = 2 + 3x of y =

x3, draw the gr aph

1 . 3 2 + 3x-x

11 Given the graph of y = x2 + 2x + 2, draw the graph

ofy=

1 x + 2x + 2 2

)'

y

7

4

6 5 4

3 -4 - 3 - 2

-1 0 -1

1

3

4

X

2

-2 -3

- 4 -3 - 2 - 1 O -1

-4

l

2

3

4

X

-2

12 Given the graph ofy graph of y =

=- x2 + 2x- 2, draw the

1 . 2 -x + 2x-2 y 4

3

2

1

-4 -3 - 2 - 1 O -1

l

2

3

4

X

-2 -3

3.2

SQUARE ROOT FUNCTIONS 2

Given the graph of y = fix), it is often useful or necessary to draw the graph of y = ..Jf (x ) and the graph of y = fix). To graph these functions, it is important to find out \Vhere fix)< 0, as ..J f(x) is not defined for these values of x 2 (because the square root of a negative number is not a real nu1nber). Siinilarly, y =fix) \vill not be defined for these 2 values of x , because y cannot be negative. Reme1nber that Jo =0 and ✓l = 1. Therefore, for O fix); for fix) > l,f(x) > ..}J( x). Graphically this n1eans that for O 1, the graph of

y = ..Jf (x ) is below y = fix).

Chapter 3

Graphing functions

41

YEAR 11

Example 4 In each part, use the graph of th e given function to dra\v the graphs of y = (a) Given y = x, draw y= ✓ x an d y 2 = x.

(b) Given y

= 2x + 1, draw

y = ✓2x + l andy2 = 2x +

)'

2

= j(x).

(c) Given y = 1 - x, draw 2 y = ✓l - x an d y = 1 - x.

1.

)'

4

4

y =x

3

3

2

2

1

I

I

-4 - 3 - 2 - 1 l

.Jf (x ) and y

2

3

4

X

y=2x+ l

y= l - x

I

-4 - 3 - 2 -

2

3

4

X

-4 - 3 -2 - 1 O

-2

3

I

- 1

4

X

-2

-3

-3

-3

-4

-4

-4

Solution (a) The graph ofy

y

=✓x is undefined for x < 0.

y =x,, ,

4

x=✓ x at x = 0, 1. Graphs intersect at (O, 0) and ( 1, 1). For O< x < 1, th e graph of y = ✓ xis above the graph of y = x.

3

,,

2

- 4 - 3 - 2 - 1,

, , , 2

,,

, ,,

,

- 1

2

3

4

X

-2 -3

2

)'

y=x,,,

4

3

2

y = x an d y = x intersect at (O, O) and ( 1, 1).

,

, , ,

,

,

2

I

_, ,

- 4 - 3 - 2 - 1, 0

,,

New Senior Mathematics Extension 1 for Years 11 & 12

I

-4

The graph ofy = x is undefined for x < 0 as y cannot be negative. It has two branches, y = ✓ x and y = - ✓ x.

42

,

,

I

, ,,

, ,,

,

, ,,

, , ,

, , -2 , -3 -4

,, X

YEAR 11

(b) The graph of y =.J2x + 1 is undefined for x < - ~.

,,

y

4

2x + 1 = ✓2x + 1 at x = - ~, 0. Graphs intersect at (- ~ , 0) and (O, 1).

3

For - ~ < x < 0, the graph of y = .J2x + 1 is above the graph of y = 2x + 1.

2

,, ,

,, , y= 2x +l

I

I

-4 -3 -2

- 1,

,o

r l

3

2

I

X

4

I

,I - 2 ,I

, ,

The graph of

-3

I

-4

I

y- =2x + 1 is undefined for x < - .!.. 2

y

, I

4

, y= 2x +l I

It has t\vo branches, y = .J2x + 1 and y = -.J2x + 1.

3

Graphs intersect at (- ~ , 0) and (O, 1).

2

I I I I I

I

-4 -3 -2

- 1,

I

2

3

4

I ',2

3

4

I

r l

X

I

,I - 2 I

, I

'

-3 -4

I

(c) The graph of y

=✓I -

x is undefined for x > 1.

1 -x = ✓1 - x at x = 0, 1. Graphs intersect at (O, 1) and (1, O).

',y=1 - x

''

Y 4

'

3

For O< x < 1, the graph of y = ✓I - x is above the graph ofy = I - x.

'' -4 - 3 -2 - I O - I

,

''

-2

''

-3

''

-4

The graph of y

2

=1-

X

''

'

x is undefined for x > 1.

It has two branches, y = ✓I - x and y = -.JI - x. Graphs intersect at (O, 1) and (1, 0).

''

3

1 ',2

-4 - 3 -2

'' -3 -4

Chapter 3

3

''

''

4

'

'

X

''

Graphing functions

43

YEAR 11

Example 5 (a) Given y = x2, draw

y= N

2

(b) Given y

2 •

= (x -

/

and y =x

(c) Given y = (x + 2)(1 - x), draw

2)2, draw

2

2

2

y= \J(x-2) andy =(x - 2) .

y= ✓(x+ 2)( 1 -x ) and y 2 =(x+ 2)(1-x).

y

)

,

4

4

y=x-

3

3

2

2

1

I

- 4 - 3 -2 - 1 0

l

- 1

2

3

y

4

X

_,

- 4 -3 -2 - I O

y

= (x - 2) 2

4

3

,-,-,.. y = (x + 2)(1 - x) I

I

2

3

4

X

_, 0

_,

-4 - 3

-2

-2

-2

-3

-3

-3

-4

-4

-4

X

Solution (a) The graph of y =

N

is defined for all x . The function is never negative. The graphs intersect at (- 1, 1), (O, 0) and (1, 1). The resulting graph is the same as y =lxl.

.

)'

'' '

I I

4

''

'' '

I I

3 I

2

'

I

1

''

,,

-4 -3 -2 - 1 O - 1

I

2

3

4

X

-2 -3 -4

The graph ofy

2

= x2 is defined for all x.

The graphs intersect at (-1, 1), (O, 0) and (1, 1). The resultin g graph is the same as the graph of y =+x.

'

'' ' '

y 4

3

y=i','

,I

X

-2 -3 -4

44

New Senior Mathematics Extension 1 for Years 11 & 12

YEAR 11

(b) The graph of y = ✓(x - 2) is defined for all x. The function is never negative. The graphs intersect at (1, 1), (2, O) an d (3, 1). The resulting graph is the sa1ne as the graph of y 2

)'

)' = (X - 2) 2 II

4

.

I I I

=Ix- 21.

2

'

1

,

I

'

,,

'' -4 - 3 -2 - 1 0 -1

3

2

I

X

4

-2

-3 -4

The graph of y2 = (x - 2)2 is defined for all x.

)'

'

4

The graphs intersect at (1, 1), (2, O) and (3, 1). r=(x - 2) 2

The resulting graph is the same as the graph of y =+(x - 2).

y = (x - 2) 2 :

3

.

' '

2

,

'

1

''

,

-4 -3 -2 - 1 0

,

2

- 1

4

X

-2

3 -4

(c) The graph of y = ,J( x + 2)(1- x) is undefined for x < -2 an d x > 1. The maxin1um value of (x + 2) ( 1 - x ) is : and occurs at x The greatest value of J(x + 2)(1- x) is ; and occurs at x

!=- !. =-

y 4

3 y= (x+2)( 1 - x), t

,

'

1

-2 < x < 1: 0 < y < 1.5. Further algebraic investigation sho\vs that the graph is a sen1icircle in the upper half plane, centre (-0.5, 0), radius 1.5.

-4 - 3 .!2 I

'

_, 0

_,

I

I

-2

I

-3 I

2

= (x + 2)( 1 -

2

'

3

4

3

4

X

'

'

y

x ) is undefined for x 1.

4

-2 2.

)'

The maxin1um turning point of y = (2 - x)(x + 1)2 is (1, 4), so the 1naxin1u1n turning point of y = ✓(2- x)(x + 1) is (1, 2). 2

y = (2 - x)(x + 1)2

4

'

'

3

,

, '

y= V(2 - x)(x+ 1) 2

'

(-1 , O) is not called a turning point of y = ✓(2 - x )(x + 1) 2 because at this point the curve changes sharply, not sn1oothly. Instead this

-4 - 3 -2 - 1 O

I

-1

point is called a cusp.

J

3

4

X

-2 -3 -4

The graph of y 2 = (2 - x)(x + 1)2 is undefined for x > 2.

)'

4

The maxin1um turning point of y = (2 -x)(x + 1)2 is (1, 4), so the maxiinum turning point of y

2

= (2 -

3

x )(x + 1)2 is (1, 2)

2

y= (2 - x)(x + 1)2 ,

,I ' ' ,, ' _.,-, ',

and the n1inin1u1n turning point is (1, -2). -4 - 3 - 2

y' = (2 - x)(x+ 1) 2

I

3

4

X

-2 -3 -4

Chapter 3 Graphing functions

47

YEAR 11

J

y

(d) The graph of y = x 3 + 2x 2 + 1 is undefined for x < -2.2 The graphs intersect at (-2.206, 0), (-2, 1) and (O, 1).

4

3

y=x3+2x2+ I

,

' , - 4 -3 . 2

'

y= ✓x'+i?+ I

'

, - ,2

0

- L1

I

2

3

4

X

-2 -3 -4

The graph of y2 = x3 + 2x2 + 1 is undefined for x < -2.2.

)'

4

The graphs intersect at (-2.206, 0), (-2, 1) and (O, 1).

'

y= x3 +2r ' +l 3 I - , 2

,

,'

'

-4 - 3

-2

=x3+2x'+ I

-3 -4

. I

EXERCISE 3.2

SQUARE ROOT FUNCTIONS y

1 The graph of y = x - 2 is shown.

4

3

2

I

- L10

- 4 -3 -2

-2 3

-4

Which of the following represents the graph of y = '1x - 2? y

A

y

B

3

C

3

2

,

,

I

3

2

,

,

I,' 2

,

3

X

-3 -2 - I O -I

-2

-2 / ,

J,'2

,

/

48

/

/

/

/

,

/ /

,

/

3

X

- 3 - 2 - IO

-3 -2

J,'2 ,,

-I

/

-2/

/

/

, ,.:'3

/

/

/

/

/

,

,

/

/

- I

3

,

I

/

- 3 -2 - I O

y

D

)'

, /

New Senior Mathematics Extension 1 for Years 11 & 12

,

/

, ,.:'3

/ /

/

3

X

YEAR 11

2 The graph of y = 1 -

x2 is shown.

y 4

3 2

I

-4

Which of the follo\ving represents the graph of y

A

y 3

8

'

I

'

0

- 3 - 2 rl I

'

'

'

-

'

I '

-2 -3

2

3

,'

'' '' '1

,'

3

2

2

2

,

'

' I ''

2

3

X

'

'

, o ,J

-3 - 2

'

'

-2

''

'

'

-3

,'

)'

3

,o rl

- 3 -2

D

)'

3

I '

X

2

x? C

I

'

=1 -

)'

2

,

2

I

'

I

I ''

-2

2

X

3

'

'

' I'

10

- 3 -2

-I

2

3

X

, , -2

'' '

'

-3

-3

1

'

3 The graph of y = x(2 -x)(x + 1) is shown.

)'

4

3 2 I

- 4 - 3 -2

- Li

3

I

4

X

-2 -3 -4

Which of the follo\ving represents the graph of y 2 =x(2 - x)(x + 1)?

A

8

y 3

' I' I

C

)'

'

'

3

2

-2

2

' '1

3

X

-3

-

I ,,

3

X

-3

4 Given the graph ofy

= x + 2, draw:

(a) y= ✓x+2

2

(b) y =x+2

3

X

\O

- 3 -2

- I

''

' -2

3

X

'

' '

'

= (x + 1)2, dra\v:

y= ✓(x+ 1) 2

y

'

I

-3

5 Given the graph ofy (a)

'

-2

' ' -3 '

-3

I

I

'

-

-2

-3

2

'

I I

-I

'

2

I

I

I

)'

3

I

- 3 - 2 - 1'0'

D

)'

2

(b) y =x+ 1

)'

4

4

3

3

2

2

I

- 2 - IO -I

-2 -3 -4

I 2 3 4

X

-4 -3 -2

-1P

I

2

3 4 x

-2 -3 -4

Chapter 3

Graphing functions

49

YEAR 11

6 Given the graph of y = (x + 2)(2 - x), draw:

7 Given the graph of y = x(x - l )(x + 3), draw:

2

(a) y= -./(x+2)(2-x)

(b) y =(x+2)(2-x)

2

(a) y=-./x(x- l)(x+3)

y

(b) y =x(x- l)(x + 3)

y

4

6

3

5

2

4

I

-t,0

--4 - 3

3

I

4

X

I

-2 -3

-4 -

-4

2

3

8 Given the graph of y = x + 3x + 1, draw: (a) y =

✓x 3 + 3x 2 + 1

2

9 Given the graph of y = ( 1 -x)(x + 2)2, dra\v: 2

3

(a) y = .J(1 - x )(x + 2)

(b) y = x + 3x + 1

y

3 -2

5

-t,0

- 4 - 3 -2

-t?

-2

-2

-3

-3

✓2 + 3x-x 3

2

3

4

X

2

3

10 Given the graph of y = 2 + 3x - x , draw: (a) y=

2

(b) y = (1 - x)(x + 2)2

y

5

--4

2

11 Given the graph of y =x + 2x + 2, draw:

2

3

(b) y =2+3x-x

(a)

y= ✓x 2 + 2x+2

)'

2

)'

6

7

4

6

2

5

4 -6

-4

-2

0 -2

4

6

X

3

2

-4 -6 -4 - 3 - 2

2

12 Given the graph of y =-x + 2x - 2, draw: (a)

+ 2x -

2

4

y= ✓-x +2x-2 2

I

)'

2

(b) y =-x

-t 10

3

2

2

I

-4 - 3 - 2 - 1 O -1

-2 -3

SO

New Senior Mathematics Extension 1 for Years 11 & 12

I

2

3

4

X

2

(b) y =x +2x+2

3

4

X

YEAR 11

3.3 ABSOLUTE VALUE FUNCTIONS Given the graph of y =f(x), it is often useful or necessary to dra,v the graphs of the absolute value functions y =If( x )I and y = J(lxl) . S01netimes these graphs ,viii be the san1e, but not always.

=If( x )I is defined ,vherever fix) exists. If( x )I>0 if -oo < fix) < If( x )I>0 wherever fix) < 0. y = J(I xi) is the same as y = fix) for x > 0, but different for x < 0. y

00•

Example 7 In each part, use the graph of the given function to draw the graph of y (a) Given the graph of y = x, draw

=If( x )I and the graph of y = J(I xi).

(b) Given the graph ofy = 2x + 1, (c) Given the graph of y = 1 - x, draw y =l2x + 11and y = 2lxl + 1. dra,v y =1 1- xi and y = 1- lxl.

Y =lxl.

y

y 4

y=x

3

)'

4

4

3

3

2

2

I

I

- 4 - 3 -2 - I

I

2

3

4

-4 - 3 - 2 - 1 O

X

-1

y=l - x I

3

4

X

-2

-2 -3

-3

-3

-4

-4

-4

Solution y

(a) The graphs are the sa1ne for x > 0. lxl > 0 for all x.

y = lxl

4

3 2

I

-4 - 3 -2

,, //

,

-lf ,

I

2

3

4

X

-2 -3

I

// y =x

-4

,I

(b) The graphs of y = 2x + 1an d y I2x + ll > 0 for all x. The graphs ofy =2x + 1and y

=l2x + 11are the sa1ne for x >- 21.

= 2lxl + 1are the sa1ne for x >0.

I2x + ll > 1 for all x. )' = 2lxl + I

,, ,o I

-4 -3 -2

,o

- J,

,-1 y =2x + l' ,, - 2

,

, ,

I

,

,

I

2

3

4

X

- 4 - 3 - 2 - ,~1 y =2x+l'

,

-3

,I

,I

-4

,

I

2

3

4

X

,, - 2

-3 -4

I

Chapter 3 Graphing functions

51

YEAR 11

(c) The graphs of y = I - x and y = II -xi are the same for x < 1. I1 - x i > 0 for all x. The graphs of y = I - x and y = I - lx l are the san1e for x > 0. 1- lx l< l y

'

4

''

y 4

''

y= l - x ' ,

3 2

3

' ',2

I

- 4 - 3 -2 - 1 0 - 1

I ' ,2

'

-2

3

''

X

4

4

X

-2

''

y= l - x' ' ,

-3

3

I

-3

'

-4

'

Y=I -

lxl

-4

Example 8 (a) Given the graph of y = x2, 2 draw y = lx and y = lxf.

I

-4 - 3 - 2

)I

y= (l xl-2J2.

)'

4

(c) Given the graph of y = (x + 2)(1 - x) , dra\vy = l(x + 2)(1-x and y=(lxl+ 2)(1- lxl).

(b) Given the graph ofy = (x-2)2, 2 dra\vy = l(x-2 ) 1and

y=x2

y

)

3

4

2

3

I

2

4 3

y

-4 - 3 -2

-2

-t?

-3

-2

-4

-3

x)

I

I

-t,0

= (x + 2)( I -

-4 -3

2

0

-t i

I

2

3

4

X

-2

-3 -4

-4

Solution

I

xf are the srune as

(a) The graphs of y = lx 2 and y = I the original graph y = x2•

By the definition, lx l= ✓ x2 so lxf = x

2 •

)'

4

y=x2

3 2

(b) The graphs of y = (x - 2) 2 and y = l(x - 2)2 1are th e sa1ne. The graphs of y = (x - 2) 2 and y = (l x l- 2) 2 are the same for x > 0. There is a cusp at (O, 4). x = 0 is an axis of syn1n1etry. 0 y

= (x - 2) 2 t 5

I

y = 0.

2

)'

4

3

(lx l+ 2) (1 - lx l) < 2. (O, 2) is not a turning point.

)' =

(x + 2 )(1 -

x = 0 is an axis of sy1n1netry.

'

x) , ,i'

)' =

(lxl + 2)( I - lxl)

- 4 - 3 .2

'' ' '

-2 -3

'' '' '

-4

Example 9

I

3

3

(a) Given the graph of y = x , draw y = lx 3 and y = Ix 1 .

(b) Given the graph of y = x(x + l)(x- 2), draw

y = lx(x + l)(x - 2)1and y = lx l(lx l + 1)(lx l - 2 ).

)'

4 )'

3

4

2

3

I

2 I

-4 - 3 - 2 -

-2

- 4 -3 -2

10 -I

-3

-2

-4

-3

3

X

4

-4

(c) Given the graph of y = (2 - x)(x + 1)2, dra\v

y= 1 -2.2. (-2.2, O) is a cusp. (O, 1) is a n1inin1u1n turning point. (-1.33, 2.19) is a n1aximun1 turning point.

3 2 I

- 4 - 3 ~2 - I 0 -I

,'

2

X

-2

,, 3

I

-3 -4

2

For y= lxl +2lxl +1: The graphs are the sa1ne for x > 0.

)'

4

lxl3 +2lx 2 +1> 1

3

1

Y =x'+2.x2+ 1, ,

(O, 1) is a minimum turning point.

Y = lxl' + 2lxl2 + I

, ' - 4 - 3 ~2 - I O ,

-I

I

2

3

4

X

-2 -3 -4

EXERCISE 3.3

ABSOLUTE VALUE FUNCTIONS

1 The graph of y = x + 2 is shown.

)'

4

3

2

I

-4 -3 - 2 - I O -I

I

2

3

4

X

-2 -3 -4

Chapter 3

Graphing functions

55

YEAR 11

Which of the following represents the graph of y = Ix + 21?

A

B

y

C

y

)'

3

2 I

'

'

' '1

1

2

3

X

I

- 3 ✓ -2 I I

I

I

'

'

I

3

''

I I

''

2

'

, '1 I I

-1

'

y

1

I ' - 3, - 2 - 1 0 I

3

2

'

D

I I

-2

'

-1 0 -1

1

2

X

3

2

3

-3

- 3... ,,- 2 - 1 0

'

-2

I

I

X

-3

-1

''

3

X

-2

-3

2 The graph of y =(x - 1)2 is shown.

2

1

-3

y 4

I

- 4 - 3 -2

-t,0 -2 -3 -4

Which of the following represents the graph of y

·,r

A

B

·~

= (lxl- 1)2? C

>'

,'

2''

I

_,

1

2

3

X

-3 - 2

I

1

_, 0

_,

I

2

X

3

- 3 -2 - 1 0

' ''

I

1

- 1

1

,'

''

2

,

, , ,, '

2,'

I

_, 0

-3 -2

)'

'';i

,

'

I

.

D

3

X

- 3 -2 - 1 0 -1

-2

-2

-2

-2

-3

-3

-3

-3

,

I

'

I

2

3

1

2

3

X

y

3 The graph of y =x(x + 2)2 is sho,vn.

4

3 2 I

-4 -3 -2

2

I

3

X

4

-2 -3 -4

Which of the following represents the graph of y

A

B

Y, 3

'

2

,'

= lxl(Ixi + 2 )2? C

Y,

,

'

0

,

,

,'

-2 -3

3

2

,'

2

1

2

3

X

, '

- 3 ,' - 2

~~/

,

-2

,

,'

2

I

1

2

3

-3

X

I

- 3 ,'- 2

,' '

' 56

y

'

I

- 3 1- 2 ~ I _ 1 , 'v'

D

)'

