Lectures in Mathematics for 1st Class Students

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Lectures in

MATHEMATICS st

For 1 class students Mechanical Engineering Department Technology University Prepared by

Assistant Professor

Dr. Nabel George Nacy and

Lecture

Dr. Laith Jaafer Habeeb References: 1) Thomas & Finney " Calculus and Analytic Geometry " (1988) , 7th edition , Addison Wesley. 2) Ford , S.R. and Ford , J.R. " Calculus " , (1963) McGraw-Hill. 3) J.K.Back house and S.P.T. Houldsworth " Pure Mathematics a First Course " (1979) , S1 Edition , Longman Group .

Chapter – one The Rate of Change of a Function 1-1- Coordinates for the plane : Cartesian Coordinate- Two number lines , one of them horizontal (called x-axis ) and the other vertical ( called y-axis ). The point where the lines cross is the origin . Each line is assumed to represent the real number . On the x-axis , the positive number a lies a units to the right of the origin , and the negative number –a lies a units to the left of the origin . On the y-axis , the positive number b lies b units above the origin while the negative where –b lies b units below the origin . With the axes in place , we assign a pair (a,b) of real number to each point P in the plane . The number a is the number at the foot of the perpendicular from P to the x-axis (called x-coordinate of P). The number b is the number at the foot of the perpendicular from P to the y-axis ( called y-coordinate of P ). y-axis

P(a,b)

b

x-axis -a

0

a

-b

1-2- The Slope of a line : Increments – When a particle moves from one position in the plane to another , the net changes in the particle's coordinates are calculated by subtracting the coordinates of the starting point ( x1 , y1 ) from the coordinates of the stopping point (x2 , y2 ) , i.e. ∆x = x2 –x1 , ∆y = y2 – y1 . Slopes of nonvertical lines : Let L be a nonvertical line in the plane , Let P1(x1 , y1 ) and P2 ( x2 , y2 ) be two points on L. Then the slope m is : ١

m

y y2  y1  x x2  x1

where x  0

- A line that goes uphill as x increases has a positive slope . A line that goes downhill as x increases has a negative slope . - A horizontal line has slope zero because ∆y = 0 . - The slope of a vertical line is undefined because ∆x = 0 . - Parallel lines have same slope . - If neither of two perpendicular lines L1 and L2 is vertical , their slopes m1 and m2 are related by the equation : m1 . m2 = -1 . Angles of Inclination: The angle of inclination of a line that crosses the xaxis is the smallest angle we get when we measure counter clock from the x-axis around the point of intersection . The slope of a line is the tangent of the line angle of inclination . m = tan Ф where Ф is the angle of inclination . - The angle of inclination of a horizontal line is taken to be 0 o . - Parallel lines have equal angle of inclination . L P2(x2 ,y2) y-axis

∆y P1(x1,y1)

Ф

∆x

Q(x2,y1) x-axis

EX-1- Find the slope of the line determined by two points A(2,1) and B(-1,3) and find the common slope of the line perpendicular to AB. y  y1 31 2 Sol.- Slope of AB is:   m AB  2 x 2  x1  1  2 3 1 3 Slope of line perpendicular to AB is :   m AB 2 EX-2- Use slopes to determine in each case whether the points are collinear (lie on a common straight line ) : a) A(1,0) , B(0,1) , C(2,1) . b) A(-3,-2) , B(-2,0) , C(-1,2) , D(1,6) . ٢

Sol. –

10 11  1 and m BC   0  m AB 01 20 The points A , B and C are not lie on a common straight line . 0  ( 2 ) 20 62  2 , m BC   2 , mCD  2 b) m AB   2  ( 3 )  1  ( 2 ) 1  ( 1 ) Since mAB = mBC = mCD Hence the points A , B , C , and D are collinear .

a) m AB 

1-3- Equations for lines : An equation for a line is an equation that is satisfied by the coordinates of the points that lies on the line and is not satisfied by the coordinates of the points that lie elsewhere . Vertical lines : Every vertical line L has to cross the x-axis at some point (a,0). The other points on L lie directly above or below (a,0) . This mean x  a ( x , y ) that : Nonvertical lines : That point – slope equation of the line through the point ( x1 , y1 ) with slope m is :

y – y1 = m ( x – x 1 ) Horizontal lines : The standard equation for the horizontal line through the point ( a , b ) is : y = b . The distance from a point to a line : To calculate the distance d between the point P(x1 , y1 ) and Q( x2 , y2 ) is : d  ( x2  x1 )2  ( y2  y1 )2 We use this formula when the coordinate axes are scaled in a common unit . To find the distance from the point P( x1 , y1 ) to the line L , we follow : 1. Find an equation for the line L' through P perpendicular to L : y – y1 = m' ( x – x1 ) where m' = -1 / m 2. Find the point Q( x2 , y2 ) by solving the equation for L and L' simultaneously . 3. Calculate the distance between P and Q . The general linear equation : Ax + By = C where A and B not both zero. EX-3 – Write an equation for the line that passes through point : a) P( -1 , 3 ) with slope m = -2 . b) P1( -2 , 0 ) and P2 ( 2 , -2 ). Sol. - a) y – y1 = m ( x – x1 ) → y – 3 = -2 ( x – (-1)) → y + 2x = 1 b) y  y1 1 20 m 2   x 2  x 1 2  ( 2 ) 2 1 y  y1  m ( x  x 1 )  y  0   ( x  ( 2 ))  2 y  x  2  0 2

٣

EX-4 - Find the slope of the line : 3x + 4y = 12 . 3 3 Sol. - y   x  3  the slope is m   4 4 EX-5- Find : a) an equation for the line through P( 2 ,1 ) parallel to L: y = x + 2 . b) an equation for the line through P perpendicular to L . c) the distance from P to L . Sol.a) sin ce L2 // L1  m L 2  m L 1  1  y  1  1( x  2 )  y  x  1 b) Since L1 and L3 are perpendicular lines then : m L 3  1  y  1   ( x  2 )  y  x  3 c)

y x2 y x  3



x

1 2

and

y

5  2

P ( 2 ,1 ) and

1 5 Q ,  2 2

 d  ( xQ  x P )2  ( yQ  y P )2  4.5

EX-6 – Find the angle of inclination of the line :

3 x  y  3

Sol.y   3x  3



m 3

m  tan   3    120 o

EX-7- Find the line through the point P(1, 4) with the angle of inclination Ф=60 o . Sol.m  tan  tan 60  3 y  4  3( x  1 )  y  3 x  4  3 EX-8- The pressure P experienced by a diver under water is related to the diver's depth d by an equation of the form P = k d + 1 where k a constant . When d = 0 meters , the pressure is 1 atmosphere . The pressure at 100 meters is about 10.94 atmosphere . Find the pressure at 50 meters. Sol.- At P = 10.94 and d = 100 → 10.94 = k(100)+1 → k = 0.0994 P = 0.0994 d + 1 , at d = 50 → P = 0.0994 * 50 + 1 = 5.97 atmo.

٤

1-4- Functions : Function is any rule that assigns to each element in one set some element from another set : y = f( x ) The inputs make up the domain of the function , and the outputs make up the function's range. The variable x is called independent variable of the function , and the variable y whose value depends on x is called the dependent variable of the function . We must keep two restrictions in mind when we define functions : 1. We never divide by zero . 2. We will deal with real – valued functions only. Intervals : - The open interval is the set of all real numbers that be strictly between two fixed numbers a and b :

( a ,b )  a  x  b

- The closed interval is the set of all real numbers that contain both endpoints :

[ a ,b ]  a  x  b

- Half open interval is the set of all real numbers that contain one endpoint but not both :

[ a ,b )  a  x  b ( a ,b ]  a  x  b

Composition of functions : suppose that the outputs of a function f can be used as inputs of a function g . We can then hook f and g together to form a new function whose inputs are the inputs of f and whose outputs are the numbers :

( go f )( x )  g( f ( x ))

EX-9- Find the domain and range of each function : a)

y

x4

c)

y  9  x2

,

, d)

Sol. - a ) x  4  0  b)

c)

b)

1 x2 y  2 x y

x  4 

x20 1 y  x2

D x : x   4

x2 1 x  2 y

9  x2  0  y  9  x2  sin ce

9  y2  0 

sin ce

y0

, R y : y  0

D x : x  2 R y : y  0

 3 x 3 x   9  y2

D x : 3  x  3

3 y3 

٥

Ry : 0  y  3

d) if if

2 x 0 x0 y 2 x4 y0

0 x4

Dx : 0  x  4



Ry : 0  y  2

x 1 and g ( x )  1  . x1 x Find (gof)(x) and (fog)(x) .

EX-10- Let

f( x )

Sol. x  ( g o f )( x )  g ( f ( x ))  g  1 x 1   

1 2x  1  x x x1

1

1 x

1   x1 ( f o g )( x )  f ( g ( x ))  f  1    1 x  1 1 x

EX-11- Let ( g o f )( x )  x and  1 Sol.- ( g o f )( x )  g    x   x

f( x) g( x ) 

1 . Find g(x). x

1 x

1-5- Limits and continuity : Limits : The limit of F( t ) as t approaches C is the number L if : Given any radius ε > 0 about L there exists a radius δ > 0 about C such that for all t , 0  t  C   implies F ( t )  L   and we can write it as : lim F ( t )  L t C

The limit of a function F( t ) as t→C never depend on what happens when t = C . lim F ( t )  L Right hand limit : t C

The limit of the function F( t ) as t →C from the right equals L if : Given any ε > 0 ( radius about L ) there exists a δ > 0 ( radius to the right of C ) such that for all t : C  t  C    F( t )  L   Left hand limit : lim F ( t )  L t C

The limit of the function F( t ) as t →C from the left equal L if : Given any ε > 0 there exists a δ > 0 such that for all t : C    t  C  F( t )  L   ٦