New Senior Mathematics Extension 1 for Years 11 & 12

'

~~~' -2 -3

I

2

3

X

- 3 1'- 2 ' ~J -0 /

,,

, ,'

'v'

-2 -3

X

YEAR 11

4 Given the graph of y = x - 2, draw: (a)

y = lx-21

(b)

y= lxl-2

5 Given the graph of y = (x + 2)2, dra,v: (a)

y = l(x + 2)2 1

y

y =(lxl+ 2}2

(b)

y

4

6

3

5

2

4

I -4 - 3 - 2 - I 0 -I

I

2

3

4

X

-2 -4 -3 -2 - I O

3

-I

-4

y= l(x- 1)(2-x)I

(b)

y=(lxl-1)(2- lxD

2

3

X

4

-2

6 Given the graph of y = (x- 1)(2 -x) , dra,v: (a)

I

7 Given the graph of y = x(x - l)(x + 2), draw:

y = lx(x- 1)(x + 2)1 (b) y= lxl(lxl-l)~xl+ 2) (a)

y

y

4

4

3

3

2

2

I

I -4 - 3

-4 - 3 - 2 - I O -I

-

-I 0 -I

I

3

2

X

4

-2

-2

-3

-3

-4

8 Given the graph of y = (x - 3)(x- 2)(x + 1), draw:

y = l(x -3)(x-2)(x + l )I (b) y=(lxl-3)~xl-2)(1x + l) (a)

3

2

9 Given the graph ofy = x + x + 1, draw: (a)

y = lx 3 + x 2 + 11

(b)

y= lxi3 +1xf + l

y 4

y 6

3

5

2

-4 -3 -2 - I O -I

2

I

2

3

4

X

-2

I

-3 -4 - 3 - 2 - - 10

4

X

-4

-2

Chapter 3

Graphing functions

57

YEAR 11

10 Given the graph of y = -'2 + 2:x!- + 1, draw: 3

2

(a) y = l-x + 2x + 11

(b)

11 Given the graph of y =:x!- + 2x + 2, draw:

y= - lxr + 21xF + 1

(a) y= lx 2 +2x+21

y= lxf + 2lxl+2

(b)

y

y 4

6

3

5

4

3

-4 - 3 -2 - 1 O -1

1

2

3

4

X

2

-2 - 4 - 3 -2 - 1 O

-3

- 1

l

2

3

4

X

-2

-4

12 Given the graph of y = -x2 + 2x - 2, draw: (a)

y = l-x 2 +2x-21

(b) y= -l xf +2lxl-2

y 4

3 2

1

-4 - 3 -2 - 1 O - 1

1

2

3

4

X

-2

3.4 GRAPHING POLYNOMIALS BY ADDING ORDINATES Given the graphs of two polynonlial functions y = fix) and y = g(x), the graph of a ne\v function y =fix)+ g(x) can be obtained by a process of adding the ordinates for each x value. This process is den1onstrated in the following exan1ples.

Example 10

)'

The graphs of y = fix) and y = g(x) are sho\vn.

20

By dra\ving vertical lines and adding ordinates, draw the graph

16

of y =fix)+ g(x). Comment on the ne\v curve. 12 8 4

1

-4

58

New Senior Mathematics Extension 1 for Years 11 & 12

2

3

4

5

X

YEAR 11

Solution On the diagram, vertical lines are dra\vn through iinportan t points such as turning points, points where a curve cuts the axes, an d points where the curves intersect.

)'

20

'.

16

"

On each vertical line, the intercepts of the two curves are added to find the position of a new point, which is 1narked on the line. These new points are then joined to obtain y =fix)+ g(x).

' ' ',,' ,' . ,,'' ,' ', ' ', '

t

~' )' =g(x)

12 8

The solid curve is the graph ofy =fix) + g(x).

'

This curve is above y = g(x) to the right of x = - 1, and below y = g(x) to the left of x = - 1. They intersect at x = -1. y = f(x) + g(x) cuts the x-axis at (-1, O) an d (O, 0).

\

' I

-s

1

. ''

-..:l- ~i

;

,, /

,,

,-

-i

-

"(}

I

J.,.. _.. I,.. ., I )'

:/

4

'

I

I

' ',,

-4 _

,l'I

.,.

I I

I

I

I

I

-

..,

R

=J\X )

.,

- I

-4

Example 11 The graphs of y = f(x) and y = g(x) are shown.

)'

By drawing vertical lines an d addin g ordin ates, draw the graph of y =fix) + g(x) . Co1n1nent on the new curve.

y= f(x)

X

Solution On the diagram, vertical lines are dra\vn through iinportant points such as turning points, points where a curve cuts the axes and points \Vhere the curves intersect. On each vertical line, the intercepts of the two curves are added to find the position of a new point, which is marked on the line. These new points are then joined to obtain y =f(x) + g(x). The solid curve is the graph ofy =fix) + g(x). This curve is above y =fix) to the right of its turning point, and belo\v y =g(x) to the left of its turning point.

' '' '' '' ''•

y

,,

'

\

'

' '

,,' ,

"" '

' '

'

' y = fix ) + g(_.j , , ,, ,, ,

\

'

\

y= fix)''

,

\

t

'

I\ I / \

\

t

" t t t

' "' ,, ' ,' ' '' '' '' ''

,

I 11' '

''

'

,_, ,

'

Chapter 3

'

-

,'y = g(x) , , X

Graphing functions

59

YEAR 11

Example 12 The graphs of y = fix) and y = g(x) are sho\vn.

)'

20

By dra\ving vertical lines and adding ordinates, draw the graph of y =f(x) + g(x). Comm ent on the ne\v curve.

15

10

-10

-4 -3

Solution O n the diagran1, vertical lines are dra\vn through important points, such as turning points, points \Vhere a curve cuts the axes and points where the curves intersect.

)'

20

15

O n each vertical line, th e intercepts of the two curves are added to find the position of a ne\v point, which is 1narked o n the line. These new points are then jo ined to obtain y =fix ) + g(x).

J' = fix) +.g(.r)Jo

The solid curve is the graph of y =f(x) + g(x).

' I I

.., -.

I I I

._ .._ ..,

1

'' I

I

I

y =f(x) - ,,, ,5

The ne\v curve is above y =g(x) for x < 4, p assing through (O, 6). They intersect at x = 4, and y = fix) + g(x) is just below y = g(x) for x > 4.

,f... ,.,. ~

I I /

I

I

II

I

1

' '

-4 - 3

Graph ing polynomials by adding ord inates

I

~2 ~ I O

I . 2

3-

4' -x

= = = = = = =10

Use graphing software to verify the add ition of ordinates for combined polynomial graphs.

EXERCISE 3.4

GRAPHING POLYNOMIALS BY ADDING ORDINATES

1 The graphs o f y = fix ) and y = g(x) are sho\vn . By drawing vertical lin es and adding ordinates, dra\v the graph of y =f(x) + g(x). Comment on th e ne\v curve. ~

~

(~ y

)'

14 12 10

14 12

10

)'

y= g(x )

10 8

8

4

6

2

4

-I 0

-2

2 -4 - 3 -2

I

2

3

4

X

-4

60

New Senior Mathematics Extension 1 for Years 11 & 12

-6

-8

4

X

YEAR 11

(d)

(e)

(f) y

)'

8

6

6

y = g(x)

y = g(x)

4 2

X

-1 0

-2

- 4 - 3 -2

-4 -6

-8

y= fix) -8

- 10

2

2

2 On the same diagra1n, sketch the graphs of y = x + 2x and y = I - 2x. Use these graphs to sketchy= x + 1. 3 On the same diagra1n, sketch the graphs of y = x 4 On the san1e diagran1, sketch the graphs of y = x

2

2

3

3x and y = x

-

3

2x and y = x . Use these graphs to sketch y = x + x

-

3

-

3

2

2x.

-

3x. Use these graphs to sketchy= x + x

5 The graphs of y = x 2 - 2x + 4 and y = x3 - 2x are sho\vn.

2

-

6x.

y

4

3

2

2

-3 - 2 - 1 O -1

3

X

-2 3

Which diagra1n represents the graph of y = x + x A >' 8 >'

2

-

4x + 4?

C

8

5

D

y 5

4

4

3

-3 - 2

0 -1

1

2

3

3

3

5

2

2

4

1

1

3

X

-3

2

-

-1 0 -1

1

2

3

X

-3

-

-1 0 -1

1

2

3

X

1

-

3.5

-2 - 1 0 - 1

1

2

3

X

GRAPHING POLYNOMIALS BY MULTIPLYING ORDINATES

Given the graphs of two polynon1ial functions, y = f(x) and y = g(x), the graph of a new function y = f(x)g(x) can be obtained by a process of multiplying the ordinates for each x value. This process is den1onstrated in the follo\ving examples.

Chapter 3

Graphing functions

61

YEAR 11

y

Example 13 The graphs ofy = j(x) and y = g(x) are sho\vn. By drawing vertical lines and

II

10

n1ultiplying ordinates, dra\v the graph ofy =j(x)g(x). Co1n1nent on the new curve.

y= g(x)

9 8 7 6 5

-3

Solution On the diagran1, vertical lines are dra\vn through important points, such as turning points, points \Vhere a curve cuts the axes, points where the curves intersect, and points where the function value is 1 or -1. On each vertical line, the intercepts of the two curves are 1nultiplied to find the position of a ne\v point, which is 1narked on the line.

y

'I I

II

I

y =f(x) x) ;

10 ', )' =g(x) 9 '. 8 I

I I I

I I

'

7

I

I

I

'

6

'I

,,

,

I' • x= -2: Ordinates are 3 and-1, so 3 x (-1) =-3. The point onj(x)g(x) 5 I ' ,, I , R I 4 I,," y=)\X) is (-2, -3). 'I 3 • x = -1: Ordinates are Oand 0, so the point on j(x)g(x) is (-1, 0). ': I 2 , ,. ,'•:/'.. :: 'I ,, : ' • x= 0: Ordinates are -1 and 1, so (-1) x 1 = -1. The point onfix)g(x) is (O, -1). : ' 1... .' . . ', , This 1nakes (-1, O) a local maxiinum turning point. ' ' , I 2 3 4 X - 4 - 3 -:v - 1 , ' • x = 1: Ordinates are Oand 2, so the point on j(x)g(x) is (1, 0). ,, ,, ' -2 , ' , , • x = 2: Ordinates are 3 and 3, so 3 x 3 = 9. The point on fix)g(x) is (2, 9). -3 ,, • x = 0.5: Ordinates are approximately-0.8 and 1.5 (-0.8) x 1.5 = -1.2. A point on j(x)g(x) is (-0.5, -1.2). The n1inimun1 turning point of y =f(x) g(x) will be near this point. The ne\v points are joined to obtain y =f(x)g(x). /

• I

:

The solid curve is the graph of y =j(x)g(x).

As x

➔

oo,fix)g(x) ➔ oo. As x

➔

-oo,f(x)g(x)

➔ -oo.

Example 14 The graphs of y = fix) and y = g(x) are sho\vn. By drawing vertical lines

)'

8

and multiplying ordinates, dra\v the graph of y =j(x)g(x). Co1n1nent on the new curve.

y= g(x)

2

y= f(x) 3

-2 -4

62

New Senior Mathematics Extension 1 for Years 11 & 12

4

5

X

YEAR 11

Solution On the diagram, dra,v vertical lines through in1portant points, such as turning points, points where a curve cuts the axes, where the curves intersect, and points where the function value is 1 or - 1.

I 0

I

I

I

8

I

Where the horizontal line y =-1 intersects a curve it shows the x value for which the product of the two functions has the opposite sign to the other function value.

I

I I

I

6

I

I

,..

I

\

•

': ,...' '.

:,4 : ,: :: ~ '

\

\ \

:

:

:\

'

•

'

,\ .:' .:i·' • \, • '

\

I

: ,: : ,, ...

:, ' : : :

- ,p :' lL. -'-

. . ,\

'

i

• x =-1: Ordinates are Oand 4.5, so the point onf(x)g(x) is (-1, O). • x= 0: Ordinates are-2 and 2, so (-2) x 2 =-4. The point onf(x)g(x) is (O, -4). • x = 1: Ordinates are -2 and 0.5, so (-2) x 0.5 = - 1.5. The point on f(x)g(x) is(l,-1.5).

'

\

I

'

I

'

'

I

~ !, ____ _ _ ~{

. ,,. ' :

2

1

: :/

,

,:

'

'

I \

_ _ _ __ __1 · ~ - - .. ... - --1

-3 -2

'

I•

I

I

Where the horizontal line y = 1 intersects a curve it shows the x value for which the product of the two functions is the san1e as the other function value.

'

' ' '

Multiply the intercepts on these vertical lines of the hvo curves and mark a ne,v point.

y

I

J. I

3

4

5

X

y =/(x) gl.x)

• x =2: Ordinates are Oand 0, so the point on f(x) g(x) is (2, O). • x = 3: Ordinates are 4 and 0.5, so 4 x 0.5 = 2. The point on f(x)g(x) is (3, 2).

Join the points to obtain y =f(x)g(x). The solid curve is the graph ofy =f(x)g(x). The new curve is a quartic, the product of t\vo quadratic functions. It is positive where the original functions both have the same sign, zero where at least one of them is zero, and negative where only one of the functions is negative. At (2, O), where both f(x) and g(x) are zero,f(x) g(x) has a horizontal point of inflection.

Example 15 The graphs of y = f(x) and y = g(x) are shown. By dra,ving vertical lines

)'

6

and 1nultiplying ordin ates, draw the graph of y =f(x) g(x) . Comn1ent on the new curve.

5

4

y =g(x)

I

- 3 - 2 _,

-

I

2

4

5

X

2 3 4

-s -6

Chapter 3

Graphing functions

63

YEAR 11

Solution On the diagran1, draw vertical lines through iinportant points, such as turnmg points, points ,vhere a curve cuts the axes, ,vhere the curves mtersect, and points where the function value is 1 or - 1.

)'

I

I I

6 5

Multiply the intercepts on these vertical lines of the hvo curves and n1ark a new point Where the horiwntal lme y = 1 intersects a curve it sho,vs the x value for which the product of the two functions is the san1e as the other function value.

y =gl.x),:,. ' ' .)'=fix) ' ' '' y = f(x) gl.x) '' 4 '' ' : ' ' ,: ,3 : ' : ' .: 2 : ' I '' : ' ' ...: ' -''' : '

.

--

Where the horizontal line y =-1 intersects a curve it sho,vs the x value for - 3 - 2 - ir ... which the product of the two functions has the opposite sign to the other .. function value. :~' • x = - 1: Ordinates are 3 and -4, so the point on J{x)g(x) is (-1,-12). 1 ~ • x = 0: Ordinates are 2 and 2, so 2 x 2 = 4. The point on J{x)g(x) is (O, 4). ' • x = 1: g( l ) = 0, so J{x)g(x) = 0 and cuts the x-axis at ( 1, O). • x = 2: }{2) = 0, so f(x)g(x) = 0 and cuts the x -axis at (2, O). • x = 3: Ordinates are (-1) and (-4), so (-1) x (-4) = 4. The point onj{x)g(x) is (3, 4). • x = 4: Ordinates are (-2) and 6, so (-2) x 6 = - 12. The point on J{x)g(x) is (4, -12). g(x) cuts the x -axis again at 0.6 and 3.6, so J{x)g(x)

X

' ' ' ' ' ' ' '

'

''

., l

'' : ':t

''

'

''

',' '

I

= 0 and cuts the x -axis at (-0.6, O) and (3.6, 0).

Drawing x = 1.5 will help locate the turning point which is at approximately ( 1.5, - 1). Join the points to obtain y =f(x)g(x). The solid curve is the graph of y =J{x)g(x) . The ne,v curve is a quartic, the product of a linear and a cubic function. It is positive where the original functions both have the san1e sign, zero where one of them is zero, and negative where only one of the functions is negative. As x ➔ +oo,f(x)g(x) ➔ -. It has a greatest value of 4, which occurs at x = 0 and x = 3.

- - - - - - - - = = = = = = =10

Graphing polynomials by multiplying ordinates

Use graphing software to verify the multiplication of ordinates for combined polyn omial graphs.

EXERCISE 3.5

GRAPHING POLYNOMIALS BY MULTIPLYING ORDINATES

1 The graphs of y = J{x) and y = g(x) are sho,vn. By drawing vertical lines and n1ultiplying ordinates, draw the graph ofy =J{x) g(x). Con1n1ent on the new curve.

(a)

(b)

)'

6

5

4

4

3

3

2

2

1

1

- 1

1

2

)'=fix)

3

X

-3

-

- 1 0 - 1

)'

10

6

)' = g(x)

5

- 3 -2 - 1 0

(c)

)'

)'=fix) 5

3

X

-s - 10

64

-4

-4

-5

-s

New Senior Mathematics Extension 1 for Years 11 & 12

- 15

YEAR 11 y

(d)

y 10

(e)

20

8 6 4

10

y 5

(f)

y = g(x)

4

y = g(x)

3

2

y = /(x)

X

-4 - 3 -2 -

0 - t2

-4 - 3

- 10

-2

-I I0

2

4

5

X

2

- 20

-8

3

- 10

4

- 12

-5

- 14

-6 -7

2 On the same diagra1n, sketch the graphs of y = x?- and y = I - x. Use these graphs to sketch y= x 2 3

3 On the same diagra1n, sketch the graphs of y = x - 2 and y = x • Use these graphs to sketch y= x 4 On the same diagra1n, sketch the graphs of y = x 2 2 y = (x - 3x)(x - 1).

5 The graphs of y = x + 4 and y = x

2

-

2

-

3x and y = x

2

-

4

-

x3. 3 •

2x

1. Use these graphs to sketch

2x are shown.

-4

2

Which diagra1n represents the graph of y = (x + 4)(x A

B

)'

-

C

y

8

8

6

6

2x)? D

)'

y

8

8 6 4

2

-2

0 -2

2 2

X

2

X

X

-2

-4

-4

-6

-6

-8

-8

-2

0 -2

2

X

-8

Chapter 3

Graphing functions

65

YEAR 11

3.6

PARAMETRIC FORM OF A FUNCTION OR RELATION

It is often useful in m athem atics to express two related variables (e.g. x and y) in terms of a third variable (e.g. tor 0) , so that, for exa1nple: x = f(t) , y = g(t) or x = f(0), y = g(0) Equations like these are called parametric equations and the third variable ( e.g. tor 0) is called the parameter. For exa1nple, recall that the function s cosine and sine can b e defined as the x - and y -coordinates respectively of a point on the unit circle x2 + y 2 = 1. Thus the unit circle can be represented by the paran1etric equations:

x = cos 0, y = sin 0 2

2

where 0 is the para1neter. When the unit circle is described by the equation x + y = 1, it is said to b e in Cartesian forn1.

Example 16 Find the Cartesian equ ation of the curve and describe it in word s, given the p aran1etric equations:

(a)

X

(b) x = 2t - 1, y = 3t + 2.

= t, y = t + 1

Solution (a) x= t,y= t + 1 Make t the subject of the equation in x: t = x Substitute in the equ ation for y: y = x + 1 The parametric equation s represent a straight line \Vith gradien t 1 and y -in tercept 1. (b) x = 2t - 1, y = 3t + 2 Make t the subject of the equation in x:

Substitute in the equ ation for y: y = 3 x

2t = x + 1 t = x+ l

x;

2

1

+2

2y=3x + 3 + 4

3x-2y+7=0 The parametric equation s represent a straight line \Vith gradien t 1.5 and y -intercept 3.5.

Example 17 Find the Cartesian equ ation of the curve whose p ara1netric equation s are x = 1 + t, y =

Solution x= l +t y=r Fro1n [ 1): t = x - 1 Substitute in to [2): y = (x- 1)2

l

[ 1l [2)

Hence the Cartesian equation is y = (x and the graph is th e parabola sh own.

t2.

s 4

1)2

I

- 4 - 3 -2 - I O -I

I

2

3

4

X

Example 18 Find the Cartesian equ ation of the curve whose p ara1netric equation s are x = 1 + t, y =

66

New Senior Mathematics Extension 1 for Years 11 & 12

t2, t > 0.

YEAR 11

Solution y

As in Exan1ple 17, these parametric equations give the Cartesian equation y = (x - 1)2, but there is now also the condition t > 0.

6 5

4

x = 1 + t and t > 0, so the condition is equivalent to x > 1.

3

Hence the Cartesian equation is y = (x - 1)2 with the domain restricted to x > 1, as shown.

2

I

-4 - 3 - 2 - I O

I

-I

2

3

4

X

Example 19 Find the Cartesian equation of the curve \Vhose paran1etric equations are given by x = 2 sin 0, y = 2 cos 0. Describe the curve in \Vords and sketch its graph.

Solution Recall the Pythagorean identity: sin 2 0+ cos 2 0= 1.

y 3

2

· 0 X , 20 Sill = - SO Sill = -X 2' 4

2 2

cos0 = Y2' so cos 0 = L4 . xi

Hence, using the identity: or

-

4

2 1 +-

4

I

=l

-3 -

-I 0 - I

I

X

3

x2 + y2=4

The curve is a circle with centre at the origin and radius 2.

-3

Example 20 Write each Cartesian equation in para1netric form. (a) 3x + y - 3 = 0

(b) x2 =4(y-3)

Solution There 1nay be n1ore that one set of paran1etric equations for each Cartesian equation, depending on how the parameter is defined for x and y.

(a) Method 1 Re\vrite the equation: y = 3 - 3x Let t = x: y = 3 - 3t The paran1etric equations are x = t and y = 3 - 3t.

Method 2 Rewrite the equation: y = 3( 1 - x) Let t = 1 - x: y = 3t The parametric equations are x = 1 - t and y = 3t.

2

(b) Re\vrite the equation:

(;) =y-3

Method 1 Let t = x=2t

X

2

:

t

2

=y - 3

y= t2 + 3

The parrunetric equations are x = 2t and y = t2 + 3.

Method 2 Rewrite the equation: Let t= x:

, x-, =4(y-3)

C=4(y-3) t2 y= - +3 4

2

The parametric equations are x = t and y = t + 3. 4

Chapter 3

G raphing functions

67

YEAR 11

(c) The equation is the sum of l\vo squares, ,vhich suggests that the identity sin 2 0 + cos 2 0 = 1 may be useful. Rewrite the equation:

( x-1)2

9

(y + 2) + 9

2

=1

Method 1

Y + 2 = cos 0 Let x - 1 = sin 0 and -'--3 3 x = 3 sin 0 + 1 y = 3 cos 0- 2 The paran1etric equations are x = 3 sin 0 + 1 and y = 3 cos 0- 2. Method2 Swapping the position of sin 0 and cos 0 would give different paran1etric equations.

Example 21 Sketch the graph of each curve fron1 its panunetric equations. (a) x = t + 2, y = 2t

(b) x = 2t, y = 2t2

(c) x = 2 sin 0, y = 2 cos 0

Solution Either use graphing soft,vare or dra,v up a table of values and plot points. (a)

)'

-2

t

-1

0 -2

X

0

1

2

y

-4

-2

0

-I

0

2

I

X

-I

-2

-3 -4

(b)

t

-2

-1

0

1

2

X

-4

-2

0

2

4

y

8

2

0

2

8

)'

9

8 7

6 5 4

3

2 I -4 - 3

(c)

68

-2 - I 0

0

0

-11:

-11:

-TC

-TC

-2rr:

-311:

-Srr:

11:

x=2sin0

0

1

✓ 2

✓ 3

2

✓ 3

✓ 2

0.5

0

y= 2cos0

2

✓ 3

✓ 2

1

0

- 1

-✓ 2

-✓3

- 1

6

4

3

2

New Senior Mathematics Extension 1 for Years 11 & 12

3

4

6

I

2

3

4

X

YEAR 11 )'

This table gives the right half of the graph of the relation. By changing the signs on values as the quadrant for 8 changes, the left half n1ay be graphed

3

I

-3 -

-10 -1

3

I

X

-3

Parametric equations of the parabola 2

The parabola x = 4ay can be represented by the paran1etric equations: x = 2at, y = at2 This can be verified by eli1ninating the parameter:

x = 2at y=at2

From [l):

t= 2a

[ l) [2)

X

Substituteinto[2): y=a(;af

,

x- =4ay The point P(2at, at2) on the parabola is the variable point that depends on the value oft, so it is frequently called 'the point t'.

EXERCISE 3.6

PARAMETRIC FORM OF A FUNCTION OR RELATION

For questions 1 to 14, find the Cartesian equation of the curves with the parametric equations given. 1 x = 2t, y = t + 2

2 x = t, y = t2

5 x=2cos8,y=2sin8,0)

20 2 sin 2A cos 4A

21 2cos3A sin 7A

A; B) cos ( A; B)

22 cos 5x + cos x may be written as:

A

B

2 sin3x cos2x

2cos3x sin2x

C

-2 sin 3x sin 2x

D

2cos3xcos2x

Express the follo,ving as products:

23 sin 3x- sinx

24 sin (x + a) - sinx

26 sin(0 + a) + sin (0- a)

27 cos ( 8 ~ a ) + cos ( 8

25 cos(x+h)-cosx 28 cos (A + B + C) - cos (A - B + C)

a)

2 30 sin (90° - A)+ sin (90° - B)

29 cos (2x + y) + cos (x + 2y)

31 sin (90° -A)+ cos 3A

32 cosx + cos (x + 120°) + cos (x + 240°) 33 cos(2A + B)- cos(A + 2B)

34 cos(A + B + C) + cos(A-B- C)

35 sin 2x + sin 3x

36 sin (2A + 2B) - sin (2A - 2B) 37 sin 165° - sin 105°

38 sin 8- cos q'>

39 cos75° -cos45°

41 sin (A -B)-sinA

40 sin50°+ cos20°

42 cos 80° - sin 50° Prove the following results.

y)

sin50-sin30 43 - ~ ~ - ~ = tan 8 cos58 + cos30

sinx+sin y (x + 44 - - - ~ =tan cosx+cosy 2

45 cos x - c~s 3x = -tan 2x s111x-s1113x

46

47 cos(8+a)-cos(8-a) =-tan a sin(0 +a) + sin(0- a) 49 sin 58+ sin 30- 2sin 20cos 8 = 2sin20cos30

sinx + sin(x + y) + sin(x+2y) ( ) 48 - - - - - ~ - - - - ~ =tan x+ y cosx + cos(x + y) + cos(x + 2y) 50 sin2x + sin4x + sin6x = 4cosx cos2x sin3x

51 sili50-sin 2 30=sin80sin20

52 sin 35° - sin 25° = ✓ 3 sin 5°

sin 2A - sin 2B _ tan (A - B) sin 2A + sin 2B tan (A + B)

53 cos35° + cos45° + cos75° + cos85° = 2 cos5° cos20° 54 2 cos 37.5° sin 7.5° =

✓ 2- 1

2

56 sin 10° + cos 40° = sin 70°

84

New Senior Mathematics Extension 1 for Years 11 & 12

55 sin 25° sin 35° - sin 20° sin 10° =

57

sin 48° + sin 12° _ ✓ 3 cos48° +cos l2° - 3

✓34-

l

YEAR 11

58 sin 3x - sinx = cot2x cosx -cos3x

59 cos(A + B) cos(A -B) = cos 2 A -sin 2 B

60 sin8 + sin78 =2cos28- l

sin(0 + )- sin(8-) 61 = -cot 0 cos (0 + ) - cos (0 - )

sin 38 + sin58

62 cos75°+ cosl5° =

✓ 3

63

sin 75° - sin 15° sin(n + l)8 + 2sinn8+sin(n-1)8 8 64 - - - ~ - ~ - - ~ - ~ - - =cot cos(n-1)8-cos(n + l)8 2

sinA+sin (A+B)+sin(A +2B) = tan(A+B) cos A + cos(A + B )+ cos(A + 2B)

65 If a + /3 + r= rr, sho\v that sin 2a + sin 2/3 + sin 2y = 4 sin a sin /3 sin r

4.6

OVERVIEW OF TRIGONOMETRIC EQUATIONS

Equations of the form sin0=sina Example 14 Find all values of 0 for which sin 8 = ~. y

Solution

p

Q

sin0= l

2

·

. 8 Sll1

. tc , SIU . ( tc - tc) , SIU . (21C + tr) , ... =Sll1 6 6 6

It

Jt -

6 It

Consider a coordinate diagra1n.

6 0

X

LXOP=: LXOQ=tc- : The ray OP defines an infinite number of angles in the first quadrant. If you rotate OP about the origin (either clock\vise or anticlockwise), then during each revolution it is along the original ray OP once. Each full rotation increases the angle by 2rr, so you find that OP is the tenninal ray defining the angles: c antic · 1oc",v1se '- · rotation · • 6tr, 2 7r: + 6tr , 4 7r: + rr: , ... 1or 6

• -2rr: + : , -4,r + : , -6rr: + : , ... for clockwise rotation.

This result can be sun1n1arised as: or:

ntr +: \Vhere n = 0, ±2, +4,...

[ 1)

n x 180° + 30° (in degrees)

Si1nilarly, the tern1inal ray OQ defines an infinite number of angles: c an t 1c ' 1ockwise . ro tat"1011 rr:- tr, 3rr:- tr , 5rr:- tr , ... ,or

6

6

6

-rr:- :, -3,r- : , -5,r- :, ... for clock\vise rotation.

This result can be sun1n1arised as: or:

ntr-: \Vhere n = +1, +3, +5, ...

[2)

n x 180° - 30° (in degrees)

Staten1ents [ l ) and [2) can be written together as:

8=ntr+(- l )" : or:

0= n x 180° + (-1)" x 30° (in degrees)

Note: (-1 )" is 1 when n is zero or even, an d is - 1 when n is odd.

Follo\ving Exainple 14 above, the general solution to the equation sin 0 = sin a can be written as:

8 = ntr + (-l)"a

(in radians)

8 = n x 180° + (-l)"a 0 (in degrees)

for any in teger n. Chapter 4

Further trigonometric identities

85

YEAR 11

The general solution of trigono1netric equations has not been included in this course. This 1naterial is included to show why you have to be very careful to consider all possible results in th e given don1ain when solving trigonometric equations. The pattern for this general solution can also be seen by considering the value of 0 at the points of intersection of the curves y = sin 0 and y = (fro1n Example 14), as shown in the following diagram.

1

y I

y=½

y=sin8

-3

3;r

From syn1n1etry, you can observe that the line y =

8

1intersects the sine curve at values of 0 that are :

units to the

right of nn where n = 0, +2, +4, .. . and that are : units to the left of nn when n = + 1, +3, ... Both of these solutions are contained in the state1nent: 0=nn +(- l)" :

nn +:, n even or zero or: 0=

nn- :, n odd

Example 15 Solve sin( 0 + )= -

1

Fi

for O < 0< 2n.

Solution

sin(0+1)=-}i

0

sin < 0 in third and fourth quadrants: sin

(0 + n) = sin Sn, sin ?n 4 4 4 0 + n =Sn, 7n 4

4

4

0= n, 3n 2

Equations of the form cos8=cosa Example 16 Find all angles 0 for which cos 0 =

1· p

Solution

)'

cos0=l

2

-3

1C

1.e. cos 0 = cos 1, cos ( 2n- 1 ), cos ( 2n + 1 ), ... In the diagra1n, OP defines the angles: and OQ defines the angles:

1, 2n + 1, 4n + 1, ...

-1, 2n -1, 4n -1, ...

0 = 2nn + 1 where n is any integer or: 0= n x 360° + 60° (in degrees)

0

1C -3

These results can be su1n1narised as:

86

New Senior Mathematics Extension 1 for Years 11 & 12

Q

X

YEAR 11

Follo\ving Exainple 16, above, the general solution to the equation cos 8 =cos a can be \Vritten as:

8= 2 nn: + a 8= n x 360° + ct'

(in radians) (in degrees) for any integer n. The pattern for this general solution can also be seen by considering the value of 8 at the points of intersection of the curves y = cos 8 ai1d y = (from Exan1ple 16), as shown in the follo\ving diagra1n.

i

)'

I

y=½ - 31
-2 and the range is y < 1. 2

/(x)\

,

, , ,

, , ,

, , ,

,

,,

,

( I, - 2)

1

Chapter 5

Inverse functions

99

YEAR 11

Makey the subject:

Method l (con1plete the square)

Method 2 (quadratic form ula)

,

y- -2y-x-l=O

2

, y -2y=x+ 1 y- - 2y + 1 = X + 2

This is a quadratic in y ,vith a = 1, b = - 2 andc=-(x + l).

(y-1)2=x + 2 y - 1 = + .,/,-x-+-2

y=

2± .J(- 2) 2 - 4(1)(- [x + 11) 2

y= 1 + ../x+2 W hich one?

y=

The range of 1- is y ~ 1, so it n1ust be: 1

y=

y=l - ✓x+2

2± ../4 + 4x+4 2

2(1± ../x+2) 2

y= l +.Jx+2 Which one? The range of 1-1 is y ~ 1, so it 1nust be:

y= l - ✓x + 2 (d) To find the point of intersection, you ,vould usually solve simultaneously the equations of the two curves.

However, for these hvo equations you need to solve x 2 - 2x - 1 = 1 - ✓x + 2, ,vhich is a fornlidable task. A better way is to re1nen1ber that a function and its inverse intersect on the line y = x. 2 2 Now you can find the solution of x - 2x - 1 = x, i.e. x - 3x - 1 = 0. 3+ ✓ 13 The solution is x =- - -. 2

Ho,vever,l and 1- intersect at one point only (as sho,vn in the diagram on p. 99). The second solution is a different poin t of intersection that would exist if the entire parabola ,vere intersected with the line y =x (instead of only half the parabola). 1

Fron1 the diagra1n, you can see that the required value in this case is the smaller of the two solutions, . . . . 3 - ✓13 so the point of 1ntersect1011 1s at x =- - -. 2

(e)

, = x- - 2x - 1 :. :i=2x - 2 At x = - 1, the gradient of the tangent is - 4. y

At correspondin g points in the reflection in y = x, the product of the gradients of a function and its inverse n1ust be 1, so 1-1 has a gradient of -0.25 at the point (2, - 1).

Exponential and logarithmic functions Exponential and logarithmic functions, ,vhich you have ,net in the Mathen1atics Advanced course, provide a very good exan1ple of inverse functions. )'

An exponential function has the forn1 l(x) = ax. The base a is any positive real nun1ber except l ; the do1nain is Rand the range is R+ (the set of positive real nun1bers). It is a strictly n1onotonic increasing function, i.e. a one-to-one function, without having to restrict its do1nain. Let Interchanging x and y:

f(x) = a•

,

, ,,

.

g(x) = los..x

,>, , ------'--

x = er

(a, I)

(definition of a logarith1n)

X

,

100

, , ,

( l , a ), ' , ,'

y = l(x) = ax y= log. x

y=x

New Senior Mathematics Extension 1 for Years 11 & 12

YEAR 11

The function g, where g(x) = log. x, is called a logarith1nic function. Its don1ain is R+(the range off) and its range is R (the don1ain off). It is also a strictly monotonic increasing function. Hence log. ax = x for all real x (i.e. for all x in the do1nain of the exponential function), and a1og"x = x for all x > 0 (i.e. for all x in the domain of the logarithn1ic function). It should be noted th at for any pair of inverse functions f and g, f is the inverse of g and g is the inverse off MAKING CONNECTIONS

Inverse functions Use technology to explore the graphs of inverse functions and their properties.

EXERCISE 5.1

INVERSE FUNCTIONS

1 Which of the follo,ving are one-to-one functions? A f(x)=x-2

B

f(x)=x2 -2x+I,x>l

D

f(x)=9-x

E

H

f(x)

f(x)= 9-x f(x) = cosx

1

= cosx, 0 < x < 1

C

F

f(x)= ✓4-x 2 G

f(x)= l9-xl J

f(x)

=x 3 -

f(x)=9-x 2

4x, -2 < x < 2

2 Find the largest possible don1ain for ,vhich the follo,ving are one-to -one increasing functions. (a)

(d)

f(x)= ✓4 -x 2 f (x) = 3x - x 2

(b)

(e)

f(x)= ✓x 2 - 4 f (x) = x2 + 6x + 8

(c) f(x)=-

1

x+2

3 For each of the follo,ving, find the inverse function and state the don1ain and range of the inverse. 2

(a) f(x)=2x-4 (e) f(x)=x (i)

3

(b) f(x)=x -I,x>O

(c)

(d)

f(x)= ✓9-x 2 , -3 -1 x+ l

1 2•

l +x 1 (b) State t\vo don1ains for which the function f(x) = is one-to-one. (State the largest possible don1ain in 2 each case.) 1+ x (c) Find and graph the two functions that are the inverse of the functions defined in (b).

5 Explain why the following functions do not have an inverse function. Suggest suitable restrictions to their do1nain so that the restricted functions will have an inverse. (a) f(x)

=✓a 2 -x 2

(b) f (x)

=4 -

x

2

(c) f(x)

= - 12 X

6 Show that the following pairs of functions are inverses by sho,ving thatf(g(x)) = g(f(x)) = x. ex (a) f(x) = 2x-l and g(x) = (x + 1) (b) f(x) = and g(x) = log,2x

1

(c) f(x)

2

= ✓16-x 2 , -4 - andg(x)= - - ,x>O

7 f(x)= ✓4 -x 2 , -2 < x < 0. The inverse function is given by: 2

2

A x +y =4

B

y= ✓4 -x 2 ,-220 -k(t - C) = log, (T- 20)

T-20

=e-• 0

IO

- -- -- --- -- -- --- -- -- --- -- -- --

Note that the asyn1ptote of the graph is x = IO, approached fro1n below. This is consistent with the restriction O< x < IOfort> 0. When t=2,x=4:

4= 10(1-e-2.k)

0

0.4 = 1- e- 2k Reciprocal of both sides:

e- 2k =0.6 e2k =2.

or

3

2

I

5

inverse functions:

-2k = log, 0.6

-2k=-0.51

2k=log,(~)

k= 0.255

k= 11og,(~)

When t= 5: But:

X

= 10(1 - e-5")

e- 2.k

x = 10(1 - e-Sk)

1.e.

= 0.6 .5.

X

= 10(1 - e- 1275)

X

= 10(1 - 0.279)

5

e-sk =(e-2k)2 =(~)2 =0.279 x= 10( 1 -0.279) x= I0-2.79

x= 10-2.79

10-x=2.79

10-x =2.79 Hence the concentration is 2. 79 units after 5 n1inutes. (Ren1ember that ( 10 - x) is the concentration remaining.)

Fro1n the previous exa1nples, it can be seen that if k < 0 then as t

➔ oo, N ➔ P.

dN Given -d = k(N - P), where k and Pare constants, if k < 0 then li1n N = P. t I➔•

Chapter 7

Rates of change and their application

165

YEAR 11

EXERCISE 7.4

HARDER EXPONENTIAL GROWTH AND DECAY

1 N is decreasing according to the equation

~~ = - 0.4(N -

30). If N = 60 \Vhen t = 0:

(a) sho\v that N =30 + Ae-OAt is a solution of this equation, where A is a constant (b) calculate the value of N when t =5.

clJ:

2 N is increasing according to the equation = 0.2(N - 40). If N = 50 when t = 0: (a) sho\v that N = 40 + Ae0·21 is a solution to this equation, where A is a constant (b) calculate the value of N when t = 10.

3 The original ten1perature of a body is 120°C, the ten1perature of its surroundings is 50°C and the body cools to 70°C in 10 n1inutes. Assun1ing Ne\vton's law of cooling, i.e. ~~ =- k(T - 50) where Tis the ten1perature of the body at tin1e t, find: (a) the te1nperature after 20 n1inutes (b) the tin1e taken to cool to 60°C. 4 If N = 70 \Vhen t = 0, which expression is the correct solution to

A

N = 20 + 50e0·5'

B

N

= 20 + 50e--0.s,

C

N

clJ: = - 0.S(N - 20)?

= 20 -

50e05'

D

N = 20- 50e--0.s,

5 A metal bar has a te1nperature of 1230°C and cools to 1030°C in lOminutes when the surrounding ten1perature is 30°C. Assun1e Newton's law of cooling, i.e. body at tin1e t.

~~ =- k(T - 30) where Tis the temperature of the

(a) Show that T =30 + 1200e-kt satisfies both Newton's la\v of cooling and the initial conditions. (b) Find the te1nperature after 20 minutes. (c) Find the tin1e taken to cool fron1 1230°C to 80°C.

6 Water at 20°C is placed in a freezer where the air is at a constant te1nperature of -10°C. The temperature of the water falls to 15°C in 5 minutes. Assu1ne Ne\vton's law of cooling, i.e. ten1perature of the body at time t.

~~ =- k(T + 10) where Tis the

(a) Show that T =-10 + 30e-k' satisfies both Newton's law of cooling and the initial conditions. (b) Find the temperature of the \Vater after another 5 n1inutes (when t = 10).

7 A body whose temperature is 180°C is iin1nersed in a liquid that is at 60°C. In 1 minute the temperature of the

body has fallen to 120°C. Assun1e Newton's la\v of cooling, i.e. ~~ =- k(T - 60) where T is the te1nperature of the body at tiine t. (a) Show that T = 60 + 120e-k' satisfies both Ne\vton's law of cooling and the initial conditions. (b) At \Vhat tin1e would the te1nperature of the body have fallen to 90°C?

8 A current of i amperes (or 'a1nps') flows through a coil of inductance L henrys and resistance R oh1ns. The current at any_ti1ne is given by i = ~ ( 1- e-f'

), \Vhere Eis the electron1otive force (i.e. the voltage) in volts.

Sho\v that L ~; + Ri = E.

9 A vessel is filled at a variable rate so that the volu1ne of liquid in the vessel at any time tis given by V = A(l - e-k'). dV (a) Show that dt

=k(A - V ).

(c) Show that liin V t ➔~

= A.

(b)

If a quarter of the vessel is filled in the first 51ninutes, what fraction is filled in the next 5 minutes?

10 A rectangular vessel is divided into two equal co1npartn1ents by a vertical porous 1nen1brane. Liquid in one compartment, initially at a depth of 20 cm, flows into the other co1npart1nent, initially empty, at a rate proportional to the difference between the levels in each co1npart1nent. The differential equation for this

=k(20 - 2x ), where xc1n is the depth of the liquid in one of the vessels at any tiine tn1inutes. Show that x = 10( 1 - e-2.k'). (b) If the level in the second con1partn1ent rises 2 c1n in the first

process is : (a)

5 n1inutes, at what tiine will the difference in levels be 2cm?

166

New Senior Mathematics Extension 1 for Years 11 & 12

YEAR 11

11 In a certain che1nical process, the amount y grams of a certain substance at tin1e t hours is given by the forn1ula y = 3 + e-k,_

';l; = - k(y-3).

(a) Show that

(b) If initially y decreases at a rate of 0.08 gran1s per hour, find the value of k. (c) Find the rate of change \vhen y = 3.5.

7.5

(d) What values can y take?

RELATED RATES OF CHANGE

Rates of change were introduced earlier in this chapter. You will now look at more sophisticated applications of rates of change. Problems with related rates arise when there is a function that relates two variables, e.g. x and y, where both variables are also functions of another variable, e.g. tin1e t. For example, you may need to detern1ine

7t Z

7t when ! is

known. In such cases it is necessary to use the chain rule, = x ~; . Re1ne1nber that in a context like this, 'increasing' means a positive rate of change while 'decreasing' 1neans a negative rate of change.

Example 13 (a) If V =

4

3

1!'r

3

dr

.

dV

and dt = 5, find the expression for dt .

dr

2

dS

(b) If S = 4,rr an d dt = 5, find the expression for dt.

(c) If V =

dV

4

3 nr3 and S = 4nr2, find the expression for dS .

Solution (a)

dV dr dV dr dV dt dr dt

-

(b) -dS =Birr

4 , = - ,rx3r3

dr dS = dS x dr dt dr dt

=4irr2 = dV x dr

dr

dr =S dt -dS =8m-x5 dt

dt

=5

dV =41l'r2X5 dt

As

1;

dV dr dV dS -dr dS dV dS -dV dS

(c)

dS =401l'r dt

= 4 1rr2, dS = Bll'r

dr = dV x dr dr dS =- 1 81rr

= 4 1rr2 x _ l_

Birr

= -r 2

dV = 20nr2 dt is the san1e in parts (a) an d (b), these results can be used to find the answer to part (c) in this case.

Example 14 If x = St cos a and y = St sin a- ~gr, where a and g are constants, find: (a) the expression for .

(c) if a=

,r

4

Z

as a function oft

(b) the expression for

Z

when t = 2

dy and g = 9.8, find the value of dx \Vhen t = 2.

Chapter 7

Rates of change and their application

167

YEAR 11

Solution (a)

dx dt = 5 cos a

(b) t=2 2

. = 5 sin a - gt

dy dx

dy dt = dt X dx

dy

(t)

a=

dy - 5sina-gx2 dx 5cosa dy - 5sina-2g dx 5cosa

y=5tsina-.!.gt2 dy dt

(c)

dy dx

1C

4 ,g =9.8 5sin¾-2x9.8

=

4

5

✓ 2-19.6

dy dx

5cos

,r

=

5 ✓ 2

c1x=(:)

7x

=-4.544

dy 5sina-gt dx = 5cosa

Example 15 Given x =

t2 -

1 and y =

(a)

t3, find as functions oft:

d2y (b) dx2

dy dx

Solution (a) :

(b)

=2t

dy = 1!_ dx = 2t dx 2 ' dt 2

y=t3

d y = _!!_(dy ) dx 2 dx dx

dy =3t2 dt dy dy dt dx = dt X dx

= .E._(dy)x dt dt dx

dx

it(!)= ;t(~)=; dt

I

dy = 3t2 X _!__ = 3t dx

2t

2

Example 16 A spherical balloon is being inflated so that its radius increases at the constant rate of 3 cn1/n1in. At what rate is its volu1ne increasing when the radius of the balloon is 5 c1n?

Solution

j

If r is the radius of the balloon, its volu1ne is V = nr3 • Given ~; = 3, need to find By the chain rule:

dt - dr

But:

V=4nr3

So:

dV 2 dr =4nr

Thus: For r

dV dV dr - -x -

= 5:

dt

3

dV = dV x dr = 4 n, 2 x 3 dt dr dt dV 2 dt =41C X5 X3 = 3001rcn13/ min

168

New Senior Mathematics Extension 1 for Years 11 & 12

~~ for r = 5.

YEAR 11

Example 17 A vessel containing water has the shape of an inverted right circular cone with base radius 2 m and height 5111. The water flo\vs out of the apex of the cone at a constant rate of 0.21113/min. Find the rate at which the \Vater level is dropping when the depth of the \Vater is 4 m.

Solution

2

Let the depth of the \Vater be h 111, the radius of the cone at the water level be rm and the volume of the water be Vn13 at tin1e tminutes. The volun1e of the water at any time t is V = ~ n:r 2h.

T

..-.J·- .. '

'

dV dh Given dt = - 0.2, need to find dt when h = 4. .

r '

I,

'

r

=52

2

T

2h

r=s Volu1ne:

2h . Su bst1tute r = :

5

'

' ''

out of the vessel. To find the link between rand h, use si1nilar triangles.

h

'

5'

dV · negative: · th e vo1un1e 1s · decreas1ng, · because th e \Vater 1s . flo\v1ng . dt 1s

Fro1n proportional sides:

' ''

' '

' '' 5'

V=l11:,2h 3

2

2h ) x h V= ~n: x ( 5

4n:h 3

Hence: Chain rule:

When h =4:

V= 75 dV 4n:h 2 dh = 25 dV dV dh dt = dh xdt 4n:h 2 dh -0.2 = 25 xdt dh 5 dt= - 411:h2 dh 5 5 dt= - 4n:x16 = - 6411:

' ..L

= -0.0249 in/min Thus the water level is decreasing at a rate of 0.02491n/inin.

Alternatively: V and h are both dependent on time, so you can differentiate both sides of the volu1ne equation with respect to time.

4n:h 3

V= 75

dV = 411: !!.._(h3) dt 75 x dt Chain rule:

= ~: x~(h3 )x ~; = 411: X 3h2 X dh 75

dV dt =-0.2, so:

4n:h 2 dh - 0.2 = x dt 25

dt as before.

Chapter 7

Rates of change and their application

169

YEAR 11

Example 18

j

The volun1e of water in a hen1ispherical bo,vl of radius 10 cn1 is V = n x 2 (30- x ), where x cn1 is the depth of water at tiine t. The bowl is being filled at a constant rate of2nc,n3 /min. 10 cm At what rate is the depth mcreasing ,vhen the depth is 2 cn1? •-------

Solution

•'x

Given ~~ = 2:IZ', need to find : when x = 2. The related variables are V, x and t.

V

=

inx (30 - x) 2

= I0nx 2 -

l nx 3 3

Method 1

Method2

dV dV dx dt = dx x dt

dx dx dV dt = dVx dt dx I - = - - - - X2:IZ' dt nx(20 - x ) dx 2 dt = x(20 - x)

dx 2n= nx(20 - x) x dt

dV 2 dx =20nx - nx =nx(20- x)

dx 2 dt= x(20- x) When x = 2: :

= / c1n/s 8 As an extension to this question, you n1ight ask: 'At what depth is the depth increasing at a minimun1 rate?'

This is the sa1ne as askmg: 'For what value of xis

';u a n1inin1u1n?'

Note that the fonnula for the volu1ne of water in the he1nispherical bowl at any depth a can be calculated by 2 2 finding the volu1ne generated by rotating the circle ,vith equation x + y = 100 between y = IO- a and y = IO.

Example 19 A ladder 101n long has its upper end against a vertical wall and its lower end on a horizontal floor. The lower end is slipping away fro1n the wall at a constant speed of 4m/ s. Find the rate at which the upper end of the ladder is slipping down the wall when the lo,ver end is 6 n1 fro1n the wall. What is this rate when the upper end is very close to the ground?

Solution At any tiine t, the lo,ver end of the ladder is x m fron1 the wall and the upper end is y n1 above the ground.

Yt

,vhen x = 6. = 4, need to find 2 2 By Pyth agoras'theore1n: x + y = 100

Given :

y (O,y)

y= ✓IOO -x 2 ,0 1. In step 2 it is proven that there is the potential for a 'chain reaction' to occur: if it is true for any one case then it is also true for the next case, ,vhich 1neans that it is also true for the case after that, and so on forever. In step 1 it is proven that it is true in the first case: this is like flicking the switch to start a chain reaction or pushing over the first domino of an infinite line of dominoes. The infinite cascade of falling do1ninoes 1neans that the statement is true for all appropriate values of n fron1 the starting value to infinity. You will now look at several different types of problems that require proof by induction.

9.1

MATHEMATICAL INDUCTION INVOLVING SERIES

Note the following pattern:

1 =1

1 + 3=4 1 + 3 + 5=9 1 +3+ 5 + 7=16 Observe that the su1n of n consecutive odd nu1nbers starting from 1 is always n Can you prove this? Yes, this can be proved by induction. (See Exan1ple 1.)

186

New Senior Mathematics Extension 1 for Years 11 & 12

2 •

YEAR 12

Example 1 2

Prove that 1 + 3 + 5 + ... + (2n - 1) = n for all integers n > 1.

Solution The two methods shown below are equivalent, but Method 2 introduces a syn1bolic representation of the staten1ent being proved. When you are fan1iliar with Method 2 you will find that it requires less writing than with Method 1. No ,natter ,vhich 1nethod you use, you n1ust ahvays write all three steps and use the appropriate words e1nphasised belo,v (or their equivalent) in each step.

Method 1

Step 1 Prove that the state1nent is true for n = 1. When n = 1:

2

LHS = 1

RHS= 1 = 1

LHS =RHS

:. the staten1ent is true for n = 1

Step 2 Assun1e the statement is true for n = k, ,vhere k is any integer greater than or equal to 1. i.e. assun1e that 11 + 3 + 5 + ... + (2k- 1)1= k 2

[a]

No,v prove that the statement will be true for n = k + 1 if it is true for n = k. i.e. prove that

1 +3+5+ ... + (2[k+ 1)-1) =(k+ 1)2 LHS = 1 + 3 + 5 + ... + (2[k + 1) - 1) = 11 +3 + 5+ ... +(2k- l )I+ (2[k+ 1) -1)

using [a]: (You can only prove this is true when n = k + 1 if it is true when n = k.)

= (k+ 1)2 = RHS

Step 3 Conclusion The staten1ent is true for n = k + 1 if it is true for n = k

(Step 2)

The staten1ent is true for n = 1

(Step 1)

:. by induction, the statement is true for all integers n > 1.

Method2 2

Let S(n) be the state1nent that 1 + 3 + 5 + ... + (2n - 1) = n ,vhere n is a positive integer.

Step 1 Prove that S( 1) is true. 2

LHS = 1

RHS = 1 = 1

LHS =RHS

:. S( l ) is true

Chapter 9

Proof by mathematical induction

187

YEAR12 Step 2 Assun1e S(k) is true.

i.e. assume that 11 + 3 + 5 + ... + (2k- 1)1= k2

[a]

No\v prove that S(k + 1) is true if S(k) is true. i.e. prove that

using [a]:

,

1 + 3+5+ ... + (2[k+ 1)- 1) = (k+ It LHS = 1 + 3 + 5 + ... + (2[k + l] - 1)

= 11 + 3+5+ ... +(2k- l)I + (2[k+ 1)- 1) =k2 +(2k+l) = (k + 1)2 =RHS

Step 3 Conclusion S(k + 1) is true if S(k) is true

(Step 2) S(l) is true (Step 1) : . by in duction, S(n) is true for all integers n > 1.

Example 2 1 2 Prove by induction that 12 + 22 + 32 + ... + n 2 = n(n + ~ n + l) for all positive integers n.

Solution 2 2 2 2 n(n + 1)(2n + 1) h . .. . Let S( n ) be th e statement th at 1 + 2 + 3 + ... + n = , w ere n 1s a positive integer. 6 Step 1 Prove that S( 1) is true. LHS = 12= 1 LHS = RHS

RHS = 1(1 + 1)(2 x 1 + 1) = 1(2)(3) = 1 6 6 : . S( l ) is true

Step 2 Assun1e S(k) is true for a positive integer k · assu1ne th at 1.e.

li2 k(k+l)(2k+l) _ + 22+ 32+ . . . + k2 I__ --'---'--'------'6

No\v, prove that S(k + 1) is true if S(k) is true. i.e. prove that

12 + 22 + 32 + ... + (k + 1)2 = (k + I)(k +62)(2k + 3) 2 2 2 2 2 LHS = 11 + 2 + 3 + ... + k I+ (k + 1)

using [a]:

= k(k + I)(2k+ I) + (k+ I)2 6

(k+l) as a factor. 6 (k+l) LHS= (k(2k+ I)+6(k+I)) 6

Now look at the RHS: you \Vant

= (k;I)(2k 2 +7k+6) = (k+l)(k+2)(2k+3) =RHS 6

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New Senior Mathematics Extension 1 for Years 11 & 12

[a]

YEAR 12 Step 3 Conclusion S(k + 1) is tru e if S(k) is true

(Step 2)

S( l ) is true

(Step 1)

:. by induction, S(n) is tru e for all integers n > 1.

Example 3 3 Prove by indu ction that n + (n + 1) + (n + 2) + ... + 2n = n(;+ l) for all integers n > 1.

Solution 3 Let S(n) be the statement that n + (n + 1) + (n + 2) + ... + 2n = n(; + l) for positive integer n.

Step 1 Prove that S( 1) is true. LHS = 1 + (2 x 1) (Note that the sun1 on the LHS starts \Vith n and fin ish es \Vith 2n.) =3 RHS =3x l(l + l )= 3 2

LHS = RHS

: . S(l) is tru e

Step 2 Assume S(k) is tru e for an integer k;,: 1.

.1.e. assu1ne that Ik + (k + 1) + (k + 2) + .. . + 2k I= 3k(k+ l ) 2

[a]

Now prove that S(k + 1) is tru e if S(k) is true. i.e. prove that

(k + 1) + (k+2} + ... +2(k + 1) = 3 (k + l )(~k+l) + l)

Note that when n = k + 1, the sun1 on the LHS starts \Vith (k + 1) and finish es \Vith 2(k + 1). LHS =(k+ 1) + (k+2} + ... +2k+ (2k+ 1) +(2k + 2) Now you h ave a problen1! At this poin t you need to u se the substitution of line [a], but you are 1nissin g the first tern1 ( the k). In goin g from S(k) to S(k + 1) you have lost the first term bu t gain ed two extra tern1s on the LBS.The solution is to break up the extra two terms (2k + 1) + (2k + 2) as a k tern1 and a (3k + 3) term. LHS = l k + (k+ 1) +(k+2) + ... + 2k l+(3k+3} using [a]:

= 3k(k + 1) + 3 (k + l) 2

= 3(k + l)[k + 2] 2 = RHS

taking the con1n1on factor of 3 (k + l) 2

Step 3 Conclu sion S(k + 1) is true if S(k) is true

(Step 2)

S(l) is true

(Step 1)

:. by induction, S(n) is tru e for all integers n > 1.

Chapter 9

Proof by mathematical induction

189

YEAR12 EXERCISE 9.1

MATHEMATICAL INDUCTION INVOLVING SERIES

. n 2 (n+ 1) 2 1 If S(n) 1s the staten1ent that n + 2n + 3n + ... + n = , then S(5) represents the statement: 2 5X6 25X26 B 1 + 2 + 3 + ... + 25 = A 1 + 2 + 3 + ... + 5 = 2 2 25x6 25 6 C 1 + 2 + 3 + ... + 25 = D 5+ 10+ 15+20+25= ; 2

Prove each of the following by in duction for all positive integers n. n(n+ l) 2 n(3n + 1) 4 2 + 5 + 8 + ... + (3n - 1) = 2 11 2 11- 1 a(r -l) 6 a + ar + ar + ... + ar = r- 1

3 1 +2+4+ ... +211 - 1=2"-1

2 1 + 2 + 3 + ... +n=

5 a+ (a+ d) + (a+2d) + ... + (a+ (n - l)d) = ~ (2a+(n-l)d] 2

3

n- 1

7 l+r+r+r + ... +r

1 -r11 = ~- 1- r

n(n + l)(n + 2) 3 n(n + 1)(2n + 7) 9 1 x 3 + 2 x 4 + 3 x 5 + ... + n(n + 2) = 6 3 3 1 1 1 1 n 3 5 10 4 + 104 + 756 + ... + (n + 3n ) = n (n +l) 11 1X2 + 2X3 + 3X4 + ... + n(n+l)=n+l 2 1 1 1 1 n 12 2x3 + 3x4 + 4x5 + ... + (n+l)(n+2) = 2(n+2) 12 22 32 n2 n(n+ l ) 14 1 1 1 1 n 13 lx3 + 3x 5 + 5x7 + . .. + (2n - 1)(2n+ 1) = 2(2n + l) lx3 + 3x5 + 5x7 + ... + (2n- 1)(2n+l)- 2n+ l

8 1 x 2 + 2 x 3 + 3 x 4 + ... + n(n + 1) =

find )· ( 13+23+ 33+ . .. +n3) 3 23 33 3_ n2(n + l)2 H . ence m1 15 (a) 1 + + + ... + n 4 n4 3 3 3 3 (b) Hence show that 1 + 2 + 3 + ... + n = (1 + 2 + 3 + ... + n)2. n➔oo

n(3n + l) 17 1 x l ! + 2 x 2! + 3 x 3! + ... + n x n! = (n + l )! -1 2 18 1 X 2° + 2 X 2 1+ 3 X 22+ ... + n X 211 - 1= 1 + (n - 1) X 2".

16 (n + 1) + (n + 2) + ... + 2n =

19 12+ 32+ 52+ ... +(2n- 1)2=n(4n;-1)

20 12-22+ 32_42+ . .. +(-l )11- 1n2=(-1)"- 1;(n + l)

21 1 X 2 + 4 X 22 + 9 X 23 + ... + n 2 X 211 = (n 2 - 2n + 3) X 2" +I 1 3 7 211 -1 1 22 - + - + - + ... + - 11- =n - 1+- 11 2 4 8 2 2

24 14 + 24 + 34 + .. .+ n 4 =

9.2

-

6

23 2x2 1+3x2 2+ 4x23 + . .. +(n+ l )x2"= nx2 11 +1

n(n + l)(2n + l )(3n 2 +3n-l) 30

(For those who like an algebraic challenge.)

PROVING DIVISIBILITY BY INDUCTION

The following points are essential for understanding a proof by induction where the assertion involves divisibility. • If an integer N is divisible by an integer d, then N = d x q, where q is an integer. For exan1ple, 18 is divisible by 3 because 18 = 3 x 6 (where 6 is an integer). But 19 is not divisible by 3 even

!

though 19 =3 x 61 , because 6 is not an integer. • The integers are closed under addition, subtraction an d 1nultiplication. This means that whenever you add, subtract or m ultiply integers, the ans,ver ,viii always be another integer. This is not true for division: son1e divisions of integers give integer answers, but others do not.

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Example 4 Prove by induction that 4

11 -

1 is divisible by 3 for all integers n > 1.

Solution Let S(n) be the statement that 4" - 1 is divisible by 3 for integer n.

Step 1 Prove that S( 1) is true. 4 1 - 1 = 3, ,vhich is divisible by 3.

(Note: The LHS- RHS style of proof can't be used here. )

: . S(l) is true.

Step 2 Assu1ne S( k ) is true for an integer k > 1. Le. assume that 4k- 1 = 3M

where Mis an integer

4k=3M + l

[a]

Now prove that S(k + 1) is true if S(k) is true. i.e. prove that

1

4k+ -

1 is divisible by 3

4k+ 1 - 1 = 4X4k- l =4(3M + 1) -1

using [a]:

= 12M + 3 = 3(4M + 1)

,vhich is divisible by 3, because it is a multiple of 3 and 4M + 1 is an integer (as integers are closed under 1nultiplication and addition).

Step 3 Conclusion S(k + 1) is true if S(k) is true

(Step 2)

S(l) is true

(Step 1)

:. by induction, S(n ) is true for all integers n > 1.

Example 5 Prove by induction that 3" + 7" is divisible by 10 for all positive odd integers n.

Solution 11

Let S(n) be the statement that 3" + 7 is divisible by 10 for positive odd integer n.

Step 1 Prove that S( 1) is true. 3 1 + 7 1 = 10, which is divisible by 10. :. S( 1) is true.

Step 2 Assu1ne S(k) is true for an odd in teger k > 1. i.e. assun1e that

3k + i = l OM

where Mis an integer

3k = lOM - 7k

[a]

(Note: i could have been made the subject instead of3k.)

Now prove that S(k + 2) is true if S(k) is true. Chapter 9

Proof by mathematical induction

191

YEAR12

(Note the S(k + 2) instead of S(k + 1), as this is \Vorking \Vith odd integer values of n.) i.e. prove that 3k+2+ 7k+2 is divisible by 10 3k+2+ 7k+2= 32 X 3k + 72 X ~ using [a]:

=9(10M-7k) +49x 7k = 90M - 9 X 7k + 49 X 7k =90M +40x7k = 10(9M +4x 7k)

This is divisible by 10, because it is a 1nultiple of 10 and 9M + 4 x 7k is an integer. Step 3 Conclusion

S(k + 2) is true if S(k) is true

(Step 2)

S( l ) is true

(Step 1)

: . by induction, S(n) is true for all odd integers n > 1.

EXERCISE 9.2

PROVING DIVISIBILITY BY INDUCTION

1 If k and Mare integers, \Vhich of the following expressions does not always generate an integer? A 9M+4x7k B 9M-4x7k C 9M+4x7k D 9Mx4x7k Prove the following by induction.

2 5" + 3 is divisible by 4 for all positive integers n.

3 3 211 - 1 is divisible by 8 for all positive in tegers n.

4 3" + 2" is divisible by 5 for all odd integers n > 1.

5 5 + 2(11") is a 1nultiple of3 for all positive integers n.

11

6 (a) Factorise k(k + l)(k + 2) + 3(k + l)(k + 2). (b) Hence prove that n(n + l)(n + 2) is divisible by 3 for all positive integers n. 7 3 311 + 2"+2 is divisible by 5 for all positive integers n. 8 7" - 2" is divisible by 9 for n > 2.

411 9 3 - 1 is divisible by 80 for all positive integers n.

10 5" + 2 x 11" is divisible by 3 for all positive integers n.

11 2311 - 1 is divisible by 7 for n > 1 11

12 6 + 10n - 6 is divisible by 5 for all positive integers n. 13 3 211 + 1 + 2"- 1 is divisible by 7 for all positive integers n. 3 3 2 14 (a) Show that (k + 3) = k + 9k + 27k + 27. (b) Hence prove that the sum of the cubes of three consecutive positive integers is divisible by 3. 15 Prove that the polynon1ial (x - 1)"+2 + x 2" + 1 is divisible by x 2 - x + 1 for all positive integers n. (Note: In step 2, you can't say (x- 1t+2+ ?'+ 1 = (x2-x + l)M where Mis an integer. You n1ust say (x - l)k+2+ x2k+i = (x2 -x + l)M(x), \vhere M(x) is a polynomial, and continue this through

the rest of the proof.) . d"IVISI . "ble by x - 1,:,or all positive . . mtegers . n. (Usexk+I - 1 = xk+l - x k + x k - 1) . 16 Prove th at x 11 - 1 1s

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9.3

WHEN INDUCTION DOESN'T WORK

A single counter-example is enough to disprove a result. Some proofs by 1nathe1natical induction seem to be correct, but turn out to be incorrect. The most likely reason is that you have failed to prove step 1, Le. that the result is true at the beginning (usually for n = 1). This means that the asswned result (n = k) is false, so when you prove the result true for n = k + 1, you are proving it from an incorrect asswnption.

Example 6 Let S(n) be the statement: n 2 - n is an odd integer, for all positive integers n. (a) Sho\v that if S(k) is true, then S(k + 1) is true.

(b) Is S( l ) true?

(c) Is S(n) true for any n?

Solution (a) Let S(k) be that k

2

k is an odd integer. Statement: S(k + 1): (k + 1)2- (k + 1) is an odd integer. (k + 1)2-(k+ 1) =(k + l )(k+ 1-1) =k(k+ 1) 2 =k -k+2k =S(k) + 2k =Odd + Even = Odd Hence the result is true for n = k + 1 if it is true for n = k. 2

(b) S( l ) is that 1

-

-

1 is an odd integer and since this result is 0, then S( l ) is false.

(c) S(n) is never true because when n is even, n

2

-

n is even, and when n is odd, n 2 - n is even.

Example 7 Use 1nathen1atical induction to prove that 2" - 1 is prin1e if n is prin1e.

Solution Using a table of values to explore this for son1e values of n: This suggests that the result \Vorks, although finding a pattern for the priine nwnbers for n is not possible. The next priine values for n are 11, 13, 17, 19, 23, ... Any non-prin1e value for 2" - 1 is enough to say the original staten1ent is false. Testing these values finds for n = 11: 211 - 1 =2047. As 2047 =23 x 89, this is not priine, so the initial assun1ption is false. The result can not be proved by mathe1natical induction.

EXERCISE 9.3

WHEN INDUCTION DOESN'T WORK

1 Let S(n ) be the state1nent: n 2 + n is an odd integer. (a) Sho\v that if S(k ) is true, then (k + 1) is true. (b) Is S(l) true? (c) Is S(n) true for any n? (d) If the staten1ent is not true, what change do you need to 1nake to 1nake it true? Prove your new staten1ent. . 2 Given 12 + 42 + 72 + ... + (3n - 2) 2 =

n( 6n 2 -

3n - 1) .

2 (a) Sho\v that if S(k ) is true then S(k + 1) is true.

(b) Is S(l) true?

(c) Is S(n) true for any n?

Chapter 9

Proof by mathematical induction

193

YEAR12

3 It is stated that n 2 - n + 41 is prime for n > 1. (a) Is this state1n ent true for n = 1? (b) Is this statement true for n = 2? (c) Is this staten1ent true for n = 5? (d) Is it possible to find a value of n for which this expression does not give a pri1ne nun1ber? Justify your ai1s\ver. (e) Is the given staten1ent true or false? 2

4 Let S(n) be the statement: n - n is an odd integer. (a) Show that if S(k) is true then S(k + 1) is true. (b) Is S(l) true? (c) Is S(n) true for any n?

CHAPTER REVIEW 9 1 Prove by induction that -2 + 2 + 6 + ... + (4n - 6) = 2n(n - 2) for all integers n > 1. 2 Prove by induction that 4"+1 + 52" - 1 is divisible by 21 for all positive integers n. 3 Prove by induction that 7" - 1 is divisible by 3 for all positive integers n. 4 Prove by induction that 6 + 24 + 60 + ... + n(n + l)(n + 2) = n(n + l)(n: 2 )(n + 3) for all positive integers n. 1 5 Prove by induction that l + 1 + 5 + ... + n + 2 = 1 for all integers n > 1. 4 6 96 n(n + 1)2" (n + 1)2" 6 Prove by induction that 5" + 12n - 1 is divisible by 16 for all positive integers n.

7 Prove by induction that 7" + 6" is divisible by 13 for all odd positive integers n. 2

8 Prove by induction that 2 x 1! + 5 x 2! + 10 x 3! + .. . + (n + l)n!

=n(n + l)! for all positive integers n.

1 1 1 . d . th 1 9 Prove byin uct1on at 4 + 28 + 70 + ... + ( 3n- 2 )( 3n + l ) - 3 nn+ 1 for all integers n > 1.

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New Senior Mathematics Extension 1 for Years 11 & 12