Note that – A function F( t ) has a limit at point C if and only if the right hand and the left hand limits at C exist and equal . In symbols : lim F ( t )  L  lim F ( t )  L and lim F ( t )  L t C

t C 

t C

The limit combinations theorems : 1 ) lim F1 ( t )  F2 ( t )  lim F1 ( t )  lim F2 ( t ) 2 ) limF1 ( t ) * F2 ( t )  lim F1 ( t ) * lim F2 ( t ) F ( t ) lim F1 ( t )  3 ) lim 1 where lim F2 ( t )  0 F2 ( t ) lim F2 ( t ) 4 ) lim k * F1 ( t )  k * lim F1 ( t ) k Sin 1 5 ) lim  0  provided that  is measured in radius The limits ( in 1 – 4 ) are all to be taken as t→C and F1( t ) and F2( t ) are to be real functions . Thm. -1 : The sandwich theorem : Suppose that f ( t )  g ( t )  h( t ) for all t  C in some interval about C and that f( t ) and h( t ) approaches the same limit L as t→C , then : lim g ( t )  L t C

Infinity as a limit : 1.The limit of the function f( x ) as x approaches infinity is the number L: lim f ( x )  L . If , given any ε > 0 there exists a number M such that x

 f( x) L  . for all x : M  x 2. The limit of f( x ) as x approaches negative infinity is the number L : lim f ( x )  L . If , given any ε > 0 there exists a number N such that x  

for all x : x  N  f( x ) L  . The following facts are some times abbreviated by saying : a) As x approaches 0 from the right , 1/x tends to ∞ . b) As x approaches 0 from the left , 1/x tends to -∞ . c) As x tends to ∞ , 1/x approaches 0 . d) As x tends to -∞ , 1/x approaches 0 . Continuity : Continuity at an interior point : A function y = f( x ) is continuous at an interior point C of its domain if : lim f ( x )  f ( C ) . x C

Continuity at an endpoint : A function y = f( x ) is continuous at a left endpoint a of its domain if : lim f ( x )  f ( a ) . xa

A function y = f( x ) is continuous at a right endpoint b of its domain if: lim f ( x )  f ( b ) . t b 

٧

Continuous function : A function is continuous if it is continuous at each point of its domain . Discontinuity at a point : If a function f is not continuous at a point C , we say that f is discontinuous at C , and call C a point of discontinuity of f . The continuity test : The function y = f ( x ) is continuous at x = C if and only if all three of the following statements are true : 1) f ( C ) exist ( C is in the domain of f ) . 2) lim f ( x ) exists ( f has a limit as x→C ) . x C

3)

lim f ( x )  f ( C ) ( the limit equals the function value ) .

x C

Thm.-2 : The limit combination theorem for continuous function : If the function f and g are continuous at x = C , then all of the following combinations are continuous at x = C : 1 ) f  g 2 ) f .g 3 ) k .g k

4 ) go f , f o g 5 ) f / g

provided g ( C )  0 Thm.-3 : A function is continuous at every point at which it has a derivative . That is , if y = f ( x ) has a derivative f ' ( C ) at x = C , then f is continuous at x = C . EX-12 – Find : 5x3  8x2 x3  a3 1) lim 4 , 2 ) lim x 0 3 x  16 x 2 xa x 4  a 4 tan 2 y Sin5 x , 4 ) lim 3) lim x  0 Sin 3 x y 0 3y Sin 2 x 1  5) lim 2 , 6 ) lim  1  Cos  x 0 2 x  x x  x  3 2 3x  5 x 7 3y 7 7 ) lim , 8 ) lim x   10 x 3  11 x  5 y y 2  2 x3  1 1 9) lim , 10 ) lim 2 x  2 x  7 x  5 x  1 x  1 Sinx     11 ) lim Cos 1   , 12 ) lim Sin Cos(tan x )  x 0 x 0 x   2  S0l.5 x3  8x2 5x  8 08 1  1 ) lim 4  lim 2  2 x  0 3 x  16 x x  0 3 x  16 0  16 2 3 3 2 2 x a ( x  a )( x  ax  a ) a2  a2  a2 3 2 ) lim 4 lim    x a x  a 4 x  a ( x  a )( x  a )( x 2  a 2 ) ( a  a )( a 2  a 2 ) 4 a Sin5 x Sin5 x 5 lim 5 x  5 . 5 x 0 5 x  5 3 ) lim x 0 Sin 3 x 3 Sin 3 x 3 3 lim 3 x 0 3x 3x

٨

tan 2 y 2 Sin 2 y 1 2 .lim  . lim  y 0 3y 3 2 y0 2 y y0 Cos 2 y 3 Sin 2 x 1 Sin 2 x  2 lim . lim 2 lim 2 x 0 2 x  x 2 x 0 2 x x 0 2 x  1 1  lim  1  Cos   1  Cos0  2 x x  5 7 3  3 3 2 3x  5x  7 x x  3 lim  lim 3 x   10 x  11 x  5 x 11 5 10 10  2  3 x x 3 7  3y 7 y y2 0  lim  0 lim 2 y  y  2 y  2 1 1 2 y 1 1 3 3 x 1 1 x lim lim    x 2 x 2  7 x  5 x 2 7 5 0   x x2 x3 1 1    lim x  1 1 1 x   1 Sinx       Cos 1  lim Sinx   Cos0  1 lim Cos 1  x 0 x x  0   x          lim Sin Cos(tan x )  Sin Cos(tan 0 )  Sin Cos0   Sin  1 x 0 2 2  2  2 

4 ) lim 5)

6)

7)

8)

9)

10 ) 11 )

12 )