10.1

INTRODUCTION TO VECTORS

Vectors give you a way to 1nove around the Cartesian plane by moving a certain distance in a particular direction, by considering the direction to be split into horizontal and vertical parts. For example, if you \Valk 40 m east, then 30 m north, how far are you from your starting point, and in what direction? This concept has many applications in Physics and is expanded to three diinensions in the Mathematics Extension 2 course. Vectors are used in kinen1atics and dynalnics (,notion), hydrodynan1ics (n1otion of a fluid), heat flo\v iI1 solids, electron1agnetism and 1nany other real-life scientific applications.

Scalar quantities and vector quantities A scalar quantity is one that is completely specified by its magnitude (size) and, if appropriate, its unit of 1neasurement. Exan1ples of scalar qual1tities include n1ass, length, speed, temperature and tiine. Note that ten1perature is a scalar quantity, even though it ,nay have a negative value (dependmg on the n1easurement scale bemg used). A negative ten1perature value indicates its relative magnitude and does not in1ply that te1nperature is oriented in any 'direction'. A vector quantity is one that is con1pletely described by its magnitude and direction (and, if appropriate, its unit of 1neasurement). Exalnples of vector quantities include: • displacement (change in position of an object) • velocity (tim e rate of chal1ge of displace1nent or position) • acceleration ( tin1e rate of change of velocity) • force (a push or a pull that can affect the ,notion of an object) • weight (the gravitational force acting on an object- this is not the sa1ne as the 1nass of all object, which is scalar) • n1on1entu1n (the product of an object's mass and velocity). Consider the vector quantity of displace1nent. In order to fully specify the displacement of all object, it is necessary to ki10\v how far it is fro1n its starting pomt (magnitude) and the direction in which it has n1oved.

Example 1 Sarall walks from point O to point C, via points A and B. (a) What distance has she walked? (b) What is her position relative to her starting point?

N 200 m c .-----------. B

80m

80m

+

Solution (a) Sarall walks a total distan ce of 480 m. (b) Sarall is 80 m directly north of her starting point at 0. This is her displacen1ent fro1n 0.

0

Chapter 10

200 m

A

Vectors in two dimensions

195

YEAR12

Vectors and vector notation A vector quantity can be represented by a directed straight line segn1ent with an arro\vhead (on the line seg1nent) indicating its direction and the length of the line indicating its magnitude. The beginning of the vector is called its tail and the end is called its head.

B

head

a

a

The vect~ with a tail at point A and head at point B is tail A denoted AB. Alternatively, it is also co1n1non to use a bold lo\ver-case letter to denote a vector, e.g. a, or an italic lower-case letter with a tilde underneath, e.g. '!· You will find it easiest to use AB or a in handwritten text.

-

Magnitude of a vector The n1agnitude of a vector is a measure of its size or length and is a scalar quantity. A

N

I

The 1nagnitude of vector AB is written AB~the 1nagnitude of vector!! is written I':! I, and the 1nagnitude of vector a is \Vritten Ia 1·

10km

4g• - - - - I " - --

The diagran1 at right shows a vector OA of magnitude 10 km in a direction N48°E.

0

E

Vector algebra Equality of vectors Two vectors are equal if and only if they have the same magnitude and the saine direction, regardless of their positions. In the exa1nple below, !! = ~ = £.

Negative of a vector B

B

The negative of vector AB= !! is the vector -AB = BA = -g. -a

a

-g has the san1e n1agnitude as!! but \Vith opposite direction. BA is the vector drawn from B to A, \Vhereas AB is the vector drawn from A to B. Therefore, BA = -AB and

-

the direction is reversed.

-

-

A

The zero vector The zero vector, denoted Q, is a vector of zero n1agnitude. Its direction cannot be defined. Adding a vector to its negative will produce the zero vector.

g +(-g) = Q

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New Senior Mathematics Extension 1 for Years 11 & 12

A

YEAR 12

Scalar multiplication of vectors Multiplying a vector by a scalar (a nu1nber) k, where k E R, the set of real nun1bers, k k tim es the magnitude and parallel to the original vector.

* 0, results in a vector with

• If k > 0, then kg has the same direction as g, but has k tin1es the n1agnitude. • If k = 0, then kg = Q. • If k < 0, then k'! is in the opposite direction to'! and has k times the magnitude.

-

la

Addition of vectors The triangle rule for the addition of vectors To add hvo vectors g and ~' the vectors are placed with the head of g at the tail of~The resultant vector or vector sum is a vector joining the tail of g to the head of~- This vector addition fonns a triangle, hence the triangle rule for vector addition, sometin1es called the head-to-tail rule.

Also, for L1ABC , AB + BC= AC or g + ~ = ~-

-a

b

-

-a

-b a+b

B

-a

-b

A

C

C

The parallelogram rule for the addition of vectors If hvo vectors to be added have the san1e initial point, they can be added by dra,ving a parallelogram rather than 1noving one vector to the head of the other vector. The diagonal of the parallelogra1n, drawn fron1 the initial point to the final point, represents the vector sun1. This works because adding~ tog is the same as adding g to~The diagram sho,vs that '! + ~ = ~ + '! = ~ and this rule is kno,vn as the parallelogram rule for addition of vectors. The parallelogram rule for addition of vectors illustrates that vector addition is con1n1utative.

A

For parallelogram OABC, you can also write OA +AB= OB and OC +CB= OB.

MAKING CONNECTIONS

C

Addition of vectors Explore the parallelogram rule for the addition of vectors.

Chapter 10 Vectors in two dimensions

197

YEAR12

Example 2 Draw diagrains to show: (b)

(a) (~ + ~)+£

a +(b+c)

- - -

Solution (a)

(b)

(a+ b) + c

a+ (b + c) a+b

-a

a

-

(a+ b) + c =a+ (b + c) =a+ b + c

Vector addition is also associative. If three vectors~,~ and£ are added, then ~ + (~ + £) = (~ + ~) + £ = ~ + ~ + £· This n1ethod can be extended to any nun1ber of vectors (so1netimes called the polygon rule for vector addition).

Example 3 Draw a diagram to sho,v ~ + ~ + £ + rJ.

Solution

C

--

...

,✓

-a , , , , , , , ---

-

----

-- ~

-d

,,, ..... a+b+c

~

~

a +b+c+d

Subtraction of vectors Subtraction of a vector is defined as addition of its negative. That is, to subtract ~ from~' add-~ to ~' and so~ - ~ = ~ + (-~).

-b

-

-b

MAKING CONNECTIONS

a- b

- - = = = = = = = = = = = =10

Subtraction of vectors Use technology to explore the subtraction and addition of vectors.

If a vector is subtracted fron1 itself, the result is the zero vector: ~ - ~ = Q.

198

-

a

a

New Senior Mathematics Extension 1 for Years 11 & 12

YEAR 12

Example 4 Given the three vectors~,~ and£, as shown, construct the follo,ving: (a) i:!+~

-b

-a

(b) f + i:!

C

(c) ~ -£

Solution (b) To add t\vo vectors£ and~, the vectors are placed ,vith the head the vectors are placed with the of cat the tail of a. head of a at the tail of b. The resultant vector or vector The resultant vector or vector su1n is a vector joining the tail sun1 is a vector joining the of c to the head of a. tail of a to the head of b. -

(a) To add two vectors I:! and ~'

-

b

-

(c) To subtract vector£ fro1n

vector~, the vector-~ is placed ,vith the head of~ at the tail of-c.

-

The resultant vector or vector su1n is a vector joining the tail of b to the head of -c.

-

-

a+ b

-

b

b- c

_,

-

Example 5 ABCD is a parallelogran1. If AB =I:! and BC= ~, express each of the following vectors in tenns of I:! and~-

(al DA

(b)

AC

(c)

b

-

BJ5

Solution

D

(a) As a parallelogran1 has sides

equal in 1nagnitude and direction, the other sides can also be labelled.

(c) Start at B and get to D by

(b) Start at A and get to C by

following a path around the edges. These edges add to give AC.

-

Either b

-

-BD=BA+AD - -

(or BJ5 =BC+ CD) BD=-a+b

AC= AB+ BC =a + b

or

b

following a path around th e edges. These edges add to giveBD.

=b-a

AC= AD + DC D

=b + a

DA has the same 1nagnitude as ~ in the opposite direction.

DA =-b-

Chapter 10

Vectors in two dimensions

199

YEAR12 EXERCISE 10.1

INTRODUCTION TO VECTORS

1 Given three vectors 'i!, ~ and £, as shown, construct the follo\ving: (a) ti!+£

(b) c+b

(c) 'i! - ~

(d) £-':!

-a

-b

B.---~£=----::7C

2 ABCD is a parallelognun. If AB= 'i! and BC = ~' express each of the following vectors in tenns of 'i! and~. (a) CD

(b) AD

(c) CA

(d) DB

-a D A

B

3 ABCD is a quadrilateral. If AB = 'i!, BC = ~ and CD = £, express each of the following vectors in terms of 'i!, ~ and £· (a) AC

(b) AD

(c) DA

(d) DB

-b C

a

C

A-------_j D

4 ABCD is a trapeziu1n with DC parallel to AB and one-and-a-half tiines the length of AB. If AB = 'i! and BC = ~' express each of the following vectors in tenns of a and b.

-

A

-

-

(a) CD

(b) CA

(c) AD

(d) DB

-

B

D

-b C

I

5 If all the short line seg1nents shown are the san1e length, express the following in tern1s of 'i! and~.

(a) OP

(b) OG

(c)

(e) AB

(t)

(g) FQ

DI

OQ

H

F

G

(d) CE

E

(h) DE+EO 0

a

C

A

-

D

6 BC is parallel to OA and twice its length. Express the following in tern1s of a- and -b. (a)

AB

(b) AC

7 Fron1 the diagra1n, find the following in tern1s of 'i!, ~. ~ and g.

(al

w

(b)

VZ

w -b

(c) WZ

a

V

y

8 In MBC, AB = 'i!, BC = ~ and CA = £· Which one of the follo\ving statements is true? A

a+c=b

B

a +b+c=O

C

a+b-c=O

D

b+c=a

-a C

200

New Senior Mathematics Extension 1 for Years 11 & 12

YEAR 12 B

9 In the quadrilateral ABCD, which one of the follo,ving statements is true?

A C

-AB+BC=CD-DA - - -AB-BC=CD-DA - - -

D

AB + BC= -CD-DA

B

C

AB+BC=CD +'BA

A

D

10 In the parallelogran1 ABCD sho,vn, the point of intersection of the diagonals is 0, ,vhere O is the 1nidpoint of both AC and BD. The vector OC is equal to:

A

-1 (a-b) 2 -

B

-

.!. (a+b) 2 -

-

C

-1 a-b 2- -

-1 b-a

D

2- -

B

11 MBC is a triangle with AB= q and AC=£· D and E are the n1idpoints of

F

AB and AC respectively.Fis a point on BC such that FC = 2 X BF.

D

(a) Express the vectors BC and DE in tenns of q and£·

A..::._- - - - : ~ - - ~

(b) Con1pare the vectors BC and DE.

E

C

(c) What geometric property of a triangle does the answer to part (b) de1nonstrate? (d) Express the vectors BF and FC in tenns of g and£· 1 (e) Show that AF= (2g+£).

3

12 ABCD is a parallelogran1 in ,vhich AB=~ and AD=~ and Eis the n1idpoint of BC. B

(a) Express AC in terms of~ and rJ. (b) Express AE in terms of~ and rJ.

E

(c) Express DE in terms of~ and ~2(d) If Fis a point on DE and DF = DE, express DF in terms 3 of band d.

-

A

-

(e) Find AF in tenns of band d, and hence sho,v that Flies on AC.

(f)

10.2

--

C

d

-

Find the ratio AF : FC.

D

VECTORS IN TWO DIMENSIONS

Position vectors on the Cartesian plane

)'

P(x,y )

A position vector is a vector drawn with its tail at 0, the origin. Position vectors are used to represent points by a distance and a direction rather than two nun1bers (the coordinates). You can draw position vectors on the Cartesian plane and represent them with ordered pairs as ,veil as a length and a direction.

0

X

The position vector of any point P(x, y) relative to a fixed origin O on the Cartesian plane is uniquely specified by the vector OP, as sho,vn in the diagran1. That is, the vector OP represents the position vector of point P(x, y) relative to 0.

Chapter 10

Vectors in two dimensions

201

YEAR12 y

For exa1nple, the diagra1n at right shows the position vector of point A(3, 2) relative to 0. The position vector is OA, where A has coordinates (3, 2).

-

3

A(3, 2)

•

2 I

-3 -2 - I 0 -I

X

3

2

I

-2 -3

EXPLORING FURTHER

-=============~10

Position vectors

Use techno logy to explore position vectors on the Cartesian plane.

Example 6 Draw the following position vectors on the Cartesian plane. (a) QA, where A is (-3, 2)

(b) ~, the position vector of (1, -3)

Solution (a) Vector is dra,vn fro1n the initial point at the origin O to the given coordinates.

(b) Vector is drawn fro1n the initial point at the origin O to the given coordinates.

)'

)'

3

OA

3

2

2

I

-3 -2 - I O - I

I

I

2

3

X

- 3 -2 - I O - I

-2

-2

-3

-3

I

2

3

X

a

A position vector can be represented by a coordinate pair (a, b). This represents the vector that is a units fron1 0 in the positive x direction, and b units fron1 0 in the positive y direction. This can be represented as a colun1n vector (: ), [:] or by the coordinates (a, b). Any vector that is equivalent to a translation of a units in the positive x-direction and b units in the positive y -direction can be represented in this ,vay. If A has coordinates (3, 2), then the position vector OA can be represented by the colu1nn vector (

The magnitude of a position vector The 1nagnitude of a position vector OA =(:)can be calculated using Pythagoras' theore1n.

l™I= Ja2 +b2. 202

New Senior Mathematics Extension 1 for Years 11 & 12

!} !) OA = (

YEAR 12

For exan1ple, the n1agnitude of the position vector OA = (~)is

loAI= J3 + 2 2

2

=

.Jo units.

Example 7 Draw the following position vectors on the Cartesian plane.

~

(a) !!• the position vector of ( ).

OB = ( ~ )·

(b)

Solution (a) !! represents a vector which is 1 unit from O in the positive x direction an d 2 units from O in the positive y direction.

Vector is drawn fron1 the origin to the point (1, 2).

(b) OB represents a vector which is - 4 units fro1n 0 in the positive x direction (4 units in the negative x direction) and 3 units from O in the positive y direction.

Vector is drawn from the origin to the point (- 4, 3).

y

y

4

3

4

B(- 4, 3)

•

A( l.2)

2

3

2 I

-4

-3

-2

-I

0

OB I

2

3

4

I

X

_,

- 4 - 3 -2 - 1 0

- I

-2

-2

-3

-3

-4

-4

I

Equal vectors

2

3

4

X

y 4

Equal vectors do not have to start at the san1e point. To be equal, they only need to have the san1e n1agnitude and direction.

3

Although the position vector of the point A(3, 1) starts at the origin, the coordinate pair (3, 1) can be used to represent any vector \Vhose head is three units across and one unit up fron1 its tail. Ho\vever, none of the other vectors can be called position vectors.

C

I~ 2

-4 -3

Vectors \viii also be equal if and only if they are expressed using the sa1ne

3

4

X

: ___.,-2

coordin:tes. For exam ple, the vector [ : ] will only be equal to the

-------

-3 --4

vector [ d] if a= band c =d.

Example 8 Given q = (2, 5), specify an ordered pair for each of the follo\ving.

(a)

2q

(b) -d

Solution (a) As this vector has twice the 1nagnitude, each coordin ate will be 1nultiplied by 2. 2q = (4, 10). (b) As this vector has the sa1ne 1nagnitude but in the opposite direction to negated (n1ultiplied by - 1).

1, each coordinate n1ust also be

The vector -q is represented by the ordered pair (-2, -5). Chapter 10

Vectors in two dimensions

203

YEAR12

Example 9 Given !: =( ~ ), specify a colun1n vector for each of the following.

(b) -3!:

Solution (a) As this vector has half the n1agnitude, each coordinate is nutltiplied by ~.

The vector ~!:is represented by the colun1n vector ( ~ ). 3 (b) As this vector has three tin1es the n1agnitude but in the opposite direction to !c, each coordinate n1ust also be negated (n1ultiplied by -3).

The vector -3!c is represented by the colun1n vector (

EXERCISE 10.2

2

~~ )·

VECTORS IN TWO DIMENSIONS

1 If vector Ii! is represented by the ordered pair (2, -6), specify an ordered pair for each of the following vectors. (a) 3t;!

(b)

~ ':!

(c) -':!

2 If vector~ is represented by the colu1nn vector ( (a) -2b

(b) 5~

(d) 0.4':!

~ ) , specify a colu1nn vector for each of the following vectors.

(c)

!~

(d) - -5 b

4-

3 If£ is the position vector of (6, -3), represent each of the following as an ordered pair. (a) -£

(b) 2~

(c)

!

£

-

(d) 1.5£

4 If~ is the position vector of (6, -3), represent each of the vectors as a colun1n vector. (c) - -1 c

(b) 2£

(a) -~

(d) 1.5£

3-

5 Represent each of the vectors in the plane shown as an ordered pair. (a) a (b) b (c) ~ (d)

4.

(e)

!:

(f)

y JO

-b

f

5

-e - JO

-s

C

0

- JO

204

New Senior Mathematics Extension 1 for Years 11 & 12

5

JOX

YEAR 12 y 10

6 Represent each of the vectors in the plane shown as a colun1n vector. (a) a

(b)

(d) d

(e)

-

b

-e

(c)

C

(f)

f

-

-

5

a

'f.

C

- JO

0

-5

5

-5

- JO

7 If vector ~ is represented by the ordered pair (3, - 5), then which of the following represents the vector -2~? A

B

(- 6, 10)

C

(6,-10)

D

(10, - 6)

(-10, 6)

8 Which of the follo,ving represents the vector from the point (2, 6) to the point (- 1, 8)?

9 Draw the follo,ving vectors on the Cartesian plane. (a) ~' the position vector of (2, 4)

(b) ~' the position vector of (- 4, O)

(c) £, the position vector of (- 3, - 5)

(d) OD, ,vhere D is (- 1, 3)

(e)

OE, ,vhere Eis (2, - 4)

(f)

OF, where F is (O, - 3)

10 Draw the follo,ving vectors on the Cartesian plane. 2

(a) 1, the position vector of ( ~)

(b) ~' the position vector of ( ~ )

(d) OD, where D is ( ~ ) 2

(e)

OE, ,vhere E is ( ~

5

)

(c) £, the position vector of ( _:)

(f)

OF, where Fis

(

=~)

11 (a) Find the colunm vector forn1 of the vector g, which has an initial point (- 3, 4) and finishes at the point (5, 9). (b) Find the colunm vector forn1 of the vector ~, which has an initial point (5, -1) and finishes at the point (- 6, -17). (c) If the vector £ = ( ~~) has an initial point (2, 1), detennine the coordinates of its tenninal point.

(d) If the vector~ = ( ~;) has a tern1inal point ( 12, - 2), detern1ine the coordinates of its initial point.

~

(e) Sketch representations of the vector = ( ~ ) ,vith initial points at: 2

(i)

(O, O)

(ii)

(3, 1)

(iii) (5, - 3)

(iv) (- 5, 5)

(v)

(- 2, - 2).

Chapter 10 Vectors in two dimensions

205

YEAR12

12 The points A, Band C have coordinates (-2, -3), (2, 3) and (8, -1) respectively.

--

(a) Find the vectors AB, BC and AC and express then1 in colu1nn vector fonn. (b) Find

IAN~1oc1and IRI.

(c) Use Pythagoras' theorem to prove that MBC is a right-angled triangle. (d) Find the coordinates of a point D such that ABCD forn1s a square. (e) Find the coordinates of the point of intersection of the diagonals of the square ABCD.

13 OABC is a parallelogra1n \Vith OA =':!and OC =£·Xis the n1idpoint of AB as shown.

cr-------. B

(a) Find the vectors OB and OX in tern1s of':! and£· (b) Find the vector CX in tern1s of g and£·

C

2(c) If Yis a point on CX, such that CY= cx, find CY in 3 tern1s of a and c.

-

X

o...___-..

-

- --.1

-

(d) Find OY and hence show that Y lies on OB.

A

(e) Find the ratio OY: YB.

14 (a) The position vectors of points A and Bare (-3, 6) and (2, 9) A1---~n~1--:;:-.,!..P_~n'..-.-.. B respectively. Find the position vector of the n1idpoint of AB. a (b) Let the position vectors of points A and B be g and~ respectively. Find the position vector of the midpoint of AB. (c) The position vectors of points D and E are (1, 7) and (5, 3) 0 respectively. Find the position vectors of the points of trisection of DE. (d) Let the position vectors of points D and Ebe

-

loci

-

9 In parallelogran1 ABCD, AB = 3f + 4 j and AD = -2i - 5 j. Find: (a)

ci5

(b)

CA

(c)

I5B

(d)

IXBI

(e)

151

10 Given Ii!= -13f + 20j and~= 2f + 15 j, find: (b) the value of x so that the vector Xf! + 4~ is parallel to the x-axis.

11 Given vectors Ii!= Sf+ 4

z,~ = -4f + 7 j and£ = -i - 9l, with 3f! + 2~ - v~ parallel to the y -axis, find the value

of V.

212

New Senior Mathematics Extension 1 for Years 11 & 12

X

YEAR 12

12 Which one of the follo\ving vectors is parallel to the vector [ = 14f-6[?

A

g=28f + 12j

B

-

~= 14f +6j

C

f=-14f-6j

D

q = -28f + 12j

13 Which one of the follo\ving vectors is parallel to the vector~ = Sf - 3j and has a 1nagnitude of 2.,/f,i?

A

B

10f-6j

C

-

D

-Sf + 3j

-

- 10f-6j

-

14 For position vectors OF = 2f - 7 zand OG = --4 i + 5l, what is the value of IPG I?

A

6✓ 3

6✓ 5

B

C

2✓ 10

D 2✓37

15 Find the unit vector parallel to each of the following vectors. (a) 8f + 6j

(b) 3f-6j

-

(c)

-

-2f + 9j

(d)

-f-12j

-

16 For~=3f-9j:

-

(a) find~

(b) find vector fin the direction of~ \Vith a magnitude of 15.

17 Given~= 2f - 3 j and~ = 6£ + j, find f for f = 2~ -3~.

18 Given~= i -3land~= -3f + 2z, find the vector fin the direction of~ \Vhere 1£1= 1~1· 19 What is the unit vector in the direction of g = -2[ + 5 j is? A

.!.(-2i + Si) 7 -

J9(-2i

B

D

C

+ sz)

Jh(-21 +sz)

20 Which of the follo\ving vectors is parallel to ~=-Sf + 5j and has a magnitude of 5?

A

~ (-sf +sz)

s(-si +sz)

B

C

1

(-sf + sz)

21 OABC is a parallelogran1 in \Vhich OA = 3f + 2j and OC = 2f -3 j. (a) Find

AB and CB.

(b) Find the diagonals OB and CA.

Let M, N and P be the midpoints of AC, OB and OA respectively. (c) Find ON and OM.

(d) Find CP and BP.

22 MBC has vertices defined by the position vectors OA = -2i + 3j, OB= 4i - 5l and OC = 6f - 9 j. (a) State the coordinates of the vertices of MBC. (b) Find the vectors AB,

oc and AC.

(c) Find

IAB~lociand IAc1.

23 t.OAB is a triangle in which OA = 6£ and OB= 4 j . The point M with position vector OM = xf + yj is equidistant from 0, A and B.

-

(a) Find the values of x and y.

(b) Find the vectors AM, MB and OM.

(c) Find the values of IAM I, IMB I and 10M 1.

24 OABC is a parallelogran1 in which vectors OA = 2i - 4 j and OC = 3f + 2 j.

-

(a) Find vectors AB and CB.

-

(b) Find the vectors OB and AC, the diagonals of the parallelogra1n.

(c) Find the vectors OP and OQ, \vhere Pis the 1nidpoint of OB and Q is the n1idpoint of AC. What can you

say about the points P and Q? (d) Find the vectors OR and CR, where R is the n1idpoint of AB.

-

-

Chapter 10 Vectors in two dimensions

213

YEAR12

25 OABC is a square in which vectors OA = 3£ -2} and OC = 2£ + 3 j. Mis the 111idpoint of AB and N divides CB internally in the ratio 1 : 2. (a) Find the vectors OB, AC, OM, ON and NB.

(b) Find the length of the diagonals, IOBI and IAcl.

' 26 (a) If!! = 3i - 4 j and ~ = -4i + 3j, find ~ and b.

-

-

-

'

(b) Hence find the exact values of x and y, such that 2x~ + 3y~ = i + j. (c) If£ = 4i - 8 j and 4 = 8i + 4 j, find _.

_.

-

' £and d. -

...

i + j.

(d) Hence find the exact values of v and w, such that 2v~ + 3wq =

27 (a) If!! = 3Pi+ 4 PL, p > 0 and li:! I = 2, find the exact value of p.

(b) Hence find~.

(c) Find the vector~ \Vhich is parallel to~' ifl~I = 10. (e) Hence find f

(d) If£= 7 qi + 24qj, q > 0, and 1£1= 4, find the exact value of q.

Find the vector 4 in the direction of£ where 14 1= 50. (g) Find the vector \vith n1agnitude 10 that is parallel to the vector~ + q.

(f)

10.4

SCALAR PRODUCT OF VECTORS

The scalar product (also called the dot product or the inner product) is a way of multiplying hvo vectors. The result of this 1nultiplication is a scalar quantity (with magnitude but no direction). The scalar product of hvo vectors !! and~ is written as!! • ~ (read as'!! dot~'). There is also another product operation on vectors, the vector product or cross product, which is \Vritten as!! x

~

(read as 'i:! cross~'). The result of this multiplication is a vector quantity. This is beyond the scope of this course.

If 8 is the angle between the positive directions of two vectors ':! and ~' then the scalar product is defined to

-a

-b

be!! • ~= l!!ll ~Icos 8.

0

a

-

For a straight angle, 8= tr, so!! • ~= - l!!ll ~Ias cos 7r= -1.

Example 18 Given li:! I = 5 and l~I = 6, find the scalar product of!! and~, correct to hvo deciinal places, in each of the following.

(a)

(b)

-b

-a b

40" a

140°

-

-

Solution (a)

214

i;! • ~= li:! ll ~lcos8

(b)

I;! • ~= li:! ll ~Icos8

=5x6xcos40°

= 5 x 6 x cos140°

= 22.98

= -22.98

New Senior Mathematics Extension 1 for Years 11 & 12

YEAR 12

Special cases of the scalar product Parallel vectors If~ and~ are parallel vectors in the san1e direction, then~ • ~=

-a

1~1~1cos 0°. b

Ascos0°= 1, this 1neans: i;! • q = li:! llql If':! and~ are parallel vectors but in opposite directions, then: ~ • ~ =

1~11~1cos,r

-a -b

Equal vectors If~ = ~' then 8 = 0 (just as for any parallel vectors), so the scalar product is the 1nagnitude squared: ':! • ':! = li:! ll i:! lcosoo

- - ~ --

-b

= li:! ll i:! I

Perpendicular vectors Two vectors are said to be perpendicular or orthogonal if the angle behveen their directions is a right angle (90°). If':! and qare perpendicular vectors, then:':! • q= li:! llq lcos

1

= l~ll ~lx 0

-a

=0

-b

Also, if i;! • ~ = li:! llqlcos8 = 0, then li:! I = 0 or lql = 0 or 8 = ; .

This leads to an important property of perpendicular vectors: if the scalar product of two non-zero vectors is zero, then the vectors are perpendicular. If~ • ~= 0 for non-zero vectors~ and~' then~ an d~ are perpendicular. An in1portant property of the unit vectors i and j is that they are perpendicular, so that i • j = j • i = 0. Also, using the property of parallel vectors: i

-

•i = j • j

- -

= 1.

Scalar product for vectors in component form In co1nponent fonn, for~= x 1f + y 1j and~ = x 2f + y2 j, the scalar product is:

-

I;! •

-

q= (Xii+ Yi l) • ( X2i + Y2l) = (x1X2)(£ • £) + ( X1Y2)(£ • = X1X2 + Y1Y2 as

f•f = j

l) + (y1x2)(1 • £) + (Y1 Y2 )(1• l)

• j = 1 and

f • j = j • £= 0

- -

Consider hvo vectors, ':!= x 1f + y 1j making an angle 81 with the x-axis and q= x 2 £+ y 1j making an angle 82 with the x-axis. The angle behveen the vectors is e= el - 82. ~ . ~= Iall bl cos e = IalIbl cos ( el - 82) = Iall bl(cos81 cos82 + sin 8 1sin 8 2 ) = Ial cos81lblcos 82 + lalsin81lblsin 8;

= X1X2 + Y1 Y2 Chapter 10

Vectors in two dimensions

215

YEAR12

For':!= x,i, + y,j and~= x 2 f + y 2 j the scalar product of the vectors in con1ponent forn1 is f:! • ~ = x 1x 2 + y1 y 2• This means there are two basic expressions for the scalar product:':! • ~= lt:! II ~Icos0 = X1X2 + Y1Y2•

Geometric interpretation of the scalar product Three vectors forn1 the triangle MOB and the length of each side is the magnitude of the vector forming that side. 2

2

B

2

The cosine rule states that It:!- ~1 = lt:!1 + 1~1 - 211:!ll ~Icos 0.

-b

Using the properties of the dot product, the left-hand side can be ,vritten as:

a- b

2

It:!- ~1 = (':!- ~) • (':! - ~)

8

o ~----;;a- - - -~

=a• a-a • b-b • a+b • b

~

2

Rewriting the cosine rule: 2

l1:!l

-

21:! .

2

2

It:! - ~1 = lt:!1 + 1~1

-

211:!II~ Icos0

~ + 1~1 = l1:!l

-

2 11:!II~ Icose

2

2

2

+ 1~1

-2q • ~ = -211:!ll ~Icos0 f:! •

~ = lt:! ll ~Icos0

Example 19 Find the scalar product f:! • ~, given':!= 2i +SL and~= -3i + 4l-

Solution

q • ~ = (2{ + 5

t) •(-3f + 4t)

Multiply the coefficients of the like components and su1n together: q • ~ = 2 x (-3) + 5 x 4 = 14

Algebraic properties of the scalar product The scalar product has the follo,ving algebraic properties. 1 The commutative la,v: a• b = b • a

This property follows immediately fro1n the definition q • ~= It:! 11 ~1cos 0, since lt:! Il~I = l~ ll t:! I· 2 The associative la,v, including multiplication by a scalar:

(mq) • ~ = m( q • ~).where m is a real nun1ber. This property also follo,vs directly fron1 the definition, although it is necessary to distinguish the cases m < 0 and m > 0. 3 The distributive law:

':! . (~+~) =':! . ~+q• ~ The distributive law can be verified using the geo1netric interpretation of the scalar product and assuming that the projections of~ and ~ on q are both positive.

216

New Senior Mathematics Extension 1 for Years 11 & 12

A

YEAR 12

Fron1 the diagram: q • (~ + £) = lqll ~ + £1cos0 = lql x l5NI as l~ + £lcos0= 15NI Also: q • ~+q • £ = lq l x lc5MI + lql x iMN I = l~l x l5NI 0

2

-a

M

2

N

4 An important result involving the distributive law is (q + ~)(q-~) = lq l - 1~1 :

(~+ ~) •(~-~) = ~- ~- ~- ~+~• ~-~ - ~

Finding the angle between two vectors The scalar product can be used to find the angle between two vectors. Let 0 be the angle behveen the directions of vectors ~ and ~ as shown. q • ~= lq ll ~lcos0

-b 0

•

.. 0 = cos

_1

(a•b)

-a

1; 11~1

Note: It is usual to consider 0 to be the s1naller of the two possible angles between the two vectors. Either way, the sn1aller angle and the larger (reflex) angle will be equivalent for any scalar product, because cos ( 0 ) = cos (2:7r - 0 ).

Example 20 Find the angle in degrees, correct to hvo deciinal places, between vectors ~ = 3f - 2 j and ~ = 4 f + j.

-

-

Solution

Scalar product of the vectors: ~ • ~: q •

~ = ( 3f -

2f )( 4f +

L)

=3x4+(-2)xl

= 10 Magnitudes of the vectors: lq l = J3 2+ (-2) 2 =✓ 13

1~1 = ✓42 + 12 =✓ 17

- 1(

= cos

✓ 13

10 X

✓ 17

)

= 47.73° (correct to two deciinal places) EXPLORING FURTHER

Angle between two vectors Use technology to explore the angles between vectors.

Chapter 10

Vectors in two dimensions

21 7

YEAR12 EXERCISE 10.4

SCALAR PRODUCT OF VECTORS

1 Consider the vectors g, ~, £, ,:J and e as shown.

-b

Find the angle behveen the following pairs of vectors. (a) g and~

(f)

~

and ,:J

.... - - -30°- -

(b) g and£

(c) g and ,:J

(d) g and g

(g) b ande

(h) £ and ,:J

(i)

-

-

-e

d

-

C

(e) ~and £

0)

-cand -e

,:J and g

2 Given li:!I = 8 and l~I = 7, find the scalar product of':! and~ for each of the following, correct to hvo decimal places \Vhere necessary.

(b)

(a)

-a

(c)

-b a -

b

(d)

(e)

b

-

-b (f)

-b -a

3 Find the scalar product g • ~for the following: (a) i:!=3f+2jand~=Sf+3j

(b) i:!=-3f + 4j and~=2f+Sj

(d) g=-4f-3j and~=7f-6j (e) g=-2f-9jand~=-7f-4j Sho\V that the vectors ':! = 3i + 7 l and ~ = 7i - 3zare perpendicular. -

-

(c) i:!=4f-Sjand~=-3f+8j

(f)

g=2f-12jand ~=6f + j

-

,.,

4

-

_,

-

5 Find the vector ,:J that is perpendicular to £ = 4 i - 3j and has a n1agnitude of 10.

6 If the vectors g = 7i - 5j and f = xi - 3j are perpendicular, find the value of x. 7 If i:!=-6i + 2j,find:

(a) ':! • ':!

(b) li:!I

8 For any vector ':!, find the value of each of the follo\ving, in tern1s of jaj where necessary. (b) ~ • ~

(a) g • g

(c) ':! • (-':!)

9 If':! is parallel to ~' express g • ~ in tenns of jaj and jbj. 10 Find the angle, correct to the nearest degree, between each of the following pairs of vectors ':! and ~: (a) i:!=3f+2jand~=3f+Sj

-

-

-

-

-

-

(e) ':! = -Sf-6j and~= -7[-2j

(d) i:!=-2f-4jand~=4f + 3j

-

(c) g=4f-jand ~=3f + 4j

(b) ':!=-3f + 2zand~=Sf+6z

-

(f)

g=8f + 4jand~=-3[+6j

-

11 Which vector is perpendicular to f =-Si + 2 j with magnitude 12?

C

C

-

=

12

✓ 29

(2i- -SJ·) -

D

12 Vectors g = xf - 2 j and ~ = -6i + y j are perpendicular. What are possible values of x and y?

-

-

A

x = 1 and y = 3

B

x = 1 and y = - 3

C

x = - 2 and y = -6

13 The angle behveen the vectors ':! = -5i - 2l and ~ = 3i + l is closest to: A 3° B 40° C 140°

218

New Senior Mathematics Extension 1 for Years 11 & 12

D

x = 2 and y = 6

D

177°

-

-

YEAR 12

14 The points A, Band C have position vectors OA = -2£ - 3!•OB= 2i + 3land OC = 8f -

z.

(a) Find the vectors AB, BC and AC in con1ponent forn1.

(b) Find

1::rn1,1oc1and IA°cl.

(c) Show that MBC is a right-angled triangle.

(d) Find the position vector of a point D such that ABCD fonns a square. (e) Find the vector BD, the other diagonal of the square ABCD.

(f)

C

Show that the diagonals of the square ABCD bisect at right angles.

-

-

-

-

15 ABCD is a kite ,vith AB= AD and CB= CD, as shown. (a) Show that the quadrilateral ABCD ,vhose vertices have position vectors

D

8

OA = 2f + _.j, OB= 2f + 4j, OC = 8f + 7 _,j and OD= Sf+ _,j respectively, is a kite. ,., (b) Find AC and BD and hence show that the diagonals of this kite cross at

right angles. (c) Express BD, BC and DC in tern1s of p, q and!:, where OB= p, QC= q and OD=!:· _. .... _, (d) Explain why BC • BC = DC • DC and use this equation to sho,v

A

thatp • q=q • r.

-

(e) Hence show that the diagonals of a kite cross at right angles.

10.5

PROJECTIONS OF VECTORS

Any vector can be resolved into a sun1 of two vectors that are perpendicular to each other. It is usual to resolve the vector into two perpendicular components ,vith one in a specified direction (e.g. parallel to a given vector) and the other perpendicular to that specified direction.

Scalar projection A scalar projection of a vector can be thought of as like the shadow of a wire when the sun is overhead. (The original vector is the wire.) In the diagrams, vector ~ is the vector projection of vector'.:! in the direction of vector b.

-a

a

-

8 8

C

-b

C

b

The magnitude of£, 1£1, is the scalar projection of vector '.i! onto vector~Using trigonon1etry, cos 0 = By definition, f! •

I!

II"

~ = lt.i! II ~Icos 0, so '.:! • ~= It.:! II ~ I/!I = 1~11£1

a• b

:. kl=1~1 • b • so le I= a • bas -"- = b - l~I •

•

a• b

The scalar projection of '.i! onto~ is f! • ~.where'.:! • ~= l ~I- •

Chapter 10

Vectors in two dimensions

219

YEAR12

Example 21 Consider the vectors!!= 5£- j and~= 3£ + 4j.

-

-

(b) Find the scalar projection of~ onto !!·

(a) Find the scalar projection of!! onto~.

Solution (b) Find~= ; by first finding l!!I·

(a) Find~= Ii i by first finding l~I·

11 ~=3f+4j

!!=st-t l!!I = .Js2+(-1)2

= Ju,

l~I = .J32+42 a !! = 1; 1=

~ b 1( . ·) ~= 1f 1=5 3!+4l

1

Scalar projection ~ • ~:

Scalar projection !! • ~ :

!! • ~ = (5£ -

(s. ·) Ju, ! - z

,

=5

~ • ~=(3£ +4t) • ,k-(5£-t)

t) •!(3£ + 4 t)

15

4

= Ju,- Ju,

15-4

11

5

- Ju,

= -11 5

_ 1 1✓ 26

26

The scalar proJ· ection of a onto b is !.!..

-

-

-f.[6.

5

The scalar projection of~ onto !! is 11

In general, the scalar projection of!! onto ~ does not equal the scalar projection of~ onto !!·

Vector projection In each of the diagra1ns above, vector £ is the vector projection of vector !! onto~. The vector projection of!! onto~ is a fraction of~' for example, : ~

-a

where m and n are real numbers. Now,£= 1£1~

8

=(!! · ~)~ The vector projection of!! perpendicular to ~ is: !! - £

The vector projection of!! onto ~ is (!! • ~) ~. The vector projection of!! onto ~ can also be expressed as The vector projection of!! perpendicular to~ is!!-(!! •

a• b b • b~.

- -

f )f

a• b The vector projection of!! perpendicular to~ can also be expressed as!!- b• b~.

220

New Senior Mathematics Extension 1 for Years 11 & 12

YEAR12

Example 22 Consider the vectors'!= 2£ -5 j and lz = -2£ + 3j.

-

(a) Find the vector projection of q onto lz.

Solution

,

-

(b) Find the vector projection of q perpendicular to lz.

b

(a) Find lz = l~I by first finding llzl-

(b) Find':!-(':! • ~)~:

19 - (-2; +31·) 13 38 . 57 . = 2 I. - S.J- 1+ - J - 13 - 13 12 . 8 . =-- 1- - 1 13- 13 -

a-(a• b)b = 2i-SJ' - - -

lz = -2£ + 3L: llz l = ✓(-2)2 + 32 = ✓13

, b lz= 1~1 =

1 ( . ·) Jo -2!_+3z '

Scalar projection: q • lz:

= -{; (3f + 21)

':! • ~ = (2f-st) • ffi(-2i +31)

The vector projection of q = 2i - 5l perpendicular

-4 - 15 = ~= + ~=

to b = -2i + 3]. is _ _±_(3; + 21·). 13 -

Jo Jo - 19

- Jo Vector projection: ( t;! • ~)~:

(a• b)b = - - -

ffi x ffi 13

13

(-2i + 31·) -

-19( 2· 3z·) =o-!.+

The vector projection of q = 2f -5 l onto · 3 )IS · · -- 19 ( -2i +3j", ) b =-2 1+

-

-

-

13

-

-

Note: The scalar product of the parallel and perpendicular vector projections should be equal to 0. This can be used to check the projections obtained. EXERCISE 10.5

PROJECTION OF VECTORS

1 For each of the follo\ving pairs of vectors, find the scalar projection of q onto lz. (a) t;!=4f-jandlz=3f+4j

(b) t;!=4f+3jandlz=3£+2j

(c) t;!=8f+3jandlz=-3£ + 8j

(d) q=-3i + 2j andlz=Sf + 6j

(e) q=-2f-3jandlz=4f+Sj

(f)

-

-

-

-

-

-

-

-

-

q = -Sf-6} and lz = - 7f-2j

-

-

2 For each of the follo\ving pairs of vectors, find the vector projections of'! onto lz. (a) q=4f+3jandlz=3f+2j

(b) q=4f-jandlz=3f + 4j

-

-

-

(d) q = -3i + 21 andlz=Si +6j

(c) q=8f+4jandlz=-3f + 6j

-

-

(e) q=-2f-3l andlz=4i +7j

(f)

_,

q = -Sf-61 andlz = -7f-2j

3 For each of the follo\ving pairs of vectors, find the vector projections of g perpendicular to lz. (a) g=4f+3zandlz=3f + 2l

(b) q=4f-Landlz=3f + 4j

(d) I;!= -3£ + 2} and lz = Sf +6}

-

-

(c) g=8f+4Landlz=-3f + 6j

(e) '! =-2£-3} andlz=4i +7}

-

-

(f)

'! = -5£-6} and lz= -7£-2}

-

_,

4 For the follo\ving vectors, find the scalar projection of lz onto g. (a) t;!=4i +3jandlz =3£+2j

-

-

(b) t;!=4f-jandlz=3f+4j

-

_,

Chapter 10

Vectors in two dimensions

221

YEAR12

5 For the following vectors, find the vector projection of~ onto q. (a) g=4f+3jand~=3f+2j

-

(b) g=4f-jand~=3f + 4j

_,

-

_.

6 For g = -2i - 3l and~= -2i + 2 j, the scalar projection of g onto~ is: A

✓ 2 2

B

-✓ 2

-2✓ 13 13

C

2

7 The vector projection of 3i + 2! onto -f + 2! is ; ( -i + 2

D

2✓ 13 13

t} What is the vector projection of 3f + 2!

perpendicular to -f + 2 j ? A

~ (3i +2t)

B

~(-i + 21)

D

C

2f + j

-

8 Consider two vectors g = -i + 7land~= Sf + 4j. (a) Find the scalar projection of q onto~(b) Find the vector projection of q onto~(c) Find the scalar projection of~ onto q.

(d) Find the vector projection of~ onto q.

9 Consider two vectors q = 3£ -4} and~= 2f -2j. (a) Find the scalar projection of g onto~-

(b) Find the vector projection of g onto~-

(c) Find the vector projection of g perpendicular to the direction of~-

(d) Hence, express the vector q = 3£ -4 j in tenns of projections onto an d perpendicular to~= 2f - 2 j.

-

-

10 For any two vectors q and ~' state the 1neaning of each of the following. ~

(a) g • ~

(b) ~ • ~

~

(c)

a• b

11 g, ~ and £ are unit vectors in the Cartesian plane. (a) Show that q = cosaf + sinaj.

-j

-a

(b) Derive similar expressions for ~ and !:· (c) Finda • banda • c.

(d) Hence deduce the co1npound angle forn1ulas

cos(a- {J) = cosacos/3 + sinasin/3 and cos(a + /3) = cosacos/3-sinasin/3.

10.6

VECTORS IN GEOMETRIC PROOFS

Many geon1etry theoren1s can be solved using vector 1nethods.

Background knowledge It is useful to recall the properties of triangles and special quadrilaterals, in particular: • a trapezium is a quadrilateral with one pair of opposite sides parallel • a parallelogram is a quadrilateral with both pairs of opposite sides parallel. It has the properties that both pairs of opposite sides are equal, both pairs of opposite angles are equal and the diagonals bisect each other • a rectangle is a parallelogran1 \Vhere one angle is a right angle. It has all the properties of the parallelograin, as well as that each angle is 90°, the diagonals are equal, and each diagonal divides the rectangle into a pair

222

New Senior Mathematics Extension 1 for Years 11 & 12

YEAR 12

of congruent triangles • a square is a rectangle \Vith a pair of equal adjacent sides. It has all the properties of a rectangle, as \Veil as that all sides are of equal length, the diagonals bisect the angles of the square, the diagonals bisect each other at right angles, and the diagonals divide the square into four congruent right-angled triangles • a rhombus is a parallelogran1 with a pair of adjacent sides equal. It has all the properties of the parallelognun, as \Veil as that all sides are of equal length, the diagonals bisect the angles of the rho1nbus, the diagonals bisect each other at right angles, and the diagonals divide the rhombus into four congruent right-angled triangles • a square can also be defined as a rho1nbus with one angle a right angle • a kite is a quadrilateral \Vith two pairs of adjacent sides equal. The opposite angles between the pairs of non-equal sides are equal.

Vector proof involving sides of a triangle Example 23 Use vector n1ethods to prove that the line seg1nent joining the n1idpoints of two sides of a triangle is parallel to the third side and half its length.

Solution Consider MBC, where P and Qare the n1idpoints of sides AB and BC respectively.

B

Let AB= a and BC= b.

-

-

P'r

Pis the 1nidpoint of AB and Q is the midpoint of BC, so -PB= -1 a and -BQ= -1 b. 22Express vectors required in tenns of g and ~: AC= AB+ BC

=a+b

-----\Q

A L.._ _ _ _ ___,. C

PQ=PB + BQ = .!.(a+b) 2 -

-

:. PQ= .!.AE 2

Thus, PQ is parallel to AC and half the length of AC.

Vector proof involving sides of a right-angled triangle Pythagoras' theore1n is used extensively throughout Mathe1natics, particularly in measuren1ent and trigonon1etry. You have used it throughout this topic on vectors to calculate the 1nagnitude of a vector. Pythagoras' theore1n can be proved using vector 1nethods.

Example 24 Use vector n1ethods to prove Pythagoras' theoren1: in any right-angled triangle, the square on the hypotenuse is equal to the su1n of the squares on the other t\vo sides.

Solution Consider MBC. Let AB= -a and AC= -c.

C

:. BC= BA + AC =-a+c

C

=c-a

-

A

-a

B

Chapter 10 Vectors in two dimensions

223

YEAR12 2 2 2 2 2 By Pythagoras' theoren1: -AC = -AB + _BC-, or 1-AC 1 = 1-AB 1 + 1-BC 1 2 2 Use scalar product of vectors to find IAc l : IAc l = AC • AC =c • c

If g and ~ are perpendicular vectors, then g • ~ = 0. As AB and BC are perpendicular, AB • BC= 0. That is, ~ • (~-~)=(£-~) • ~= 0. No,v: ~• (£-~) = ~ • ~-~ • ~ = 0 so,g • £=g • g If g and ~ are parallel vectors, then g • ~ = lg

II~1-

Now: g • g= lgll g l 2 = lgl

=c • c-c • a-a • c+a • a

--------

= c • c - c • a as a • c = a • a from above

=c • c-a • a as c• a=a • a 2 2 = 1£1 - 1~1 Thus li3c l2 = 1£12-1~12 So: IAC 12 = 1£ 12 2 2 2 = 1~1 +(1£1 - 1~1 ) 2 2 = IAB l + li3cl

EXERCISE 10.6

VECTORS IN GEOMETRIC PROOFS b ~-------- 8

1 Consider the parallelogran1 OABC where OA =~and OC = ~- Express OB and CA in terms of~ and~- Hence show that the diagonals of a parallelogram 1neet at right angles if and only if it is a rho1nbus.

0

-b

C

2 Consider the parallelogran1 ABCD, where AB=~ and AD= 4.. Prove that the 1nidpoints of the sides of a parallelogram join to forn1 a parallelogra1n. A L-- - -~ " ' -- - - '

E

-a

224

New Senior Mathematics Extension 1 for Years 11 & 12

B

YEAR 12

3 Consider the quadrilateral ABCD, as shown. Let AB=~, BC=~, CD=!: and DA= 4-

C

-a

Which one of the follo\ving statements is correct? A a-c=b-d B a+b=c +d C

a+c=b-d

D

a- + c= -b-d _, _. _.

A

d

4 Consider the circle with centre O and radius OA = ~- B and Care points on the circle and BC = b.

-

Which one of the follo\ving staten1ents must be true?

A

c

1

a= - b - 2a•b = b•b

B D

0

Ak---->-----:,!C

1

a=- - b 22a • b = -b • b

B

5 Use vector methods to prove that the 1nidpoint of the hypotenuse of a right-angled triangle is equidistant fron1 all vertices.

A

M

a

-

-b

B 2

2

2

A

6 Use vector methods to prove that, ifj!,1 = 1~1 + 1~1 for MBC, then LACB is a right angle.

C

b

-a

C

7 ABCD is a rectangle. (a) Prove that the diagonals of a rectangle bisect each other.

C

B

-a

A

B

(b) Prove that the diagonals of a rectangle are equal in length.

-b

-b

D

8 BDEF is a parallelogram contained within a triangle ABC, as shown.

F

A

B

Let AF=~' BD =~and F be the midpoint of AB. D

(a) Find the vector AE in terms of~ and~(b) Use vector methods to prove that BD =

~ BC.

C

9 ABC is a triangle with AB=~ and AC = ~- Xis the 1nidpoint of BC as sho\vn. (a) Find BX and AX.

A

(b) Find 2(BX • BX+ AX • AX).

Apollonius' theore1n relates to the length of a n1edian of a triangle to the lengths of its sides. In any triangle, the su1n of th e squares on any hvo sides is equal to hvice the square on half the third side together \vith twice the square on the median \Vhich bisects the third side. 2

2

2

2

(c) Prove this, i.e. prove that lAB l + IA"c l = 2(IA"x l + IID?l

).

Chapter 10

Vectors in two dimensions

225

YEAR12 b A ~ - - --------::,-, 8

10 Consider the parallelogram OABC \Vhere OA = g and OC = ~(a) Find the diagonals OB and AC in terms of g and~-

(b) Find the sun1 of the squares of the lengths of the sides of the parallelogram in tern1s of I;! and ~. (c) Find the sun1 of the squares of the lengths of the diagonals OB 0 b and AC in terms of a and b. - (d) Hence prove that the sum of the squares of the lengths of the diagonals of a parallelogra1n is equal to the sum of the squares of the lengths of the sides.

C

CHAPTER REVIEW 10 1 Four vectors, g, ~' £ and g, are shown in the diagran1. Which one of the

-b

following staten1ents is true?

A

a +c=b+d

B

a+b=c +d

C

a+b+c +d =0

D

b +c=a +d

C

d

C

2 In the parallelogra1n ABCD shown, the point of intersection of the diagonals is 0 . The vector OD is equal to:

A C

-1 (a-b)

B

2 - -1 b-a 2- -

D

.!. (a +b) 2 - -1 (b-a) 2 - -

A

3 If vector I;! is represented by the ordered pair (- 2, 3), then the vector -3(;! is represented by the ordered pair:

A

B

(-6, 9)

(- 6, -9)

C

D

(6, -9)

(6, 9)

4 The vector that runs from the point (- 3, 1) to the point (3, - 2) can be represented by the column vector: 6

A

(:)

B

( ~3 )

C

(~ )

5 Which one of the following vectors is parallel to th e vector f = A

g=24f-16j

-

B

C

~=3f+2j

-

D

( ~ 3)

D

g=-3f-2j

-6i + 4 j?

£=-24f-16j

-

6 Which one of the following vectors is parallel to the vector I;!= -3f + 7 j and has a magnitude of 2Jss? A

-24f +28j -

B

_l;+Z. 1· 2-

C

2-

3i-7j -

D

-6f + l4j -

I I

7 Given position vectors OA = -3f + 4 Land OB = 4i + 3f, \Vhat is the value of AB ? A

✓ 2

B

5✓ 2

z,

8 If g = -4i + 2 j and~ = i - 4 then 2g A

-Sf-21

B

-of +lOj

C

2✓ 5

D

7✓ 2

C

-9f +Sj

D

-10f-4L

2✓ 5

D

20

~ is:

9 What is the magnitude of the vector g = 4 f- 2 j?

A

B

2

2✓ 3

C

10 Which of the following vectors is parallel to the vector 2£ + 3 j and has a n1agnitude of 2Jo? A

226

-4f+6j

-

B

4f-6j

-

C

6f+9j

New Senior Mathematics Extension 1 for Years 11 & 12

,.,

D

-4f-6j _.

YEAR 12

11 What is the unit vector in the direction of':! = -3f - j?

A

}o(-3i-

l)

B

l~(-3i- L)

C

}o(-3i +

l)

D

12 What vector is perpendicular to f = 4£ - j and has 1nagnitude 5?

A

(:!= .k-(i+4L) 8

~= .k-(4i-L) C

£= ffi(i+4L)

13 The vectors ':! = 3i - xj and q= 2i + 5 j are perpendicular. What is the value of x?

A

- -15

8

2

- -6

C

5

-6 5

14 The angle between the vectors':!= 2i + 3j and q= 3i - j is closest to: A 75° 8 g90 C 91°

D

-15

D

105°

D

3 17

2

15 If':! = 2i + 5 j and q= 4i - j, the scalar projection of':! on to q is:

A

3

8

✓ 17

3

C

✓ 15

16 The vector projection of i -3 j onto i + 5 j is tof + 5j? A

- -7 (.I

5 -

8

+ 3]·) -

-4 (5.!. - ]·) 13

-

3 ✓ 13

_1._(; + 5j). What is the vector projection of i -3 j perpendicular 13 C

D

-20 (.!. + 5]·)

13

-

-12 (.!. - 3]·) 5

-

17 ABCD is a trapeziun1 ,vhere DC is parallel to and t\vice the length of AB. Let AB=':! and BC = Therefore, DA is equal to: A a+b 8 -a-b C b-a D a-b -----

-

-

-

18 ABCD is a trapezium with AB parallel to and three tim es the length of DC. If AB = ':! and BC = q, express each of the following vectors in terms of a and b. - (a) CD (b) AC (c) AD

q.

A

,_____ B

(d) DB C

19 Label the following vectors that have been drawn on the Cartesian plane:

y 4

3

• ':! the position vector of (- 3, 2)

2

I

• OB where B is ( ~) I

- 4 - 3 -2

• £ the position vector of (:)

2

3

4

X

-2 -3

• OD where Dis (1, -4)

-4

20 Given':!= -2i - 5 j and q= 3f - j, find the follo,ving vectors in component form. (a) ':! + q

-

(b) 3~ - 2(:!

(c) -2':! - 7~

21 Given ':! = -2i - 5 j and q= 3f - j, calculate the n1agnitude of the following vectors. (a) It:! + ~I

(b) 13~ - 2(:! I

(c) 1-2(:!- 7~1

Chapter 10

Vectors in two dimensions

227

YEAR12

22 Given vectors £ = ( ~) and

q = (-:),find the following vectors in colu1nn vector form. (b) 7q-8~

(a) 4£-34

23 The position vectors of points A and B are OA = 15i - 7 j and OB = -6i - 19j, respectively.

-

(a) Find the vector AB in co1nponent fonn.

(b) Calculate IABI.

24 Find the magnitude of each of the following vectors. (a) q=lOf +24j

(b) ~=-6f-3j

-

-

25 Resolve the follo\ving vectors into con1ponent forn1 xf + yj. Give answers correct to t\vo deciinal places. (a) q has a n1agnitude of 16 units and 1nakes an angle of 48° to the positive x -axis. (b)

~

has a n1agnitude of 24 units and n1akes an angle of 148° to the positive x-axis.

26 Find the exact values of the unknown pronu1nerals in the follo\ving vector equations. (a) (2a-3b)£-2bL =5f-12L (c) ( a

2

-

(b) (2f + 5)£ +(8-7g)f = f(3f-2f)+2g(f + 4L)

9a) f + ( 2b3 + 1) l = 10.!, - 5l (list n1ultiple solutions)

27 Consider the vector q = -9£ - 3 j.

-

(b) Find the vector~ in the direction of q \Vith a magnitude of 5.

(a) Find~-

28 Find the scalar product q • ~, given the follo\ving pairs of vectors. (b) q = 3£ - 7 j and~= 6£ - j

(a) q=-4f + jand~=2£ + 7j

-

-

-

-

29 Calculate the scalar product and hence show that the vectors q = -3f + 5zand~= l Of + 6zare perpendicular. 30 For each of the following pairs of vectors, find the scalar projection of q onto~(a) q=3f-4zand~=6f+3z

(b) q=-5f + 2zand~=f-7z

31 Forq=2f-5jand~=4f+j,find:

-

-

(a) the vector projection of q onto ~

(b) the vector projection of g perpendicular to ~-

32 The points A, Band C have coordinates (2, -5), (5, 9) and (-9, 12) respectively. (a) Find the vectors AB, BC and AC in column vector form.

(b) Find IABI, IBCI and IACI.

(c) Show that MBC is an isosceles triangle. (d) Find the coordinates of a point D such that ABCD forn1s a rho1nbus. (e) Find the coordinates of the pomt of mtersection of the diagonals of the rho1nbus ABCD. 2

33 (a) If q = -4e£ + 2ef, e > 0 and lq l = 40, find the exact value of e.

(b) Hence, find ~-

(c) Find the vector~ that is parallel to~ with I~ I= 10. 2

(d) If£= 4Ji - 3fj, f > 0 and 1£1 = 250, find the exact value off. (f)

2

Find the vect;r q in the direction of~ where 14 1 = 20.

(e) Hence find c. (g) Find b-d.

34 Consider two vectors q = 2£ -5 j and~= -3£ - j. (a)

Find the scalar projection of q in the direction of~-

(b) Fmd the vector projection of q onto~-

(c) Find the vector projection of g perpendicular to the direction of~(d) Hence, express the vector q = 2£ -5l in tenns of projections parallel to and perpendicular to~= -3.!, -

228

New Senior Mathematics Extension 1 for Years 11 & 12

l·

YEAR 12

35 MBC is right-angled with M being the n1idpoint of the hypotenuse AC, as shown. Let AM = a and BM = b.

-

C

-

A

(a) Find AB and BC in tenns of g and~-

(b) Prove that Mis equidistant from the three vertices of MBC.

B

36 OABC is a parallelogram where OA = ~ and OC = £·Mand N are the midpoints of AB and BC respectively. (a) Draw a diagra1n of parallelogran1 OABC, sho,ving the given vectors and midpoints. (b) Find the vectors OM and ON in tenns of g and £ and show the1n on your diagran1. (c) Hence find the vector MN in tern1s of g and f. (d) Find vector AC in terms of g and f and show this on your diagra1n. 2(e) Pis a point on OM such that OP =- OM. Find the vector OP in tern1s of a and c. 3 2(f) Q is a point on ON such that OQ = 0N. Find the vector OQ in terms of g and f.

3

(g) Show that vector MN is parallel to and half the n1agnitude of AC. (h) Find vectors AP, PQ and QC, and hence prove that the diagonal AC is trisected at P and Q.

Chapter 10

Vectors in two dimensions

229

11.1

VOLUMES OF SOLIDS OF REVOLUTION

You have seen that the area of a region bounded by a line y = r, the x-axis and the ordinates x = 0 and x = h can be found by adding up the areas of all the rectangles of \vidth 8x and height r bet\veen x = 0 and x = h, as 8x becomes vanishingly

y r -+-~,~ .--~--

h

s1nall: A= lim

''

' ' ' '

L, f(x)8x .

' '

ox➔O O

I

This area is given by the definite integral A = oh r dx, \Vhich is A=

s:· rdx = [ rx ]~ = rh.

0

8x

,,

X

You should recognise this as the area of a rectangle of sides rand h. Consider what happens when the area bounded by y = r, the x-axis and the ordinates x = 0 and x = h is rotated about the x-axis to form a solid of revolution, as shown in the diagram below to the left. The solid of revolution forn1ed is a cylinder of radius rand height h. The rectangles of side rand width 8x have become circular disks of radius rand thickness 8x. The volu1ne of this disk is given by flV = n(f ( x) )2 8x. Adding all the disks as

)'

r ,11----,-,----..,.-1' \ I \

' '

I I IJ I I

h

I I I I I I

I I

I

8x gets sn1aller gives V

I

' " ' '' ' I

I II I

)

' '' •'•

.." '' " •" • •

' '

h

V

X

ii ,

I

,-

8x➔O

dx.

0

0

' '' ' ""...__ I

= 1C

= lim L, 1C (f (x))2 8x, \Vhich is given by the definite integral

Thus the volume is V =

_ _,.,__

1C

J

hr O

2

dx = 1C [ r 2x ]

When the arc CD of the curve y =f (x) on the interval a < x < b is rotated about the x-axis, the volume of the solid of revolution formed is given by: b

230

= nr 2h, which you should recognise as the

volume of a cylinder of radius rand height h.

lix

V

'' O

f

= 1C • (f(x))

2

dx

or

b

f

)'

y= f(x) -

71\D

C

2

V=n .Y dx

New Senior Mathematics Extension 1 for Years 11 & 12

0

b

X

YEAR 12

Example 1 Calculate the volu1ne of the solid forn1ed when the portion of the line y = 2x between x = 0 an d x = 3 is rotated about the x-axis. What is the naine of the kind of solid formed?

Solution

b

2

Volun1e = n: • y dx

J

Draw a diagrain: )'

3

2

J

= n: (2x) dx

y= 2x

0

=4n:

J:x dx 2

X

3

=4n:(9 - 0)

= 36n: units3 The solid is a right circular cone of base radius 6 and height 3.

Volumes of solids of revolution-formal development Consider a continuous function fin the interval a< x < b. If y the plane section ABDC is rotated about the x-axis then a solid is generated with circular vertical cross-sections, as shown in the diagram on the right. This solid is called a solid of revolution. P(x, y) is a point on the curve y =f(x) an d Q( x + 8x, y + 8y) is a point close to P. The ordinate PM describes a circle of area n:y2 ---+-0 a11d QN describes a circle of area n:(y + 8y)2. The typical lower rectangle PRNM describes a cylinder of volume n:y2 8x and the typical upper rectangle describes a cylin der of volun1e n:(y + 8y)2 8x. If a typical layer PQNM describes a solid of volun1e 8V, then:

B Q A

.t "----------',

a

' '

p

M

8x

R D

-N- - -

X

b

n:y2 8x < 8V < n: (y + 8y)28x b

Thus:

b

I, n:y 2 8x 1, is rotated about the x -axis. Find V, the volume generated. Hence find Jim V. a➔~

21 The area bounded by the parabola y = 2x volu1ne generated.

x2, the y -axis and the line y = 1 is rotated about the x-axis. Find the

22 Find the volu1ne of the solid generated by rotating the region bounded by the parabola y = 1 - x 2 and the lines x = 1, y = 1 about:

(a) the x -axis

(b) the y -axis.

23 Find the volu1ne of the cone fonned by rotating the seg1nent of the line x + 2y =4 that is cut off by the axes about:

(a) the x-axis

(b) the y -axis.

24 Use the trapezoidal rule with five function values to estiinate the volu1ne of the solid fonned by rotating the curve y =

1 , about the x -axis benveen x = -2 and x = 2. l+x·

25 The area under the curve y = e-x between x = 0 and x = 1 is rotated about the x-axis. Find the volume of the solid of revolution.

=e', 0.5 < x < 1.5, is rotated about the x -axis. Find the volu1ne generated when the curve y =e-o.sx, -2 < x < 2, is rotated about the x-axis.

26 Find the volu1ne generated when the curve y 27

238

New Senior Mathematics Extension 1 for Years 11 & 12

YEAR 12

28 Find the volun1e generated when the curve y = ex + e-x between x = - 1 and x = 1 is rotated about the x -axis. 29 (a) Find the area of the region bounded by the curve y = e-x, the coordinate axes and the line x = a, a> 0. (b) Find the limit of this area as a ➔ oo.

(c) Find the volun1e of the solid generated by rotating the region in (a) about the x-axis and find the lin1it of this volu1ne as a ➔ oo.

30 Find the volun1e of the solid generated by rotating about the x -axis the region enclosed by the curve y 2 = ~, the x -axis and the ordinates x = 1 and x = 3.

31 Find the volun1e of the solid generated by rotating about the x -axis the area beneath the curve y = benveen x = 6 and x = 11.

1

.J

x- 2

32 (a) Given a> 1, sketch the curve y = log,x for 1 < x < a. Find the area enclosed by the curve and the lines y=O andx= a.

(b) The region enclosed by the curve y = log,x and the lines x = 0, y = log, a and y = 0 is rotated about the y -axis to forn1 a solid of revolution. Find the volun1e of this solid.

33 Sketch the curve y = : 2 for values of x fron1 x =

1to x =

1. This part of the curve is rotated about the y -axis to

forn1 a solid of revolution. Find its volume. 1 fron1 x = 0 to x = 5. The region enclosed by the curve, the x -axis and the 4+x ordinates x = 0 and x = 5 is rotated about the x -axis. Find the volume of the solid fonned.

34 Sketch the curve y =

.J

35 The region enclosed by the curve y =

.J 2

x-7

and the x -axis between x = 8 and x = 10 is rotated about the

x -axis. Find the volume of the solid fonned.

36 The region enclosed by the curve y =

.Jxx+ 1 and the x -axis between x = 3 and x = 5 is rotated about the x -axis.

Find the volun1e of the solid formed. 2

2

37 (a) Sketch the region bounded by the curves y = 2(x - 1) and y = 1 -x • (b) Calculate the area of the shaded region. 2 2 (c) The region bounded by the y -axis and the curves y = 2(x - 1) and y = 1 - x for x > 0, is rotated about the y -axis. Calculate the volu1ne of the solid of revolution generated. 38 The curve y =

✓2 cos( 1x) n1eets the line y = xat the point A(l, 1), as sho,vn in the diagram.

(a) Find the exact value of the shaded area. (b) The shaded area is rotated about the x -axis. Calculate the volun1e of the

solid of revolution forn1ed. (c) The shaded area is rotated about the y -axis. Write the integral for this volu1ne. (d) By using a combination of exact integration and the trapezoidal rule, as appropriate, calculate the volun1e of the solid in (c).

39 In the diagran1 on the right, the parabola y = 4x - x 2 and the line y = 2x intersect

>' 2

y= x

A(l , I)

I

y= V2cos({ x) 0

I

2

X

at the points (O, 0) and (2, 4). (a) Calculate the area of the region bet\veen the curves. (b) The shaded region between the curves is rotated about the x -axis to form a solid

of revolution. Calculate the exact volun1e of this solid. (c) The shaded region between the curves is rotated about the y -axis to forn1 a solid of revolution. Calculate the exact volun1e of this solid. (0, 0)

(0, 4) X

Chapter 11

Applications of calculus

239

YEAR12

40 (a) Sketch the region bounded by the curve y = 1 + ✓ x and the lines y = 1 and x = 4. (b) Calculate the area of this region. (c) This region is rotated around the x-axis to form a solid. Calculate the volu1ne of this solid. (d) Calculate the volume of the solid fonned if this region is rotated about the y -axis. 2

2

41 (a) On the san1e diagran1 sketch the graphs of x + y = 1 for - 1 < x < 0 and :

2

+ y 2 = 1 for O ~ x < 2.

(b) An egg is modelled by rotating about the x-axis the curves x 2 + y 2 = 1 for -1 < x < 0 and£ + y 2 = 1 4 for O < x < 2 to form a solid of revolution. Find the exact value of the volu1ne of the egg.

42 A bowl is formed by rotating the curve y = Slog, (x - 1) about the y-axis for O< y < 4. Y (a) Calculate the volume of the bowl (capacity), giving your Y =810s, (-x - I) 4 ans,ver correct to one decimal place. (b) This bowl is to be 1noulded out of plastic with vertical sides and a solid base 0.5 units thick. The cross-section of the bo,vl is shown in the diagra1n, right. Calculate the volume of plastic used to make the bo,vl.

11.2

y=81og, (x - l)

3

2

I

-3

_,

3

X

INDEFINITE INTEGRALS AND SUBSTITUTION

Some integrals can only be solved using particular substitutions for the variables. In this Mathematics Extension 1 course, any substitutions needed to find an integral are given. Integration using a substitution can be considered as the converse of the method of differentiating a con1posite function- it's like using the chain rule backwards. The aim of a substitution is to transfonn an integral into one that involves a standard result,

f

e.g. u" du= n ~ u"+' + C. Variable substitution works as follows: 1 Let

.

f

y = f(u)du whereu=g(x) Yu=J(u)

But dy dx

= dy x du du

dx du

= f(u)x dx du y= f(u)x dxdx

f JJ(u) x ~~ dx = Jf (u)du

This 'backwards' fonn of the chain rule is convenient when the substitution of u = g(x) allo,vs a function to be expressed as the product of: and a function of u. For exan1ple: 2

3

• Iff(x) = 2x (x

-

1)4 then you can substitute u = x

3

-

1. As

~~ = 3x2, you can ,vrite 2x2 as ~ x 3x 2, so that

f(x) is written: f(x)= ~(x 3 -1) 4 x(3x 2 ) = 2.u4 du where u = x 3 -1 3

dx

2

• If f (x) = x ../1 + x 2 , you can see that 2x is the derivative of 1 + x , so if you n1ake the substitution u = 1 + x

1

and write x as (2x) , then f (x) =

240

1u½ ~~ where u = 1 + x2 and~~ = 2x.

New Senior Mathematics Extension 1 for Years 11 & 12

2

YEAR 12

• If f (x) =

1 x + , you can see that 2x + 2 is the derivative of x2 + 2x, so if you make the substitution 3 2 (x + 2x)

u = x2 + 2x and write x + 1 as

1(2x + 2), then f ( x) = 1u- ~~3

• If f ( x ) = x../1- x , then you can 1nake the substitution u = 1 -x. As x = 1 - u,

~~ = - 1, so:

f(x)= -x✓l - x x (- 1) l du = -(l - u)u 2 x dx

.:!.) x du dx

( 1

_

-- u 2 - u 2

i)

1 du = ( u 2 - u 2 x dx

du

= f(u) dx Example 10 Find:

r

(a)

I

(b)

Jx✓l + x 2 dx using the substitution u = 1 + x2

3x 2 ( x 3 -1

dx using the substitution u = x3 - 1

· th e subst1tut1on . . u =x ()c J( x + l ) dx using x + 2x 2

3

2

+ 2 x.

Solution 3 du 2 u = x - l , dx = 3x

(a)

I3x (x - 1)4 I 2

3

dx=

u

(b )

4

X

, du u = 1 + x-, dx = 2x

~~dx

= Ju4 du =lus +C 5

= ~(x 3 - 1)5 +C

(c)

2 du u = x + 2x, dx = 2x + 2

J

(x

3 x+l 3 dx= 1 J (2x+2 )(x 2 +2x)- dx 2 + 2x) = lJ u-3 xdudx 2 dx 3 =1 J u- du =1 x ( 12)u-2 +C =

- 1

2 +C 4( x + 2x) 2

A quick way to check your answer is to differentiate it to see that it gives the integrand.

Chapter 11

Applications of calculus

241

YEAR12

Example 11 Find:

(a)

f

(c)

f(3x - 5 )4 dx using the substitution u = 3x - 5.

x .J~l---x dx using the substitution u = 1 - x

f✓l + t

(b)

t

dt using the substitution u = 1 + t

Solution du (a) u = l -x,dx= - 1,x = l -u

fx✓l - xdx= f(1- u)u½x:dx = -f(ut - ut ) c- 1)dx (note that = - f( u½ - u½) du

du= ( -l)dx)

1 2 .2.) 2 = - ( 3u2- 5u2 + C

.2.

2

1

2

= - (l - x) 2 - - (l - x) 2 +C 5 3 du (b) u = 1 + t, dt = 1, t = u - 1

(c)

f x = f(u½ - u-½

t dt = u - 1 x du f ✓l + t ✓ u dt

du u = 3x - 5, dx = 3 f (3x - 5)4 dx = i f 3(3x - 5)4 dx

dt

= lfu4x dudx 3 dx

) 0, of the abscissa OS at the point P? The tenn abscissa is the fonnal name for the x-coordinate of point P. In other words, Bernoulli is asking for the curves \Vith ST = aOS. T Bernoulli's question can be rewritten using a more modern notation.

'f = f(x)g(y) x

)'

)' =

f(x)

()

It follo\vs fron1 the right-angled t.SPT that ST = SP . tan 0 However, TP is a tangent line to the curve y = j(x) at point P, given by tan 0 =

0

s

X

N

Z.

SP is the ordinate (they-value) of point P. Therefore: ST =

t:; t) 0 = (

ST = aOS :. y = ax

(Z)

Solving for the derivative in this equation gives ddy X

= L. ax

In so1ne \vays, this problen1 is slightly n1ore co1nplicated than the equations solved so far because it has the forn1

t

= f (x )g (y ). Both the independent and the dependent variables appear as separate factors in the derivative

Z.

In this section you \vill learn how to answer Bernoulli's question. You will learn how to solve any first-order differential equation of the forn1

Z=

f (x )g( y ). The key to solving these equations is the separation of the t\vo

variables onto either side of the equality. Chapter 12

Differential equations

287

YEAR12

Z Z

= f (x )g(y ), transpose the equation so that all tenns involving the dependent variable

In other words, to solve

are on the left side of the equality and all tern1s involving the independent variable are on the right side. This would give gty)

= f(x).

In the table belo\v, the original equation is in the left-hand colu1nn \vhile the equation \Vith variables separated is in the right-hand colu1nn. Original differential equation

Variables

Rewritten with variables separated

cosx+ y Z =0

xandy

yZ=-cosx

.

dw -COSV = 0 dv

vandw

0z'=z + l

0andz

Sll1V -

d~v - =cotv dv 1 dz = -1 z+ l d0 0

Finding the general solution of a first-order differential equation by the method of separation of variables Example 19 Find the general solution of the differential equation

z

= x(y-1).

Solution The dependent an d independent variables are separated onto either side of the equality, with the dependent on 1 the left and the independent variable on the right: ddy = x. y- 1 X Both sides of the equality are integrated \Vith respect to the independent variable, using the change of variable

z

dx = dy on the LHS:

Jy-1 dxdy dx= f xdx 1

f y~ldy= f xdx

I

I

log, y - 1 =

~x2+c

Represent the dependent variable as an explicit function of the independent variable.

\ \ \ , ~f-,,., - I 1I 1I

Exponentiating both sides: l x 2 +c

\ \ "-i T.. ., , / ''~, ---///

IY- l l =e2

=

-I x 2 e' e 2 I 2

Removing the absolute value: y = 1+ Ae2x

,

\Vhere A= +e 0 experiencing exponential decay, the half-life, t1 is the tiine it takes the value of y to h alve. The fonnula for the half-life is 2 1 (t2 -t1) 1 t = - log, 2 = ( ) log, 2. "i" r I y(t1) og, y(t2)

)'

Half- life I~

Yo

'

yof2 yof4 yofS 0

y =Yoe-11 r

2r

3r

t

First-order decay reactions In a si1nple first-order decay reaction, the rate of decrease \Vith the value of the concentration [A] of that reactant.

d~: ] of the concentration of a reactant A varies directly

That is, the concentration [A] of the reactant A is modelled by the differential equation relative decay rate r is usually called the reaction rate.

d~:] = -r( A], where the

Exa1nples of so1ne first-order decay reactions are included in the table belo\v. Reaction rater (s- 1)

Reactant

Half-life t 1 2

23su 92

4.87 X 10- 18

4.51 x 10 years

9

1: c

3.83 X 10- 12

5.73 x 10 years

32 p 15

5.61 X 10- 7

14.3 days

3

Newton's law of cooling Ne\vton's law of cooling states that the time rate of decrease in the te1nperature T of an object varies directly with the difference between the object's ten1perature and the a1nbient temperature ( the ten1perature of the surrounding 1nediu1n, T,,).

dT

Therefore: dt = -r(T - T. ),

r > 0,

T (O) = T0

If T0 > T., the body's mitial te1nperature is higher than the te1nperature of its surrounding, so the body is cooling. The model predicts that the body will cool quickly at first, reducing the excess of its ten1perature to that of the surrounding environment. This cooling reduces the excess ten1perature of the body, thus continually slowing its rate of cooling. However, if T 0 < T., then the body's initial temperature is lower than th e temperature of its surroundings, so the body is wanning. Again, this change in te1nperature reduces the te1nperature difference \vith the surroundmg 1nediu1n, thus slowing the rate of warn1ing.

Chapter 12

Differential equations

295

YEAR12

Example 23 Newton's la,v of cooling can be used to model the ten1perature of a cup of coffee cooling on a kitchen bench. If the ten1perature of the kitchen is a constant 20°C and the initial ten1perature of the coffee is 95°C, after 20 minutes the coffee will have cooled to 65°C. (a) Find the temperature of the coffee after an additional 20 1ninutes. (b) Sketch a graph of the ten1perature of the coffee for the first 2 hours.

Solution (a) This problem models the process of cooling, so Newton's la,v of cooling is: ~T = -r(T- 20), r > 0, no)= 95. r

The two variables are separated: (T ~ ~; = -r 20 Both sides of the equation are integrate~ ,vith respect to t and a change of variable is applied on the LHS of the equation: f (T ~ 20 ) ~; dt =

-frdt

f(T ~ 20) dT =- f r dt log, IT- 201=-rt+ c Both sides of the equation are exponentiated and solved for the dependent variable: T - 20 = Ae-, ', where A = +e'. The initial condition is substituted and the constant of integration is determined: T(O) = 95: 95 - 20 = A

:. T=20 +75e -''

t = 20, T = 65, is substituted to detennine the constant of proportionality: log, ( ~~) = -20r r = 1

i

Olog, ( ~)

los,(1) 3 The constant of proportionality is substituted in the general solution: T = 20 + 75e 20 - -

t

No,v: e

_.!._log,(1) 20

3

.!._log,(1)

= e 20

s =e

log,(1)20

s

=

( 3 );O

5

t

Simplify:T(t) = 20+75( ~)

20 2

After an additional 20 minutes, t = 40: T ( 40) = 20 + 75( ~) = 20 + 27 = 47°C The ten1perature of the coffee is 47°C. (b) Using the 1nodel fro1n part (a) to plot the changing value of T:

n·c>

95

20 - - - - - - - - - - 0

306090 t(min)

296

New Senior Mathematics Extension 1 for Years 11 & 12

YEAR 12

Models of modified growth and decay The uninhibited growth model \vith its constant relative gro\vth rate r = .!_ ddy does not take into account the y t inherent lin1itations on the gro\vth of a population. In practice, n1ost populations have a size (the carrying capacity C) beyond which their enviro111nent can no longer sustain the1n. To account for the finite carrying capacity inherent in 1nost systen1s, the Belgian 1nathe1natician Pierre Fran 0 e

3,,

C

log, .!_ = log, 10- 2t, t > 0 y

D

log,(y) = log,( 10) + t, t > 0

8 A can of soft drink at a temperature of 24°C is placed in a freezer. The temperature inside the freezer is maintained at -4°C. When the can has been in the freezer fort minutes, the temperature of the can is T'C. The rate of decrease in the ten1perature of the can is proportional to the excess of its te1nperature over the temperature inside the freezer. If k is a positive constant, the temperature of the can is modelled by the solution of: dT dT A dt =-k(T-24); t=O,T=-4 8 dt=-k(T-24); t=O,T=-4 C

dT = -k(T + 4); dt

t = 0, T = 24

D

dT = -k(T - 4); dt

t = 0, T = 24

9 The rate at ,vhich a ru1nour spreads throughout a population of 1000 students is proportional to the product of the nun1ber N of students ,vho kno,v the run1our and the nu1nber of students ,vho haven't yet heard the ru1nour after t hours. If hvo students decide to start a ru1nour, the model that best decribes the spread of the ru1nour t hours later is: dN A dN =k(lOOO-N) N(O)=O dt = k(N - 2)(1000- N), N(O) = 2 8 dt 1000 ' dN C dt = kN( l OOO- N), N(O) = 0 = kN(lOOO- N), N(O) = 2 D

ft

10 A chemical dissolves in a pool at a rate equal to 10% of the a1nount of undissolved chen1ical. Initially the ainount of undissolved chen1ical is 5 kg and after t hours x kilogra1ns has dissolved. The differential equation that 1nodels this process is: x-5 dx x dx 5-x -dx =5- -x C D A dt = 10 B dt = 10 dt 10 10 11 The rate of increase in the nu1nber of bacteria in a laboratory is directly proportional to the nun1ber present. If the nu1nber of bacteria triples every 2hours, after how many hours will the number of bacteria be quadruple its initial value?

A

2log,

!

8

2log, 4 log, 3

D

log, 3 log, 2

12 In a simple ,node! for the body mass m of an adult, the rate of change of body 1nass in kg/day varies directly with the difference benveen the total energy (food) intake C per day and the total energy expended per day. The energy expenditure per day depends upon the metabolic rate of the individual, but 165 kJ/kg of body n1ass is a realistic average value. Assun1ing t is n1easured in days, the differential equation 1nodelling the rate of change of body 1nass in kg/day could be: dC 8 dt = k(l65m-C), k>O A ~7=k(l65-C), k>O C

dm dt =k(C- 165m),

k>O

D

dC dt =k(C- 165m),

k>O

Chapter 12

Differential equations

301

YEAR12

13 A population P(t) of an ani1nal satisfies';/;= / P( 2- ~ ), P(O) = 100 \Vith t n1easured in years. What is P(t) 5 0 0 ?Yi . th 1 _ 1 (1 1 ) as t ➔ =. ou are given at P(lOOO-P) - 1000 p - 1000-P .

A

25

B

250

C

D

500

1000

14 The growth rate of a tree varies jointly with the product of the current height h and the difference between the current height and the 1naxin1u1n height H. The differential equation modelling this growth could be: A

-dH = r(h dt

C

-dh =rh(h-H )

H)

dt

' '

r >0

B

-dH =rh(H dt

r>O

D

dh - =rh(H-h)

h)

dt

'

'

r>0 r>O

15 At any tin1e t > 0 (in days), the rate of gro\vth in the number of bacteria in a laboratory is directly proportional to the number N currently present. The initial population of bacteria is 1000. (a) Asstuning this gro\vth rate continues indefinitely, \Vrite a differential equation to model the nun1ber of bacteria present in the dish after t > 0 days. The initial population of 1000 bacteria triples during the first 2 days. t

(b) Hence, sho\v that N(t) =ax 3b for a suitable choice of the positive integers a and b. (c) By what factor will the population have increased in the first 4 days? (d) How n1uch time will it take for the population to gro\v to 10 tiines its initial value? Express your answer correct to the nearest hour.

Challenging 16 Ahnost all carbon in the \Vorld is carbon- 12, \Vhich is the 1nost common stable 'isotope' (nuclear form) of carbon. In the late 1940s the A1nerican scientist Willard Libby studied carbon-14, which is not stable: it radioactively decays according to the reaction 14 C ➔ 14 N + e- + v,, in \Vhich a neutron spontaneously transforn1s into a proton (thus changing the ato1n fron1 carbon C to nitrogen N) as it emits an electron and an antineutrino. In the upper attnosphere, carbon- 12 son1etiines transforn1s back into carbon-14 due to interactions with cosmic rays, so the proportion of both isotopes in the atmosphere stays relatively constant. But whenever carbon is absorbed by plants to becon1e part of living organis1ns in the \Vorld the carbon- 12 is n1ostly shielded fron1 transfonning into carbon- 14. This 1neans that when an organisn1 dies, its concentration 12

14

of carbon- 12 { [ C ] ) ren1ains relatively constant, but its concentration of carbon- 14 {[ C ]) radioactively

d [ 14 c ] decaysat therate dt = -r[ 14 C], r=l.2097xl0--4 years- 1• (a) Find the half-life

(t½) of carbon-14, correct to the nearest year.

The ratio of carbon- 12 to carbon-14 remains relatively constant in living organis1ns, roughly

[14c]

R = [ 12C] "'1.3 x 10- 12, but this ratio changes after the organisn1 dies (because it stops absorbing new carbon-14 atoms from the at1nosphere, while any carbon-14 present is still decaying). Consequently, you can detennine the length of tiine since an organis1n's death by measuring how much this ratio has changed. (b) Half the original carbon- 14 has radioactively decayed. How 1nany years ago did the tree die? [ 14c]

(c) Find a differential equation for the ratio R = [ 12C]. (Hint: [12C] can be considered a constant.)

302

New Senior Mathematics Extension 1 for Years 11 & 12

YEAR 12 t

(d) Hence find a formula for the ratio R after t years in the form R(t) = nun1ber a and integer n.

a(.!.);; for a suitable choice of real 2

(e) The skeleton of an extinct n1ega-marsupial is found to have a carbon ratio years ago, to the nearest year, did the anin1al die?

~ ::c~ = 0.9 x 10""". How many C

17 Bubonic plague (known as the Black Death) ravaged Europe most severely between the years 1347 and 1351. It is estiinated that the plague killed n1ore than one in three people living in Europe at that tiine. The disease was survivable only in about 5% of cases. Epide1niologists com1nonly use S to represent the fraction of a population that have survived an epide1nic disease, t days after its arrival. The change in this fraction over tiine can be modelled by the differential equation: ~; = -r(S-I), S(O) = 1, ,vhere I is the fraction of the population that ultimately recover (and hence survive). (a) Find a fonnula for the survivability fraction Sas a function of the tin1e t days. 2 d S (b) Show that - 2 = r 2 ( S - I) dt

After l n1onth, the survivability fraction S approaches a steady state value of 0.05. (c) Find I. (d) Find the value of r if after 14 days, only 6% of the population has survived. (e) Find the tin1e when the death rate reaches its n1aximun1 value and state this death rate. (f) Plot the survivability fraction over the first 2 ,veeks.

18 Two types of bacteria, type A and type B, coexist in a biological system. Assu1ne that each population grows exponentially. The proportion of the total bacteria population belonging to type A is given by A(t) p(t) = A(t) + B(t). (a) Express the growth rate ddp in tern1s of A(t), B(t), A'(t) and B'(t). t d (b) Given that A'(t) = rAA(t), rA> 0 and B'(t) = r ii3(t), r 8 > 0, express the gro,vth rate in terms r 11 , r 8, A and B.

ft

(c) Hence write the growth rate dp in tern1s rA' r 8 and p. dt

t

and r,1 - r8 =

hour- 1 , and given that p (l ~ p) =

2

~ + 1 P' find p(lO). 0 1 0 19 Toxins that enter an ecosystem are generally observed to be 1nore concentrated per unit of biomass as you n1ove up the food chain. This phenon1enon is known as bioaccun1ulation. (d) If p (O) = /

Following one gro,vth rate model, the length L (111) of a species of tuna at age t (years) is given by ddL = .!.(3-L), L(O) =0, 0 < t < 20. t 5 (a) Find L(t). 3 The weight W (kg) of a tuna is related to its length L by the equation W(t) = 16L • Let H (t) (mg) be the accu1nulated mass of mercury (in n1illigrams) in a tuna after t years. The rate at which mercury is added to the tissue of the tuna is ddH = 1 W, H(O) = 0. 100 (b) Find H(t). t (c) Find a fonnula for the concentration C of n1ercury in units of milligra1ns of mercury per kilograin of tuna, for a tuna of age t years. (d) Assun1ing the tuna has a lifespan of 20 years, plot the 1nercury concentration in the tuna over its lifetime, 0 < t < 20.

Chapter 12

Differential equations

303

YEAR12

20 The British actuary and 1nathe111atician Benja1nin Gompertz (1779- 1865) proposed the following growth rate n1odel: dd~ = rW log, ( ~)

[1]

Nun1erous experimental studies have den1onstrated that the growth rate dd~ of tu1nours is modelled by dW = dt

1 W ( 10-log, W ) ,

W ( 0) = e

[2] 20 \Vhere Wis the weight of the tun1our (in n1illigrams) and tis the tin1e in days. (a) Given that W grows according to the Gon1pertz growth rate ,node! [ 1], find the Gon1pertz parameters r

and C from [2] . =.l.

(b) Verify that W =

20 e 10- 9'

is the solution of the differential equation [2]. (c) Find the equilibriu1n weight of the tumour, i.e. the weight of the tun1our when it stops gro\ving. (d) Find the maxin1um growth rate for this tumour. Treatn1ent of many tun1ours is n1ost effective when the tun1our is growing at its fastest rate. (e) After ho\v long is the tumour growing at the fastest rate? (f) Sketch the graph of W for the first 3 n1onths, sho\ving any equilibriu1n solutions and the point of maxiinum gro\vth rate. 21 Oil is pumped fro1n a Bass Strait oil well at a rate proportional to the volwne V of oil (in w1its of 'barrels') remainmg in the well after t years. Initially, the \veil had 1 000 000 barrels of oil, but 5 years later, only 600 000 barrels of oil remam. It \viii not be profitable to continue pumping oil \Vhen fewer than 1000 barrels of oil ren1ain. t

(a) Show that the volume of oil V re1naining after t years is given by V(t) = a ( b )dfor a suitable choice of positive integers a, b, c and d. c (b) At \Vhat rate is the remaining nu1nber of barrels of oil decreasing after 5 years?

(c) For how many years \viii the oil well remain profitable? Express your answer to the nearest 1nonth.

22 A nature conservation group releases 21 Tasmanian devils onto a remote island off the coast ofTas1nania. The group believes the island can support at most 588 Tas1nanian devils. The growth rate of the Tas1nanian devil population pis (a) Show that

1 111( =

p

(

1 l-

p( 0) = 21, with r > 0 and t n1easured in years.

1- : ), 5 8

P ) 588

1 =- -

p

p

_

1 . 588

(b) State the ,node! for the Tasn1anian devil population p in terms of rand tiine t.

Three years after the beginning of the breeding progran1, the population is 294. (c) Find r. (d) Use the model fro1n part (c) to estiinate the devil population after 6 years. Six years after the Tasn1anian devils are first taken to the island, the nature conservation group decides it is tin1e to repopulate the 1nainland with the island's devils. A decision is made to take 140 devils from the island each year. (e) Find a differential equation = f (P) for a suitable choice of the function f(P) that n1odels the modified

t

(f)

growth rate in the island population P(t), t years after the devils \Vere first taken from the island back to the 1nainland. Assume that the value of r and the carrying capacity of devils on the island both re1nain unchanged dP . 1 1 ( 1 1 ) Ifdt = -K(P-a)(P-b),P(O) =P0, findP(t),g1ven that (P-a)(P-b ) = b-a P-b - P-a ·

More challenging question parts:

i8

10 3 (g) Express the functionfiP) found in part (e) in the forn1 f (P) = (P-a)(P-b), a< b, for a suitable 5 choice of real nu1nbers a and b. (h) Hence find a solution curve for the island population P(t), t years after the decision to repopulate the mainland. (ij Plot the Tas1nanian devil population over the first 14 years of the breeding prograin. Hence, comment on the effectiveness of the breeding progra1n.

304

New Senior Mathematics Extension 1 for Years 11 & 12

YEAR 12

CHAPTER REVIEW 12 1 A can1era at ground level is 400 n1etres a,vay fro1n a hot air balloon just prior to the balloon lifting off. The balloon lifts off and the can1era records the balloon rising into the sky at a constant rate of 10 n1etres per second. (a) If 0 is the angle of elevation of the balloon, express the height h of the balloon in terms of this angle. (b) Ho,v fast is the angle of elevation 0 radians changing when the balloon is 300 m above the ground? 2

2 Verify that y = e-•cos x is a solution of d ; + 2 ddxy + 2 y = 0. dx 3 The gradient of the tangent to a curve at any point (x, y) is x I' x > -1. If the curve passes through the point (1, 1), find the equation of the curve. x+ 4 (a) If f'(x) =cos2x-3sinx andf'(O) = 3, findf'(x). (b) Hence if f

(

1) = 3, find fix).

5 Solve the differential equation 6 (a) Show that

4 y 2 -4

1

Z

= 1- 2 y, given that y = - 1 where x = 0. 1

+ . y 2

y- 2 d 2 (b) Find y as a function of x if y satisfies the differential equation = y - 4 and y = 0 where x = 0.

Jx

7 A species of tuna is declining so that T, the nu1nber of tuna at a time t years fro1n no,v, satisfies the differential . dT equation dt = - O.1T .

(a) Write the general solution to this differential equation, where T(O) = A > 0 is the initial population. (b) Find the tin1e it ,viii take for the nun1bers to fall to one-quarter of their present value.

8 Consider the initial value problen1 equation. 9 If sin x = if, 0 < x < n, ,vhat is

A -cot x

B

Z

= 2x (1 + y

2

),

y(O) = 1. Find the exact solution to the differential

Z

in tern1s of x?

C

tan x

- tan x

D

cotx

10 Which one of the follo,ving differential equations is not satisfied by x = e- 31?

(-dx) dt

A

2 2

-9x =0

B

2 d - x -9x=O dt2

11 Ify = e'"' satisfies the differential equation

land-6

A

12 If

A

f

= 2-

B

d2 y

dx

2

C

2 d - x +9x=O dt2

D

dx - +3x=O dt

dy + dx - 6 y = 0, then the possible values of k are:

-land6

C

-2and3

D

2and-3

D

6✓ 3-5+2x-2x ✓ x

3✓x and y = 3 ,vhere x = 1, then y is given by the rule:

1+3x(l- ✓ x)

B

3+2x(l- ✓ x)

C

3+2x + 2x ✓ x

Chapter 12

Differential equations

305

YEAR12 2

13 What is the slope field of y' = x 2 ? y y A , ,

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= -3 y, an d y = 1 where x = 2, then which of the following is true? B C D y= 2e3-3x y=e6-3x y= 2e3x-3 y=e3x-6

15 The general solution of A

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z

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Z=

-(2 + y) is:

y = e-x + A, A is a real nu1nber y = Ae' + 2, A is a real nun1ber

16 Consider the differential equation

y = Ae-x + 2, A is a real number D y = Ae-x - 2, A is a real nun1ber

B

ix =

x - 2 y , for \Vhich the solution is g(x). Which of the follo\ving

staten1ents about the particular solution that contains the point (O, - 1) is true at x = O?

A C

B D

the graph is increasing and concave up the graph is decreasing and concave up

the graph is increasing and concave do\vn the graph is decreasing and concave down

Questions 17 and 18 refer to the following information.

Consider the differential equation

ix =

ysinx, for which the solution is y = f(x). LetfiO) = 1.

17 Which of the following staten1ents about the graph off(x) are true? The slope off(x) at the point ( ~ ,

(i)

1) is 1.

(ii) fix) has a horizontal tangent where x = 0. (iii) fix) has a vertical tangent where y = 0.

A

i only

B

ii only

C

i and ii only

C

y=e

D

ii and iii only

D

y=e

18 The particular solution is: A

306

y=e

1 - cosx

B y=ecosx -

I

New Senior Mathematics Extension 1 for Years 11 & 12

- sinx

sinx

YEAR 12

19 When added to water, 5 grams of a substance dissolves at a rate equal to 10% of the ainount of undissolved chemical per hour. If xis the nu1nber of grains of undissolved chen1ical after t hours, then x satisfies the differential equation: dx = -1 (10-x) dx 1 dx =--1 x dx = _!._(5-x) 8 C D A dt=-1ox dt 5 dt 10 dt 5

20 The population P(t) of a certain species satisfies the differential equation clJ: = P( 2-

~ ) with an initial 10 00 population P(O) = 4000, \Vhere tis the time in years. What is the population as t approaches infinity? 1 1 20000 th Note at P + 20000- P - P(20000-P)" A

4000

8

5000

C

10000

D

20000

21 A quantity of sugar is dissolved in a tank containing 100 litres of pure \Vater. At tin1e t = 0 n1inutes, pure water is poured into the tank at a rate of 4 litres per 1ninute. The tank is kept well stirred at all times. At the sa1ne time, the sugar solution is drained from a tap at the bottom of the tai1k at a rate of 6 litres per minute. A differential equation for the mass m gran1s of sugar in the tai1k is: A

dm dt = -6m

8

dm dt

=4 _ 3m

C

50

dm _ _ 3m dt - 50- t

D

dm dt

=4 -

3m

50-t

22 According to Fourier's law of heat conduction, the rate of heat transfer~~ through an ice sheet in Antarctica is given by the differential equation

~~ = k(Twh- T. ), \Vhere k is the thern1al conductivity of the ice, h is

the thickness of the ice sheet and Twand T. are the te1nperatures at the ice/water boundary and the ice/air boundary respectively. As the \Vater loses Q joules of heat through the ice sheet, the rate of increase in ice thickness h is given by

f3 =

Llp' where Lis the latent heat of sea water (in other words, the amount of heat loss required to freeze

1 kilogran1 of it) and p is the density of the ice.

(a) Find the rate of increase of the ice sheet thickness ~;. (b) If h(O) =h0 , find h(t), assu1ning that k(TL~ T. ) is a positive constai1t.

23 The quantity q sold of a new product is a function of the selling price p. The revenue R(p) =p q(p) from selling

q units at price p is also a function of the selling price. It can be sho\vn that ~R d p

= q ( 1 + P 7" )· q p

Econon1ists call E = P dq the price elasticity of de1nand q with respect to price p. This price elasticity of

q p

den1and measures the relative change of the quantity den1anded in response to given relative change in price. (a) Show that revenue is a 1naxin1u1n \Vhere E =-1. The inverse den1and curve of a product n1easures the selling price p(q) in tern1s of the available supply q. The la\v of de1nand states that, 'when the price of a good rises, and everything else re1nains the san1e, the quantity of the good den1anded will fall'. In other words, selling price varies inversely \Vith the de1nanded supply. Suppose that a business has the exclusive right to iinport a new product into a large city. Market research

J25 -

indicates that the selling price $p for a new product is modelled by p = q 2 , \vhere the supply q is n1easured in units of 1000. (b) State the don1ain of the n1odel. (c) Find the rate of change of the den1anded quantity q \vith respect to the selling price p. (Express your ai1swer in tenns of q.) (d) Hence find the selling price that will n1axi1nise revenue.

Chapter 12

Differential equations

307

YEAR12

24 The relationship behveen air temperature T Kand wind speed V m s- 1 is 1nodelled by the differential equation dT dV

=-

V , where CP is a constant kI10\vn as the specific heat of the air mass.

Cp

(a) Find the general solution for the wind speed V(T) as a function ofte1nperature.

Near a particular tropical cyclone the wind speed is zero and the ten1perature is 308 K. (b) Find the particular solution for the wind speed V(T) as a function of ten1perature. (c) If CP = 1004.6, find the wind speed near the centre of a tropical cyclone if the te1nperature there is 307 K.

25 A mathematical ,node! for the relationship between x the number of predators (in hundreds) and y the nu1nber of prey (in thousands), in a particular environn1ent at tin1e t years, gives the follo\ving pair of a: .al equations. . . -d dy = - - 1 ( x - 2) dx = - 1 (y - 1) diuerent1 -d t 400 t 100 d (a) Use the chain rule to obtain a differential equation involving x and y only.

fx,

(b) At so1ne point in tiine there are 100 predators (x = 1) and 3000 prey (y = 3). Hence show that a2(x- h) 2 + b2(y- k) 2 = 1 for real numbers a, b, hand k. (c) The graph of the solution in part (b) is an ellipse with a centre at (h, k). Sketch your solution from part (b). (d) Fron1 your previous answers, find the maxin1u1n and 1ninin1um nun1ber of predators the environn1ent can support.

26 This question uses data fro1n the table below. V(h) = l ,rh2 (3R-h)

Volu1ne of spherical cap (m

A 1(h) = nh(2R - h)

Free surface area of a spherical cap (n/)

A/h) =2nhR

Curved surface area of a spherical cap (1n

£

Evaporation coefficient for lake surface (m 3/in2 day)

a

Seepage coefficient for lake bottom (n1 /m day)

Qm

Net rate of inflow from rivers (m 3/day)

3

3 )

2

3

)

2

The Lake Eyre basin is the lo\vest point in Australia and usually dry. However, the southern lake of the Lake Eyre co1nplex fills to a depth of about 3 n1etres via inflo\v fro1n flooding rivers, a fe\v tin1es each century. This southern lake can be n1odelled as the cap of a sphere of radius R (1n), with a centre at 0, as sho\vn. Assume the inflow from the flooding rivers has stopped so 3 that Q111 "" 0 (1n /day).

'

A possible balance for the volun1e of \Vater in the southern lake of the Lake Eyre con1plex during a flood year is

~~ =Q.,, - £A1( h )- a A2 ( h ).

(a) Use the information from the table above to express (b) Show that

!r =

~~

as an explicit function of h.

nh (2R - h) and hence find ~; in simplest forn1.

The depth of the \Vater in the lake can now be n1odelled iinmediately after an inflow of 1 gigalitre (1000000 n1 3), which fills the lake to a depth of 31n above its centre so that h0 = 3 1n. (Assume no further inflows or precipitation.) (c) Find the radius R of the hen1ispherical cap.

27 The coroner arrives at Frog1norton Manor on a cold winter evening. Lady Frogmorton has just been found dead in her clin1atically controlled greenhouse by the butler. The temperature of the body is taken, and is recorded as 29°C at precisely 7:20 pn1. The coroner checks the progran1n1able thermostat and detern1ines the greenhouse has been kept at a constant 20°C for the past 12 months. At precisely 8:20pm, the coroner takes the te1nperature of the body once more, and it is recorded as 27.4°C.

308

New Senior Mathematics Extension 1 for Years 11 & 12

YEAR 12

Sn1iggins is Lady Frogn1orton's chauffer. Under questioning by the police, he claims that he delivered Lady Frogmorton to the manor at around 6 pm that evening, after a day at the Chelsea Flower Show. (a) Assun1ing a normal body ten1perature of 37°C, propose a differential equation to n1odel the te1nperature T(t) of the deceased at a tin1e t hours since death. (b) Solve the initial value problem in part (a). Recall that the coroner arrived on the scene at precisely 7:20 pm, which is r hours since death. (c) Show that at r hours after death e'' = 1;.

(d) Find an expression, similar to your answer in part (c), for the ten1perature at 8:20 pn1. (e) Hence find r. (f) The coroner inforn1s the police that the chauffer's story may need further investigation. Explain why this is so.

Challenging 28 Bob's credit card bill Bis initially $15 000 and he pays 18% interest on this debt per year, co1npounded continuously. He decides to pay it off by transferring 1noney fro1n his savings account continuously at the rate of $300 per n1011th . (a) Find and solve a differential equation to n1odel the credit card balance B after t years. (b) Ho\v n1uch tiine \viii it take to pay off the credit card bill (to the nearest day)? (c) What is the sum total of Bob's repayments?

Assun1e Bob has $40 000 in a savings account that accu1nulates interest at an annual rate of 6%, also co1npounded continuously. (d) Find and solve a differential equation to n1odel the balance S of Bob's savings account. (e) Ho\v n1uch n1011ey \viii Bob have in his savings account when the debt is finally paid off (assu1ning no other transactions)?

29 An abandoned open-cut ,nine just outside a large city has been purchased as a landfill for solid waste by a city council. When purchased, the open-cut ,nine had a volume of 1 1nillion cubic 1netres. At the beginning of 2015, the landfill already had 100 000 cubic metres of solid \Vaste. The volu1ne of solid \Vaste W in the landfill (n1easured in units of 100000 cubic metres) t years after the beginning of2015 is modelled by the solution of the differential equation dd~ = ~ (10- W) , W(O) = 1. 1 (a) Find the volun1e of solid waste in the landfill t years after 2015. (b) Hence detennine the volun1e of solid waste in the landfill at the beginning of 2035. (Express your answer in cubic n1etres, correct to the nearest cubic n1etre.)

30 The population P(t) of penguins on an island in the Southern Ocean t years after the beginning of2015 grows at a rate directly proportional to 1000 - P(t), where the constant of proportionality is k. (a) If the population at the beginning of 2015 is 200, express the penguin population t years after the beginning of2015 in tenns oft and k. (b) If the population after 2 years is 300, find k. (c) Hence detennine the long-tenn population of penguins on the island.

31 While on an unauthorised trip to the local fast food restaurant during their study period, a pair of Year 12 students are convinced that they have just seen the Priine Minister buying a hamburger. On returning to the school,

f=

~ p(l- p), where p is the 1 proportion of the school con1n1unity that has already heard the ru1nour, t minutes after their return to school. their ainazing discovery spreads throughout the school community at the rate

(a) What proportion of the school community has heard the rumour \vhen it is spreading most rapidly? By the beginning of the afternoon period, 20% of the school co1n1nunity had already heard the rumour. 1 . (b) Find p(t), at time t 1ninutes since the beginning of the afternoon period, given (l l ) =.!. _ p -p p p- 1 (c) At \Vhat time (correct to the nearest 1ninute) is the run1our spreading most rapidly?

Chapter 12

Differential equations

309

CHAPTER13 Motion, forces and projectiles 13.1

PROBLEMS INVOLVING DISPLACEMENT AND VELOCITY

Displacement When considering the ,notion of an object, you need to consider the position of the object, how fast it is travelling and the cause of the motion. The position of an object n1ust be defined relative to a reference point, \Vhich is usually its starting point. The position can be described by both its distance and its direction from the starting point. This is a vector quantity, called displacement. On the other hand, if you are only concerned with how far the object has travelled then the direction can be ignored, so you can consider the distance travelled by the object. The international syste1n of units (SI) uses the metre (n1) as the standard unit for displace1nent and for distance. Other common SI units include the centiinetre (c1n), n1illi1netre (mm) and kilometre (kn1). For exa1nple, if you walk around a square park of side length 500 m, the distance you travel is 2000111. Ho\vever, as you arrive back at your starting point, your displace1nent is zero.

Velocity The rate at \Vhich the displacement of an object changes \Vith respect to tin1e can be described by the vector quantity velocity. As a vector quantity, velocity is defined by its 1nagnitude and direction. The 1nagnitude of velocity is the scalar quantity speed. The SI standard unit for velocity and speed is 1netres per second (n1/s or ,ns- 1). Another co1n1non unit is kilometres per hour (km/h or kn1 h - I). The average velocity of an object between two positions is defined as: . change in position displacen1ent 1 average ve ocity = ti1ne taken = ti1ne On the other hand, the average speed between two points is defined as: average speed = dist~nce t~velled t1n1e ta en Note: Speed ,nay need to be converted from ms-1 tokinh- 1 or fro1n kmh- 1 to n1s-1: For exa1nple: 90kinh- 1 = 90 x 1000 + (60 x 60) = 90 + 3.6 = 25 n1s-1 and 15n1s-1 = 15 x 3.6 = 54km h- 1•

--------+

3.6

X

3.6

• Displacement is the change in position of an object relative to its starting point. It is a vector quantity \Vith both magnitude an d direction. • Distance is ho\v far an object has travelled. It is a scalar quantity with 1nagnitude only. • Velocity is the rate at \Vhich the displacement of an object is changing with respect to tin1e. It is a vector quantity with both magnitude an d direction. • Speed is the magnitude of the object's velocity. It is a scalar quantity \Vith 1nagnitude only.

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New Senior Mathematics Extension 1 for Years 11 & 12

YEAR 12

Example 1 Marli is going for a \Valk around the block. She starts at 0 and 1Tioves to point A and then to point B, where she stops to talk to a friend.

c- - - - - - - - - - - - - B

Detern1ine:

O'----------------' 750m

(a) the distance Marli travels fron1 0 to B

250 m A

N

+

(b) Marli's displacement fron1 0 to B, correct to one deciITial place.

If Marli takes 10 minutes to reach point B, determine: (c) her average speed in 1Tietres per second, correct to one decin1al place (d) her average velocity in 1Tietres per second, correct to one decin1al place.

Solution (a) Distance travelled fro1TI Oto point A to point B = 750 + 250 = 1000 !TI (b) Using Pythagoras' theorem to find the distance c bet\veen the startmg and finishing points:

,--------------=~

1001= ✓750

2

2 + 250

= 790.61TI . . o f n1ot1on: . 0 = tan -1(250) D1rect1on 750 = 18.4° Bearing is N(90 - 18.4) 0 E = N71.6°E

B

250 m

0 ~=--'"----- - - - - - - - - - ~ A

N

+

750 m

Marli's displace1Tient is 790.6n1 fro1TI O in a direction ofN71.6°E. (c) 101Tiin=600s average speed = dist~nce t~velled tln1e ta en 1000 600 = l.7 n1s- 1

. displacement (d) average velocity = - ~- . - - tln1e - 790.6 600 = 1.3 m s- 1 in the direction N7 l.6°E.

Finding resultant velocity Example 2 Magayi can swi1TI in still water at a rate of 3.0 !TI s- 1• If she swiITis in a river that is flo\ving at 4.0 n1s- 1 and keeps her direction (\vith respect to the water) perpendicular to the flo\v, find the n1agnitude of her velocity with respect to the riverbank.

Solution Vector diagra1TI to illustrate the situation: Flow4ms- l

Using Pythagoras' theore1TI to find the 1Tiagnitude of the resultant velocity !'.:

11'. I = .J32 +42 Magayi 3m s-l

Resultant velocity vms- 1

=5n1s- 1

-

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Finding resultant velocity using component form Example 3 Let .£ and j be unit vectors in the directions of east and north, respectively. Pravat is s\vimming in the ocean and his velocity relative to the water 1'.'i 1n s-I is given by the vector 1:'.i = 2.5 .!, + 1.0 j. The ocean's current has a velocity ]'.' 21n s- 1 where ]'.'2 = 0.5.£ - 3.0 j. Find the 1nagnitude and direction of Pravat's-resultant velocity):'. 1n s- 1, correct to one decin1al place. -

Solution Vector diagram to illustrate the situation: N

Adding vectors to find the resultant velocity:

j

-

3

L~

2

I

v,

-

3 E

-I

= 2.Sf + l.Oj + 0.Sf-3.0j = 3.0f- 2.0 j 11'.'I = ✓(3.0)2 +c-2.0)2 =JD.a = 3.6n1s- 1

-1

0 is an acute angle in the diagran1, so use positive values for side lengths.

-2

0

-3

The direction bearing: 90 + 33.7 = 123.7°T. Pravat is swimming at 3.6 m s-1 in a direction of

20 · ) = tan- 1 ( 3.0 = 33.7°

123.7°T.

EXPLORING FURTHER

0

Velocity vectors Use technology to explore combinations of velocity vectors.

EXERCISE 13.1

PROBLEMS INVOLVING D ISPLACEMENT AND VELOCITY

1 Jaide is going for a walk around the block. She starts at 0 and 1noves to point A and then to point B, \vhere she stops to have a rest.

650m

N

A~ - - - - - - - - - - - ~ B

+

200m

(a) What is the distance Jaide has travelled fro1n Oto B? O ' - - - - -- - - - - - ' C (b) What is Jaide's displacen1ent from O to B, correct to one decin1al place, and \Vhat is her direction? Jaide has taken 12 1ninutes to reach point B. (c) What is Jaide's average speed in 1netres per second, correct to one decin1al place? (d) What is Jaide's average velocity in 1netres per second, correct to one decin1al place, and what is her direction? N

2 Courtney \valks 2001n south, then 330 n1 east, and finally 190 m south (diagran1 shown is not to scale). What is Courtney's displacement for her entire walk, correct to one decin1al place, and what is the direction?

+

200m

330m

190m

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YEAR 12

3 A 1nountain-clin1bing expedition establishes its base camp and hvo intern1ediate ca1nps at positions A and B. Camp A is 8400 m west of and 18001n above base can1p. Ca1np Bis 5900 m west of and 8501n higher than Camp A. What is the displace1nent of can1p B fro1n the base can1p and what is its angle of elevation? Give answers correct to one decin1al place. N ,.__ 130m _ _ Start 4 Kenneth is taking his dog for a walk along the path in the local park, as shown in the diagran1 on the right (not to scale).

Finish

210m

What is the 1nagnitude and direction of Kenneth's resultant displace1nent, correct to the nearest \Vhole nun1ber?

t

30° 160m

5 While exploring a recently discovered cave system, a spelunker (cave explorer) starts at the entrance and makes the follo\ving movements: 85 m north , 1901n east, 250 m N45°E, and 1001n south. What is the spelunker's final displacen1ent fro1n the cave entrance and what is their direction? 6 Brianna can swiin in still \Vater at a rate of 6.0kin h- 1• If she S\Vin1s in a river flowing at 4.5 kin h- 1 and keeps her direction (with respect to the water) perpendicular to the flo\v, then \Vhat is the 1nagnitude of her velocity (correct to one decin1al place) with respect to the riverbank? 7 (a) A river flows at 5 kin h- 1 and Rhani rows at 10 kin h- 1• In what direction should Rhani row to go straight across the river? A Angle of 30° to the riverbank downstrea1n B Angle of 60° to the riverbank do\vnstream D Angle of 60° to the riverbank upstream C Angle of30° to the riverbank upstrean1 (b) If the river flo\vs at l Okn1h- 1 and Rhani rows at 5kinh- 1, then in \Vhat direction should she row to go straight across the river?

A C

Angle of 30° to the riverbank downstream Angle of 30° to the riverbank upstrean1

B D

Angle of 60° to the riverbank do\vnstream It is not possible

8 Youlin and Nick are riding in a boat that has a speed relative to the water of3.00ms-1• The boat points at an angle of 35° to the shore, moving upstream on a river that is flo\ving at 0. 75 n1s-1• Ho\v n1uch tiine does it take for the boat to reach the opposite shore, correct to two decin1al places, if the river is 50 1n \vide?

9 (a) Raphaela is \Vatching an aircraft that is flying at a velocity of l OOms-1 N with respect to the flow of air. If the velocity of the \Vind is 101n s- 1 north, what is the resultant velocity of the aircraft relative to Raphaela? (b) Later, she observes a second aircraft flying with a velocity of 125 1n s- 1 N \Vith respect to the flow of the air. If the flow of the air has a velocity of 10 m s-1 south, then the resultant velocity of this aircraft is 115 n1s-1. What is the direction of this resultant velocity, relative to Raphaela? (c) The next day, Raphaela observes another aircraft flying with a velocity of 1001n s- 1 N \Vhich encounters a \Vmd coining fro1n the side at a rate of25 n1s- 1 W What is the resultant velocity of this aircraft relative to Raphaela, correct to one decimal place, and \vhat is its direction? 10 Let !. and j be unit vectors in the directions of east and north respectively. Mitchell is swi1n1ning in the ocean and his velocity !'.i 1n s- 1 is given by the vector !'.i = I.Of - 2.0 j. The ocean's current has a velocity !'.2 n1s-1 \Vhere !'.2 = -0.5i + 2.5 j. What is the 1nagnitude and direction of Mitchell's resultant velocity!'. m s-1, correct to one decimal place, and \Vhat is its direction? 11 Let i and j be unit vectors in the directions of east and north respectively. Give answers correct to two decin1al places \vhere required. (a) Express each vector in the forn1 xi +

-

~) OA = 5.0 mat 053°T

-

yz.

(ii) AB = 6.0 mat 315°T

(iii) BC = 4.0 n1 at 240°T

(iv) CD= 3.0 n1 at 150°T.

(b) Express the swn of the four displacen1ent vectors in the fonn xi+ y l, with values correct to two decm1al places. (c) What is the 1nagnitude and direction of the resultant vector OD, correct to hvo decimal places, and what is its direction? Chapter 13

Motion, forces and projectiles

313

YEAR12

12 Let f and j be unit vectors in the directions of east and north respectively. Morgan and Eilish finish their last day of school and in1n1ediately decide to take a road trip across the desert in Eilish's ne\v car. Their trip involves the following moven1ents: 43kr11 at 337°T, 65kI11 at 270°T and 22kin at 93°T. Eilish's car breaks down after the last leg of the trip. At this tiI11e, how far are Morgan and Eilish fron1 the school and in what direction?

13 Johanna is driving along the free\vay at 108 kin h- I when it begins to rain. She observes the raindrops running do\vn the driver's side windo\v. Calculate the angle that the raindrops 111ake \Vith the vertical windo\v, as seen by Johanna, if the raindrops have a speed of 10 111 s-I relative to the Earth's surface. Assu111e that the raindrops fall vertically down, relative to the Earth's surface.

14 Bus # 1 is moving at a speed of 50kn1h-1, while Bus #2 is 111oving at 30kinh-1 in the opposite direction. What is the relative velocity of Bus # 1 with respect to Bus #2? A 20kn1h- 1 in the direction of Bus #1 8 20kn1h-1 in the direction of Bus #2 C 80kn1h-1 in the direction of Bus #1 D 80kI11h- 1 in the direction of Bus #2 15 On her \Vay hon1e fron1 school, Teluila \valks along three streets after exiting the school gate. She walks 240 m east, 720 n1 north and 75111 east. What is the magnitude ofTeluila's resultant displacement? A

555111

8

763n1

C

786m

D

1035n1

16 Freddie rows his boat in a direction perpendicular to the riverbanks. The river is 200 n1 wide. The boat's speed in this direction is 2 m s-1. The speed of the river's flow is 1 n1s - I and the riverbanks are parallel straight lines. (a) Calculate the velocity of the boat relative to the riverbank. (b) At \Vhat distance, in the direction downstrean1 fro111 his starting point, will Freddie get to the other riverbank? (c) Calculate Freddie's final displacen1ent from his starting point. (d) Rohanne is going to cross the river in another ro\vboat. She can ro\v at the speed of 2 111 s- I relative to the river's flow. At what direction should Rohanne ro\v to get to the other riverbank at the san1e point as Freddie, and how n1uch tiI11e will this take? 17 The pilot of an aircraft flying due south is notified by the flight controller that there is a second aircraft flying due north in the same general area and at the san1e altitude. The pilot is told that the northbound aircraft is currently located at a position that is 12.6 kin, 170°T \vith respect to the pilot's aircraft. (a) Ho\v n1any kilon1etres to the south is the second aircraft? (b) Ho\v n1any kilon1etres to the east is the second aircraft? (c) If the hvo aircrafts both have an airspeed of 300 kin h-1, how much tiI11e (in seconds) will elapse before they are side by side? 18 An aircraft drops a package of emergency rations to a fan1ily stranded in the floods. The aircraft is travelling horizontally at 45.0n1s-1 and is 100111 above the ground. A parachute allo\vs the package to fall with constant speed and hit the ground 10 s after release. (a) Find where the package hits the ground, relative to the point fro111 where it was dropped, to the nearest n1etre. (b) Find the velocity of the package just before it hits the ground, correct to one decimal place.

13.2

PROBLEMS INVOLVING FORCES

Forces exist everywhere and are funda111ental to the structure of the universe and to the nature of 111atter. A force can be thought of as a push or a pull acting on an object For most siI11ple objects, it is reasonable to consider the forces to be acting through a single point at the object's centre of n1ass. (This 111eans the object is being considered as a particle, as defined earlier in Chapter 7, section 7.2 Velocity and acceleration as rates of change.) The action of a force can affect an object by changing the speed or direction of its n1otion, or by deforming the object. The amount of force acting on an object is measured using the SI standard unit called the ne\vton (N). A force of 1 ne\vton will accelerate an object of 111ass 1 kilogran1 at a rate of 1 metre per second, so 1 N = 1 kg111s-2 .

314

New Senior Mathematics Extension 1 for Years 11 & 12

YEAR 12

Every object near the Earth's surface is subject to a force called gravity. This force is called the weight of the object and it acts vertically downwards on every object, to\vards the centre of n1ass of the Earth. The force due to gravity (the weight force) of a body of mass 1 kilogram is 9.8 N. For exa1nple, a person whose n1ass is 80 kg has a weight equal to their n1ass multiplied by g, where g= 9.8 n1 s-2, so an 80 kg person weighs 80 x 9.8 = 784 N. Force is a vector quantity, so it needs both magnitude and direction to fully describe it. If more than one force is acting on an object, the su1n of all the forces is the resultant force or net force. The object will n1ove as if the net force is the only force that is actually acting on the object. For exan1ple, in the diagran1 on the right, the resultant force acting on the object is f, the sun1 of the two forces f 1and f 2 acting on the object: f = f 1 + f 2• This principle can be extended to involve the addition of many forces.

F,

-

The resultant force or net force f is the vector sum of all the forces acting on an object. If forces f 1, f 2 , f 3 , ••• f,, act on an object, then f = f 1 + f 2 + f 3 + ... f,,

Calculating resultant forces Example 4 Forces are acting on an object. Calculate the resultant force on the object for each of the following. (a) F1 = 130N east, F2 = 210N \Vest

(b) f1 = 240 N north, f 2 = 165 N west

-

-

Solution North

(a) Vector diagran1 to show the forces acting on the object:

f = f1 + f2

Fi ,.

210N

I.

•

I 130N., F, -

= 130N east + 2 10N west= 80N \Vest (b) Vector diagran1 to sho\v the forces acting on the object: F,

Vector diagram to add the forces:

-

240 N 165 N r-t--, F,---._.

-

+

165 N

240 N

F

North

0

+

Using Pythagoras' theoren1: If I=

✓2402 + 165 2

= 291.25 N Direction of resultan t force: tan 0 =