EX-13- Test continuity for the following function :  x 2  1  1  x  0   0 x1   2 x  f( x)  1 x1   2 x  4 1  x  2     0 2  x  3  Sol.- We test the continuity at midpoints x = 0 , 1 , 2 and endpoints x = -1 , 3 . At x  0  f ( 0 )  2* 0  0 lim f ( x )  lim( x 2  1 )  1 x 0 x 0  lim f ( x )  lim 2 x  0  lim f ( x ) x 0

x 0

x 0

Since lim f ( x ) doesn' t exist x 0 Hence the function discontinuous at x  0

٩

At

x  1

f (1) 1 lim f ( x )  lim 2 x  2 x1

x1

lim f ( x )  lim ( 2 x  4 )  2  lim f ( x )  lim f ( x )

x  1

x 1

x1

Since lim f ( x )  f ( 1 )

x  1

x1

At

x 2

Hence the function is discontinuous at x  1 f ( 2 )  2 * 2  4  0 lim f ( x )  lim( 2 x  4 )  0 x 2 x 2 lim f ( x )  lim 0  0  lim f ( x )  lim f ( x ) x 2

x2

x2

Since lim f ( x )  f ( 2 )  0

x  2

x 2

Hence the function is continuous at x  2 At

x  1 

f ( 1 )  ( 1 ) 2  1  0 lim f ( x )  lim ( x 2  1 )  0  f ( 1 )

x   1

x  1

Hence the function is continuous at x  1

At

x  3

f(3)0 lim f ( x )  lim 0  0  f ( 3 )

x  3

x3

Hence the function is continuous at x  3

EX-14- What value should be assigned to a to make the function :  x 2  1 f(x)   2 ax

x  3  

x  3 

continuous at x = 3 ?

Sol. – lim f ( x )  lim f ( x )  lim ( x 2  1 )  lim 2 ax  8  6 a  a 

x3

x3

x3

x3

١٠

4 3

Problems – 1 1. The steel in railroad track expands when heated . For the track temperature encountered in normal outdoor use , the length S of a piece by a linear equation . An of track is related to its temperature t experiment with a piece of track gave the following measurements : t 1  65 o F , S 1  35 ft t 2  135 o F , S 2  35.16 ft Write a linear equation for the relation between S and t . (ans.: S=0.0023t+34.85) 2. Three of the following four points lie on a circle center the origin . Which are they , and what is the radius of the circle ? A(-1.7) , B(5,-5) , C(-7,5) and D(7,-1). (ans.: A,B,D;√50) 3. A and B are the points (3,4) and (7,1) respectively . Use Pythagoras theorem to prove that OA is perpendicular to AB . Calculate the slopes of OA and AB , and find their product . (ans.: 4/3, -3/4;-1) 4. P(-2,-4) , Q(-5,-2) , R(2,1) and S are the vertices of a parallelogram . Find the coordinates of M , the point of intersection of the diagonals and of S. (ans.: M(0,-3/2) , S(5,-1)) 5. Calculate the area of the triangle formed by the line 3x-7y+4 =0 , and the (ans.: 8/21) axes . 6. Find the equation of the straight line through P(7,5) perpendicular to the straight line AB whose equation is 3x + 4y -16 = 0 . Calculate the length of (ans.: 3y-4x+13=0;5) the perpendicular from P and AB. 7. L(-1,0) , M(3,7) and N(5,-2) are the mid-points of the sides BC , CA and AB respectively of the triangle ABC. Find the equation of AB. (ans.:4y=7x-43) 8. The straight line x – y – 6 = 0 cuts the curve y2 = 8x at P and Q . Calculate (ans.:16√2) the length of PQ . 9. A line is drawn through the point (2,3) making an angle of 45o with the positive direction of the x-axis and it meets the line x = 6 at P . Find the distance of P from the origin O , and the equation of the line through P perpendicular to OP. (ans.: √85,7y+6x-85=0) 10. The vertices of a quadrilateral ABCD are A(4,0) , B(14,11) , C(0,6) and D(-10,-5) . Prove that the diagonals AC and BD bisect each other at right angles , and that the length of BD is four times that of AC . ١١

11. The coordinates of the vertices A, B and C of the triangle ABC are (-3,7) , (2,19) and (10,7) respectively : a) Prove that the triangle is isosceles. b) Calculate the length of the perpendicular from B to AC , and use it to (ans.:12,78) find the area of the triangle . 12. Find the equations of the lines which pass through the point of intersection of the lines x - 3y = 4 and 3x + y = 2 and are respectively parallel and perpendicular to the line 3x + 4y = 0 . (ans.:4y+3x+1=0;3y-4x+7=0) 13. Through the point A(1,5) is drawn a line parallel to the x-axis to meet at B the line PQ whose equation is 3y = 2x - 5 . Find the length of AB and the sine of the angle between PQ and AB ; hence show that the length of the perpendicular from A to PQ is 18/√13 . Calculate the area of the triangle (ans.:9,2/√13,25/12) formed by PQ and the axes . x2  2 14. Let y  2 , express x in terms of y and find the values of y for x 1 y2 which x is real . (ans.: x   ; y  2 or y  1 ) y1

15. Find the domain and range of each function : 1 1 1 a) y , b ) y  , c ) y  1  x2 1 x 3 x ( ans . : a )x ,0  y  1 ; b ) x  0 , y  0 ; c ) x  3 , y  0 ) 16. Find the points of intersection of x2 = 4y and y = 4x . (ans.:(0,0),(16,64)) 17. Find the coordinates of the points at which the curves cut the axes : a ) y  x 3  9 x 2 , b ) y  ( x 2  1 )( x 2  9 ) , c ) y  ( x  1 )( x  5 )2 (ans.:a)(0,0);(0,0),(9,0);b)(0,9);(1,0),(-1,0),(3,0),(-3,0);c)(0,25);(-1,0),(5,0)) 18. Let f(x) = ax + b and g(x) = cx + d . What condition must be satisfied by the constants a , b , c and d to make f(g(x)) and g(f(x)) identical ? (ans.:ad+b=bc+d) 19. A particle moves in the plane from (-2,5) to the y-axis in such away that ∆y = 3*∆x . Find its new coordinates . (ans.:(0,11),(0,-1)) 20. If f(x) = 1/x and g(x)=1/√x , what are the domain of f , g , f+g , f-g , f.g , f/g , g/f , fog and gof ? What is the domain of h(x) = g(x+4) ? ( ans . : x  0 ,x  0 ,x  0 ,x  0 ,x  0 ,x  0 ,x  0 ,x  0 ,x  0 ; x  4 ) ١٢

21. Discuss the continuity of : 1    x for x0   x   x3 for 0  x  1  f( x )  1 for 1  x  2    for x2   1  0 for x  2   (ans.: discontinuous at x=0,2 ; continuous at x=1) 22. Evaluate the following limits : x  Sinx 1  Sinx a ) lim b ) lim x 2 x  5 x  x x x .Sinx c ) lim d ) lim x  0 tan 3 x x   ( x  Sinx ) 2 1 x x  1  2x e ) lim f ) lim x 1 1  x x1 x2  x g ) lim ( n 2  1  n ) n

(ans.:a)1/2, b)0, c)1/3, d)0, e)1/2, f)-1/2√2, g)0)  f(x)   for x  3  23. Suppose that : f(x) = x – 3x -4x +12 and h( x )   x  3  .  k for x  3  Find : a) all zeros of f . b) the value of k that makes h continuous at x=3 . ( ans . : a ) x  2 ,3 ; b )k  5 ) 3

2

١٣

Chapter two Functions 2-1- Exponential and Logarithm functions : Exponential functions : If a is a positive number and x is any number , we define the exponential function as : y = ax with domain : -∞ < x < ∞ Range : y > 0 The properties of the exponential functions are : 1. If a > 0 ↔ ax > 0 . 2. ax . ay = ax + y . 3. ax / ay = ax - y . 4. ( ax )y = ax.y . 5. ( a . b )x = ax . bx . x

6. a y  a x  ( a ) x . 7. a-x = 1 / ax and ax = 1 / a-x . 8. ax = ay ↔ x = y . 9. a0 = 1 , a∞ = ∞ , a-∞ = 0 , where a > 1 . a∞ = 0 , a-∞ = ∞ , where a < 1 . The graph of the exponential function y = ax is : y

y

Logarithm function : If a is any positive number other than 1 , then the logarithm of x to the base a denoted by : y = logax where x > 0 At a = e = 2.7182828… , we get the natural logarithm and denoted by : y = ln x Let x , y > 0 then the properties of logarithm functions are : 1. y = ax ↔ x = logay and y = ex ↔ x = ln y . 2. logex = ln x . 3. logax = ln x / ln a . ١

4. 5. 6. 7. 8. 9.

ln (x.y) = ln x + ln y . ln ( x / y ) = ln x – ln y . ln xn = n. ln x . ln e = logaa = 1 and ln 1 = loga1 = 0 . ax = ex. ln a . eln x = x .

The graph of the function y = ln x is : y2 1.5 1 0.5 0 -0.5 0

1

2

e

X

3

4

5

-1 -1.5 -2

Application of exponential and logarithm functions : We take Newton's law of cooling :

T – TS = ( T0 – TS ) et k where T is the temperature of the object at time t . TS is the surrounding temperature . T0 is the initial temperature of the object . k is a constant . EX-1- The temperature of an ingot of metal is 80 oC and the room temperature is 20 oC . After twenty minutes, it was 70 oC . a) What is the temperature will the metal be after 30 minutes? b) What is the temperature will the metal be after two hours? c) When will the metal be 30 oC? Sol. : T  T S  ( T0  T S )e tk  50  60 e 20 k  k 

ln 5  ln 6   0 .0091 20

a)

T  20  60 e 30 ( 0.0091 )  60 * 0.761  45 .6 o C  T  65.6 o C

b)

T  TS  60 e 120 ( 0.0091 )  60 * 0.335  20.1 o C  T  40 .1 o C

c)

10  60 e 0.0091 t  0.0091 t   ln 6  t  3.3 hrs . ٢

2-2- Trigonometric functions : When an angle of measure θ is placed in standard position at the center of a circle of radius r , the trigonometric functions of θ are defined by the equations : y y 1 x 1 1 Sin , Cos    Sin   , tan    r csc  r sec  x Cot Cos

r

y

θ o

x

The following are some properties of these functions :