~~~

1 0 = tan- ( ~~~)

0 = 34.5 1° The resultant force is 291.25 N in the direction N34.5 l 0 W.

If the forces acting on an object are in the sa1ne plane, then each of the forces and the resultant force can be expressed in component form as a su1n of two perpendicular vectors, using horizontal(£) and vertical components. The force f can be resolved into the hvo perpendicular con1ponents as f = xf + For exan1ple, fron1 the diagram: f = If Icos 0{ + If Isin 0;_·.

(z)

~

yz.

1D sin (0)j

0

_ _.____..,___

Chapter 13

J

l~icos(0)f

Motion, forces and projectiles

315

YEAR12

Calcu lating resultant forces using vector components Example 5 Two forces are acting on an object: f 1 = 120 N acting at N60°E and f 2 = 150 N acting at N50°W Calculate the resultant force f acting on the object.

Solution Vector diagram to show the forces acting on the object: N

Resolving forces into horizontal (i) and vertical components:

LL

N

F,

•

F,

-

F,

!SON

!SON

120N

120N

150sin (40°)

w-----0~1-----E

(1)

F1

-i

120sin (30°) 40°

\V - - -

150cos (400)

'

120cos(30°)

-- E

'

s s

f 1 =120cos30°f+ l 20sin30°z

f 2 =-150cos40°f+ l 50sin40°z

= 103.92.i, + 60.00 j

= -1 14.91£ + 96.42}

-

f = f1 +f2 = 103.92.i, + 60.00j- 114.9 1{ + 96.42}

-

-

156.42

= (103.92-l 14.91)f + (60.00 + 96.42)} = -10.98£ + 156.421 10.99

Using Pythagoras' theorem:

IEI= -JO The particle hits the water after 2 seconds.

Note: If there is no vertical co1nponent of projection, then the velocity of projection does not affect the tiine taken to reach the water. If you drop an object and at the sa1ne tiine throw another object horizontally, you will notice that both objects reach the ground at the san1e tiine (assuming that air resistance is not significantly different for the

objects). In both cases the vertical n1otion is governed by y in the previous exa1nple.

= - ~ gt 2 , independent of any horizontal ,notion, just as

Example 13 A particle is projected fron1 ground level and 1.5 seconds later it just clears a \Vall 3. 75 1netres high at a horizontal distance of 36 1netres. If the initial velocity is 0) = cos af + sin al n1 s-l and

~(t) =l~(O)Itcosa i + (l~(O)It sin a-5t

2

)

I~( I I

t

I

(a) Calculate the initial velocity and the angle of projection.

(b) Find expressions for r(t) and ~(t) in component form.

(c) Find the range.

)'

I!'. (OJI

(36, 3¾) 0 ............................ 36m ····························--I

328

New Senior Mathematics Extension 1 for Years 11 & 12

X

YEAR 12

Solution (a) t = 1.5, x = 36, y = 3.75: r(t) = l~(O) Itcosa i + (l~(O) It sina-5t Hence x = l~(O) Itcosa and y = l~(O) Itsina-5t

2

)

l

2

Solve simultaneously to find the angle of projection a.

t = 1.5, x = 36: 36 = 1.51~(0)1cos a

(1)

t = 1.5, y = 3.75: 3.75 = 1.51~(0)1sin a - 5 x 1.52 15 = L5l~(O)lsina

(2)

15 L5l~(O) lsina 36 = L5l~(O) icosa

(2) + (1):

tana=212 a= 22°37' To calculate the velocity of projection lv(O) Isubstitute into [ 1): cos(a) = ~;, as tan a= 36 =I~(O) IX l3 12 1.5

5 12

360 x .Ll. = 26,ns- 1 v(O) I = l15 12

i

(b) r(t) = 26tcosa + ( 26t sin a - 5t

= 26t X

2

)z

(c) y=O: 10t-5t2=0

5t(2 - t) = 0 t= 2s x=24x2=48 m The range is 48 metres.

12 i + (26t X 2_ - 5t 2 ) j 1313 -

= 24t i + ( l Ot - 5t

2

)

z

Differentiate r(t) ,vith respect to t: ~(t) = 24f + (10-lOt)j

Example 14 A stone is thro,vn to hit a s1nall object sitting on top of a wall that is 20 metres horizontally fro1n the point of projection and 5 n1etres high. If the stone is thrown from ground level with a speed of 401ns- 1, show that there are hvo angles of projection that will allow the object to be hit. The position vector of the particle is given by 2 r(t) = l~(O) It cosaf + o~(O) It sin a-5t ) f where l~(O) I

)'

40 ms- 1

, P (20, 5)

X

O

is the initial velocity and a the angle of projection fron1 th e horizontal.

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Solution l~(O) I= 40, i:(t) = 40tcosaf + ( 40tsina-5t 2 )L, x = 20, y = 5. x = 20: 40tcos a= 20 1 t= ---'c...._ 2cosa

y= 5: 40tsin a-Sr= 5

[l]

1 1 Substitute t = into [l]: 40sinax 2 cos a 2 cos a

-s(

2

1 ) =5 2 cos a 5

20tan a- cos 2 a= 5 4 80tan a-Ssec 2 a= 20

[2]

Substitute the identity sec2 a= I + tan 2 a into [2] to solve for a: 80 tan a - 5 ( 1+ tan 2 a) = 20 5tan2 a-80tana+25 = 0 tan 2 a-16tan a+5 = 0

20 Quadratic fonnula·. tan a= 1 6 ± ✓256 2 _ 16± ✓ 236 2

=8 + ✓ 59

tan a= 15.68 or 0.3 189 a= 86°21' or 17°41'

There are two angles of projection that allow the object to be hit, 17° 41' and 86° 21'.

EXERCISE 13.3

PROJECTILE MOTION

1 A particle is projected with a speed of 100 n1etres per second from a horizontal plane at an angle of 60°. Given ~(t) = 100cos60°f + (lOOsin 60° - I Ot)l, find: (a) when the particle reaches its greatest height (b) its position vector r(t) (c) the greatest height reached (d) the time of flight (e) the horizontal range (f) the equation of the trajectory.

Y

1oo ms- 1

60° 1

2 A particle is projected with a speed of801ns- fron1 a horizontal plane at an angle of 30°. (a) Write the vector for v(O)·

0

X

2

(b) Given that ~(t) = l~(O) Icosaf + (l~(O) Isin a - gt) L, find i:(t), using g = IO m s-

(c) Hence find the tiine of flight and the range of the projectile. (d) Find the maxin1um height reached by the projectile. 2

(e) Show that the equation of the trajectory is y = ~ - ;;, . 0

330

New Senior Mathematics Extension 1 for Years 11 & 12

•

YEAR 12 y

3 An object is projected horizontally from the top of a building 125111 high at a speed of 10 n1s-1. Usingg = 10111 s-2 , find:

!Oms- 1

(a) !'.(t) and r(t)

. ... ..

(b) the tiI11e \Vhen the object hits the ground and its distance

125m :

fron1 the base of the building (c) the n1axi111un1 height reached by the object.

0

4 A cricket ball is hit with a velocity of 12.5 m s-1 at an angle with the horizontal whose tangent is (a) If ):'.(t) = l!'.(O)Icosaf + (l!'.(O)Isma -

i·

z,

gt) find ):'.(t) and r(t), given that g = IO m s-2 and tan a=

X

i·

(b) Find the greatest height reached. (c) Find the tin1e of flight and the horizontal distance travelled.

5 A particle is projected so that at time tits position is given by r(t) = 36tf + ( 15t in metres and time is in seconds. If follo\ving staten1ents is correct? A

l!'.(O)I =36 n1s-1 and a = sin- •

~ gt 2 )f, \Vhere distances are

a is the angle of projection and l!'.(O)Iis the initial velocity, which of the

(i53 )

8

l!'.(O)I = 15111s- 1 and a = cos- •( ~;)

D

l!'.(O)I = 39111s- 1 and a = tan- 1 (

5 ) 12 2

6 The equation of the trajectory of a particle projected fron1 the origin is given by (a) where it hits the ground

y= ; - :Oo. Find:

(b) the greatest height reached.

7 A particle is projected at an angle a to the horizontal with a velocity of 50 111 s-l. It passes through the point (200, 25) where distance is measured in n1etres and g = IO 111 s- 2 . (a) Write the expression for ):'.(O).

(c) Find

(b) Find the vectors ):'.(t) and r(t).

a given that the particle passes through the point (200, 25).

8 A ball is projected so that its horizontal range is 45 metres. It passes through a point l l .25111etres vertically above and 22.5 metres horizontally fro111 the pomt of projection. (a) Ifl):'.(O) I is the initial velocity and a the angle of projection, find expressions for ):'.(t) and r(t).

(b) Giveng= lOms-2 , find the angle of projection and the speed of projection. y

(22.5, 11.25)

0 - ···························45m················································· - I

x

9 A stone is projected fro111 the edge of a cliff \Vith a speed of 30 n1s-1 to hit an object 120 n1 horizontally fro111 the edge and 35111 belo\v. Giveng= l On1s-2 , find: (a) expressions for ):'.(t) and r(t)

(b) the angle of projection.

10 A ball is thro\vn horizontally with speed v 111 s-l fron1 a point h n1etres above the ground and lands at a horizontal distanced 111etres fron1 the point of release. Useg = lOms-2 . (a) Find expressions for ):'.(t) and r(t). (c) Find h given v = IO and d = 20.

(b) Find v given d = 3 and h = 1.25.

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11 A particle is projected fron1 a point \Vhose coordinates are (O, O) \Vith a velocity 20 Ill s- 1 horizontally and l OOms-1 vertically. Assumeg= 101lls-2 . (a) What is the angle of projection? (b) Find expressions for !'.(t) and r (t ). (c) Find the tin1e when the projectile passes through the point whose horizontal distance from O is 120 Ill. (d) What is the vertical height at this time in (c)?

12 A particle is projected so that at any tin1e tits position is given by r(t ) = 36tf + (15 t -5t 2 ) f, \Vhere distances are llleasured in metres. Find: (a) the velocity of projection

(b) the angle of projection

(c) the greatest height reached

(d) the range.

13 A particle is projected so that at any tin1e tits position is given by r(t ) = 15 tf + ( 20 t - 5t 2 ) l, where distances are llleasured in metres. Find: (a) the point of projection (b) the angle of projection (c) the speed of projection (d) the coordinates of the highest point on the path. 14 A shell is fired at 200 1lletres per second to strike a target 2 lull a\vay. Using g = 10 ms- 2 , find: (a) expressions for y(t) and r(t)

(b) the angle of projection.

15 An aircraft travelling at 630lunh-1 drops a package fro1ll a height of lOOOn1. Usingg = lOn1s-2, find: (a) expressions for y(t) and r(t)

(b) the tiille taken for the package to hit the ground

(c) the horizontal distance travelled by the package.

16 A particle is projected to just clear two walls that are each 7 Ill tall, and 7 m and 14 Ill respectively a\vay from the point of projection. It is given that r(t) = Vt cosaf + ( Vt sin a -5 t 2 ) L(a) If a is the angle of projection, prove that tan a= 1.5. (b) Show that if the walls are h llletres high and are respectively b metres and c n1etres distant fron1 the point . . th h(b + c) of proJectlon, en tan a = be .

17 A particle is projected with a speed of 25 ill s- 1 in the direction of a point P that is 7 n1 vertically above and 24 m horizontally fron1 the point of projection. Use g = 10 Ill s-2 . (a) Draw a diagram to sho\v this infonllation.

(b) What is the angle of projection?

(c) Find expressions for y(t ) and r(t). (d) Find the coordinates of the pomt on the trajectory directly below P.

(e) At the mstant of the particle's projection, a second particle is dropped from P. Find the position vector for this second particle. (f) Prove that the two particles will collide (i.e. will be in the same place at the san1e tiille). 2

18 A particle projected at an angle of 15° has a horizontal range of 80 Ill. Use g= 10 n1 s- • (a) Find expressions for y(t ) and ~(t). (b) What is the velocity of projection? (c) What would be the range if the angle of projection were increased to 45° \vith the sa1lle speed of projection?

19 A stone is projected horizontally with a velocity of 15 ms- I fro1ll the top of a building 30 m high. -2 Useg = lOn1s . (a) Find expressions for y ,(t) and!'. 1(t ), the velocity and position of the stone. At the same instant another stone is projected fron1 the base of the same building \vith a velocity of 30 m s-1 at an angle of 60° to the horizontal. The two stones collide. (b) Find expressions for !'. 2 (t) and r 2 (t ), the velocity and position of the second stone.

(c) If the coordinates of the base of the building are (O, 0), find the coordmates of the point of collision.

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New Senior Mathematics Extension 1 for Years 11 & 12

YEAR 12

20 A golfer hits a ball fron1 a point on a flat golf course. Two seconds later, it strikes the ground 50 m a\vay. Usingg= 101ns-2 , find: (a) expressions for ~(t) and r(t)

(b) the velocity and angle of projection

(c) the n1axi1nun1 height of the ball above the golf course. 21 A particle is projected \Vith velocity 20n1s-1 to hit a target at a horizontal distance 20m fron1 the point of projection and a vertical height of 10111. Usingg= 101ns-2 , find: (a) expressions for ~(t) and r(t)

(b) t\vo possible angles of projection.

22 A particle is projected so that its position vector is given by r(t) = 30t i + ( 40t -5t 2 ) j, where distances are measured in n1etres and tin1e in seconds. Find: (a) the expression for ~(t) (b) the velocity of projection

(c) the angle of projection

(d) the Cartesian coordinates of th e highest point on the path.

23 A particle projected from a point 1neets the horizontal plane through the point of projection after travelling a horizontal distance a. In the course of its trajectory it reaches a greatest height b above the point of projection. (a) Find expressions for ~(t) and r(t). (b) Find the horizontal and vertical con1ponents of the velocity of projection in tern1s of a and b. 4 (c) Show that \Vhen the particle has travelled a horizontal distance x, it has reached a height of bx(~ - x). a1 24 A football kicked at 16ms- just passes over a crossbar 4n1 high and l6n1 away. (a) If ais the angle of projection, finds expressions for ~(t) and r(t). (b) Usingg= 101ns-2, show that a satisfies 5 tan2 a- 16tan a+ 9 = 0.

CHAPTER REVIEW 13 1 A car is moving at a speed of 100 kn1 h-1 and a truck is n1oving at 90 km h- 1in the opposite direction. The relative velocity of the car with respect to the truck is: A 10 km h- 1 in the direction of the truck B C 190 kin h_, in the direction of the truck D

10 km h- 1in the direction of the car 190 kin h-1in the direction of the car

2 On his way home from training, Mitch walks along three streets. He \valks 1501n north, 300 n1 east and 250 n1 north. The n1agnitude of Mitch's resultant displace1nent is: A

200111

B

400 m

C

500111

D

700 n1

3 Three forces are acting sin1ultaneously in the san1e plane on an object. The resultant force is f = 520 N due east. Which of the following combinations of forces will give this resultant force? A f 1 = 350 N west, f 2 = 270 N east, f 3 = 440 N west B

f 1 = 350 N east, f 2 = 270 N west, f 3 = 440 N east

C

f 1 = 350 N east, f 2 = 270 N west, f 3 = 440 N \Vest f 1 = 350 N west, f 2 = 270 N east, f 3 = 440 N east

D

4 A soccer ball is struck with a force of 420 N in the direction N22°E. The northerly and easterly components of this force are f 1 and f 2 respectively. The 1nagnitudes of these forces are: A f 1 =87N,f2 =411N B f 1 =411N,f2 =87N C

f 1 = 157N, f 2 = 389N

D

f 1 = 389N, f 2 = 157N

5 Forces are acting on an object. Calculate the resultant force acting on the object for each of the following: (a) f 1 = 250 N north, f 2 = 450 N south, f 3 = 125 N north

(b) f 1 = 100 N east, f 2 = 100 N south

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6 A particle of mass 50 kg is suspended by two strings attached to two points in the same horizontal plane. If the t\vo strings each 111ake angles of 30° respectively to the horizontal, find the n1agnitude of the tension in each string, in newtons, given that sin 30° = ~.

7 Jaide paddles her canoe in a direction perpendicular to the riverbanks. The river is 120 n1 wide. The canoe's speed in this direction is l.5ms-1. The water in the river is flowing at 0.8ms-1 and you can assume that the riverbanks are parallel straight lines. (a) Calculate the velocity of the canoe relative to the bank. Give the angle correct to t\vo deci111al places. (b) At \Vhat distance, in the direction downstrean1 fro111 the starting point, will Jaide get to the other riverbank? (c) Calculate Jaide's final displacen1ent fro111 the starting point. (d) Jake is going to cross the river in another canoe. He can paddle at the speed of l.8111 s-I relative to the water's flow. At what direction should Jake row to get to the sai11e end-point as Jaide, and how 111uch tin1e will this take? 8 A particle is projected with a velocity whose horizontal and vertical components are 6 m s-1and 4 m s-1 respectively. Use g = l On1s- 2 . (a) Draw a diagram to sho\v this infon11ation. (c) Find expressions for ~(t) and r(t). (e) Find the horizontal range.

(b) What is the angle of projection? (d) Find the greatest height reached.

9 A cricketer hits a cricket ball off the ground towards a fielder \Vho is 65 m away. The ball reaches a maxin1u111 height of 4.9 n1 and the horizontal component of the velocity is 28 n1s-1. Use g= 9.8 n1 s-2 . (a) (b) (c) (d)

Find expressions for ~(t) and r(t). How n1uch time does it take for the ball to reach its greatest height? Ho\v far has the ball travelled horizontally when it has descended to a height of 1.3111? Find the constant speed with which the fielder must run forward, starting at the instant the ball is hit, in order to catch the ball at a height of 1.3 m above the ground.

10 A particle is projected fron1 a point 15111 above horizontal ground. At its highest point it just clears the top of a wall 26.25m high and 30m away. Use g= 10 ms-2 . (a) Draw a diagram to sho\v this infon11ation. (b) Find expressions for ~(t) and r(t). (c) Find the speed ai1d the ai1gle of projection of the particle.

11 A tennis ball is struck \Vith a force of 300 N in the direction S40°W The southerly and \Vesterly con1ponents of this force are f 1 ai1d f 2 respectively. The n1agnitudes of these forces are: A C

334

f 1 =230N,f2 =230N f 1 =230N,f2 = 193N

B D

f 1 =193N,f2 =230N f 1 = 150N,f2 =260N

New Senior Mathematics Extension 1 for Years 11 & 12

14.1

BERNOULLI TRIALS

When a coin is flipped, there are hvo outcomes possible: heads or tails. When a standard die is rolled, there are six outcomes possible: 1, 2, 3, 4, 5 or 6. If you are interested in \Vhether or not you roll a particular number, like 6 for exainple, there are only two outco1nes of interest in this scenario: 'getting a 6' ai1d 'not getting a 6'. The saine argu1nent can be applied to n1ai1y practical situations: aJ1 archer hits the target or 1nisses the target; a footballer scores a goal or 1nisses; a 1nedical test indicates the presence or absence of a disease. In all of these cases, you need to consider success aJ1d failure. When you attach a probability ofp to success, then the probability of failure will be 1 - p. Bernoulli trials are a \Vay of an alysing situations like these where there are exactly two possible outcomes: success or failure. The trials are independent, so the outcon1e of one trial has no influence over the outco1ne of the next trial, aJ1d the number of trials is fixed. Bernoulli trials are nan1ed after Jacob Bernoulli ( 1654-1705).

Example 1 Decide \Vhether each state1nent could represent a Bernoulli trial. (a) (b) (c) (d)

You You You You

can either pass or fail an exa1nination. can either buy a particular braJ1d of phone or not buy it. can either \Valk, ride your bike or catch the bus to school. can either get the job that you applied for or not get that job.

Solution (a) Only two outcon1es, so it is a Bernoulli trial. (c) Three possible outcomes, so it is not a Bernoulli trial.

(b) Only two outco1nes, so it is a Bernoulli trial. (d) O nly t\vo outco1nes, so it is a Bern oulli trial.

Bernoulli random variables Associated \Vith Bernoulli trials are Bernoulli raJ1do1n variables. Bernoulli randon1 variables are often encoded using the convention that the nu1nber 1 is success, and Ois failure. If X represents a Bernoulli randon1 variable, aJ1d a probability of p is associated to success, where O< p < 1, then: P(X = x )=

p ifx= l 1-p ifx = O

The probability distribution table for this situation is:

The expected value is given by: E(X) =0 X (1 -

p) + 1 X p

=p

0

1

1-p

p

The variaJ1ce is given by: Var(X) =E(X2)

-

= 02 X ( 1 -

[E(X)]2

p) + 12 X p- p2

2

For a Bernoulli random variable X:

=p -p =p(l-p)

E(X) =p Var(X) =p(l - p)

Chapter 14 The binomial distribution

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YEAR12 EXERCISE 14.1

BERNOUL LI TRIALS

1 In an exa1nination there are 10 multiple-choice questions \vith four possible answers, only one of which is correct. (a) If you just guess your ans\vers, does this form a sequence of Bernoulli trials? (b) Use Pascal's triangle (or the binomial expansion) to calculate the number of different ways you could get 6 out of the 10 questions correct.

2 At the recent local government elections, 20 people are selected at random after leaving the polling booth and asked if they have voted for a particular candidate. Would the list of responses represent a Bernoulli trial?

3 The babies born at a hospital in one 111011th have their sex recorded as '1nale' or 'female'. (a) Would the list of sexes represent a Bernoulli trial? (b) In 20 births at the hospital during the 111onth, 11 are male. Use Pascal's triangle (or the bino1nial expansion) to calculate the number of different ways this result could occur.

4 A coin is tossed four tin1es. (a) Would the list of outco1nes represent a Bernoulli trial? (b) If the possible outcomes are recorded as H (heads) or T (tails), list all possible sets of outcomes \Vith at least three heads.

5 Two standard dice are rolled together and the sun1 of the numbers rolled is noted. The result is recorded as either 7 or not 7. (a) Does this experiment represent a Bernoulli trial? (b) If pis the probability of a sun1 of 7, find the value of p. (c) In 36 rolls of the dice, how many tin1es would you expect the su1n to be 7?

14.2

BINOMIAL DISTRIBUTION

Bernoulli trials can help you to understand the 1nost in1portant of the discrete probability distributions, the binomial distribution. Suppose you are conducting an experiinent, consisting of n trials, where:

• n is detennined before the experiinent begins • all n trials are identical Bernoulli trials, \vith probability of success p and probability of failure

q=

1-

p

• all the trials are independent, so that the outco1ne fro1n any one trial has no effect on the outco1ne of any other trial. In these cases there is a binomial random variable, which in turn has a binomial probability distribution. There is a shorthand notation to indicate a binon1ial distribution: X - B(n, p) indicates a random variable X that has a binomial distribution \vith n identical trials and a

probability of success of p.

n and p are called the parameters of the distribution. X - B(n, p) is read as 'X is distributed as a bino1nial variable \vith para1neters n and p'.

Consider the experi1nent of dra\ving three cards, one at a tiine \Vith replace1nent, fron1 a standard pack of 52 playing cards. If interested in the nun1ber X of hearts cards selected, then Xis a binomial variable. Consider dra\ving a heart card to be a success. In tern1s of notation, n = 3 (the nun1ber of trials) and p = .!. (the probability of success on any 4 particular trial, as P(heart) = .!.. 4

So,x-B(3,:). 336

New Senior Mathematics Extension 1 for Years 11 & 12

YEAR 12

If you were drawing three cards fron1 a pack without replace1nent, then X would not be a bino1nial variable because the probability of success p would change with each draw. Even if an experin1ent has only two possible outcomes, it is not auto1natically bino1nial.

Example 2 Find the probability of obtaining exactly hvo hearts in a selection of three cards if the card is replaced after each selection.

Solution There are three ways to obtain exactly hvo hearts: HHN, HNH, NHH, ,vhere H represents 'getting a heart' and N represents 'not getting a heart'. In each case, the probability is equal to .!.x .!.xi. 4 4 4 2

P(exactlytwohearts)=3x ( -1 ) x -3 = - 9 4 4 64

Binomial theorem In Chapter 6 Pern1utations and combinations, you developed Pascal's triangle and the general expansion of (a + b)" using the binomial theore1n. This is ,vritten as: (a + b)" = "C0a" + "C1a''- 1b + "C2a" - 2b2 + .... + "C,a" - ' b' + ... + "C,,b"

n'

where "C = · ' (n-r)!r! MAKING CONNECTIONS

Pascal's triangle and the binomial theorem Use tech nology to construct Pascal's triangle and explore its relationship to the binomial theorem .

Example 3 Calculate, using technology or by hand, the coefficients in the expansion of (a + b) 6•

Solution 6

C0 = 1, 6C1 = 6, 6C2 = 15, 6 C3 = 20, 6C4 = 15, 6 C5 = 6, 6C6 = 1

Thus (a+ b)6 = a6 + 6a 5b + 15a4b 2 + 20a3 b3 + 15a2b4 + 6ab 5 + b6 .

It is usual to replace "C, by (~) when working with binon1ial probabilities. Do not n1ix up this notation with the colu1nn vector! It is easy to ,vork out sin1ple exan1ples like Exan1ple 2, but when finding the probability of dra,ving 13 hearts 3 from 30 draws, it is not as easy to list all the possible outco1nes. Instead you can use C13 to find the total nun1ber

°

of possible outcon1es. With the binomial distribution, use the alternative notation (;) or in this case, ( :~}

Chapter 14 The binomial distribution

337

YEAR12 Consider the probability of obtaining any one of these outco1nes, such as 13H followed by 17N. The probability of 3

7

thisis ( 1 ) ' x 4

(

3) '

4

When you consider all possible outcomes: 3

7

P(exactly 13 hearts fro1n 30 draws)= ( 30) x ( 1 ) ' x (3)' 13 4 4 Using technology to evaluate, the expression is equal to 0.013414448 8 ... which can be rounded to 0.0134.

A general expression for the probability that X takes a particular value x is as follo,vs:

P(X = x) = (:) px(l - p)" - x

Recalling that the probability of failure is so1netimes ,vritten as q, ,vhere p + q = 1, this can also be ,vritten as:

Example 4 A variable x follo,vs the distribu tion X - B( 10, 0.6). Find P(X = 5) for this distribution, expressing your answer correct to four decin1al places.

Solution Identify n (the n un1ber of trials), x (the nu1nber of successes), p (the probability of success), and 1 - p (the probability of failure):

p = 0.6

X=5

n = 10

1-

p = 0.4

Substitu te into the forn1ula: P(X = x) = (:) px( 1 - p )" - x

P(X = 5) =

(lsO) (0.6)

5

(0.4) 5

P(X = 5) = 0.2007 There is a clear link between the binomial probability distribution and the binon1ial theorem, although the order of the terms has been reversed. You can use this link to show that the sun1 of the bino1nial probabilities equals one. Reme1nber, it is a condition for a probability distribution that the su1n of the probabilities equals one. Using the expansion of the binomial theore1n:

L(n). p" - i (l-p)i =(p + l -p)" II

i=O I

= l" =l You can also think of a binomial distribution in tern1s of the expansion of (p + q)", where p + probability of success.

338

New Senior Mathematics Extension 1 for Years 11 & 12

q=

1, and pis the

YEAR 12

Example 5 The probability of any particular egg being cracked in a carton containing a dozen eggs is 0.05. Find the probability that exactly three eggs are cracked, stating your ans\ver correct to three deciinal places.

Solution Binomial probability is appropriate because there are two outcon1es, a fixed probability of success and a defined nu1nber of trials. Identify n (the number of trials), x (the nun1ber of successes, i.e. cracked eggs), p (the probability of success), and 1 - p (the probability of failure):

n = 12, X = 3, p = 0.05, 1 - p = 0.95 Substitute into the forn1ula: P(X = x ) = (:) px( I - p) ,, _x 1 P(X = 3) = ( : )(0.05) 3 (0.95) 9 P(X=3) = 0.017

Example 6 A particular medical test correctly identifies whether or not a person has an illness 98% of the tin1e. If 10 people are tested, find the following probabilities, correct to three deci1nal places: (a) No people are incorrectly diagnosed. (b) At least one person is incorrectly diagnosed.

Solution (a) Identify n, x and p: n = IO, x = IO, p = 0.98

Substitute into the fonnula: P(X = x) = (: )px(l - p)" - x

P(X = 10) = (~~) (0.98) 10 (0.02) 0 P(X= 10)= 0.8 17 (b) P(at least one incorrectly diagnosed) = 1 - p (none incorrectly diagnosed) (None incorrectly diagnosed is the same as all correctly diagnosed) P(X < 10) = 1 - P(X = 10) = 1-0.817 = 0.183

Note that P(X > a) = 1 - P(X < a). This can be a useful result in many situations.

Chapter 14 The binomial distribution

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Example 7 Accurate Andy is a darts player whose favourite shot is the triple-hventy. Andy is successful with this shot 80% of the time. In each of the following situations, An dy has five shots. In each case state your answer correct to four decin1al places where necessary. (a) Find the probability that Andy makes exactly two triple-hventies. (b) Find the probability that Andy 1nakes n1ore than three triple-twenties.

(c) Find the probability that Andy 1nakes triple-hventies on his first and fifth attempts only.

Solution (a) n = 5, x = 2, p = 0.8

Substitute into the formula: P(X = x) = (: )px(l -

p)" - x

P(X=2l= (~)co.8)2(0.2)

3

P(X = 2) = 0.0512

(b) P (more than three successes)= P(four successes)+ P(five successes) P(X > 3) = P(X = 4) + P(X = 5) =

(!)

1

5

0

(0.8)4(0.2) + (:) (0.8) (0.2)

= 0.4096 + 0.32768 ""0.7373 (c) Certain conditions have been specified, therefore you cannot use a complete tern1 fro1n the binon1ial distribution. In this case, consider only the sequence SFFFS, not the 1nore general case of hvo successes anywhere within the five trials. P(triple-twenty on first and fifth only)= 0.8 x 0.2 x 0.2 x 0.2 x 0.8 = 0.00512 "'0.005 1

Graph of the binomial distribution Consider again the experiment where you draw three cards, \Vith replacement, fron1 a standard pack of 52 playing cards. Reme1nber, you are interested in the nun1ber of hearts that occur in the three cards drawn, so X - B(3, 0.25). The following table shows the distribution of the rando1n variable: X

0

1

2

3

P(X=x)

0.422

0.422

0. 141

0.016

Here, the su1n of the probabilities is 1.001. This is due to the rounding of the probabilities and can therefore be regarded as 1. A graph of this distribution is shown:

P(X= x)

0.45

0.422 0.422

0.40 + 0.35 + 0.30

·•

0.25

"

0.20 0. 14 1

0.15 + 0.10 + 0.05

·•

0.0 16

0

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New Senior Mathematics Extension 1 for Years 11 & 12

1

2

3

'

X

YEAR12

No,v, consider the effect that the values n and p have on the graph of the distribution. The following diagrams ,vill help you understand the effect each of these variables has on the graph. Set n = 10. B( lO, 0.25)

B(lO, 0.5)

B(lO, 0.75) P(X=x)

P(X= x) P(X =x)

0.30 0.25

II= 10, p = 0.25

0.30 0_25 _

II= 10, p = 0.5

0.25

0.20

0.20 ••

0.20 -

0.15

0. 15 +

0. 15

0.10

0. 10 ••

0.10

0.05

0.05

0.05 -

.n. 0 I 2 3 4 5 6 7 8 9 10

0

X

.n

.n .• ..•

I 2 3 4 5 6 7 8 9 10

t1=10, p=0.75

X

0 . 1 . 2 3 " 4 5 6 7 8 9 10

X

MAKING CONNECTIONS

Graphing binomial distributions Move the sliders to explore the effects of changing n and p on the graph of a binomial distri bution.

For graphs of the binomial distribution, a,vhere X - B(n, p): • Ifp < 0.5 the graph is skewed to the right (positively skewed). • Ifp = 0.5 the graph is symmetric about the mean . • Ifp > 0.5 the graph is skewed to the left (negatively skewed). • As n increases, the graph clusters n1ore tightly about the n1ode but retains the sa1ne shape as other distributions with the same value of p.