1)

Sin 2  Cos 2  1

2 ) 1  tan 2   sec 2  and 1  Cot 2  csc 2  3 ) Sin(    )  Sin .Cos  Cos .Sin 4 ) Cos(    )  Cos .Cos  Sin .Sin tan   tan  5 ) tan(    )  1  tan  . tan  6 ) Sin 2  2 Sin .Cos and Cos 2  Cos 2   Sin 2 1  Cos 2 1  Cos 2 7 ) Cos 2  and Sin 2  2 2   8 ) Sin(   )  Cos and Cos(   )   Sin 2 2 9) Sin(  )   Sin and Cos(  )  Cos and tan(  )   tan  1 10 ) Sin .Sin  [ Cos(    )  Cos(    )] 2 1 Cos .Cos  [ Cos(    )  Cos(    )] 2 1 Sin .Cos  [ Sin(    )  Sin(    )] 2

٣

   .Cos 2 2    Sin  Sin  2Cos .Sin 2 2    12 ) Cos  Cos  2Cos .Cos 2 2    Cos  Cos  2 Sin .Sin 2 2 11 )

Sin  Sin  2 Sin

θ Sinθ Cosθ tanθ

0 0 1 0

Π/6 Π/4 1/2 1/√2 √3/2 1/√2 1/√2 1

Π/3 Π/2 √3/2 1 1/2 0 √3 ∞

Π 0 -1 0

Graphs of the trigonometric functions are :

1.5 1 0.5 -2Л

0

-Л

Л

-0.5 -1 -1.5 y  Sinx

D x : x R y : 1  y  1

٤

2Л

1.5 1 0.5 -2Л

0

-Л

Л

2Л

-0.5 -1 -1.5 y  Cosx

-2π

-π

D x : x R y : 1  y  1

0

y  tan x

D x : x  R y : y

٥

π

2n  1  2

2π

-2π

-π

π

0

y  Cotx

2π

D x : x  n  R y : y

1

-2π

-π

π

0

-1

y  Secx

2n  1  2 R y : y  1 or y  1

D x : x 

٦

2π

1 -2π

-π

π

0

2π

-1

y  csc x

D x : x  n  R y : y  1 or y  1

Where n  0 ,1 ,2 ,3 ,...... EX-2 - Solve the following equations , for values of θ from 0o to 360o inclusive . a) tan θ = 2 Sin θ b) 1 + Cos θ = 2 Sin2 θ Sol.Sin  2 Sin a ) tan  2 Sin  Cos  Sin ( 1  2Cos )  0 either Sin  0    0 o ,180 o ,360 o 1 or Cos     60 o ,300 o 2 Therefore the required values of θ are 0o,60o,180o,300o,360o . b ) 1  Cos  2.Sin 2   1  Cos  2( 1  Cos 2 )  ( 2Cos  1 )( Cos  1 )  0 1 either Cos     60 o ,300 o 2 or Cos  1    180 o There the roots of the equation between 0o and 360o are 60o,180o and 300o . ٧

EX-3- If tan θ = 7/24, find without using tables the values of Secθ and Sinθ. Sol.y 7  r  7 2  24 2  25 tan    x 24 ٧ y 7 r 25 Sec   and Sin    x 24 r 25 ٢٤

EX-4- Prove the following identities : a ) Csc  tan  .Sec  Csc .Sec 2 b ) Cos 4   Sin 4   Cos 2  Sin 2 Sec  Csc tan   Cot c)  tan   Cot Sec  Csc Sol.-

1 Sin 1 .  Sin Cos Cos Cos 2  Sin 2 1 1 .    Csc .Sec 2  R .H .S . 2 2 Sin Cos  Sin .Cos  4 4 b ) L .H .S .  Cos   Sin   ( Cos 2  Sin 2 ).( Cos 2  Sin 2 )

a)



c)

L .H .S .  Csc  tan  .Sec 

 Cos 2  Sin 2  R .H .S . 1 1  Sec  Csc Cos Sin 1 L .H .S .    Sin Cos Sin  Cos tan   Cot  Cos Sin 1 2 2 Sin   Cos  Sin .Cos tan   Cot .    R .H .S . 1 Sin  Cos Sec  Csc Sin .Cos

1

EX-5- Simplify Sol.-

1 x2  a2

x a 2



2

x  a .Csc

when

1 a 2 Csc 2   a 2



1 a Cot 2

EX-6- Eliminate θ from the equations : i) x = a Sinθ and y = b tanθ ii) x = 2 Secθ and y = Cos2θ Sol.-

٨



. 1 tan  . a

i)

x a  Csc  a x y b y  b tan   tan    Cot  b y 2 2 a b Since Csc 2  Cot 2  1  2  2  1 x y x  a .Sin  Sin 

x

ii )