EXERCISE 14.2

BINOMIAL DISTRIBUTION

1 Find the stated probability for the following bi1101nial distributions. Give your answers correct to four decin1al places (d.p.). (a) P(X = 4) if X - B( l 2, 0.7) (c) P(X = 2) if X - B(B, 0.3)

(b) P(X = 6) if X - B(20, 0.45) (d) P(X = 30) if X - B(50, 0.6)

2 A spinner is divided into eight equal sections, of which three are shaded. Find the probability, correct to four decimal places, of the spinner landing on a shaded section exactly five tiines out of 10 spms. 3 A stan dard six-sided die is rolled 50 tin1es. Find the probability, correct to five deciinal places, of the follo,ving outcomes: (a) exactly 30 even nun1bers are rolled

(b) at least one 6 is rolled.

4 Peter, a keen gardener, knows fro1n past experience that only 60% of his tulip bulbs will flower. He plants 20 bulbs. Fmd the probability, correct to five decin1al places, of the follo,ving outcomes: (a) exactly 15 bulbs flower

(b) n1ore than 15 bulbs flo,ver.

Chapter 14 The binomial distribution

341

YEAR12

5 For each of the following situations, state \vhether or not it is suitable for binomial 1nodelling. For those that are not, explain why the binon1ial model is not suitable. (a) You are playing a gan1e where you are required to roll a 6 before you can start. The number of rolls

required for the eight players is recorded. 1 (b) A die is biased in such a way that P(4) = - .The die is rolled 40 tiines and the nu1nber of 4s rolled

is recorded.

3

6 If 100 cards are dra\vn with replacen1ent fro1n a standard pack of 52 playing cards, find the probability of the following events, correct to four deciinal places: (a) exactly 25 hearts are dra\vn

(c) exactly 55 red cards are dra\vn

(b) exactly 40 cards are clubs (d) exactly 60 cards are black. 4

7 The probability that an archer will shoot a bullseye \Vith any particular arrow is - . The archer shoots 5

10 arrows. Which of the follo\ving expressions best represents the probability that the archer misses the bullseye \Vith exactly two of the arrows?

8 In Amazonia, 62% of all births are girls. Find the probability, correct to four decin1al places, that of eight births exactly half are girls. 9 In the town of Sinistraville, 35% of the population is left-handed. Find the probability, correct to four decimal places, that exactly 35 people out of a randon1 san1ple of 100 will be left-handed. 10 A coin is biased in such a \Vay that P(heads) = 3 x P( tails). Find the probability, correct to three deciinal places, that exactly six tosses out of 10 land on heads. 11 For the graph shown, choose the best representation for its binon1ial distribution.

A

X - B( 12, 0.5)

B

C

X - B( l 2, 0.3)

D

P(X=x)

0.25

X - B(20, 0.3) 0.20 X - B(20, 0.5) 0.15 0. 10

0.05 0 1

2

3

4

5

6 7

8

9 10 11 12 13 14 15 16 17 18 19 20

x

12 Cards are dealt from a standard pack of 52 playing cards without replacen1ent. Let X be the nun1ber of hearts dealt and assu1ne five cards are dealt. Can this experiinent be described as X - B ( 5,:)?

13 On the way to work, Philo1nena 1nust drive through six sets of traffic lights. The lights are independent of each other and the probability that Philon1ena must stop at any particular set is 0.7. Find the probability, correct to four decin1al places, that Philo1nena stops at the following sets of lights: (a) exactly five sets of lights (b) fe\ver than two sets of lights (c) more than five sets of lights (d) the first three sets of lights (e) the second and sixth sets of lights only ( f) four sets of lights, including the first t\vo.

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New Senior Mathematics Extension 1 for Years 11 & 12

YEAR 12

14 Boxes of matches are supposed to contain 47 ,natches. Production records indicate that 80% of boxes contain 47 ,natches. A batch of20 boxes is sa1npled. If 1nore than four boxes do not contain exactly 47 ,natches, production is stopped. (a) Find the probability, correct to four deci1nal places, that of the 20 boxes selected, the nu1nber of boxes that do not contain 47 ,natches is the following: ~) 0 boxes (ii) 1 box (iii) 2 boxes (iv) 3 boxes (v) 4 boxes (b) Find the probability that the nun1ber of boxes that do not have 47 matches is no n1ore than four. (c) Find the probability that production is stopped. 15 The local baker kno\vs that, on average, 8% of the loaves of bread baked each day are slightly burnt and cannot be sold at full price. The likelihood of a particular loaf being slightly burnt is independent of any other loaf being slightly burnt. For each of the follo\ving, give your answers correct to four decimal places \Vhere necessary. (a) If a random sample of 12 loaves is selected, what is the probability that exactly one loaf cannot be sold

at full price? (b) For the san1e rando1n sa1nple, calculate the probability that exactly three loaves are slightly burnt. (c) What is the probability that all loaves in the random sample are sold at full price? (d) To make a profit, the baker has to sell more than 40% of the loaves at full price. Calculate the probability that the baker n1akes a profit selling the 12 loaves.

16 A recording co1npany receives a large number of ne\v songs from various artists. On average, only 4% of the ne\v songs become popular hits. One of the producers decides to select a random sample of 35 new songs. (a) What is the probability that a maxin1um of three new songs fro1n the sample \vill become popular hits? State your ans\ver correct to three deciinal places. (b) What is the probability that 1nore than four but less than six songs fro1n the sample \vill become popular hits? State your ans\ver correct to three deci1nal places. (c) Detern1ine the 1nost likely number of songs fron1 this san1ple that will becon1e popular hits. Explain your answer using appropriate calculations. (d) The producer decides to select 70 songs at rando1n from the new songs received. If the probability of a new song becon1ing a popular hit has not changed, \Vhat is the probability that exactly one ne\v song will become a popular hit? (e) What \vould be the n1ost likely nu1nber of new songs fro1n this sa1nple to beco1ne popular hits? Explain your answer using appropriate calculations.

17 Black n1arket DVDs have a 0. 15 chance of being faulty. Five DVDs are purchased independently. (a) Draw a table to sho\v the probability distribution of X, the nun1ber of faulty DVDs purchased. (b) Find the probability that fe\ver than four faulty DVDs are purchased if you know there is at least one faulty DVD.

14.3

MEAN AND VARIANCE OF THE BINOMIAL DISTRIBUTION

Consider a rando1n variable X - B(5, 0.4). The probability distribution for Xis shown in the follo\ving table: X

0

1

2

3

4

5

P(X=x)

0.07776

0.2592

0.3456

0.2304

0.0768

0.0 1024

Using the 1nethod outlined earlier in this chapter, you can find the expected value of X, E(X), and the Variance of X, Var(X). 5

E(X)=

L

X;P ;

;=o = 0 X 0.07776 + 1 X 0.2592 + 2 X 0.3456 + 3 X 0.2304 + 4 X 0.0768 + 5 X 0.01024

= 0 + 0.2592 + 0.6912 + 0.6912 + 0.3072 + 0.0512 =2

So, [E(X) ]

2

=4 C hapter 14 The binomial distribution

343

YEAR12 5

2

E(X

)

='I,x; P; i=(J

= 0 X 0.07776 + 1 X 0.2592 + 4 X 0.3456 + 9 X 0.2304 + 16 X 0.0768 + 25 X 0.01024 = 0 + 0.2592 + 1.3824 + 2.0736 + 1.2288 + 0.256

= 5.2 So, Var(X) =E(X2) = 5.2 -4

[E(X)]2

= 1.2 When dealing with a binon1ial distribution, if Xis the randon1 variable representing the nu1nber of successes in n trials and p is the (constant) probability of success, then: E(X) = np = µ (1nean)

=np(l - p) a(X) = Jnp( l - p) Var(X)

You can check these results for the example given at the very beginning of this section, where X - B(5, 0.4). Using the rules: E(X)

= np = 5 X 0.4

Var(X)

=np(l -

p)

= 5 X 0.4 X 0.6 = 1.2

=2 The same answers are obtained.

Example 8 Find the follo,ving statistics for X - B(l5, 0.3). If necessary, give answers correct to hvo deciinal places. (a)

E(X)

(b)

Var(X)

(c)

a(X)

Solution (a) E(X)

= np = 15 X 0.3

(b) Var(X) = np(l - p) =4.5 X (1 - 0.3) = 3. 15

=4.5

Example 9 Given X - B(20, p) and E(X)

=5, find the value of p.

Solution Use E(X) = np: np = 5 Substitute the kI10,vn value: 20 x p = 5

5 1 Solve for the unknown: p = - = - = 0.25 20 4

344

New Senior Mathematics Extension 1 for Years 11 & 12

(c) a(X) = .Jnp(1- p)

= ✓3.15 = 1.77

YEAR 12

Example 10 Given X - B(n, p), µ = 9 and a 2 = 6.3, find the values of n and p.

Solution a 2 =np(I-p) a 2 = 6.3 :. np(I - p) = 6.3

µ=np µ =9 :.np=9

Substitute for the known value of np in the a 2 equation: 9(1 - p) = 6.3 1-p =0.7 p= 0.3 Fin d n:

n x 0.3 = 9

9 n= 0.3 n=30

n = 30, p = 0.3, so X - B(30, 0.3)

EXERCISE 14.3

MEAN AND VARIANCE OF THE BINOMIAL DISTRIBUTION

1 Find the stated values for each of the following binon1ial distributions. (a) X - B(20, 0.7). Find E(X), Var(X), a (X)

(b) X - B(IOO, 0.55). Find E(X), Var(X), a (X)

(c) X

~ B( 50, ~ ). Find E(X), Var(X), a (X)

(d) X

(e) X

~ B(25,0.8). Find E(X) , Var(X), a(X)

(f)

X

~ B( 10,0.5). Find E(X), Var(X), a (X) ~B( 40,0.4). Find E(X), Var(X)

2 For each of the follo\ving, find the value of the unknown. (a) X - B(l5, p), E(X) = 5, p =? (c) X - B(n, 0.2), E(X) = 7, n =?

(b) X - B(30, p), E(X) = 3, p =?

(d) X - B(n, 0.4), E(X) = 10, n = ?

3 For each of the follo\ving, find the values of n and p for X - B (n, p). (a) X - B(n, p), µ = 4.8, a 2 = 2.88 (c) X - B(n, p), µ = 4.5 and a 2 = 3.825 (e) X-B(n,p),µ=65anda 1 =22.75

2

(b) X - B(n, p), µ= 2 and a = 1.8

(d) X - B(n, p), µ= 11 and a 2 = 8.58 (f) X-B(n,p),µ=41.25anda 2 =29.90625.

4 Given X - B(30, 0.4), which of the following pairs of values is correct? A E(X) = 12, Var(X) = ✓ 72 B E(X) = 12, a 2 = 7.2 C E(X) = 18, a 2 = 7.2 D E(X) = 18, a = ✓ 72

5 Given X - B(n, p), µ=IO and a 2 = 8, \Vhich of the following pairs of values is correct? A

n = 80, p = 0.8

B

n = 50, p = 0.8

C

n = 80, p = 0.2

D

n = 50, p = 0.2

6 Find P(X = 3) given that E(X) = 3 and Var(X) = 0.75 and X ~ B(n, p). 7 A spinner is divided into four equal sections, one of \Vhich is coloured blue. If the spinner is spun eight times, calculate the probability of obtaining less than the expected number of blue outcomes, correct to three deci1nal places.

8 For the variable X - B(40, p), a 2 = 9.6. The correct value(s) of pis: A 0.4 B 0.6 C 0.4 and 0.6 D

none of these.

Chapter 14 The binomial distribution

345

YEAR12

9 Consider X - B(8, 0.25). (a) Construct a table showing the probability distribution. Round probabilities to four deci1nal places. (b) Use the ruleµ= np to find E(X). (c) Use the rule a= ✓Var(X) to find a(X), correct to hvo deciinal places.

10 A fan1ily has six children ,vho are all boys or girls. Assu1ne the probability of any child being a boy is 0.5 and that the probability is independent for each child. Find the probability of each of the follo,ving, roundmg ans,vers to four deci1nal places: (a) the first two children born are n1ale (c) there are n1ore girls than boys

(b) there are three boys and three girls, in any order (d) there is at least one boy, but more girls than boys

(e) there are no consecutive births of the san1e sex.

11 A researcher has calculated the n1ean and the variance for a san1ple of a given random variable that has a binon1ial distribution. (a) If the mean of the data set is 42 and the variance is six, determine the nu1nber of trials (n) and the probability of success (p). (b) A new set of data is collected and the researcher notices that the mean and variance of the new set of data are double the corresponding values of the first set of data. Co1npare the hvo sets of data using appropriate calculations involving n and p. (c) The researcher realises that the results for 10 trials have been on1itted fro1n the first set of data. When the 1nean is recalculated with the 10 additional results, the n1ean is unchanged atµ= 42. What variance should the researcher expect for this set of data? Give your answer correct to two decimal places. (d) Another researcher conducts the same experilnent and records the data for 56 trials. The probability of 6 success for this set of data is calculated to be p = . What values of the mean and variance should the 7

researcher expect to calculate for this set of data? Give your answer correct to hvo decimal places. (e) Write a statement to describe the effect on the 1nean and variance as the nu1nber of trials increases.

14.4

NORMAL APPROXIMATION FOR THE SAMPLE PROPORTION

You can make sense of the ,vorld around you by studying particular features or characteristics that occur ,vithin populations. To study a whole population can be impractical, costly, time-consun1ing and so1netimes in1possible. In real-life situations, this 1neans it is rare to kI10,v the probability of a certain characteristic occurring within a given population. This issue can be overcome by obtainmg a sample and usmg the inforn1ation from this san1ple to infer particular characteristics about the population. In other words, you study the 'part' in order to predict information about the 'whole'. This is the basis of statistical inference. In your previous studies you have analysed sets of data, sun1n1arised and then presented the mforn1ation. This is kilown as descriptive statistics. In this chapter, you will be using inferential statistics. While descriptive statistics sun1n1arises and presents infonnation, inferential statistics uses the statistics fro1n a rando1n san1ple to draw conclusions about the population. Recall that a san1ple is considered random if every unit of the population has an equally likely chance of being selected in the sainple.

Sampling distributions and variability The statistical results of rando1n sa1nples taken from a population ,viii vary. This leads to the idea that statistics associated with samples from a population can be represented by a rando1n variable ,vith its own probability distribution. This applies when ,vorking ,vith either sample 1neans or sainple proportions. To understand the behaviour of a probability distribution of results fro1n sa1nples, you need to look at what happens when n1any samples are taken. One way of doing this is to create a sinlulation and observe the behaviour of the data.

346

New Senior Mathematics Extension 1 for Years 11 & 12

YEAR 12 Probability distribution of a fair ten-sided die

The probability distribution of a fair ten-sided die is shown on the right. This is a discrete unifonn distribution.

P(X=x)

...

In theory, it is expected that the results will follow a discrete uniform distribution.

0.11

Using the formulas for the expected value and standard deviation of a unifonn distribution:

0.09 "

E(X)=

u+l 2 10+ 1

2 11 =2 =5.5

2

ando- =

(u-1)

0.10 "

... 0.07 ... 0.06 ... 0.08

2

a=JN 12

12

_ (10-1)

2

0.05

=2.60

"

0.04 "

12

0.03

...

... 0.01 ...

81 =12 =6.75

0.02

1

The graph on the right displays the results fro1n 1000 observations of rolling a fair ten-sided die. This can be considered as a very large san1ple of a random variable fron1 a discrete unifonn distribution. The sample 1nean and standard deviation were calculated.

23456

78910x M==S.36

P(X =x)

St:wcbrd d,:,,iatioo = 2.8 7

120

100

-

80 . -

Note that this sample distribution is very close to the discrete uniforn1 distribution, but not identical. This would be expected in n1ost cases. Such a random sam ple ,vould have a n1ean around 5.5 and the expected number of tin1es of each value occurring is close to one-tenth of the san1ple size, in this case 100. This sample's 1nean is 5.36 and the standard deviation is 2.87.

60

.-

40

20 l

2

3

4

5

6

7

9

8

10

X

... E 15 ...

Now consider what happens if you take 100 samples of size 50 and plot the n1ean of each sa1nple. Note in this case a histograin is used to plot the means of the sa1nples, so that similar sized n1eans are grouped within one interval, providing a better picture of the overall pattern thai1 plotting each individual sam ple 1neai1 separately.

,., 20 .£

"-

~

"' 10 0 ti E :;;

..0

5

...

z

.

4.5

. .

5.0

.

.

.

5.5

.

. I' . I

6.0

6.5

Mean of samples

Repeating the same process 1000 tin1es gives a better indication of the shape of the distribution of the san1ple 1neai1s. Here it can be seen that the distribution is beco1ning more bell-shaped and syin1netrical, 1nuch like a norn1al distribution. From the histogram it can be seen that the mean is very close to 5.5, ,vhich reflects the probability of the discrete unifonn distribution.

"'

c"..

E ~

"'0

... 80 ...

-

100

-

60

~

" z

..0

E :,

40 20

.

_r

. .

4.5

-

- .....

-.....

-

5.0

5.5 Mean of samples

- .....

- -

6.0

-

. .

6.5

.

The central limit theorem This sin1ulation de1nonstrates one of the 1nost important theore1ns in statistics- the central limit theorem (CLT). While the central lin1it theore1n is not required knowledge for this course, it is helpful to understand the significance of this concept ai1d the part this plays in inferential statistics. Informally, the CLT says that the san1pling distribution for a given statistic is essentially norn1ally distributed, regardless of the nature of the parent distribution. This means that you can use a norn1al distribution to 1nodel the sampling distribution for that statistic, and hence develop confidence intervals for estimates for a population parameter. Chapter 14 The binomial distribution

347

YEAR12

In most cases, the population para111eters are unknown, so the CLT is used in reverse. Taking samples of size n fro1n any distribution, you can calculate the 1nean for each san1ple and kno,v that the collection of sample 1neans will be approxilnately norn1ally distributed. This implies that the average of the san1ple means ,viii be approxilnately equal to the population 1nean. Consider obtainmg san1ples of size 50 ,vhen rolling a ten-sided fair die. When a s1nall nu1nber of samples of size 50 are generated, the distribution of the san1ple means does not show a norn1al distribution as clearly as when larger sa1nples are taken. The shape is n1ore randon1 and often can be ske,ved to the left or the right, or very spaced out. When a large nun1ber of sainples of size 50 are generated, the distribution of the san1ple means starts to resemble the shape of the nonnal curve 1nore closely; the larger the number of san1ples generated, the 1nore evident this is. MAKING CONNECTIONS

- - - - - - - - = = = = = = = -10

Central lim it theorem and sample proportions Explore how the shape of a distribution changes as more samples are taken.

2

If a large enough san1ple is taken fro1n any population where the n1ean JL and the variance cr are known, then the sa1nple observations have a distribution ,vhich is approximately nonnal ,vith 1nean nµ. and variance ncr 2• In n1ost cases, a san1ple size greater than or equal to 30 is usually enough to assun1e that the san1ple means will follow a normal distribution. If X 1, X 2, X3, ••• , X,, are independent identically distributed randon1 variables, each ,vith 1nean µ and varian ce er 2, then as n ➔ 00 , D= X 1 + X2 + X3 + ... + X11 ~ N(nJL,ncr 2 ). The central li1nit theore1n is often expressed in another way. Instead of dealing ,vith the sun1 of the san1ples, you can instead deal with the n1ean of the samples, X. In this case, you have X ~

NG,:

2 ) .

This n1eans that no matter what the distribution of XI> X2, X3, ••. , X11 looks like, if there are lots of distributions then the san1ple statistics will be approxilnately nonnal. In fact, 'lots' is often not a very large number, only n > 30 as stated above. If the underlying population has a norn1al distribution, then the sample size is usually not an issue. It is only when the distribution of the underlying population is son1ething other than nonnal that the sa1nple size is n1ore iinportant.

Sample proportions-notation and terminology Medical research indicates that 4% of the Australian population carry the gene for cystic fibrosis, one of the most co1n1non life-threatening genetic conditions in Australia. You could ask, 'Ho,v accurate is 4% with respect to the population?' Realistically, the exact population paran1eter is unknown . Unless you are willing to perforn1 a census, your infonnation and subsequent calculations ,viii con1e from a rando1n sa1nple. However, it can be shown that a proportion calculated fro1n a simple random san1ple can provide a reasonable estimate of the population paran1eter. For exa1nple, consider if a randon1 san1ple of 4000 people is tested for the cystic fibrosis gene. Fron1 this san1ple, it is 140 found that 140 people are carrying the gene. This gives a proportion of- - = 0.035. As this proportion is obtamed 4ooo from the random sample, it is called the sam ple proportion.

348

New Senior Mathematics Extension 1 for Years 11 & 12

YEAR 12

The table below illustrates definitions and terminology associated with sample proportions fron1 this example. Notation X

n

Definition Observation For any one san1ple, the number of people or things that possess a certain characteristic is called the observation.

Example The observation is the 140 people \Vho carry the gene.

Sample size

The san1ple size is 4000.

This is the number of people or things that n1ake up the san1ple. X

' p=n

p

Sample proportion This is the number of people or things with a characteristic that occurs \vi thin the san1ple (observation), as a proportion of the nu1nber in the sainple.

140 out of 4000 carry the gene. 140 ' p= 4000

Population proportion

=0.035 (Unknown.)

This is the true proportion of the people or things with a characteristic that occur within the whole population. The sainple proportion

p= ~ is not the population proportion; rather, it is the best estiinate of the population

n proportion fro1n a single sa1nple. Because of this, the sa1nple proportion is sometiines called the point estimate.

Example 11 Consider a san1ple of2000 18-year-olds in NSW Fron1 this sa1nple, 210 have their provisional driver's licence (P plates). Using this inforn1ation, identify: (a) the population proportion

(c) the observation value

(b) the sa1nple size (d) the san1ple proportion.

Solution (a) p is the population proportion.

The population proportion, \Vhich is unkno\vn, is the proportion of all 18-year-olds in NSW who have their provisional licence. (b) The sample size is n, the nu1nber of people in the san1ple: n = 2000 (c) The observation value is the nun1ber of people in the sample \Vho have their provisional driver's licence: x = 210 ,.,

X

,.,

X

(d) The sample proportion is given by p=- : p=-

n

n 210 2000 =0.105

Chapter 14 The binomial distribution

349

YEAR12

The CLT and proportions Probability distribution of a fair ten-sided die

Earlier in this chapter, the variability of results benveen rando1n sa1nples taken from a single population was explored. Recall that any statistic from all of the possible sa1nples is a randon1 variable and each of these rando1n variables has a probability distribution.

P(X= x) 0.11

0.10

To understand the behaviour of a sa1npling distribution for proportions, a sin1ulation has been created that will allo\v you to observe the behaviour of the data.

0.09 .

0.08 . 0.07 .

From the previous section, recall the probability distribution of a fair IO-sided die. This is a discrete unifonn distribution.

0.06

0.05

Rather than considering the san1pling distribution of the 1neans, you will now consider \Vhat happens if a particular number occurs, in this case the nun1ber '3'. This simulation \viii observe 100 samples of size 50. In each san1ple, the die has been rolled 50 tin1es and the nun1ber of threes 'recorded' for each sa1nple.

0.04 .

0.03 . 0.02 . 0.01 ' I

A histogram is used to plot the occurrences of the nu1nber '3' as it provides a better picture of the whole pattern than plotting each individual result separately.

2

3

4

5 6 7

8

9 10 11

20 .,.

.

. .,.

This n1eans that in san1ples of 50 rolls, 3 samples had one 'three', 12 samples had 2 'threes', 11 samples had 3 'threes' (and so on); up to one sample had 10 'threes'.

.

..,.

.I

Note that the distribution of the occurrence of the nun1ber 'three' does not resemble a unifonn distribution.

'

12

345678910 Number of threes rolled

Repeating the saine process 1000 times gives a better indication of the shape of the distribution of the occurrence of the nun1ber three. Rolling a IO -sided die 50 tiines, with the probability of a 'three' being 0.1 means that you would expect 50 x 0.1 = 5 'threes'. In fact, you n1ay get n1ore or less 'threes' in son1e sa1nples.

.

200 .,. "'~ 150

~

.

.,. . ~ . § 50 ,. z .I

'o 100 ~

I

2

3

4

5

6 7 8 9 Number of threes rolled

10

II

' 12

13

' 14

The n1ore sa1nples that you take, the n1ore the distribution beco1nes bell-shaped and syin1netrical, like a norn1al distribution. The tallest column on this histogran1 is above five, meaning that the expected result of 5 'threes' per sa1nple n1ore often thai1 any other result. The CLT applies equally to the proportion of a specific randon1 variable in a san1pling distribution, as it does to the sampling distribution of meai1s. Taking samples of size n, fro1n any distribution, a specific proportion of a randon1 variable can be calculated and it will be approxiinately norn1ally distributed. This implies that the proportion of a specific randon1 variable in a sa1nple will be approxin1ately equal to the proportion of the same specific random variable in the population.

350

New Senior Mathematics Extension 1 for Years 11 & 12

x

YEAR 12 MAKING CONNECTIONS

The CLT and sample proportions The activity simulates obtaining samples of size 50 for a 10-sided fair die. Observe how the shape of the d istribution of the number of threes obtained changes as more samples of size 50 are taken.

When a s1nall nu1nber of san1ples of size 50 are generated, the distribution of the nun1ber of threes does not clearly show a normal distribution. The shape is more random and often skewed to the left or the right or very spaced out. When a large number of san1ples of size 50 are generated, the distribution of the nu1nber of threes starts to rese1nble the shape of a nonnal curve. The larger the nun1ber of samples generated, the n1ore evident this is.

Sampling distribution for sample proportions It is useful to explore the relationship behveen Bernoulli randon1 variables and binomial distributions, and how these relate to san1ple proportions. Recall fro1n your work on discrete random variables that a Bernoulli rando1n variable is a discrete random variable ,vhich has hvo outco1nes: a 'success' with an associated probability of p and a 'failure' with the associated probability of 1 - p. Within a binon1ial situation, if n units are chosen at rando1n, then the variable X (the count of successes in a san1ple) has a binon1ial distribution X ~ B( n, p). The sainple proportion

p= Xn

is the bino1nial count divided by the sa1nple size, ,vhich gives a value between , , 0 and 1. This 1neans that the set of p values are related to a rando1n variable called P and follow a binomial distribution. Notice the differences in notation. The values X and pare used ,vhen observing a single random sample. However, ,vhen considering the ,vhole set of values that can be obtained fron1 all sainples, X and P' are used. The standard deviation of a binomial distribution can be estin1ated fron1 the sa1nple proportion pof one sample, rather than 1nany. This cannot be done easily for other distributions. As previously discussed, both the 1nean and the standard deviation are defined using the population para1neter p. However, the value ofp is unknown, so the exact meai1 and standard deviation cannot be calculated. To overcon1e this problem, use the sample proportion pas an estin1ate for both the expected value and standard deviation, and replace p ,vith p. When preplaces p in the standard deviation, then it is often called the standard error S ( instead. This is to distinguish between the use of a paran1eter and a statistic.

j,)

The distribution off, for the sample proportions is defined as f, ~

,vhere the expected value

N p,) p(lnP) ,

E(P)"' pand the standard deviation SD"' ✓p(ln p) =S(P).

Example 12 A social 1nedia survey found that 134 out of 265 respondents had used son1e forn1 of social n1edia within the past year. Using this infonnation: (a) calculate the expected value for the proportion of people ,vho access social n1edia, correct to

two decimal places (b) calculate the standard error, correct to three decimal places.

Chapter 14 T he binomia l distribution

351

YEAR12

Solution (a) Use the formula for the expected value: E (

P) "' p n

Substitute the given values into the fonnula:

E(fa) "' 134

265 "' 0.51 The expected value E( f>) "' 0.51, to hvo deciinal places.

(b) Use the formula for the standard error: s(f>) = ~

Substitute the given values into the fonnula: s( fa) =~ 0.5lx0.49 265 "' 0.03 1 The standard errors( f>) = 0.031, to three deci1nal places.

Example 13 A recent survey of 1500 Year 12 students sho\ved that 990 were intending to apply for university. (a) Calculate the expected value. (b) Calculate the standard error, correct to four deciinal places.

Solution (a) Find the expected value using the forn1ula:

E(i>)"' p

__

,,, 990 1500 "' 0.66 (b) Use the formula for the standard error: s(

p) = ~ 0.66(1-0.66) 1500 "' 0.012231

The standard error is s(f>) = 0.0122.

352

New Senior Mathematics Extension 1 for Years 11 & 12

YEAR 12

The size of n Fro1n the CLT, you know that where a randon1 sample of size n has X 'successes: ,vhen n is sufficiently large, the san1pling distribution of the san1ple proportion pwill be approxin1ately normal with mean p and standard deviation

✓P(1n P) .

Sampling distribution of pro portion 11=50, p=O.I

,._ V

~

5,l: §1 ~

~

0.20 0.18 0.16 0.14 0.12 0.10 0.08 0.06 0.04 0.02

" " " " "

Sampling distribution of proportion t1 = 100, p = 0.1

--

0. 14 >-

C

"'g. _g

-

!/ ·.!E=

-

"-

0.05 0. 10 0.15

~

:,

0.10

8" 0.04

J::

0.08

.~ 0.03 ~

0.06

.!E ~

0.04

' 0.20

' 0.25

' 0.30

0.02 0.0 1

0.02

-,

0.06

E o.os

1;- 0. 12

'""

-

Sampling distribution of proportion t1 = 500, p= 0.1

j, 0.05 0.10 0.15

0.20 0.25

0.04 0.06 0.08 0.10 0. 12 0.14 0. 16

Here you can see that for a value of p = 0. 1 and n = 50, the distribution ,nay look bell-shaped but it is noticeably skewed. However, as the value for n increases, the shape, ,vhile retaining the bell shape, becomes 1nore syn1n1etrical about p. This means that even if the value of pis not close to 0.5 (i.e. the population is unbalanced), the larger the san1ple, the n1ore the sampling distribution of presembles a nonnal distribution. So what value of n is considered sufficiently large? The following conditions allow you to detennine if the CLT can be applied. The following hvo conditions n1ust be n1et for the CLT to apply:

• np>5 • n(1-p)>5

You also need to approxin1ate the population proportion p with the sample proportion p. However, note that if tile population already has a norn1al distribution then the conditions are not necessary.

Example 14 A survey of 25 people has obtained a sa1nple proportion of 0.2. Detern1ine if tile distribution can be approxiinated by a normal distribution.

Solution Define the parameters: n = 25 and p= 0.2. Test these paran1eters for each condition:

np > 5 25x0.2=5

5>s

And: n(l- p)> 5 25x0.8=20 20>5

Botll staten1ents are true, so the distribution can be approximated by a norn1al distribution.

EXERCISE 14.4

NORMAL APPROXIMATION FOR THE SAMPLE PROPORTION

1 Read through tile following scenarios and identify the population proportion, san1ple size, observation value and sample proportion. Give answers correct to two decilnal places where necessary. (a) A report by an IT data con1pany in tile Northern Territory surveys 800 people and reports that

42% of then1 have received a fake en1ail clain1mg tlley have won a large ainount of 1noney. (b) A database search shows tllat approximately 215000 Tas1nanians own a Holden. A survey of2500 Holden owners in Tasn1ania sho,vs that 1893 prefer a n1ediun1 - to large-sized car over a s1naller ,node!. Chapter 14 The binomial distribution

353

YEAR12

2 A survey has found that 64 of 104 Year 12 students are not sure what to do after they leave school. Using this inforn1ation, calculate the expected value and standard error for the population proportion of Year 12 students who are not sure what to do after they leave school. Give answers correct to three deciinal places. 3 The results of a survey of 334 people give a sa1nple proportion of0.98. Do these given values allo,v the distribution to be approxin1ated by a normal distribution? 4 Calculate the expected value and standard error for the follo,ving proportion elements. Give answers to three decin1al places where necessary. (a) x = 507 and n = 1300

(b) x = 415 and n = 500

(c) x = 50 and n = 1000

5 Detennine ,vhether the following values satisfy the conditions for normality. (a) p=0.55 and n = 1000

(b) p=0.9 and n = 10

(c) p=O. l and n = 50

6 For each of the following, given the san1ple proportion and observation value, find the sa1nple size. Give your ans,vers correct to the nearest whole nu1nber. (a) p=0.7andx=9 1

(b) p=0. 15andx=300

(c) p=0.83andx= l 5

7 For each of the following, given the san1ple proportion and the sa1nple size, find the observation value. (a) p = 0.34 and n = 2500 (b) p = 0.02 and n = 100 (c) p = 0.98 and n = 5000 A

A

A

8 The 'san1pling error' is considered to be the difference between the values of the population proportion and the san1ple proportion. Why is the value of the sampling error difficult to calculate? 9 In the follo,ving cases, calculate the expected value, correct to two deciinal places. Then calculate the standard error, correct to three decin1al places. Finally, describe the shape of the sa1npling distribution. (a) Jess is running for school captain. From a simple random sample of900 students, the proportion

of students who favour Jess is 0.75. (b) A siinple rando1n sa1nple of 50 people found that five people were prepared to quit s1noking \vithin the ,veek. (c) Last year 67 000 children were hospitalised. Of these ad1nissions, 42 600 were boys. 10 Explain the difference between p and p. 11 In your own words, explain the standard error.

12 What happens to the value of the standard error as the sa1nple size increases and what does this 1nean? 13 According to a survey of 500 e1nployers, it was found that 78% of employers agreed that the de1nand for jobs has increased. This survey ,vas repeated many tiines with randon1 san1ples being taken fro1n the saine population and the san1ple proportions recorded. (a) Do you think that the san1ple proportion from each survey \viii always be 0. 78? (b) Would a sa1nple proportion value ofO be possible? What ,vould this 1nean? (c) Could some of the sa1nple proportions be lower than 0.45? What could this be caused by? (d) Is the sample proportion of0.78 close to the true population proportion? Justify your answer. (e) In what scenario could the sample proportion value not be representative of the true population proportion? 14 A randon1 san1ple of 1500 patient records showed that since the introduction of heparin injections for all patients, only 7% of patients developed blood clots. This research was repeated n1any times with random samples being taken from the san1e population and the san1ple proportions recorded. (a) Why do you think that the sainple proportion varies for each sainple? (b) Would a sa1nple proportion value of one be possible? What ,vould this 1nean? (c) If a sa1nple proportion appeared abnormally high, for example 0.23, ,vhat could this indicate? (d) A proportion of 7% is fairly s1nall. Does this indicate a proble1n with the sampling process? (e) In what scenario could the sample proportion value not be representative of the true population proportion?

354

New Senior Mathematics Extension 1 for Years 11 & 12

YEAR 12

15 A fitness club surveys a sainple of 120 of its me1nbers on \Vhether they are satisfied with the club's facilities or not. The survey results show that half of the n1e1nbers are satisfied, a third are indifferent aJ1d a sixth are dissatisfied. (a) Calculate the observation value for each type of response. (b) Calculate the point estimate for each type of response, correct to hvo deci1nal places where necessary. (c) Ho\v would the point estiinate of the dissatisfied n1embers change if the sa1nple size is doubled but the results are in the saine proportion? (d) Ho\v would the point estiinate of the dissatisfied n1embers changes if the san1ple size is the san1e but the proportion of dissatisfied custo1ners changes to one-fifth of the sample?

16 The sales 1nanager from a publishing co1npany believes that 35% of the con1pany's sales come from returning custo1ners. A randon1 san1ple of 200 sales has been chosen. (a) For this san1ple, state the expected value, correct to two decin1al places. Then calculate the standard error, correct to four decimal places. (b) Assun1mg nonnality, plot the san1pling distribution of p. (c) The sales 1naJ1ager chooses a second randon1 san1ple of 200 sales and notices that only 55 of the sales are fron1 returning customers. ~) For the new san1ple, calculate the expected value, correct to three decin1al places. Then, calculate the stai1dard error, correct to four decin1al places. (ii) Assun1ing nonnality, plot the saI11pling distribution of this saI11ple on the same axes as the first sainple. (d) If a third rando1n sa1nple is chosen and the nu1nber of sales from returning custo1ners is 80, aJ1d assun1mg normality, how \vould the new san1pling distribution co1npare to the other hvo sainpling distributions? Plot the third saI11plmg distribution on the san1e axes as the first and second san1ples.

CHAPTER REVIEW 14 1 Find the stated probability for the follo\ving bino1nial distributions. Express your answers as fractions in siinplest form. (a) P(X=3)ifX-B(5,;).

(b) P(X=4)ifX-B(6, 1:)-

2 For the variable Y - B(n, p) it is known that E(Y) = 32 and Var(Y) = 6.4. (a) Find the probability of success, p. (b) Find the nu1nber of trials, n. 3 A coin is biased in such a way that P(heads) = 3 x P(tails). The coin is tossed 100 tin1es. Let X stai1d for the nun1ber of tails obtained. Find the value of E(X). 4 When Yehudi and Carlos play racquetball, the probability that Yehudi wins a point is 0.35. (a) Ho\v n1aJ1y points would you expect Yehudi to win from the first 15 points? Give your ans\ver to the nearest whole number of points. (b) Choose the correct terms in the following staten1ent. If Yehudi won 10 out of the first 15 points, I would [not be/ be slightly/ be very) surprised as the nun1ber is [about the saine as / just above/ well above) the expected nu1nber.

5 A saI11ple of three ite1ns is selected at raJ1don1 fro1n a box containing 10 ite1ns of which three are defective. Let Y represent the nu1nber of defective ite1ns selected. (a) Con1plete the table to sho\v the probability distribution of the variable. State the probabilities in sin1plest fraction forn1. 0

3

(b) Find the expected number of defective items, E(Y).

Chapter 14 The binomial distribution

355

YEAR12

6 A jar contains seven white marbles, three green n1arbles and t\vo blue marbles. Two marbles are dra\vn, with replacen1ent, fro1n the jar. What is the probability of dra\ving exactly one white 1narble? A B C 2 D -7 x -5 12x l 2 12xii + 12xii x12x12 12 12

77

(7 5) (5 7)

75

7 Find tile value oft in the following probability distribution table.

A

10

B

0.1

C

D

6

0.6

X

0

1

2

3

P(X=x)

t

2t

3t

4t

8 A coin, which is biased so that P(heads) = 2 x P(tails), is tossed eight tilnes. The probability that the result is heads exactly three times, P(X = 3), is best represented by: A

(!)(!)5(~)3

B

(!)(!)3(~)5

C

(:)(!)3(~)2

D

80

D

(!)(~)3+(:)(~)5

9 If Y - B( l OO, 0.2), what is the value ofµ?

A

0.8

B

16

C

20

10 For a particular binon1ial rando1n variable Y, it is known that Var(Y) = 14.4. If 60 trials are conducted, what is tile probability p of success? A 0.4 B 0.6 C 0.24 or 0.76 D 0.4 or 0.6 11 A die, with sides labelled 1-6, is biased so tllat P(odd) = 3 x P(even). (a} If rolling an odd nu1nber is considered a success, find P(success). (b} The die is rolled 30 times. ~) Write this infonnation in the form X - B(n, p). (ii) What is tile expected ntunber of odd nu1nbers that will occur in the 30 rolls? (iii) Ho\v unusual would you consider it to roll 28 odd nu1nbers in the 30 rolls? Explain with reference to the 95% confidence interval. (c} ~) What is tile probability of any pair of rolls resulting in two odd nun1bers? (ii) Draw a table to sho\v the probability distribution of tile nu1nber of odd nu1nbers in the two rolls. 12 A cereal manufacturer is running a pron1otion on single-serving boxes of cereal, which states that one in five cereal boxes contains a free gift. Carol likes to win free gifts and intends to buy one cereal box every day for one week. State your answers correct to three decimal places where necessary. (a} If tile variable X represents tile nun1ber of cereal boxes with a free gift, calculate tile probability that Carol will win exactly one free gift fro1n tile seven cereal boxes she intends to buy. (b} What is tile probability that Carol \Vins no free gifts from the seven cereal boxes she intends to buy? (c} Given that Carol wins one free gift on the first day, calculate the probability that she \Vins exactly one more free gift in the next six days. (d} Given that Carol does not win a free gift in the first five days, calculate the probability tllat she \Vins more than one free gift in the next two days. (e} Carol thinks that if she buys one cereal box per day for 10 days, her chances of \Vinning 1nore than two free gifts will increase. Explain why Carol is correct in her thinking, using appropriate calculations.

356

New Senior Mathematics Extension 1 for Years 11 & 12

SUMMARY

x 2 +7x+ l 2

• - - - - , x-:1:-3

, or (x· + 7x + 12) + (x+ 3),

x+3 x -:I: -3 are two ways of writing a long division of polynon1ials.

1 FURTHER WORK WITH FUNCTIONS Properties of inequalities • Adding or subtracting the same nun1ber fro1n both sides does not alter an inequality. • Multiplying both sides of an inequality by a negative nun1ber reverses the direction of the inequality. • Multiplication by a positive number does not alter the direction of the inequality. • Taking the reciprocal of both sides of an inequality reverses its direction \Vhen both sides have the same sign, but not if the signs are different. • Squaring both sides of an inequality: this produces a positive nu1nber on both sides, so the direction of the inequality a 2 ~ b2 depends on \Vhich oflal and lbl is larger. • Square root of both sides of an inequality: this is only defined if both sides of the inequality are positive. The direction of the inequality does not change, i.e. if a > b then ✓ a> ✓b, provided a is positive and b is positive or zero.

2 POLYNOMIALS • A real polynomial P(x) is an algebraic expression of the fonn P(x) = a,,x" + a11 - 1x"- 1 + .. . + a 1x + a0, \Vhere n, n - 1, ... are all positive integers and a11 , aIf- 1 , ••• are the coefficients, \Vhich for convenience will usually be chosen as integers. • The tenn a,,x" is the leading term. If a11 -:I: 0, the polynomial is said to be of then -th degree.

• P(x ) is defined for all real x and is a continuous and differentiable function of x. • P(x) = 0 is a poly110111ial equation of degree n. Real nun1bers x that satisfy this polynomial equation are called the real roots of the equation or the real zeros

• Dividend = Divisor x Quotient + Ren1ainder Degree of the ren1ainder < degree of the divisor. Degree of the quotient< degree of the dividend. • When the divisor is a linear function (first degree) , the remainder will be a constant.

The remainder theorem If a poly110111ial P(x) is divided by (x - a) until the ren1ainder R does not contain x, then R = P(a). For any polyno111ial: P(x) = (x - a)Q(x) + R where Q(x) is another polynomial. Note:

• If P(x) is divided by x + a, as x + a= x - (-a), then R = P(-a). • If P(x) is divided by ax- b, as ax-b=a(x- b) , then

R=P(:).

a

The factor theorem When the remainder of a division is zero, the divisor is a factor of the expression being divided: • For a polynomial P(x), if P(a) = 0 then (x - a) is a factor of P(x). • Conversely: if (x - a) is a factor of P(x) then P(a) =0. By finding the zeros of a poly110111ial (i.e. values of a such that P(a) = 0), \Ve can factorise the polynon1ial. Zeros can be found by trial and error and then by long division.

Multiple roots of a polynomial equation A polynomial of degree n has n zeros, but they are not necessarily all different. You say that c is a zero of multiplicity r (r > 1) when the factor (x - c) occurs r tiines.

• The polynon1ial equation a,,x" + a11 _ 1x"- + ... + a 1x + a0 = 0 \Viii have at most n real roots.

If x = c is a zero of multiplicity r of the real polynonlial P(x), then x = c is also a zero of 1nultiplicity (r - 1) of the derived polynomial P'(x), a zero of 111ultiplicity (r - 2) of the second derived polyno111ial P''(x), and so on.

• Expressions are not polyno111ials if they include po\vers of x that are not positive integers, 1 2 X -3• e.g.x½+ 2 +x,X+

If P(x) is a polynon1ial of degree n, then P'(x) must be a polynon1ial of degree (n - 1), P''(x) a polynon1ial of degree (n - 2), and so on.

Division of polynomials 2 • If P(x) = 3x + 7x + 4, then P( l) is the value of P(x) when x = 1, which is found by substituting x = 1 into

Polynomial functions The general polynomial function is the function f where f(x) = a,,x" + a11 _ 1x"- 1+ ... + a 1x + a0 (a11 -:I: O). This function is defined for all real values of x and is

of the corresponding poly110111ial.

1

2

the polynon1ial: P(l) = 3(1) + 7(1) + 4 = 14 2 i.e. P(a) = 3a + 7a + 4.

continuous and differentiable.

Summary

357

For a ;c 0, \Ve have:

• f (x) = ax + b (general linear function)

= ax2 + bx + e (general quadratic function) 3 2 f (x) = ax + bx + ex + d (general cubic function)

• f(x) •

If the coefficient of the highest power of x is unity ( 1), the polynon1ial is said to be monic.

Graphs of polynomial functions The simplest polyno1nial function of degree n is f (x) = x". Graphs of these basic polynomials for n = 1, 3, 5 follow, and for n =2, 4 below: Odd functions y n=3 ,. = 1 f(-x) =-f(x) n=S

5 Odd functions will have an inverse function, although so1ne n1ay require a restriction on the domain. 6 Even functions will not have a single inverse function, but can be split into two parts (by restricting the don1ain) so that each part has an inverse function. (Inverse functions are covered in Chapter 5.) 7 Note that n1ost functions are neither even nor odd, e.g. J(x) = x 2 + x , f (x) = ex.

Cubic functions A general cubic function is a polynon1ial function f of 3 2 the third degree, defined by f(x) =ax + bx + ex + d, where a, b, e, dare constants and a ;c 0. Every cubic polynomial has at least one linear factor of the form (x + a ), \Vhere a is a real number.

Summary of polynomial functions If P(x) =a,,x" + a11 _ 1x"- 1 + ... + a 1x + a0 _ a11 ;c 0, then:

(I, I)

0

1 For very large lxl, P(x) ,., a,,x".

X

2 A polyno1nial of odd degree always has at least one real zero (i.e. its graph cuts the x -axis at least once).

(- 1,- 1)

3 At least one n1aximun1 or 1ninin1u1n value of P

occurs between any two distinct real zeros.

n=S

4 For a polynon1ial of odd degree, the ends of the

n=3

Even functions f(-x) =J(x)

Y

tJ

graph go in opposite directions.

=2

5 For a polynon1ial of even degree, the ends of the

graph go in the san1e direction. 6 When the graph of a polynomial function 1neets the x -axis, it may cut it (single zero), touch it (double

(I , I)

(- 1, I)

zero) or cut it at a point of inflexion (triple zero). 0

X

Polynomial graphs, standard forms y Degree O (constant) P(x) = e

C

0

Important features of f(x) =x' 1 The x -axis is a tangent to each graph at the origin (n ;c 1). 2 f (x) = x" for even values of n defines even functions. For even functions,f(-x) =f(x), so their graphs are sy1n1netrical about the y-axis. 3 f(x) = x" for odd values of n defines odd functions. For odd functions, f(-x) =-f(x). Because f(-x) and f(x) are opposite in sign, the graph offfor x < 0 can be obtained by rotating the graph off for x > 0 through 180° about the origin. 4 Recognising that a function is odd or even 1neans that you only need to draw half of the graph in detail. The other half can then be drawn using the sy1n1netry properties.

358

New Senior Mathematics Extension 1 for Years 11 & 12

X

Does not cut No zeros Degree 1 (linear )

)'

P(x) = mx + b 0

Cuts once One zero

X

Degree 2 (quadratic)

P(x)

=ax2 + bx + c

a> 0: )'

2

y

)'

6

6

4

4

2

- 3 - 2 _, 0

l

-2 - 1 O

X

Cuts twice ll > 0 Two zeros

Touches ll = 0 Double zero

)'

)'

I

2

X

Does not cut ll < 0 No real zeros

a< 0: 2 -3 I

- '- 2

)'

0

1

X

I

-4

-4

X

-6

-6

Cuts twice ll > 0 Two zeros

X

2

Touches ll = 0 Double zero

Does not cut ll < 0 No real zeros

A quadratic polynomial may have t\vo, one or no real zeros. Degree 3 (cubic)

P(x)

=ax3 + bx-, + ex + d

a> 0: y

)'

2

2

I

-2

- _,0

y

)'

2

I

1

2

X

0

-2

I

X

2

-2

- 1

-2

2

I

X

-2

-2

Cuts once Triple zero

_,

1

1

-1

-2

Cuts three times Three zeros

0

I

Cuts once and touches One zero, one double zero

X

2

Cuts once One zero

a< 0: y

y

)'

2

2

2

I

I

I

y I

-2

- 2 - 1_,0

2

X

-2

-

2

X

-2

Cuts once Triple zero

-2

I

2

- _,0

X

X

2

-2

Cuts three times Three zeros

Cuts once and touches One zero, one double zero

Cuts once One zero

A cubic polynomial ,nay have one, two or three real zeros. A cubic polynomial always has at least one real zero. Summary

359

Degree 4 (quartic)

P(x) = ax4 + bx3 + cx2 + dx + e a> 0:

)' )'

)'

3

2

2

I

I

- 2 _, 0

l

2

-2

I

0

2

1

-1

X

X

-1

I

- 1

Touches twice Two double zeros

Touches Quadruple zero

-2

X

Touches once, cuts hvice Two zeros, one double zero )'

)'

4

)'

2

I l

X

-2

-2

_, 0

I

2

X

- 1

Cuts twice Onezero,onetr~lezero

Cuts four tiines ~ ur zeros

Does not cut No real zeros

The addition of an appropriate constant to each equation can create a polynon1ial with no real zeros.

a< 0:

A negative a mverts each of the six graphs above so that they open downwards, with the properties of their zeros the sa1ne.