2 x y  Cos 2  y  Cos 2   Sin 2  4 x2  4  x2 y  8  x2 y 2  2 x x

x2  4

x  2 Sec   Cos  

2

EX-7- If tan2θ – 2 tan2β = 1 , show that 2 Cos2θ – Cos2β = 0 . Sol. – tan 2   2 tan 2   1  Sec 2   1  2( Sec 2   1 )  1 1 2  Sec 2  2 Sec 2   0   0 2 Cos  Cos 2   2Cos 2  Cos 2   0 Q .E .D . EX-8- If a Sinθ = p – b Cosθ and b Sinθ = q + a Cosθ .Show that : a2 +b2 = p2 +q2 Sol.p  a .Sin  b .Cos and q  b .Sin  a .Cos p 2  q 2  ( aSin  bCos ) 2  ( bSin  aCos ) 2  a 2 ( Sin 2  Cos 2  )  b 2 ( Cos 2  Sin 2 )  a 2  b 2

EX-9- If Sin A = 4 / 5 and Cos B = 12 / 13 ,where A is obtuse and B is acute . Find , without tables , the values of : a) Sin ( A – B ) , b) tan ( A – B ) , c) tan ( A + B ) . Sol. -

13

5 4

5

A B 12

-3

٩

a ) Sin ( A  B )  SinA .CosB  CosA .SinB 4 12 3 5 63  .  .  5 13 5 13 65 tan A  tan B b) tan( A  B )  1  tan A .tan B 4 5   63  3 12   4 5 16 1 . 3 12 tan A  tan B c) tan( A  B )  1  tan A .tan B 4 5   33  3 12  4 5 56 1 . 3 12

EX-10 – Prove the following identities: a) b) c) d)

Sin ( A  B )  Sin ( A  B )  2.SinA .CosB Sin ( A  B ) tan A  tan B  CosA .CosB SecA .SecB .CscA .CscB CscA .CscB  SecA .SecB Sin 2  Cos 2  1  Cot  Sin 2  Cos 2  1

Sec ( A  B ) 

١٠

Sol.a ) L .H .S .

 Sin ( A  B )  Sin ( A  B )  SinA .CosB  CosA .SinB  SinA .CosB  CosA .SinB

b)

c)

d)

 2 .SinA .CosB  R .H .S . Sin ( A  B ) SinA .CosB  CosA .SinB R .H .S .   CosA .CosB CosA .CosB  tan A  tan B  L .H .S . 1 1 1 1 . . . SecA .SecB .CscA .CscB R .H .S   CosA CosB SinA SinB 1 1 1 1 CscA .CscB  SecA .SecB .  . SinA SinB CosA CosB 1 1   CosA .CosB  SinA .SinB Cos ( A  B )  Sec ( A  B )  L .H .S . Sin 2  Cos 2  1 2 Sin  .Cos   ( Cos 2  Sin 2 )  1 L .H .S .   Sin 2  Cos 2  1 2 Sin  .Cos   ( Cos 2  Sin 2 )  1 2 Sin  .Cos   2 Cos 2 Cos     Cot   R .H .S . 2 2 Sin  .Cos   2 Sin  Sin 

EX-11 – Find , without using tables , the values of Sin 2θ and Cos 2θ, when: a) Sinθ = 3 / 5 , b) Cos θ = 12/13 , c) Sin θ = -√3 / 2 . Sol. – a)

٥

3

5

3

θ

θ

٤

-4

3 4 24 Sin 2  2.Sin .Cos  2. .(  )   5 5 25 4 3 7 Cos 2  Cos 2  Sin 2  (  ) 2  ( ) 2  5 5 25

١١

b)

١٣ 5

θ

١٢

θ

-5

١٣

5 12 120 ).( )   13 13 169 12 5 2 119 Cos 2  Cos 2  Sin 2   ( ) 2  (  )  13 13 169

Sin 2  2.Sin .Cos  2( 

c)

-١

1

θ

θ

-√3 2

-√3

2

3 1 3 ).(  )   2 2 2 1 3 2 1 Cos 2  Cos 2  Sin 2   (  ) 2  (  )  2 2 2

Sin 2  2 Sin .Cos  2( 

EX-12- Solve the following equations for values of θ from 0o to 360o inclusive: a) Cos 2θ + Cos θ + 1 = 0 , b) 4 tan θ . tan 2θ = 1 Sol.-

١٢

a ) Cos 2  Cos  1  0  2Cos 2  1  Cos  1  0  Cos( 2.Cos  1 )  0 either Cos  0    90 o ,270 o 1 or Cos      120 o ,240 o 2 o o o   90 ,120 ,240 ,270 o  2 tan  b) 4. tan  . tan 2  1  4. tan  . 1 1  tan 2   9 tan 2   1 1 either tan      18.4 o ,198.4 o 3 1 or tan       161.6 o ,341.6 o 3 o o   18.4 ,161.6 ,198.4 o ,341.6 o 

2-3- The inverse trigonometric functions : The inverse trigonometric functions arise in problems that require finding angles from side measurements in triangles :

y  Sinx



x  Sin 1 y

-1

1



200 150 100 50 0 -50 -100 -150

-

-200

y  Sin 1 x

D x : 1  x  1 R y : 90  y  90

١٣

-1



- y  Cos 1 x

D x : 1  x  1 R y : 0  y  180

π

0

-π

y  tan 1 x

D x : x R y : 90  y  90

١٤

1

2π

π

0

-π y  Cot 1 x

D x : x Ry : 0  y  



-1

1

-

y  Sec 1 x

Dx :  x  1 Ry : 0  y   , y 

١٥

 2

2π

π

-1

0

1

-π

-2π y  Csc 1 x

Dx :  x  1   Ry :   y  , y  0 2 2 The following are some properties of the inverse trigonometric functions : 1. Sin 1 (  x )   Sin 1 x

2. 3. 4. 5. 6. 7. 8.