Summary of quartic polynomials 1 A quartic polyno1nial n1ay have four, three, two, one or no real zeros. 2 If a quartic polynon1ial has only one real zero, then it 1nust be a quadruple zero. 3 If a quartic polynon1ial has only two distinct real zeros, then they are either a triple zero and a single zero or they

are both double zeros. 4 If a quartic polynon1ial has only three distinct real zeros, then they are a double zero and two single zeros.

5 If a quartic polynon1ial has four distinct real zeros, then it can be factorised into four real lmear factors. 6 If a quartic polynon1ial has no real zeros, then it cannot be factorised into any real linear factors.

Relationship between roots and coefficients Quadratic equations Cubic equations The , general quadratic equation is ax- + bx + c = 0, a ;t 0. The roots of this equation are a and /3, so that (x - a)(x- /3) =0. • Sum of roots:a + /3 =-

b 0

• Product of roots: a/3 = ~

2

b

• Sum of roots: a + f3 + y =-a • Sun1 of products of pairs of roots: a/3 + ay+ f3r = ~

-ad

Identities:

• Product of roots: af3r = Identity:

• a2 + /32 = (a + /3>2 -

• a 2 + /3 2 + r 2 = (a + f3 + r >2 - 2(af3 + ay+ /3r)

2af3 3 • a + f33 = (a + /3>3 - 3af3(a + /3)

360

3

The general cubic equation is ax + bx + ex + d = 0, a ;t 0. The roots of this equation are a, f3 and r, so that (x - a)(x - /3)(x - y) =0.

Note: These relationships bet\veen roots and coefficients are not enough to find the roots of an equation \Vithout so1ne additional infonnation.

New Senior Mathematics Extension 1 for Years 11 & 12

Parametric equation of the parabola , • The parabola x- = 4ay can be represented by the para1netric equations x = 2at , y = at2.

Quartic equations (4th degree)

The general quartic equation is 4 3 2 ax + bx + cx + dx + e = 0, a-:;; 0. The roots of this equation are a, /3, y and f>, so that (x - a)(x- /3)(x - y)(x - f>) = 0. b • Sum of roots: a + /3 + y+ f> = - a • Sum of products of pairs of roots:

a/3 + ay+ af> + /3r+ /3f> + yf> = ~

4 FURTHER TRIGONOMETRIC IDENTITIES Sum and difference of two angles • cos (0 + 2

- 10 -4

X

I

'

y

2

-6

''

-4 -3

r-

3 4

'

'

X

'

:' ',

'

:3 ' :4

'

'

y 4 3 , ,2

:: ,' '

''

' ''' ' ' ''' ''

2

-8 - 10 - 12 - 14

, '' 2

' - !10 ' ' l

X

0

,

' ,.

1

~I

2

- 0.5

, ,' ,'

'' ' ' ' ' '

-2 -3 -4

2 3 4

X

-6

7

y

(b)

-8

I

,

y ' 4 ' 3

'' '

- 10

0

X

2

I

(c) x< - 2, - 1 I

I

X

-3

I

8 0

X

-1

X

2

I

., I - 4 - 3 f - !10 , -2 ,' : - 3

(d) all x, x ;t o. 2

'

2

2

3

X

y 4, ,3 , , 2

:, ' ' '

y

y

0 I

2 3 4

-4

y 80

-10 - 20

,,

-4 -

y 2

- 0.5

'' '

I

I

- 30

382

'

y :4

6

0

(d)

'

''

- 4 - 3 - 2 'l 0

(c)

X

'

y 0.5

12 B 13 (a)

3 4

1:

-2 -3 -4

- 0.2

0)

3 A

New Senior Mathematics Extension 1 for Years 11 & 12

2

X

' ' ''' ' ' '

-4

'

'' I

' :' ,.' 1 3 :' ' :' '' ' ' ' ' ' ' '

4

X

9

5 (a)

)'

4 ' 3 • 2 ' :' 1

1

3 4

X

'3

)'

(b)

1'

•

0

1

--4 - 3 - 2 - !1

3 4

--4 - 3 -

'

'

+3 '

,4

'

'

'

1 2 3 4

X

9 (a)

'

-4 -3 -2 - 1 -1 ,

''

,

-f

3 4

''

-,4

,

(b) '

'' )'

,,

7 (a)

,,

'

X

1 2 3 4

X

)'

i '•

' 1

X

•

•

'• '•

-3 -4

• 10 (a)

6

' ' '• 5 '

,, 1 2 3 4

3 4

)'

'

- 4 -, J '- 2 - ! f , , -2 , -3

2 3 4

'

4

- 4 - 3 -, - ! 10

'

,

- 4 - 3 - 2 - !P

1

'

'

,,

X

)'

,'

'

4 3

3 4

-3 -4

'

EXERC ISE 3.2 1 B 2 C 3 D 4 (a)

2)

1

-3 -4

X

X

,

-2

4, '3

I 2 3 4

4

-2

(b)

X

)'

'' 1

1 2 3 4

i,, ' ••

1

- !1

'

2 1

'

0

-4 -3

)'

-2

-3 -4

'3 ,,

-2

12

-3 -4

'

- 4 - 3 -2 - ! P

-4 -2

1

''

X

)'

'

'• 4

-2

6 (a)

4 3 2,

1 2 3 4

)'

•3

'

-!1

•2 '

)'

'

'

'

X

,, '

(b)

4 3 2 1

'

,'

X

-3 -4

)'

'

1 2 3 4

-2

-3 -4

,, ' •

11

4

- 4 - 3 - 2 - !P

'

4

'•

X

1 2 3 4

• •

'

.

)'

'' ' '

('

''

10

8 (a)

,

4 3 2

,, ' ' '''

-!'f

-4 -3 -2

)'

••4

i - 4 - 3 - 2 - !P

- 4 - 3 - 2 - !10 , ' l

'

4 3

,, ' •

•

•3

1

X

)'

2 3 4

•

-2

X

-3 -4

-2

--4

(b)

)'

4 3

,

, , ,

(b)

,,

(b)

)'

6 '' ' 5 ' '••4 ' • ' il

4 3 2

' , , ,

2 3 4

,

" ' • '• 1

X

,,

--4 -

-3 -4

)'

2 3 4

X

• •

3 4

X

2

Answers

Chapter 3

383

11 (a)

y 5 4 3

I I

\

(b) 'I

(b)

)' I

'

'

I

'

- 4 - 3 - 2 - ! 10

5 4 3

1 2 3 4

'2

'1 '0

2'

''

-2

-4 -

- 4 - 3 - 2 - !10

-3

I

, ,J'

X

1 2 3 4

2 - }1

X

4 3

'

\

'

\

'

2

' '

1 2 3 4

- 4 - 3 - 2 - 10 -1

X

' 1 2'\ 3 4

-3

- 4'

12 (a) There is no graph as the original function is always negative and hence does not have a square root. (b) There is no graph as the original function is always negative, so can not be equal to a positive . y . quantity

'

(b)

' 0 - 4 - 3 - 2 .-!1

X

' ' '

I

'

' (b)

)'

3 C

4

- 4 - 3 - 2 - 10 l ' 'l 2 3 4 I

-

'

-2

'

-4

, 5 (a)

X

'

(b)

-2 -3 -4

)'

'

'I

',4

'J

\

h

-f

- !1

, ~'.l , , -4

-4 - 3 -

-2 -3 -4

6 5 4 3

0 - !1

1

3 4 X

-2 -3 -4

y 6 5

8 (a)

7

I

\

- 4 -3

)'

I

. )'

(b)

4 3 ,2

2 3 4 X

- 4 - 3 - 2 - ! 10

-3 -4

I

-4 - 3 -2

4

1

, •3 , ,

y 4 3 2 1

X

-3 -4

y

X

- ,2,

0 1 - 2,

10 (a)

4 3

2 1

(b)

'

y

7 (a)

1/"2 3 4 - 4 - 3 - 2 - 10 -1 ,,

'

-?'

y

,

- 4 - 3 - 2'- l O ' -1 -2

-2 - 3'

2 A

1 2 1I 3 4

4 3

'

1 B 4 (a)

-3 -4 )'

4 3 2 1

EXERCISE 3.3

X

' -2

\

-2 -3

1 2 3 4

4 3

3 2 1

'

X

y

9 (a)

y

6 (a)

1 2 3 4

~2 '- 3

'

)'

5

X

I

-2

(b)

4

1 2

11 (a)

y 5 4 3 2

1 - 4 - 3 - 2 - ! 10 -2

1 2 3 4 X

-4 - 3 -2 - 1O rl

~2 '- 3

'

384

New Senior Mathematics Extension 1 for Years 11 & 12

1 2' - '.l 4

X

- 4 - 3 - 2 - ! 10

-2

1 2 3 4

X

(b)

(c)

)'

'

'

2

5

y

\

'

\•

'

\

2,

\ \

:\

; I

I

2 3 4

X

'

(

-2

- 3''

I 2 ' ' '

'

3 4

X

'

' \

\

\

\'

'

\

(e)

'

\

'

12

' y=i' - 2x•'

IO

\

8

\

'\

X

I \

I

6

I

I

14 12 IO 8 6 4

''

-:,' /

...• .; - --, -:

-- - 4 - ~ -? -lz - 1 2 : ~J

.• •.... .

I

'

'

8

:,

-,,,,

6

)-

' '

'

IO

\

'

6

' '

3 4

-

''

'

'' -

'

- l ' - L20

I ''

3

4

X

'

'

y= .) - sin(8 - 4>) cos(0+ 4> )- cos(0 - 4>) _ 2cos0sin 4> - 2sin0sin4> = - cot0 62 LHS = cos75° + cos!5° sin 75° - sin !5° _ 2cos45°cos30° 2cos45° sin30° = cot 30° = ✓3

. . :. : ~ =-=-

63 LHS = sin A +sin(A + B)+ sin(A + 2B) cosA +cos(A + B)+cos(A + 2B) _ 2sin(A + B)cos(- B) + sin(A+B) 2cos(A + B)cos(- B)+cos(A + B) _ sin(A+B)(2cosB + I) - cos(A+ B)(2cosB+ I) = tan (A + B) 64 LHS = sin(n + 1)0 + 2 sinn0 + sin(n - 1)0 cos(n - 1)0 - cos(n + 1)0 2sin n0cos0+ 2sin n0 - 2sinn0 sin(-0) = co~ 0; 1 [usecos0 = 2cos'~ - I and sin0 = 2sin sin 2cos' .ll. - l + I ' 2 sin !l. cos !l. ' ' c.os 2 sin ~ = cot !l. 65 a + f3 + r = n, (a + /3) = 11: - r LHS = sin 2a + sin 2/3 + sin 2y = 2sin (a + f3) cos (a - fJ) + sin 2y = 2sin(n - y) cos (a - /3) + sin 2y = 2sin ycos (a - fJ) + 2sin ycos y = 2sin y(cos (a - f3) + cos ii

EXERCISE 4.7 ' ' (b) 1 (a) .!!.. 3,3

(9)

~ cos~)

a- ~ + y) cos ( a- ~- r ))

= 4sin ycos (

~Z/3 ) cos( 2

n

a; n)

= 4 sin a sin f3 sin y.

(e) X = (g) X -

1!. 6

=-

(c) x = ' 0 so a< 0 to all I~ 0.

- 10

\ 0 when t >I¼- The velocity is now an increasing function. 3 (a) 1= 0, v = - Sms-1

{b) v = 0: 5 = .~, I= I a = - '"-

(1+1)2

_,

l =l,a = Sms

Answers

Chapter 7

401

(c)

ll;

EXERCISE 7.4

1 (a) N = 30 + Ae...,·"'• dN = - 0 4 Ae-o.◄ t, but Ae...,·" = N - 30, dt . so dN = - 04{N- 30) dt . (b) 34

JOO

90 80

_____

70

2 (a) N = 40 + Ae0·",

-----~-~_,

~

but Ae0·" = N - 40,

',

(b) 113.9

40

3

30

20 IO 0

I

2

3

4

5

6 I

As t increases, the velocity approaches 60 km h" 1 from above. 8 (a ) x(t) = 2(1 - e_,), x(t) = 2e· 1 • x(I) = - 2e· 1

x(OJ = o, x (OJ = 2, x(O) = - 2 (b) x(I)

I

02

so ft = 0.2(N - 40)

50

2

dJ: = 0.2Ae

- - - - - · __:. · ;. · = · -

-

•

~~ = - k(T -

50),

f} = k(T-~ SO)' T ;o 20

1= - .l J I d1 = - .l log (T - SO)+ C, T>50 k T - 50 k e - k(t - C) = log, (T - 50), T - 50 = e·•'! S K 3>'f 7,r 2 II X- z, z, or x - ' ◄ '2' ◄ 'n, 4 •2• "' 1C

••)

(

jjj)

x-

SR'

It

,r

3>'! Sn- 7 1l 9Jt IJ>'f UJI' ISJf 2 ' 8 ' S' ti ' i ' 8 , i

3Jt

1t

4' 4, S' 2' i '

CHAPTER REVIEW 8

1 (a) cos8 - ./3 sin(:/ = 2cos( 0 + 1)

2 (a) 5 s in (x + 0.927)

(b) 2cos(0+ l) = I, 8 = •;

(b) 0.64 (c) nn+( - 1)" x·;-o.927

1- -j; ✓ 2 - 1 (✓2 - 1)' , , 1-cosf 3 tan•= ! +cost - l+-j; - ✓ 2 +1 I

tan!= ✓ 2 -l 4 cos3 8 = cos(28+ 0) = cos28cos 8 - s in28sin 8 = 2cos3 8 - cos8 - 2sin2 8cos 8 = 2cos3 8 - cos8 - 2cos 8 + 2cos3 8 = 4cos3 8 - 3cos 8 3 Now8cos 8 - 6cos8 + I = 0 4cos3 8 - 3cos8= - ½ Hence cos38 -

-t 39 _

\,r, 8 _ = cos (8 + I/>)+ cos(8 - 1/)) Let A =8 + ,t>and B=8 - ,f>

+ tan8

tan8= - l , tan8 = 4 +

, . • 1C (..II) 8 = 0 , 3• , T

New Senior Mathematics Extension 1 for Years 11 & 12

+2(

c.osA 9 -cosAL 9 - c.oslL 9 ) = 36 2 cos llcos .1t cos~

•

tan ; + + tan ◄; = 33 5 8=L 2x 3,r .!:!. g ix 2

4 '

tan2 29x

3 ' ◄ '3 ' ◄ ' ◄

2

•

•

6 (a) cos (2A + B) = cos 2A cos B - sin 2A sin B = (2cos2 A - l) cosB - 2sin A cosAsinB Let A = B = 8 : cos38 = 4cos' 8 - 3cos8 ¼cos38 = cos3 8 - ¾cos(:/ {b) 27:'f , n

k+ I

- I -r

CHAPTER 9 Many of the p roofs below are not complete: in most cases only the outline of the p roof of S(k + I) is given. For more complete solutions, see the Student Worked Solutions book.

1-r

8

= RHS

I: LHS = 2. RHS = I X; X 3 = 2 = LHS. Result is true for n = I

11 =

S{k) = k(k + l )(k + 2l_s(k + I) = (k + l)(k + 2)(k + 3) 3

EXERCISE 9.1 1 D 2 For S{k + I): LHS = I + 2 + 3 + ... + k + k + I = k(k+l) + k + I

S(k + I) = I x 2 + 2 X 3 + 3 x4+ ...+ k(k + I) + (k + l)(k + 2) = k(k + l)(k + 2) + (k + l ){k + 2) 3

_ k(k + l)(k + 2) + 3(k + l)(k + 2) -

2

= (k+l)x(k+2) = RHS

3

= (k+ l )(k + 2)(k + 3) =RHS

2

3 For S(k + I): LHS = I + 2 + 4 + ... +2k• l + 2• =zk - I + z• =2 X z•- I = 2k• ' - 1= RHS

3

3

9

!

I: LHS = 3. RHS = I X X 9 = 3 = LHS. Result is true for11 =I

11 =

S(k) = k(k + 1)(2k + 7), S(k + I) = (k + l )(k + 2)(2k + 9)

6

6

An swers

Ch apter 9

407

S(k + I) = I X3 + 2 x 4 + 3 X 5 + ...+ k(k + 2) + (k + l )(k + 3)

15 (a ) For S(k + I): LHS = 13 + 23 + 3' + . . . + k' + (k + 1) 3

= k( k + l )(1k +7) +(k + l)(k + 3)

2

2

4

_ k( k + 1)(1k + 7) + 6(k + l )( k + 3) 6

4

= (k + l )'(k + 2)' = RHS,limit =.!.

_ ( k + i) [k( 2k +7)+ 6( k + 3)]

4

4

(b) In question 2 it was shown that J + 2 + 3 + . . . + n = n (n + I)

6

_ ( k + l) [ 2k 2 + 7k + 6k + 1s ]

Hence ( I + 2 + 3 + ... + 11) n 3 as shown in (a).

6

2 _ 11 '

( n + l) ' _ , , , 2 I + 2 + 3 + ... + 2

- ---'- - - ' - - -

=( k + 1)( 2k' + 13k + 18]

16 For S(k): (k + I)+ (k + 2) + ... + 2k = k( 3~ + I)

6 = ( k + l)(k + 2)( 2k + 9) = RHS

For S(k + I): LHS = (k + 2)+ . . . + 2k + (2k + 1)+(2k + 2) = (k + I) + (k + 2) + . . . + 2k + {3k + 2) 2 = k (3k+ I) +( 3k + 2) = 3k + 7k + 4 = (k + 1)(3k + 4) 2 2 2 =RHS 17 ForS(k + I): LHS =I x I! + 2 x 2!+ 3 X 3! + . . . + kxk! + (k + l )X (k + I)! = (k + 1)!-1 + (k + l )X (k + I)! = (k + I)! x [I +(k + I)) - I = (k + I)! x (k + 2)- l = (k + 2)! - 1 =RHS 18 11 = I: LHS = I. RHS = I + 0 = I = LHS. Result is true for n = I S(k) = I + (k - I) X 2\ S(k + I) = I + k x 2•• 1 S(k + I) = I X 2° + 2 X 2 1 + 3 X 2 2 + ... + k X 2k-t + (k + I) X 2k = I + (k - I) X 2• + (k + I) x 2• = l +(k -l + k + l)xzk = I + 2k x 2• = I + kx2•• 1 = RHS 19 11 = I: LHS = I. RHS = I ; 3 = I = LHS. Result is true for n = I

6 3

10 11 = I : LHS = 4. RHS = I X 2 = 4 = LHS. Result is true for 2 n= I S(k) = k' (k + 1)

3

3

S(k + I) = ( k + 1) (k + 2)

,

2

3

2

S(k + I) = 4 + 104 + 756 + ... + (k 3 + 3k5 ) + [(k + 1)3 + 3(k + 1)5 ) = k' (k + I )' +((k + l )'+ 3(k + I)') 2

_ k 3 ( k + l )3 + 2(k + l )3 + 6( k + l )5 2

= ( k + 1)' ( k'+ 2 + 6(k + 1)'] 2

_ ( k + 1)

3

[

k' + 6k 2 + 12k +

8]

2

= ( k + l)'(k + 2)' =RHS

k ( 4 k' - 1) k(2k - 1)(2k + I) S(k) = --'---"- = ~-~~, 3 3 S(k + I) = ( k + 1)(2k + 1)( 2k + 3)

2

11 ForS(k + l ): I I I I I LHS = I X 2 + 2 X3 + 3 X4 + . . . + k(k + I) + (k + l )(k + 2)

3 2

S(k + I) = I + 3 + 5 + . . . + (2k - 1)2 + (2k + 1)2 k( 2k - 1)(2k + I) ( k )' = + 2 + 1 3 = ( 2k + 1)(k( 2k - 1)) + 3(2k + I) 2

=~+ I = k(k + 1) + 1 = k + I = RHS k + I ( k + l)(k + 2) (k + l )(k + 2) k + 2 12 ForS(k + I): I I I I LHS = 2 X 3 + 3 X4 + . . . + ( k+ l)(k + 2) + ( k+ 2)(k + 3)

13 ForS(k + I ): I I I I LHS = J X 3 + 3X S + ... +(2k - 1)(2k + l ) +(2k + l)(2k + 3)

2

3

= ( 2k + 1)( 2k2 + Sk + 3)

_ ~'-'k ~ + ~---,'le,.,--~ _ - k("'"k'--'+-"3_,,)-'+-" 2~ _ -" k,c. +-'-1~ 2( k+2) (k + 2)( k + 3) - 2(k + 2)(k + 3) - 2( k + 3) = RHS

3

= ( k + 1)(2k + 1)( 2k + 3) = RHS 3

20

I: LHS = I. RHS = I x I x 2 = I = LHS. 2 Result is true for n = I

11 =

_ k I _ k(2k + 3)+ 1 _ k + I - 2k + l + (2k + l )(2k + 3) - (2k + l )(2k+3) - 2k + 3

S(k) =

= RHS

S(k + I) = I 2 - 22 + 3' - 42 + ... + (-1it - 1 k' + (-1it (k + I )2

=(- 1t(k + 1)

2

= ( - ll(k + l )(k + 2) = RHS

2

New Senior Mathematics Extension 1 for Years 11 & 12

k ( k + I ) +(-1/ (k + I)'

2

_ (- ll(k + 1)( - k + 2( k + 1))

k(k + I ) (k + l ) = 2(2k + I) + ( 2k + 1)(2k + 3)

= (k + l )(k + 2)(2k + I) = (k + l )( k + 2) = RHS 2( 2k+l)(2k + 3) 2( 2k+3)

1

2

2

_ (k + l )[k( 2k + 3)+ 2( k + !) ] 2(2k + l )(2k + 3)

2

= (- it-• k ( k + I)+ 2(- ll(k + 1)2

2

+ ( 2k + 1)(2k + 3)

_ k(k + 1)(2k + 3) + 2(k + l ) 2(2k + l )( 2k + 3)

(- it-•k(k + I), S(k + I) = (- l l(k + l)(k + 2) 2

14 ForS(k + I ): 12 22 32 k' LHS = I X 3 + 3 X5 + S X7 + . . . + (2k - 1)( 2k + l )

408

2

= k'(k + l ) + (k + l )' = ( k + 1) ( k + 4k +4)

6

2

21

I: LHS = 2. RHS = 8 - 6 = 2 = LHS. Result is tru e for n = I S(k) = (k' - 2k + 3)x 2•• 1 - 6, S(k + I) = ((k + 1) 2 - 2(k + I) + 3) X 2t• 2 - 6 =(k' + 2) x2•• 2 - 6 S(k + I) = I X 2 + 4 x z' + 9 x 23 + . . . + k 2 X 2k + (k + 1) 2 X 2•• 1 11 =

= (k'-2k + 3) x 2t• 1 -6+ (k + 1) 2 x2t• 1 = 2' • 1 (k' - 2k + 3 + k' + 2k + I) - 6 = 2'• 1 (2k' + 4) - 6 = (k' +2) X2k+i_ 6 = RHS 22 For S(k + 1): I 3 7 2k - I 2k+I - I LHS = 2 +4+8+ ... + 2k + 2k•I I 2k•I - I I I = k - l + - + - - - = k - l+ - + l - 2• 2k+I 2k 2k+I 2- 1 I = k+ = k+ = RHS 2 k+l k+I 2

I: LHS = 4. RHS = I X 22 = 4 = LHS. Result is true for 11 = I. S(k) = kx2'• 1,S(k + l) = (k+ l )X2t +z S(k + I) = 2 x2 1 +3 x i + 4 x2' + ... + (k+ I) x 2t+ (k + 2) x 2t• 1 = kx2k• • +(k + 2)x2t• 1 = 2•• 1 (2k + 2) = (k+ l )X2k+i =RHS IX2X3XS 24 11 = I: LHS = I. RHS = = I = LHS. 30 Result is true for n = I

23

11 =

2

k(k + 1)(2k + IJ (3k + 3k - 1) 30 • S(k) = 2

S(k + I) = (k + l)(k + 2)(2k + 3)( 3k + 9k + 5) 30 4 S(k + l) = 1 + 2" + 3'+ ... + k'1 +(k+ l )'

+ 3k - I)

=

IJ(3k

30

_ k(k + 1)(2k +

IJ(3k

+ 3k - I)+ 30(k + 1)4

k(k + 1)(2k +

2

• + ~ + I)

2

30 _ (k+ 1)[ k(2k+ 1)(3k2 + 3k - 1) + 30(k + I)' ] 30 _ (k + 1)[(2k'+k)(3k2 +3k - 1) + 30(k+ 1)3 ] 30 4

2

_ (k+ 1)[ 6k + 9k' + k' - k+ 30(k' + 3k +3k + I)]

30 4

2

_ (k + 1)[ 6k + 39k' + 9 1k + 89k + 30]

30 2

(k + l )(k + 2)(2k + 3J(3k + 9k + 5) RHS = - - - - - -3~0~ - ' - - - -~ 2

_ (k+ l )(k + 2)(6k' +27k + 37k+ 1s)

30 4

2

(k + 1J( 6k + 39k' + 9 1k + 89k + 30)

=

30

= LHS

EXERCISE 9.2 1 C 2 S(k)is5• = 4M - 3 For S(k + I): s •• • +3 = sxs t+ 3 = 5(4M - 3) + 3 = 20M - 12 = 4(SM - 3) which is divisible by 4 as (SM - 3) is an integer 3 S(k)is3 2• = 8M + I For S(k + I): 3 2• • 2 - I = 9 X 3 2k - I = 9(8M + I) - I = 72M + 8 = 8(9M + I) which is divisible by 8 as (9M + I) is an integer 4 S(k) is 3k = SM - 2k For S(k + 2): 3k+ 2 + 2•• 2 = 9 X 3k + 4 X 2k = 9(51\1 - 2') + 4 X 2' = 45M - 5 X 2k = 5(9M - 2') which is divisible by 5 as (9M - 2') is an integer

5 S(k)is5• = 3M - 2xll• For S(k + 1): 5k+ 1 + 2x u •• 1 = 5 X 5 k + 22 X ll k = 5(3M - 2 X ll k) + 22 X ll k =ISM + 12x llk= 3(5M + 4x ll') which is divisible by 3 as (SM+ 4 x 11k) is an integer. 6 (a) (k+ l )(k + 2)(k+3) (b) S(k) is k(k + l)(k + 2) = 3M For S(k + 1): (k+ l )(k + 2)(k + 3) = k(k + l)(k+2) + 3(k+ l )(k+2) = 3M + 3(k + l )(k + 2) = 3(M + (k + l)(k + 2)) which is divisible by 3 as (M + (k + l )(k + 2)) is an integer. 7 S(k) is 33k = SM - 2h 2 For S(k + 1): 3"• 3 + 2t+ 3 = 27 X 3 3k +2 X 2k• l = 27(SM - 2• • 2 ) + 2 X 2t+ 2 = 135M - 25 X 2k+ z = 5(27M - 5 X 2k+ 2 ) which is divisible by 5 as (27M - 5 x 2k+ 2) is an integer. 8 11 = 2, Exp = 7 2 - 2 2 = 49 - 4 = 45, which is divisible by 9. 2• = 9M S(k) is i.e. i = 9M+2k For S(k + 1): Exp = t • 1 - 2k• 1 = 7 X 7k - 2X 2k = 7(9M - 2') - 2X2k = 7 x 9M - 2•(7 + 2) = 9(7M - 2•) which is divisible by 9. 9 11 = I, Exp = 3• - I = 80, which is divisible by 80. S(k) is 3◄> - I = SOM i.e. 34k = SOM + I For S(k + 1): Exp = 34C>• 1>- I = 3•x 3.,- l =3•csoM + I) - 1 = 8 1 X80M+81 - l = 80(81M + I), which is divisible by 80. 10 11 = I: Exp = 5 + 22 = 27, which is divisible by 3. S(k) is s• + 2 x llk = 3M i.e. s• = 3M - 2 x ll t ForS(k + l): Exp = st• 1 + 2x u t• 1 = Sx5k+ 22xll• = 5(3M - 2X ll') + 22X lit = ISM + l2x llk = 3(5M + 4 x 11'), which is divisible by 3. 11 11 = I: Exp = 8 - I = 7, which is divisible by 7. S(k)is23 - 1 = 7M For S(k + 1): Exp = 23 k• 3 - I = 8 X 2" - I =8(2" - I ) + 7 = 8x7M - 7 = 7(8M - 1), which is divisible by 7. 12 11 = I: Exp = 6 + 10 - 6 = 10, which is divisible by 5. S(k) is 6• + IOk - 6 = SM For S(k + 1): Exp = 6k+ 1 + IO(k + I) - 6 = 6x6k+ IOk + I0 - 6 = 6 x 6k + 6x 10k - 6x6 - 50k+40 =6 x SM - 5( I Ok - 8) = 5(6M - !Ok+ 8), which is divisible by 5. 13 11 = I: Exp = 27 + I = 28, which is divisible by 7. S(k)is3 2t• 1 +2'- 1 = 7M is 32k• 1 = 7M - 2•- 1 For S(k + 1): Exp = 31 k• 3 + 2k = 9 X 32k+ 1 + 2X2k-l = 9(7M - 2k- I)+ 2 X 2k-l = 63M - 9x2k- l +2x2k-l = 63M - 7X2k- l = 7(9M - 2•- 1), wh ich is divisible by 7.

t-

Answers

Chapter 9

409

14 (a) (k+ 3)' = (k + 3) (k + 3) 2 = k' + 9k' + 27k+27 (b) Nlust prove that 113 + (n + 1) 3 + (n + 2) 3 is divisible by 3 for all integers n 2: I. S(k) isk3 +(k + 1)3+ (k+2) 3 = 3M ForS(k+ 1): (k + 1)3 + (k+2) 3 + (k + 3)3 = (k + 1)3 + (k + 2) 3 + k' + 9k' + 27k + 27 = 3M + 9k' + 27k+ 27 = 3(M + 3k2 + 9k + 9) which is divisible by 3 as (M + 3k' + 9k + 9) is an integer 15 Prove S(l) is true: (x - 1)3+ x3 = 2x3 - 3x' + 3x - 1 = (x' - x+ 1)(2x - l ) which is divisible by x' - x + l Assume S(k) is true, i.e. (x - 1)k+ 2 = (x' - x + l )M(x) - _x'l• 1 Prove S(k + 1) is true: (x - 1it• 3+ x2>+ 3 = (x - l)(x - 1l• 2 +x'xx'k• l = (x - l)[(x' - x + l )M(x) - xll• l1 + x' x ?·' = (x - l)(x' - x+ l )M(x) - (x - 1) X x'k+ l + x' xx'k• I = (x - l)(x' - x+ l )M(x) + x'k•'(x' - x + 1) = (x' - X + l)((x - l )M(x) + x ll• 1 ) which is divisible by x' - x + l 16 S(k) is / - 1 = (x - l)M(x) For S(k + 1): / • I - 1= / • l - / + / - 1 = /(x - l ) + (x - l )M(x) = (x - 1)(/ + M(x)) which is divisible by x - l EXERCISE 9.3 1 (a) S(k) = k' + k = (2N - 1), where N is a positive integer S(k + 1) = (k + 1) 2 + (k + 1) = k'+ 3k + 2 = k' + k + 2(k + I) = Odd + Even = Odd, so true when S(k) is true (b) S( 1) = 1 + 1 = 2, so result is not true. 2 (c) No. If n is odd, then n is odd and odd+ odd = even If n is even, then ,? is even and even + even =even (d) The statement should be that n 2 + n is an even integer for n 2: I. S( l ) = 1 + 1 = 2, so result is true for n = l S(k) = k' + k = 2N, where N is a positive integer S(k + 1) = (k + 1) 2 + (k + 1) = k'+ 3k + 2 = k' + k + 2(k + I) = Even+ Even = Even so, result is true if S(k) is true

2 (a) S(k) =

~ (6k' 2

2

4

S(k + l ) = (k + l)(k + 2)(k + 3)(k + 4) 4

S(k + 1) = 6 + 24+60 + . . . +k(k+ l )(k + 2) + (k + 1) (k + 2)(k+ 3)

= k(k + l )(k+ 2)(k + 3) + (k + l )(k + 2)(k+ J) 4

4 2

3k - 1) + (3k + 1)

2

2

= 3k3 - 3k' - k +9k' + 6k + l

2

2

2

= Jk3 + 15k + llk+ l

2

2

And Exp = k; 1(6(k+ l )2 - 3(k + l) - 1) 1 = k; ( 6k 2 + 12k + 6 - 3k - 3 - 1) 1

= k; (6k' + 9k + 2)

~ (6k3 + 15k2 + Ilk+ 2) 2

= Jk3+ 15k + l l k + l

2

2

which was the expression obtained for S(k + 1) 1 (b) S( 1): LHS = 1, RHS = (6 - 3 - 1) = 1 = LHS so true 2 (c) Hence result is true for all 11 2: 1

41 0

CHAPTER REVIEW 9 1 For S(k + 1): LHS = 2k(k - 2) + 4(k + 1) - 6 = 2k2 - 2 = RHS 2 S(k) is4,. , = 2 1M - sa- , For S(k + 1): 4>+ 2 + s'k• 1 = 4 x (21M - 52t - 1) + 25 x 52t - • = 84M + 2 1 X 52l- l = 2 1(4M + 52k- l) which is divisible by 2 1 as 4M + s't - , is an integer 3 S(k) is 7k = 3M + 1 For S(k + 1): 7k• l - 1 = 7(3M + 1) - 1 = 3(7M + 2) which is divisible by 3 as 7M + 2 is an integer l x2x3x4 4 11 = l : LHS = 6. RHS = = 6 = LHS. 4 Result is true for n = l S(k) = k(k + l)(k + 2)(k + 3)

= (k + l)(k + 2)(k + 3)(k + 4) = RHS

2

~ (6k 2 -

=

= k(k+ l ) = k'+k = k' - k + 2k = odd integer+ even integer = odd integer Result is true for n = k + l if it is true for 11 = k . (b) S( l ) = 1 - 1 = 0, which is not odd (c) It is never tr ue. \.Yhen n is o ff, odd' - odd = odd - odd = even . \.Yhen n is even, even' - even = even.

3k - 1)

S(k+ l ) = 1 +4 +7 + . . . +(3k - 2) +(3k + 1)

=

3 (a) S( l ) = 1 - 1 + 4 1 = 41 which is prime (b) S(2) = 4 - 2 + 4 1 = 43 which is prime (c) S(S) = 25 - 5 + 41 = 6 1 which is prime (d) S(42) = 4 12 - 4 1 + 41 = 41 2 which has a factor of 41 so is not prime (e) Original statement is false 4 (a) S(k) = k' - k is an odd integer. Show that S(k + 1) = (k + 1) 2 - (k + 1) is an odd integer. S(k+ l ) = (k + 1) 2 - (k+ l ) = (k + l)(k + 1- 1)

New Senior Mathematics Extension 1 for Years 11 & 12

5

3 1 3 . RHS = 1 = = LHS. 4 4 4 Result is true for n = l 1 1 S(k) = 1 ) k,S(k+ l ) = 1- ( ) k+I (k+l 2 k +2 2 11 =

l : LHS =

S(k + l ) = l + .!.+ 2.. + ... + k+2 + k+ 3 4 6 96 k(k+ l )2' (k+ l )(k + 2)2k+l

=l-

1 + k+3 (k + 1)2' (k + l)(k + 2)2'•1

= l - 2(k + 2) - (k+3) (k + l )(k+ 2)2'• 1 2k+4 - k - 3 =1 - --="-'--'--'----"-----i-~ (k + l)(k + 2)2'• 1 =l-

k+ l (k + l)(k + 2)2'•1 = 1 - - -1- = RHS (k + 2)2>+ 1

6 S(k)iss• = 16M - 12k+ I For S(k + I): s•• 1 + 12(k + 1) - I = 5(16M - 12k + I) + 12k + 11 = SOM - 48k + 16 = 16(5M - 3k + I) which is divisible by 16 as SM - 3k + I is an integer 7 S(k)is7• = 13M - 6t For S(k + 2): t •2 + 6k+' = 49(13M - 6') + 36 X 6k = 637M - 13 x 6• = 13(49M - 61) which is divisible by 13 as 491\1 - 6k is an integer 8 For S(k + I): LHS = k(k + I)!+ ((k + 1) 2 + l)(k+ I)! = (k + l)!(k+k2 + 2k+2) = (k+ l)!(k+2)(k + l ) = (k+ l)(k+2)! = RHS

7 (a) 8 B

9 D

- - -

10 B

(d) BF = ½(£ - q) . FC = ½(£ -

q)

(el AF = AB+BF = (!+½(f - !'!) = a+ l c - l a -

3-

2

3-

l

= 3 q +,£ = .!.(2a + c)

' - -

12 (a) AC = ~+t!

(bl AE = ~+½4

(cl DE = £ - ½4

(d) Express the vector in terms of AD and AB. DF = ½£ - ½4

(el

AF' = AD+ i5F = d+lb - l3 d3 -

k + I 3k + I (3k+1)(3k+4) _ k(3k +4)+ 1 - (3k + 1)(3k + 4)

= .lb+.ld 3 3 = 2 (b +d)

'- -

Now AC = b +d

+4k+ I - (3k3k'+ 1)(3k + 4) _ (3k+ l )(k+ I) - (3k+ 1)(3k +4) = k+ I =RHS 3k +4

-

2 -

:. AF = 3 AC Hence Flies on AC. (f) AF : FC = 2 : l EXERCISE 10.2

CHAPTER 10

1 (a) (6, - 18)

(b) (I, - 3)

(c) (- 2, 6)

(d) (0.8, - 2.4)

8

EXERCISE 10.1 ~

(b)

C

2 (a) - 2b =( ) - 10

+b

(c) lb =

-b

C

a -.:;:-------,

-c - -a

C

(d) DB = a - b -a

(b) AD = b

3 (a) AC = q + ~

(c) CA = - a - b

(bl AD = q + !! + £ (c)

' -

(b) S!! = ( -:so)

(-f)

(d) - lb 5 ) 4- =( _2s

•

s

'

3 (a) - c = (-6, 3)

(b) 2£ = (12,-6)

(c) - ½£ = (- 2, 1)

(d) -b

a- b

(d) DB = - b -

(c) WZ = b+ c+d

(c) The line joining the m idpoints of two sides of a triangle is parallel to and half the length of the th ird side.

=

2 (a) ci5 =- a

_,

(b) DE = ½BC

I I I I I S(k + I) = 4 + 28 + 70 + ... + (3k - 2)(3k + I)+ (3k + 1)(3k + 4)

(c)

=a- +b + -c+ d-

11 (a) BC = f - q, DE = ½q- ½q = ½(£ - q)

· t11at -1 + I + - I + ... + ( J g prove by .tn d uct1on )( ) 4 28 70 3n - 2 3n + I = n for all integers n 2: I 311+ I 11 = I: LHS = .!.. RHS = _ I_ = .!. = LHS. Result is true for n = I 4 I X4 4 k k+I S(k) = 3k+ l'S(k + I) = 3k+4

1 (a)

(b) VZ

W =q + £+ £

i5A = - q -

~- q

(d) I.Sf = (9 ,-4.5)

4 (a) - £ = (~)

(b) 2£ = ( : )

2 (c) _ l e = ( - ) 'I 5 (a) q = (S,s ) (d)

9 (d) I.Sc = ( ) -4.5 -

(b) ~ = (4, - 5)

4 = (- 13,- 5)

(c) £ = (0,9)

(e) ! = (10, 2)

(f) [ = (-3,5)

C

4 (a) CD = - ½q

(b) CA = - q - ~ (c) AD = - ½q+~

(d) DB = la - b

'- -

5 (a) OP = q + ~

OQ = 2q + !! (e) AB = - q + !! (f) i5i = - 3q + 2~ (b)

0G = 2q + 2~

(c)

(d)

cr = !! + ~

(g)

FQ = - q - ~ (h) DE+ ro = - 3q

6 (a) AB = - q + ~

6 (a) q = (:)

(b) ~ = ( : )

(C) f = ( ~ )

(d) 4 =(~)

(e) ! = ( ~~)

(f) [ = ( ~ )

7 A

8 C

(b) AC = q + ~

Answers

Chapter 10

411

14 (a) The position vector of the midpoint of AB is (- ½, •; ).

9

(b) The position vector of the midpoint of AB is ½!!+½~(c) The position vectors of the points of trisection of DE are (1.3 , ll.) 3 and . (11 3 , ll) 3 · 3

4

X

(d) The position vectors of th e points of trisection of DE 2 d + 3!are3 I ' d_ + 3f• ' 3-

(e)

10

TB =Ai5 + 5B = OB - OA =b - a

)'

4

AP = m and AB = m +n AP m = = AB m +n

3

I 2 3 4

AP =

X

OD

111

{b) The column vector is

= (1-

(-ll)·

Y

2 (a) l!!I = ✓,ii

I

e

p

-2

(e) 6 7

2 3 4

X

e

- (4}- (6)- (JO) 6

l!I= 9

(b) l~I = ✓ 65

(f) 1£1=

BC = _ , AC = 4 2

4 (a) 7{ - 3j

(c) ITBI' + lac l' = ( ~)' + ( ~)' = 104

(b) - { + 7j

5 m= - 1, n= - I 5 6 (a) a = , b = - 2

3

I

-

{b) CX = q - ½f

= c.., + 2a - l.c 3 3-

,

= ~, +~a 33= 1.(a + c) 3 -

,-

= ,OB Hence, Y lies on OB (e) OY : YB = 2: I

412

New Senior Mathematics Extension 1 for Years 11 & 12

(c) 29{ +4 j

(d) - 3f - 25j

3

(b) 7 ✓ 34

8 (a) - 21{ + 35j

(e)

(b) CA = -L + L (c) DB = 5{ + 9L

1a 1= ✓2

1 O (a) 5 ✓10

(b) x = - 3

11

12 D

V

=-7

(c) 16 (al

(b)

a=.r. (3i - 6;·) -

15

-

_

(d) ij = ~ (- f - 121)

f = ~ (3f - 9L)

(b) £ = f (3f - 9i)

-

1b

-

C

= JiTo (- 3i- + 2;·) )3 -

19 C

20 C

-

14 B

a=Ji} (-2; + 9;·) _

11 ~= ~ (- 14f - 9L)

18

(f) IDBI = ✓106 13 B

1s (al ~ = .~(sr + 6L)

(c) CY = !!! - ½£ (d) OY = OC + CY

(d) ,j = - 35.86f - 3.141

7 BA = - 2{ + 31

(d) ITBI = 5

13 (a) OB = q + £, OX = !! +,£

,

(b) ~ = - l l.85{ + 19.711

(d) p = 2, q = 8 (e) x = - 3, - 2, y = 2 (f) x = 2 - ✓ 6,2 + ✓ 6, y = - I , I

= ITc: I'

-

(d) lt.! I = 5

I (b) x = - 17, J' = 7 (c) g = o, Ii = -

9 (a) CD = - 3{ - 41 and so MBC is a right- angled triangle. (d) The coordinates of Dare (4, - 7). (e) The coordinates of the point of intersection of the diagonals of the square ABCD are (3, - 2).

(c) 1£1= 25

./89

3 (al !! = 12.29L + s.60 L

13, lac l = 2 ✓13, ITc: I = 2Ju {bl ITBI = 2 ✓

-

(c) 2L - 4L (d) - 3{ - 31

(cl f = 1.01t, - 1.01L

~

12 (a) AB =

EXERCISE 10.3

L

2

-4 - 3 - 2 -1

= m n+n-a + m m+n b1 (a) 2{ + 3L (b) - 31 (el 2r (f) 1 L+ 1

4 3

-

111

)a + b m +n - m + nm

= (m +n - m)a + m b m +n - m + n-

- 16 (c) The coordinates of the terminal point are (13, - 2). (d) The coordinates of the initial point are ( I 9, - 15). e

AB

=a + (b - a) - m +n - -

11 (a) The column vector is ( : ).

(e) ~

m +n

OP = OA + AP

OE

-4

111

21 (a) AB = 2i - 3j, CB = 3i + 2j

(b) OB = Sf - j, CJ\ = i +5j

9 If q and ~ are parallel and in the same direction, then

(c) ON = 2.2 -; - .!.;· OM = 2.; - .!.;· 2_' 22_

q • ~ = l!!ll £1coso = l!!ll ~I If q and ~ are parallel and in the opposite direction, then

(d) CP = - ½i + 4L, BP = - ½i + 21

q . ~ = l!!ll £1cos isoo = - l!!ll ~I

0

22 {a) A(- 2, 3), B(4, - 5), C(6, - 9)

10 (a) 8=25° (f) 8= 90°

(bl AB = 6f - 8L, BC = 2i - 4L, AC = Si - 121 (c) IABI = io, loc i = 2.rs, IACI = 4 ✓ 13 23 (a) x= 3, y = 2 (b) OM = 3i + 2j, AM = - 3i + 2j, MB = - 3i + 2j

-

(c) IAM' I = IMBI = IBM I =

-

✓13

-

11 B

AB • BC = 4X6 + 6 x(-4 )

=0 : . LABC=90°

OB = OA + AB

(d)

2f - 4 i) + ( 3f + 21)

-

AC = AB + BC

7L

BD = 2i -

= (3i + 21) - (2f - 4i)

10 l and lm51= ✓ 104 = 2-ffe,

AC • Bi5 = 10 x 2 + 2 x (- 10)

= i + 6j

=0 :. AC .L BD let M be the midpoint of AC, then AM = MC = Si + j and so IAM I = IMCI = .Ju, and its position vector will be

21, AC = f + 6l

(c) OP = ½i - l, AQ = ½f + 3Land 0Q = ½i -

L

Thus, OP = OQ and so P and Qare coincident. Therefore, the diagonals OB and AC bisect each other.

3;·, ER =1.2-; - s;·

-

-

,VB = 2i - 1L

26 (a) ~ = ¼(3i - 4L)and f = ¼( -4f + 3L) (c) ? = ;' (i - 21) and ~ = ;' (2f + L)

-

(c) 6i + 8 j

(f) 14i + 481

= - 2i - 3L + si + L

21

Let N be the midpoint of BD, then BN = ND = f - 5j and so IBN I = INDI = .Ju, and its position vector will be

-ON = -OB + -BN

.Ju, and IACI = .Ju,

27 (a) p = ¾ (b) ¾i + f j

-OM = -OA + -AM = 3f -

25 (a) OB = Sf + L, AC = f + SL, OM = 4f - ½L, ON = 3f + iL,

-

(b) x = - ½, r = - ½ (d) v =

- f, w = ~

(d) q = ;,

(g) ••~(si + l4i)

EXERCISE 10.4

1 (a) 45° (b) 45° (c) 135° {d) 165° (g) 120° {h) 180° (Q 150° (J) 30°

(f) 90°

(e) 90°

= 2i + 31 + f - SL = 3f -

21

Since OM = ON, the midpoints of AC and BD coincide and so the diagonals bisect each other at right angles. 15 (a) Use vector addition to calculate the vectors of the kite's sides. Then compare the magnitudes of those side vectors.

-AB = -OB - -OA

= 2i + 41 - (2i +

IABI = 3

3 (a) q • ~=21 (b) q • £= 14 (e) q • ~ = so (f) !! • ~ = o

BC = OC - OB

{c) !! • ~= - 52

{d) q • ~= - 10

4 Since!! • ~= 0 vectors q = 3f +7 j and£ = 7i - 3j are

-

perpendicular.

q = +(6i +8L)

7 (a) 40

8 (a) l!!I'

6 x=- ~

2 {b) 2.JIO (c) l!!l (b) I

(c) - lq l'

1)

= 31

2 (a) q • ~ = 23.67 (b) q • £ = - 23.67 (c) q • £ = 23.67 (d) q • ~ = 56 (e) q • £ = - 56 (f) q • ~ = 0

5

(e) BD = 2f - 101

AC = !Of + 21, and lACI = ✓ 104 = 2-J26

= AB - CB

(b) l5BI =

00 = 4i -

(f) Use dot p roduct to show the diagonals are perpendicular. Let Mand N be the mid points of the diagonals, then show that they are the same point.

= Sf - 2j

OR =2.; 2-

13 177°

12 B

(bl IABI = 2-m, loci = 2-m, IAC I = 2 -Ju,

5B = 5A + AB and AC = AB + Fc':

(d)

(e) 34°

14 (a) AB = 4i + 6L,BC = 6f - 4Land AC = IOi + 21

(b) Consider vector addition to obtain vectors OB and AC.

OB = Si -

{d) 8= 153°

(c) MBC is a right- angled triangle if AB .L BC and L ABC=90°

24 (a) AB = 3i + 21,CB = 2i - 41

=(

{b) 96° (c) 8=67°

-

= 8i +11 - (2i +41)

=6f + 3j loc l = .J62+ 3' = ✓ 45 = 3✓5 CD = OD - OC = Si + L- (8i + 7L) = - 3i - 6L

Answers

Chapter 10

41 3

lci5I =

✓(-3)' +(- 6)' = +1s = 3..Js

EXERCISE 10.5

1 (a) a • i, = 8

DA = OA - OD

-

= 2 f+ 1 - (st+

L)

-

.

(b)

5

' (d) a- • .b = - --,--= - -,r.; V61 61

(c) a • b = O

= - 3i

(e) a- • b- = -

23

.r.;

4)

l™ l = 3

(el (q • ~)~ = -.',9(4f + 11)

(fl (q • ~)~= ::(- 1r -

21)

q- (q •~)f = i,(76{ - 57 L) (c) q- (q •~)f = 8{ +41 (d) q- (q •~)f = ~ (42{ - 351) (el q - (q • ~) f = ;~ ( 1i - 4 L) !fl !! - (!! • f) ~ = ;; ( 2 t - 11) (b)

=31 + 6{ +3{ =6f+6j BD = BC+CD

= 6{ + 3 j - 3f - 6 j

-

-

- 31

AC • BJ5 = 6 X 3 + 6 X (- 3) =0

4 (a) b • a = .!§. -

:. AC .L BD

Bi5 = r - p, Fc: = q - p and ix = q -

-

(b)

3 (a) q - (q • ~)f = - ,',(2r - 31)

AC=AB + BC

(c)

a•b=" -m 53

(c) (q • ~)f = Q

ITBI = 3, loc i = 3 ..Js, lci5I = 3 ..Js, l™ I = 3

= 3f

(f)

(q •~)~=:,(3i +4l) (d) (q •~)~=;;(sf+6l)

2 (a) (q • ~)~ = :~(3f+2f)

Adjacent sides AB and DA are equal in length and adjacent sides BC and CD are equal in length. Thus, ABCD is a kite. (b) Use vector addition to calculate th e vectors of the kite's diagonals, then calculate the dot product of those diagonals.

a• b= ,sJu 13

-

_,

- r-

(d) Use the distributive law applied to vectors. Now BC • BC = Inc l' and DC • DC = IDC l' Since IFc: I = loci, then IFc: I' = locf and so

s

-

s

(al ('1 • ~)~=!:(4f + 3L)

(b) b- • a- = -./17 •- = •"!7 l, (bl - - - = .!.(4; - ,·) 17 _

(b •a)a

7 C

6 B

8 (a) The scalar projection o f q onto '1 is

(b) The vector projection of q o nto '1 is ;; (5 i + 4 l ).

(c) The scalar projection o f onto q is

BC • BC = DC • DC.

";(:i.

'~'[i.

(d) The vector projection of~ onto (l is ~~ (- f + 7

Now, BC · BC= locl' = l~ - r l

=(2- e)•(q - e) =2•2- 1· e- e•2+e•e =2 • 2 - 2 2 · e+e· e

9 (a) The scalar projection of q o nto '1 is

1

l).

f.

l-

(b) The vector projection of q onto '1 is% i - %

(c) The vector projection of q perpendicular to '1 is

-½! - ½l ·

l ,· - -l ;· ) (d) a= ( -72-,· - -72_;·) + ( - -22..,

.

andOC · OC= jix l' = 12 - r l'

10 (a) q • '1 is the scalar projection of q o nto '1-

=(q - r) • (q - r)

(b)

q•'1 is the scalar projection of'1 onto q.

(c) ~ • f is the cosine of the angle between q and '1-

=q• q- 2 q• r+ r • r NO\\\ ITB I = IAJ51and so le I= k l and

e· e= [ • r-

BC • BC = DC • DC q• q- 2q • E+ E• E= q• q- 2q • r + r • r

- 2q • p = - 2q • [ since p • p = [ • r

:. q• e=q •r (e) Show that the dot product of the diagonals is 0. Tc • iID = q • (r - e) =q•r - q•

e

= 0 sinceq • p=q • r :. AC and BD are perpendicular

11 (al q = lq lcosar+ lq lsina1

Since q is a unit vector l!!I = I. q = IXcosa{ + l xsinaj = cosaf + sin aL (b) ~ = cos/3i + sin/31 and f = cosf3L - sin/31

(c) q• '1 = cos a cos /3 + sin a sin /3; q• £ = cos a cos /3 - sin a sin f3 (d) Now q • '1 = l!! ll '11cos8where 8 is the angle between q and '1-

Here 8 = a - /3and l!! I = 1'11= I. Therefore q • '1 = cos( a

- /3) Thus, cos( a - /3) = cosa cos f3 + sin a sin /3 Similarly. q • £ = l!!ll£1cosef> where ef>is the angle between q and£·

414

New Senior Mathematics Extension 1 for Years 11 & 12

Here 4> =a+ ,Band l!! I = l~I = I. Therefore I.J • £ = cos(a + ,8).

5 AC = a + b

AM = MC

Thus, cos( a + ,B) = cosacos,B - sin a sin,B

= ½(Tc)

=.!.(a + b)

'- -

,

EXERCISE 10.6

BM=BC - MC

=~ -

½(q + ~)

= - ½(!! + ~)

-

0

IANfl = IMCI = IBM I = ½1!! + £1

b

Let 0A = CB = (x - 2) 2 = 4 - y=>

x - 2 = ±J4- y => x = 2± J4 - y and since0SxS2 then 2

x = 2 - J4 - y-

X

x = l.. 2

V

=tr

J:{(; )' -(

2-

J4 - y )' }dy

Sn . , = - uruts 3

>'

40 (a)

4

(b) Area = J ~, ( I - x' - 2( x' - I)) dx .

2

= 4 units

3

(c) y = 2(x' - l) ⇒ x 2 =

i + l. - 2SySO

2

y = I - x' => x' = I - y. 0 s y s I Volume = tr

J:x' dy

Volume = n

J:,(i + I )dy +n J~(I-y)dy

= 3n units' 2

38 (a) Area =

J~( ✓2cos(~x) - x )dx

I

0

(b) V = 1r

J~( 2cos

(

~x) - x' )dx

= 2(tr3+3) Ulll'ts' 418

New Senior Mathematics Extension 1 for Years 11 & 12

3

4

5

X

23 . 2 = - units 3

=1r { (1+ ✓x}'dx = u ;n 4

(c) V 2

2

(b) A = J:(1 +✓ xdx)

= S- n units' 21f

I

(d) y = 1+ Fx ⇒ ✓ x = y - 1 =>X =(y - 1) 2

For OS y S 2 the new solid is a piece of cylindrical pipe with volume, V, = n(42 - 12 ) X 2 = 30n

For 2 S y S 3 the volume of the pipe is given by 3

V, = n s: x' dy =n J: (y - l)' dy = ~n Total volume = V 1 + V, =

181n 5

11

. 3 units

x' + /

=

X

3 -2

12 r = 1- ~ J- 150

19

14 x = 4 -

13 y = 3 (2x- 1) + ,

4

8t+I

EXERCISE 11.3

1-~- - - -T~ x' + y' = I

3 (b) -2✓ 5-

I 1 (a) 3

9 3 6 (a) ~ X

I

a

(a)

(c) 2 ( vr. 5 - I)

16 2 (a) O (b) 44..fi.15

(b) ¼ (c) 6½ 5 (a)

4 (a) 69½

3 B

0

10 f (x ) = 3.../x'+ I - I

= .../t' - 21 +4 + 8 1

y

41 (a)

3

9 y = ½(x -4)2 + ½ 2

5

(b)

f

4

-;..fi.

(b)

¼

(c) 8f 7 (a) 7# (b) 21t

(c) O

(c) o

(c) - 51

t

J:1,f = [ -2.ruJ: = 2✓3 =[-2./4 - u ] 4 = --0+ 2✓3 = 2✓3 ./4- U ~ I

(b) J • du I

{b) y' = I -

V=n

(c) Integrating .J4 - u means that the negative sign needs to be taken into account. It is easily missed. 9 ½units' 10 I unit' 11 Area = • units . 2 o.Sy

x' and y' = I - ~'

15

J:(1- x' )dx +n f 0'(1- ~, )dx

- 0.5

1

X

0.5

= 2rr units 3 - 0.5 1

42 (a) y = Slog. (x - I)=> ; = log, (x - 1) => x - 1 =e• => 12 Area = J: x .Jx - 3dx =

l.

X

3

1

= I + e•

= [¾u½ +211½ ] ~ =

f~x' dy = n J.'(l +ei ) ' dy 1

V=n

~ 66.8 units3

13 Area =

{b) Solve81og. (x - I) = 4 => x - 1 = e05 ⇒ x = I + e05 is the radius of the outer shell. Volume of plastic = Outer Volume of cylinder - Capacity

= n(I + e0·'

)'

1 (a) x +½sin 2x+C

3

(b) ¼(x + 4)'+ c

(b) ;

2

(d)

. 3

';.fi> (e) 1•

31

5

f

(f) ~

·' · 3 units

3 C

(c) ¼(x' +x + 2)6+ c

6 ~: = ½(l - cos2x1 y = ½ (l - cos2x)dx = ; - ¼sin 2x+C,

4

L12 UtlltS

-··2

J

3 D

2x - 2' y = •·•• Y = "-2 - .lsin 0, - 2 if x < 0, undefined at x = 0 (r) 0 l +x 2 l+x 2

(q)

2

11 V = .n: J (cos x - sin' x)dx 0

• =.n: J: cos2xdx • = .n:[ ~ sin2x =1C2 Wlits1

(s) tan x + tan_, x sec' x l +x 2 3 y = 2x+ 2 - ~

r

12 (a)

4 (a) dy = 2 sec' 2x = 2

dx

I + tan' 2x

(c)

(t)

I

x✓x2 -

I

(b) tan ( + ~) is undefined

)'

y

"2

I

y= tan- 1 (tan2x)

0

2 3,r

I

, . --

X

,-.

"-4

2

X

2

4 3Z

2

(b) A = J,;"(sinx+cosx)dx

•

3,T

= (- cosx + sinx

i• = I+ ✓2 units'

3K

•

3K

= area of triangle above x-axis - area of triangle below x-axis

(c) V = J,;" (sinx +cosx)' dx = .n: J,;" (l+sin2x)dx

2

3,T

J: = ¼.n:(3.n: + 2)units

3

= [ x - ½cos2x

5

3

1 (a) - ½cos x+C

(b) ½tan 'x + C

(c) - ¼cos''x + C

3

(e) - -fcos x + C

(d) fsin' x + C

(b) o (c) o.s

(d) o.s

6

(f) ½sin' x + C

(e)

7

: . y is a constant for all x in th e domain.

=0

- l

-l

t

~; = 0

:. y is a constant for all x in the domain.

(f) ;

7 (a)

X 2

+ tan -l X

l +x (b) x tan- 'x - ½log, (l + x 2 ) + C

1 8.n:, a semicircle of radius 4

6 0

11 du=sec xdx,

J ''du 3 u'' =

l.

2

./3 - 1

7

,

= 5i6 t'

\O

cosx = ✓- 1 ;.Js x = cos_, ✓-1 + .Js :.

24 (a) 0 < X < I )'

2

Jr

(b) 0.5 units' y=sin- 1 x

Jr

CHAPTER 12

2

- 3 -2 - 1

O I

2

3

4

X

Jr

2

EXERCISE 12.1 1 y = x" : y' = n x"- 1

LHS = y' - ; y (b)

=n xn-l - .!!. x xn

(; ,o)

X

n-1

(c) y = log,(sin- 1x)- log,(cos- 1x)

.

~

I I - ----.==+---,==

· · dx - sin_, x✓I - x'

cos-• x✓I - x' Over the d omain O< x < I, every factor in the expression is positive :. :

n-1

= nx - nx = O= RHS 2 (a)

(b)

)'

8

> 0 for all x in the domain

-8

Hence the function is increasing for all x in the domain. (d)

>'

(d)J ~

(c) 0

2

3

4

25 (a) log, (l + x)+ C (b) tan-'x + C (d) log, x + tx' + C

422

~r

X

(c) ½log, (l + x')+ C

(e) x - tan-•x + C

New Senior Mathematics Extension 1 for Years 11 & 12

, 3 (a) y = - 7e_,, + 10 (b) y = e-, + 5 (c) y = _ e_ e" -

5e10x

(d) y = 4

,o, +e

;

lx

e y=, e

( )

,

I

- 2- ,

l. 2

-x

1 {b) A = 2 ⇒ y = ----'-x- 2

4 {a) A = 2 ⇒ y = ----"'-

l + 2x

(c) A = ½ ⇒ y =½e-'' +½

)'

{b)

(d) a = 3,b=2 ⇒ y = e-'(3x+2)

(e) a = l ,b = 1 ⇒ y = sin2x+cos2x 1 Jx (f) a = 2l , b = Il ⇒ Y = 2l e-x + 2e 5 (a) The differential equation is first order, second degree, dependent variable: y, independent variable: x (b) The differential equation is first order, first degree, dependent variable: x, independent variable: I (c) The differential equation is second order, first degree, dependent variable: x, independent variable: t {d) The differential equation is first order, first degree, dependent variable: y, independent variable: x 6 ''' =ke" • •r y = D(l + e_{T + ['kT + e-3kT + ... + e 0.

(b) S = ATk where A = ec and A > 0 since S > 0 and T > 0.

dv gR' 14 (a) dh = - v( R + It)'

2

4'

187 _

J6 -

- = gR +c 2

37

32

'is: = 1!!~:k k --

..!!. IS

I = 14 minutes

(b) s = 10 - ! Oe-•'

(c) s(3) = JO - !Ox(¾)' = 4.88 z 5 truck loads

gR'

2 {a) RHS = (l:P) [ (l + ! - I) + ~]

R +h

2

( I + l + P- 1) - (l + P) I(l + P- I)

u'

gRh =2 - R +h , , 2gR V = U J + lt. h

I

- (1:P)(I(l ~+PP- I) ) I

(b) To escape the planet's gravity th e upwards velocity v must be non-negative ash ➔ co.

g~)

2 1+ h

=

u' -2gR ~ 0 when u ~ .J2gR.

The minimum launch velocity is u = .J2gR 15 (a) T = 288 - 0.00649h (b)

lSk X

10 _ , ) 8 32

t = r( IO- s), s(O) = O

- gR

(u' -

1K

ISk -

I Sk

JSk

1 (a) Maximum monthly sales = JO, Differen ce of current sales = JO- s.

u'

=!!.:_ _ gR' + gRh -

lim

14Jf

16

EXERCISE 12.6

v' = gR' u' 2 R + lt +2 - gR

h➔oo

16 .

h=0:

v' gR' -2 = R + lt +c

u'

{b) I = 14:ir - 2:ir (101tf - 3/tt ) 15k 15k (c) h _ 1 1 _ 187 • 187 _ ,., _ , . (

(d) t = 14 - 2( ! Oh½ - 3h½)

Jvdv = - gR' J( R dlt+ h)'

c=

-k

2

(b) V = 60: P = 7-7: ~ JO'

h = o. V = u :

!~

~~ = 0.000189 ~ ,

= I(P + i - I) = LHS (b) I -

I+ p

+

l

Pe-(J+P)r r

3 (a) Net rate of change of pollutant (g/day) = rate of inflow (g/day) - rate of outflow (g/day)

~7 = 10000 X 2 -

: = -0.0352 f.

10000 X

,.;'000

= 20000 - ;';.

.,

Initially pond is unpolluted, so m(O) = 0

= 0.000189 -;, X (-0.0352 7·)

(b) c(t )

= - 6.65 X 10-6 ~

_

dt(r)

But T = 288 - 0.006 49h dB - 6.65 x JO_. B2 So dlt = 288 - 0.00649h

m(r)

= ,00000

di =

I dn~l ) 200000 X d i

! = icx!ooo X ( 20000 - .!. _

-980

(c) h = 288 - 3990e_ 8_ 0.006 49

; )

m

-

10

20X200000

=

1 10 -

{0

where c(O) = 0

-980

(d) c(IO) = 2( 1- e-½) = 0.79 g/L

h = 44 400 - 6!5000e--.

The actual numbers obtained will depend on when you used your calculator, but should be of a similar magnitude. -980 7

(d) B = 357°K, h = 44400 - 6!5000x e "

z 4890 m z 5000 m

Answers

Chapter 12

427

4 Net rate of change of salt (kg/ min) = rate of inflow (kg/ min) - rate of o utflow (kg/ min) Net rate of change of volume (Umin) = rate of inflow (Umin) - rate of outflow (Umin) Q(O) = 1000 X 0.01 = 10kg ":, = rate of inflow - rate o f o utflow

(c) N(4) = l OOOx3' = 9000 N(◄ ) = 9 N{O)

Has increased ninefold in four days.

-

.4 -

= 0.4 -

dt

V~t)

d["c] ["c] =

t =o:

IOCIO+Sl

,.i.,

t =-

~ log, ([ " cl)

c=-

~ log, (( "c ])+~ log, (( '''c l )

l ([ C])

= - ~ log, [ '''Cl

(b) v(t) = 0.0002: t = ':;: ~ 44 minutes t when

t

6 (a) Net rate of change of balan ce ($/ year) = rate o f inflow ($/year) - rate of outflow ($/year)

= .!.

[ " cl

2

log, -1 = 5730 years 2 1.2097 X 10~ 2 (b) If half the original carbon-14 has radiated away then the tree died 5730 years ago.

= O.OS( x- 600000) where x(O) = 5000000

(b) x(t) = 1000000(6- e 0.o;')

(c)

(c) t = 20: x(20) = 1000000( 6 - e' ) = 3281 718.17

1

1

J 1

[ " c] = - r ( " c ]

Since ( 12 C] is a constant then divide the previous equation

Balanceis$3281 718.15 B y = 10 x (0.5)' 8 C D 10 B kl 2k t fls,r3 2~◄ D N =Ae , 3A =Ae ,k = ,log, 3, 4 = e ,t = C

,og,,

fi = s:Oo (1000 - P), J(~+ 100~ - P )dP = ~ + C, P = Ae} A = .!. p = 1000 - p ' 9'

[ "c]

.,=...~~

t, = -

j; = O.OSx - 300000

1000 I 1 +9e-5

14 D

15 (a) ';' ~N ';' = kN,

dt

14

5 (a) v(t ) = 0.0003( 1 - e-o.o,s, )

13 D

J

~ log, ([ 14C ] )

t +c= -

=5

Q(O) =10

7 9 11 12

~ 4 days 5 hours

= - r [ " c] , r = l.2097 X l0 4 years- '

-~J

= rate of inflow - rate of outflow = 0.04 X 10 - 0 SQ

d[ C] 1

V(t) = St + 1000

f

= 4.192 days

1og,,

14

16 (a)

= 10 - S = SUmin V(O) = lOOO L

21ost 10

d t=

( )

by it. d( [ '' c ])- [ ''c] ['' c ] _ ['' cl dt [ 12 C] - - r [ 12 C] and [ 12 C] - [ 12 C] when t = 0. [ "c] dR . [ " cl Let R = [ 12 C] so dt = - rR with R = [ 12 C] = 1.3 X 10- 12 when t = 0.

~~ = - 1.2097 X 10~1 R with R = 1.3 X 10- 12 when t = 0. (d) ~~ = - rR where r = 1.2097 X 10-
O

(b) ~~ = kN dN = kdt N

': = - rdt

log, R = - rt + c t = 0, R = l.3 x 10- 12

log, INl = kt + C

log,( t.3X l 0- 12 ) = c

:

log, R = - rt + log, ( 1.3 x 10- 12 )

N = Ae'' where A = +ec N(O) = 1000 :

A = 1000

log, ( -1.-3-x"' ~-o-- ,~, ) = - rt

3000 = lOOOe"

----"'Re.__,~ =err 1.3 X 10- 12

N = lOOOe'' N(2) = 3000 :

e2 k = 3 ek =

-Tl I

,J3

But e

/' = ( ✓ 3)' = 3½ !.

N = l000 x 3 2

428

New Senior Mathematics Extension 1 for Years 11 & 12

2 =

½i.e. e-sJ;o, = ½ I

I

so { e-S370, )5370 = (½)SJ7o

[ c.. ]

A(t) 18 (a) p(t) = A(t) + B(t)

_.

(e) R = [ c" ] = 0.9 x JO

'(t) = A'(t)(A(t) + B(t)) - A(t)(A'(t) + B'(t)) p (A(t)+B(t)) 2 I

= 0.U9 X 108

( l.)S370 2

A'(t)A(t) + A'(t)B(t) - A'(t)A(t) - A(t)B'(t)

-

(A(t)+B(t))

8

,;,. Jog,(½) = log..(~: x 10

)

5370 log, (-/r X 108 ) I= = - 13986Jyears - 1og, 2

17 (a)

~ = - r( S - I)

{b) A '(t) = rAA(t), rA> 0 and B'( t) = r,,B(t), r8 > 0 '(t) = rAA(t)B(t) - r8 A(t)B(t) p (A(t) + B(t)) 2

IS - f l = e-•l+C

(rA - r.)A(t)B(t)

(A(t)+ B(t)) 2

A(t) [A(t) + B(t) - A(t) ] = ( '• - r,) A(I)+ B(t) A(t) + B(t)

S - I = Ae-• 1 where A+ ec 1- l= A

A(t)

= ( '•

S{t) = I+ (I - I)e-"

!: = - r(S - I) . Use the Chain Rule. .d1!!..(JS) =..!!.. (- r(S 4IJ dS

!)) X dS dr

=- r X (- r( S - I))

- r,) A(t)+B(t)

(d) p(O) = ,~. '• - r8 =

JS di -

_,

p 1- p

=-

J's2 = - r(l - 1)(- re-•') dr

So rs = - r'(S - 1)

t ➔-

lim JS

!: = 0 as S has become a constant.

=lJim [- r(0.05 ➔oo

t➔- dt

I)]= 0

Sol = 0.05

S(14) = 0.06 : 0.06 = 0.05 + 0.95e-"' 0.9se-"' = 0.01 =-

~ 1

log,;. =

~ 1

r

( 9+

e

p = e';.

1~0 )

p(t) =

I

--' I + 9e ,oo

p(IO) =

I

-o t

I + 9e ·

~0.11

dL dt =3- L 5 dL =f dt f 3- L 5

log, 13 - Ll = - ~+C

(d) S{t) = 0.05 + 0.95e-•'

r

hr-•:

19 (a) ! ; = ; (3 - L)

,,,2

(c) Jim S = 0.05 so

1~0

9 p = eioo - peioo

= r'(l - l)e-•'

But S - I = (I - l)e-"

]

9

I

- r( I - l)e-•'

A(t) I - A(t)+B(t)

elOO

= r (S - I) S(t) = I + ( I - I)e-"

[

= (rA - r8 )p(t)(1- p(t))

2

OR

2

, [ A(t) B(t) ] (c) p(t) = (r. - r,) A(t) + B(t)x A(t) + B(t)

log, IS - I l = - rt+C

(b)

(A(t)+ B(1))

-

J

S{O) = l :

A'(t)B(t) - A(t)B'(t)

-

dS = - rdt S- 1 dS = - r f dt S- 1

2

log, 95 = 0.3253

(e) Death rate = - :;; and is a maximum when Sis a ma.ximum Death rate = 0.3253(S - 0.05). S(O) = I, which is the largest value o f S so the d eath rate is greatest when t = 0.

I

I

3- L = e

- L +c

3 - L = Ae L(O) = 0 :

5

_L 5

C

where A = ±e

A =3 I

L(t) = 3 - 3e-,

Maximum death rate = 0.3253(1 - 0.05) = 0.309 per day

(f)

S(I)

00 • c.. I =--=--=--=--=--_;:-;:-:::-::::-==-•

J

0

7

14 t(days)

Answers

Chapter 12

429

~7 =

(b) W(t) = J6L3,

~ 1 0

W, H(O) = 0

(d) For the maximwn growth rate find when ,2'; = 0 dr-

dl = JOO X 16L' dH

1

d2w = .1L( 411V ) dtl

_ ◄X2i ~( J - e--02, )'

H + c=

108 ~

=L( dW '"')x""' = d~v(;0W(!O 4ft

e~.6, )

J(J - 3e--0.2r + 3e-o.◄ r

1 :

-

108 25

J'w = Ogives W = e9 or W = e10 mg

.ii) 6

-

d,2

When W = e10 mg, this is the equilibriwn position

(c) C(t) = ~~:~ (mg/kg)

as t ➔ oo and 4~~ = 0.

l.!l!!.(t + lSe~ur _ ll.e~.◄ 1 +ie-0.61

C=

2s

J

2

lie ~.◄t

Hence the Greatest Growth Rate is when W = e9 .

_.ii) 6

::.!.

16X27(1-e-o2r)' t + l Se~.2r -

log, w))x •~;

= ;. ( I x(!O- log,

(0 + 15 - .!.2. + .l.) = 12 °' 5x 655 23

H -- .ffi!..(t + ie-0.61 zs + l Se ~r - lie~.◄, 2 3

4ft

w) + w X ~i )x (;. W(!O= ;0 (9 - log, w)x ;0 w(10 - log, w) = .~0 (9- log, w)(10 - log, w)

e-o·•• )dt

(t + l Se~ur - 12.2 e~-◄t + ie~..6t ) 3

H(O) = 0 so c =

dt 4

= •:: ( 1 - 3e~ur + 3e-o..◄ , -

fdH =

dt

(e) W = e

+ ie-0.61 -

9

:

20 10 -•'

9

e =e

_, 9 = JO - 9e 20 _,

.ii

- ----~'~ --~ ' --~·~ JOO( I - e-•2r )'

e 20

(d) Hg(mg/kg)

-. l.

-

t = 20 1og, 9 ~ 44 days

0.12

~V(mg)

(f)

0.08 20000

0.04

15000 0

IO

5

10000

20

JS

Age(years)

20 (a)

5000

d;,v = rW log, ( i) d,'; = rW(log, C - log, W)

(1)

d;; = ,\, W(!O- log, w)

(2)

r -(b) W

log, C = JO,

1 20 '

C=e

0

dV

where u = 10 - 9e 20

-_,

= -io e10-9t-20

V = Ae-kt where A = ±ec

_,

x

=- kdt

lvl= e-tr.c

= ..2...e-io x e'o-9e 20 20

JOO

log, IVI = - kt +C

=L

-I

80

Jd: = - k fdt

_,

du dt

60

21 (a) ".(, = - kV, V(O) = I OOOOOO, k>O

10 V

= ilW X du

40

t(daJ•)

=e10-•,-1120

dW dt

20

9e20

V(O) = 1000000 :

A = 1000000

V = JOOOOOOe-•'

Now JO - log, ( e10-•,;; ) = 10 - ( J0 - 9e;;) = 9e;~

V(SJ = 600000: 600000 = 1ooooooe-•k

10-•,;~ ( JO-

So dW dt = _I 20 e

og, ( e10-•,;~ ))

I

And W(O) = e10-• = e so the initial condition is satisfied ::.!.

W =e

20 o-9 '

1

e-kl =(e-k)'

is a solution to equation (2)

=[(¾)'.1]' =(¾)' I

I

(C)

dlV -

di -

V = JOOOOOO(f)'

0

,\, tV(!O- log, W) = O JO- log, W = O W=e

430

10

mg

New Senior Mathematics Extension 1 for Years 11 & 12

(b)

dv dt

= - .!.S logt-3lx V

V = 600 000 ··

dvdt = _ .1S logt-3.l. x 600000

= 61299 barrels per year.

log, w))

dP (f) dt = - K(P - a)( P - b),

I

(c ) V = 1000 :

I 000000(¾)' = 1000 I (s3)L= 1000

dP

5

(P - a)(P - b) = - Kdt

t log, 0.001 s = log, 0.6

I = 67.6 It will remain profitable for 67 years 7 months. I

P - b = Ae-(1'-a)K, where A = ±ec P- a Po - b = A Po - a P - b = PAe- 0 and T(O) = 37

(b) dy = - (x - 2) dx 4(y - l) 4(y - l )dy = - (x - 2)dx

f

4 (y - l )dy = -

x' 2

{b) T dT _

f(x - 2)dx

20

log, IT - 201= - rt + c

C = -9

T - 20

log,

2

x' + 2x + -9 2

17

T - 20

2

4y - By = - x' + 4x+9

T - 20 _ + - ,, 17

4(y' - 2y + 1)= - (x' - 4x +4)+ 17 2

(x - 2) + 4(y - 1) = 17

(_.Jl7!_)'

(x - 2) 2

+(-.Jl7

= - rt

- ,, =e

17

1

2

log, I 7 = c

T(O) = 37 :

2

2y 2 - 4y = - -

= - rdt

f T ~T20 = - r fdt

+ 2x + C

I J8 - 12= - - +2+C

(1,3) :

(c)

J 000000 =fn x9(3R - 3)

- (x - 2) JOO =~~x 400 y- 1

2y1 - 4y = - -

~7

- _e

T = 20+ 17e-•' Since t(O) = 37, T = 20 + 17e- •'

2

2 - ) (y - 1) 2

=I

Y

(c) Let 7:20 pm be -r hours after the death. Thus T(-r) = 29 . 29 = 20 + 17e-"

e

5

-T'r

=u9

err =!Z.

•

(d) At 8:20 pm, t = -c + I: 27.4 = 20+ 17e-d'

e

-T _

7.4.

-

)7

XQ _ 9 -

7.41

9

r = log, ( ,9,) = 0.196

(2, (2 + WJ/2)

(f) Find the value of -r : err

(0, (2 + VJ3)/2)

= l1.

•

r-c -- log 15) = 0.0510 (a) This s ituation is not suitable for binomial modelling because the number of trials was not determined beforehan d. (b) This situation is suitable for binomial modelling. (a) The probability that exactly 25 hearts are drawn is P(X = 25) = 0.0918. (b) The probability that exactly 40 cards are clubs is P(X = 40) = 3.6263 X 10-4 • (c) The probability that exactly 55 red cards are drawn is P(X = 55) = 0.0485. (d) The probability that exactly 60 cards are black is P(X = 60) = 0.0108. B The probability tl1at of eight births, exactly half are girls is P(X = 4) = 0.2157. The probability tl1at exactly 35 people out of a random sample of 100 will be left-han ded is P(X = 35) = 0.0834. The probability tl1at exactly 6 tosses out of IOland on heads is P(X = 6) = 0.146. B This experiment could not be described as X - B ( 5,¼) because the probability of success is not constant. (a) The probability that Philomena stops at exactly five sets of lights is P(X = 5) = 0.3025. (b) The probability that Philomen a stops at fewer than two sets of lights is 0.0109. (c) The probability that Philomena stops at more than five sets of lights is 0.1176. (d) The probability that Philomen a stops at tl1e first three sets of lights is 0.343. (e) The probability that Philomena stops at the second and s ixth sets of lights only is 0.0040.

New Senior Mathematics Extension 1 for Years 11 & 12

17

(f) The probability that Philomena stops at exactly four sets of lights including the first two is 0.1297. (a) (i) 0.0115 (ii) 0.0576 Oi~ 0.1369 (iv) 0.2054 (v) 0.2182 (b) 0.6296 (c) 0.3704. (a) Ifa ran dom sample of 12 loaves is selected, the probability that exactly one loaf cannot be sold at full price is 0.3837. (b) The probability that exactly three loaves are slightly burnt is 0.0532. (c ) The probability that all loaves in the random sample are sold at full price is 0.3677. (d) The probability that the baker would make a profit is 0.9999. (a) The probability that a maximum of three new songs from the sample will become popular hits is 0.950. (b) The probability that more than four but Jess than six songs from the sample will become popular h its is 0.010. (c ) It is most likely that one song will become a popular hit because it has the h ighest probability, P(X =I) = 0.349 (d) The probability that exactly one of70 new songs will become a popular hit is 0. 167. (e) 1\~o songs are most likely to become popular hits because P(X =2) = 0.241 has the highest probability value. (a) X

0

P(X=x)

0.4437

I

2

0.3915 0.1382

3

4

5

0.0244

0.0022

7.59x 10-5

(b) The probability that the boxed set will contain fewer than four faulty DVDs, if you know it contains at least one faulty DVD, is 0.996. EXERCI SE 14.3 1 (a) E(X) = 14, Var(X) = 4.2, a(X) = 2.05

(b) E(X) = 55, Var(X) = 24.75, a(X) = 4.97 (c) E(X) = 33½ , Var(X) = II ¼, a (X) = 3f (d) E(X) = 5, Var(X) = 2.5, a(X) = 1.58 (e) E(X) = 20, Var(X) = 4, a (X) = 2 (f) E(X) = 16, Var(X) = 9.6 2 (a) p = ½ (b) p = 0.1 (c) n = 35 (d) " = 25 3 (a) n = 12andp=0.4 soX - B(12, 0.4) (b) 11 = 20 and p = 0.1 so X - B(20, 0.1) (c ) 11 = 30 and p = 0.15 so X - B(30, 0.15) (d) n = 50 and p = 0.22 so X - B(50, 0.22) (e) 11 = I 00 and p = 0.65 so X - B( I 00, 0.65) (f) n = 150 and p = 0.275 so X - B(150, 0.275) 4 B 5 D 6 P(X = 3)=0.42 7 If the spinner is spun eight times, the probability of obtaining Jess than the expected num ber of blue outcomes is 0.367.

8 C 9 (a) The probability distribution is shown in the table. X

0

1

2

3

4

5

6

7

8

P(X = x) 0.1001 0.2670 0.3115 0.2076 0.0865 0.0231 0.0038 0.0004 0

(b) E(X) = 2 (c ) a (X) = 1.22 10 (a) The probability that the first two children born are male is 0.25. (b) The probability that there are th ree boys and three girls, in any order, is 0.3125. (c ) The probability that there are more girls than boys is 0.3438. (d) The probability that there is at least one boy, but more girls than boys is 0.328 I. (e) The probability that there are no consecutive births of the same gender is 0.0313.

E( P) z pz

11 (a) The number of trials is n = 49 with a probability of success of

P =2.7 · (b) \) = ✓0.4200 X 0.6 = 0 0346 .

= 0.5

Use technology to plot the sampling distribution of j, with mean 0.40 and standard de,~ation 0.0346 on the same set of axes.

Dissatisfied members: p0 = ; = ,;~ = 0. I 7 ( 2 d.p.)

(c)

x0

= ¼X240 = 40,

Po=;=2~~ = 0.17(2 d.p .)

The point estimate d oes not change because the observation value is doubled, whereas the proportion of observations doesn't change. (d) Calculate the new observation value for the dissatisfied members. x 0 = ¼x 120 = 24, j,0 = ; = i';~ = 0.20 The point estimate becomes greater because the proportion of results is greater, which produces a higher observation value. 16 (a) The expected value

E( P) z

s(f>) = ✓p( I - p) =

j,, where j, = 0.35.

/0.35 X 0.65

z

O 0337

n ~ 200 · (b) Use technology to plot the sampling distribution of j, with mean 0.35 and standard deviation 0.0337.

0.15

0.25

0.35

0.45

Sample distribution of j,

CHAPTER REVIEW 14

1 (a) P(X = 3) = !~ (b) P(X = 4) = :.•~ 2 (a) p = 0.8 (b) n = 40 3 E(X) = 25 4 (a) 5 points {b) If Yehudi won JO of the first JS points I would be quite surprised as the number is well above the expected number.

5 (a)

y

0

1

2

3

P(Y= y)

7

,."

7

_,

"

••

120

(b) E(Y) = ,~

0.25 0.35 0.45 Sample distribution of j,

(c) (i) Let x be the observation value. The expected

E( fa) z j, where j, = ; . For x = 55, n = 200: p = ;! = 0.275

value

f,)

E ( z 0.275, Calculate the standard error using the formula:

s(f>) = ✓p(l - p) =

/0.275X0.725 z 00 3 ! 6 n ~ 200 · (ii) Use technology to plot the sampling distribution of with mean 0.275 and standard deviation 0.0316 on the same set of axes as previously.

p

6 C 7B SB 9C 10 D 11 (a) P(success) = 0.75 (b) (Q X - B(30, 0.75) (ii) The expected number of odd numbers which will occur in th e 30 rolls is E(X) =22.5 (iiQTo roll 28 odd numbers in the 30 rolls would be unusual as it is outside the 95% confidence limit. (c) (Q The probability of any pair of rolls resulting in two odd num bers is 0.5625. (ii) The probability distribution of the number of odd numbers in the two rolls y

0

1

2

P(Y= y)

0.0625

0.375

0.5625

12 (a) 0.367 (b) 0.2 1 (c) 0.393 (d) 0.04 (e) Chance of winning more than two if buying one box per d ay for seven d ays is 0.148. Chance of winning more than two if buying one box per d ay for JOd ays is 0.322. Therefore, Carol is correct, her chances will increase. 0.15

0.25

0.35

0.45

Sample distribution of j,

(d) Consider whether the expected value of the new sample is a h igher or a lower value. A higher number of orders from retur ning customers will produce a higher expected value, which means that the curve will be to the right of the previous two curves.

(P) z

:~

= 0.40

Answers

Chapter 14

441

binomial coefficient

A acceleration The rate of change of velocity with respect to time: 2 .. . (t) dv d x dv d (J d d . x,v , dt' dt2 ,v dx or dx 2 v , Stan ar un1tsms .

2)

-2

amplitude

as "c, or (;) where r = 0, I, . .. , n and is given by:

n! r !(n-r )!

The amplitude of a sine or cosine function is half the difference between the greatest and least values of the function. The amplitude of y = a sin nx is a.

angle of projection The angle, measured from the horizontal, at which a particle is projected.

arc (1) Par t of the circumference of a circle. (2) Par t of a curve.

binomial distribution The binomial distribution with parameters n and p is the discrete probability distribution of the number of successes in a sequence of n independent Bernoulli trials, each of which yields success with probability p.

binomial expansion A binomial expansion describes the algebraic expansion of powers of a binomial expression.

binomial probability

arrang ements in a circle The number of ways of ar ranging n different objects in a circle, regarding clockwise and anticlockwise arrangements as different, is : ' = (n - I )!. As there is no start or finish to a circle, one object needs to be fixed as the 'starting object' for counting purposes.

asymptote A line or curve that a function approaches but never reaches.

auxiliary angle method \,Vriting asinx + bcosx in the form rsin (x + a) or rcos(x-a), where r = ✓a2 + b2 and a is in the first quadrant such that tan a =

!-

B Bernoulli distribution The Bernoulli distribution is the probability distribution of a random variable which takes the value I with 'success' probability p, and the value O with 'failure' probability q = I - p. The Bernoulli distribution is a special case of the binomial distribution, where n = I.

Bernoulli random variable A Ber noulli random var iable has two possible values, namely O representing failure and I representing success. The parameter associated with such a random variable is the probability p of obtaining a I.

Bernoulli trial A Ber noulli trial is an exper iment with only two possible outcomes, labelled 'success' and 'failure'.

442

The coefficient of the ter m x" - ' y' in the expansion of 11 (x + y ) is called a binomial coefficient. It is written

New Senior Mathematics Extension 1 for Years 11 & 12

When the same trial is repeated several times and there are only two possible outcomes in each trial ( often called 'success' and 'failure'), the probability of r successes is given by P( X = r ) = (;

)q•-rp', where

0 < r < n, where p is the probability of a 'success' and

q = I - p is the probability of a 'failure'. This expression is the corresponding ter m of the binomial expansion of (p + q)".

binomial random variable A binomial random variable X represents the number of successes in n independent Bernoulli tr ials. In each Bernoulli trial, the probability of success is p and the probability of failure is: q = I - p.

binomial theorem The sum of two terms raised to the nth power can be expanded as a series of n ter ms according to the following formula:

(a+b)" = "C0a" + "C 1a"-1b+ "C2a"- 2b2 +.. .

+

+. .+ =I

•c , a"-rbr

n

•c . b"

· c , a"-'b'

r=O

(l + x )"

=(:)+(~)x+(;)x +. .+(:)x'+... +(:)x" t(:)x' 2

=

E

C column vector notation

even function A function is even i ff(-x) = j(x) for all values of x in the domain. The function is also symmetrical about the y-axis.

A vector,~, in two dimensions can be represented in column vector notation. For example, the ordered pair a = (4,5) can be represented in column vector notation

exponential decay

as: a = (:).

Exponential decay ( or 'exponential decline') occurs when a quantity decreases by a constant percentage over time.

combination

exponential gro,vth

A combination is a selection of r distinct objects from n distinct objects, where order is not important. The

Exponential growth occurs when a quantity increases by a constant percentage over time.

(n),

number of such combinations is denoted by "C or I r r and is given by: n· r!(n-r)!

F factorial (!)

component form of a vector

The product of the first n positive integers is called the factorial of n and is denoted by n!. n! = n(n - 1) (n - 2) (n - 3) x ... X 3 x 2 x 1 By definition: O! = 1

The component form of a vector, !'. , expresses the vector in terms of unit vectors i, a unit vector in the x-direction, and j, a unit vector in they-direction. For example, the ordered vector pair !'.' = ( 4, 3) can be represented as: !'. = 4£ +3j.

factorial notation (!) The notation n! is called 'n factor ial' and means n x (n - 1)(n - 2) x .. . X 3 X 2 x 1.

cubic function or cubic polynomial A polynomial function f of the third degree, defined by 3 , j(x) = ax + bx- + ex + d where a, b, c, dare constants, a '# 0. Every cubic polynomial has at least one linear factor of the form (x + a) where a is a real number.

factor theorem The factor theorem states that a polynomial P(x) has a factor (x - k), if and only if, P(k) = O; that is, k is a root of the equation P(x) = 0. The factor theorem links the factors and zeros of a polynomial.

D

fundamental counting principle

decreasing function

The fundamental counting pr inciple states that if one event has rn possible outcomes and a second independent event has n possible outcomes, then there is a total of ,n x n possible outcomes for the two combined events.

A function f defined on the interval a < x < b is said to be strictly monotonic, decreasing when, for all x 1 and x 2 in the domain , if x 2 > x 1 then j(x2) < j(x1). In other words, the cur ve always slopes downwards to the right, so the function never has a positive gradient.

degree of a polynomial A polynomial P(x) = a. x" + a. - ,x•-• + ... + a 1x + a0 is of

H half-life

the nth degree if a. '# 0.

The time taken for half of the atoms in a radioactive substance to decay.

differential equation A differential equation is any equation containing the der ivative of an unknown function.

horizontal line test To check that any hor izontal line that can be drawn will only cut a curve once.

direction field A direction field (or slope field} is a graphical representation of the tangent lines to the solutions of a first-order differential equation.

displacement The signed (positive or negative) distance of a par ticle from the origin; x(t).

displacement vector A displacement vector represents the displacement from one point to another.

division algorithm

I increasing function A function f defined on the interval a < x < b is said to be str ictly monotonic, increasing when, for all x 1 and x 2 in the domain, if x 2 > x 1 then j(x2) > j(x 1). In other words, the curve always slopes upwards to the r ight, so the function never has a negative gradient.

indefinite integral The pr imitive of a function, written

Jj(x)dx .

For any polynomial P(x) = (x - a}Q(x) + R, where Q(x) is a polynomial.

Glossary

443

independent events

multiplication principle

Events are independent if the occurrence or nonoccurrence of one event cannot change the probability of the occur rence of another event, i.e. the events have no effect on each other.

See fundan1ental counting principle.

multiplicity of a root Given a polynomial P(x) , if P(x) = (x- a)' Q(x), Q (a)"# 0 and r is a positive integer, then the root x = a has multiplicity r.

induction (mathematical) A method used to prove that something is true for all positive integers n (or e.g. for all positive integers n greater than a specified starting value, or for a subset such as all positive odd integers n) . First, the proposition is proved tr ue for a starting value of n; next, it is proved that if the proposition is true for n = k then it is tr ue for n = k + I; by induction, it is therefore tr ue for all values of n greater than the starting value.

mutually exclusive events Two events are mutually exclusive ( or 'disjoint') if membership of one event excludes membership of the other, so that they cannot occur simultaneously.

N Newton's la,v of cooling The cooling rate of a body is proportional to the difference between the temperature of the body and the temperature of the surrounding medium: ~; = - k(T - M), where T is the temperature at any time t and M is the temperature of the surrounding medium (a constant). A solution of this differential equation is T = M + Ae-k', where A is a constant.

initially \"/hen time t = 0 (at the beginning).

integrand An integrand is a function that is to be integrated.

integration by substitution Changing the variable in an integral to make it easier to determine the integral, e.g. by substituting a new variable u = g(x).

invers e cosine function

0 odd function A function is odd if j(-x) = -f(x) for all values of x in the domain. The function has rotational symmetry about the origin.

1

y = cos- xis the inverse cosine function; it means 'that part of x = cosy for which O< y < n-'.

invers e function Iff(g(x)) = g(f(x)) = x theng(x) = f - 1(x) is the inverse

one-to-one function A function for which any vertical line can only cut it once and any horizontal line can only cut it once; every one-to-one function has an inverse function.

function ofj(x). The graph ofy = J- 1(x) is the reflection of y =j(x) in the line y = x, when the inverse function exists.

invers e sine function 1

y = sin- xis the inverse sine function; it means 'that par t

of x = sin y for which

-1< y < 1'.

invers e tangent function y = tan-' xis the inverse tangent function; it means 'that

part of x = tany for which

-1< y < 1'.

parameter (1) A parameter is a quantity that defines cer tain

character istics of a function or system. For example, 0 is a parameter in J = X COS 0. (2) A parameter can be a characteristic value of a

L leading term The term a. x• of a polynomial, where a. "# 0.

logistic equation The logistic equation is the differential equation dn - = kN ( P - N) where k, Pare constants. Thus: if dt dn N = Oor N = P, - = 0 dt

M mathematical induction Mathematical induction is a method of mathematical proof used to prove statements involving the natural numbers. It is also known as proof by induction or inductive proof. The principle of induction is an axiom and so cannot itself be proven.

444

p

New Senior Mathematics Extension 1 for Years 11 & 12

situation. For example, the time taken for a machine to produce a certain product.

parametric equations When two related variables (e.g. x and y) are expressed in terms of a third variable, the 'parameter' (e.g. tor 0), so that x = f(t),y = g(t) or x = f(0),y =g ( 0).

Pascal's triangle An arrangement of numbers that gives the coefficients of the ter ms in the binomial expansion of ( I + x)".

permutation

roots (of an equation) (1) The values of x for which j(x) = 0.

A permutation is an arrangement of r distinct objects taken from n distinct objects where order is important. The number of such per mutations is denoted by "P, and is equal to: " P,. = nn-1 ( ) .. .( n-r + I ) = ( n! ) n- r !

(2) The abscissae of the points at which the graph of y = j(x} cuts or touches the x-axis.

s sample proportion The sample proportion (j,) is the fraction of samples out of n Bernoulli trials which were successes (x), that is:

The number of permutations of n objects is n!

point of contact

•

A point where two curves touch.

polynomial An algebraic expression of the for m 1 P(x) = a. x" + a. - ,x"- + ... + a 1x + aO' where n, n - I, ... are all positive integers and a. , a. _,, ... are the coefficients (for convenience, usually chosen as integers).

scalar A scalar is a quantity with magnitude but no direction.

secant The reciprocal of the cosine function. (Abbreviated 'sec'.)

solid of revolution

position vector The position vector of a point P in the plane is the vector joining the origin to P.

A three-dimensional solid formed by rotating an area under the curve about one of the coordinate axis.

statement

projectile motion The motion of a particle when it is projected in any direction and subject only to gravitational acceleration, i.e. ignoring any air resistance.

An assertion that can be true or false, but not both.

T tformulae Expressions for sinA, cosA and tanA in terms oft where t = tan

Q quadratic function or quadratic polynomial A polynomial function f of the second degree, defined 2 by j(x) = ax + bx+ c, where a, b, care constants, a '# 0. A quadratic polynomial with no real zeros cannot be factor ised into linear factors of the form (x + a) where a is a real number.

quartic function or quartic polynomial A polynomial function f of the fourth degree, defined 4 3 2 by j(x) = ax + bx + ex + dx + e, where a, b, c, d, e are constants, a'# 0. A quartic polynomial with no real zeros cannot be factor ised into linear factors of the form (x + a) where a is a real number.

1-

trajectory The path followed by a particle.

trifecta In horse racing, when the first three horses are picked in the cor rect order.

V velocity The rate of change of displacement with respect to time:

x(t), v(t) or : , standard units ms-'.

vertical line test

quinella

To check that any vertical line that can be drawn will only cut a curve once.

In horse racing, when the first two horses are picked correctly (in any order).

R range on the horizontal plane The hor izontal distance between a particle's point of projection and the point where the particle meets the horizontal plane through the point of projection.

remainder theorem The remainder theorem states if a polynomial P(x) is divided by (x - k), the remainder is equal to P(k).

X

p=n For large n, j, has an approximately normal distribution.

If a vertical line cannot pass through a curve more than once, then the curve represents a function.

z zero factorial By definition, O! = I.

zeros (of a function) The values of x for which a function j(x} = 0.

rest When a par ticle is 'at rest; its velocity is zero.

Glossary

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