Cos  1 (  x )    Cos  1 x  Sin  1 x  Cos  1 x  2 1 1 tan (  x )   tan x  Cot  1 x   tan  1 x 2 1 Sec  1 x  Cos  1 x 1 Csc  1 x  Sin  1 x 1 Sec (  x )    Sec  1 x

and noted that ( Sinx )  1 

1  Cscx  Sin  1 x Sinx ١٦

3 , find : 2 Csc , Cos , Sec , tan  , and , Cot

EX-13- Given that   Sin  1

Sol.-

٢

3 ١ 3 3 x  Sin   r 431 2 2 y 1 1 2 , Cos  , Sec  2 , tan   3 , Cot  Csc  2 3 3

  Sin  1

EX-14 – Evaluate the following expressions : 1 a ) Sec ( Cos  1 ) b ) Sin  1 1  Sin  1 ( 1 ) 2 Sol.-

c ) Cos  1 (  Sin

1  )  Sec  2 2 3   b ) Sin  1 1  Sin  1 ( 1 )   (  )   2 2 1 2  c ) Cos  1 (  Sin )  Cos  1 (  )   6 2 3

a ) Sec( Cos  1

EX-15- Prove that : a ) Sec  1 x  Cos  1

1 x

b ) Sin  1 (  x )   Sin  1 x

Sol. 1 Cosy 1 1  y  Cos  1  Sec  1 x  Cos  1 x x b ) Let y   Sin  1 x  x  Sin(  y )  x   Siny  y  Sin  1 (  x )  Sin  1 (  x )   Sin  1 x

a)

Let y  Sec  1 x  x  Secy  x 

١٧

 ) 6

2-4- Hyperbolic functions : Hyperbolic functions are used to describe the motions of waves in elastic solids ; the shapes of electric power lines ; temperature distributions in metal fins that cool pipes …etc. The hyperbolic sine (Sinh) and hyperbolic cosine (Cosh) are defined by the following equations :

4.

1 u 1 ( e  e  u ) and Coshu  ( e u  e  u ) 2 2 Sinhu e u  e  u Coshu e u  e  u tanh u   and Cothu   Coshu e u  e u Sinhu e u  e  u 1 2 1 2 Sechu   u and Cschu   u u Coshu e  e Sinhu e  e u 2 2 Cosh u  Sinh u  1

5.

tanh 2 u  Sech 2 u  1

1. Sinhu  2. 3.

and

Coth 2 u  Csch 2 u  1

Coshu  Sinhu  e u and Coshu  Sinhu  e  u Cosh(  u )  Coshu and Sinh(  u )   Sinhu Cosh0  1 and Sinh0  0 Sinh( x  y )  Sinhx .Coshy  Coshx .Sinhy Cosh( x  y )  Coshx .Coshy  Sinhx .Sinhy Sinh 2 x  2.Sinhx .Coshx Cosh 2 x  Cosh 2 x  Sinh 2 x Cosh 2 x  1 Cosh 2 x  1 13. Cosh 2 x  and Sinh 2 x  2 2 6. 7. 8. 9. 10. 11 . 12.

y=Sinhx

y=Cschx y=Cschx

0

١٨

y

y=Coshx

1

1

1 1

1

y=Sechx

0

0

0 0

y=Cothx 1 y=tanhx 0

y=Cothx

y  Sinhx y  Coshx y  tanh x y  Cothx y  Sechx y  Cschx

-1

D x : x and D x : x and D x : x and D x : x  0 and D x : x and D x : x  0 and

R y : y Ry : y  1 R y : 1  y  1 R y : y  1 or y  1 Ry : 0  y  1 R y : y  0

EX-16- Let tanh u = - 7 / 25 , determine the values of the remaining five hyperbolic functions . Sol.-

١٩

Cothu 

1 25  tanh u 7

tanh 2 u  Sech 2 u  1 1 25  Sechu 24 Sinhu tanh u  Coshu



49 24  Sech 2 u  1  Sechu  625 25

Coshu 

Cschu 



7 Sinhu 7   Sinhu   25 25 24 24

1 24  Sinhu 7

EX-17- Rewrite the following expressions in terms of exponentials . Write the final result as simply as you can : a ) 2Cosh(ln x ) b ) tanh(ln x ) c ) Cosh5 x  Sinh5 x d ) ( Sinhx  Coshx )4 Sol.e ln x  e  ln x 1 a ) 2Cosh(ln x )  2.  x 2 x 1 x 2 e ln x  e  ln x x  x 1 b ) tanh(ln x )  ln x  1 x2  1 e  e  ln x x x 5 x 5x e e e 5 x  e 5 x c ) Cosh5 x  Sinh5 x    e5x 2 2 4 x x x e e e  ex  4   e 4 x d ) ( Sinhx  Coshx )    2 2  

EX-18- Solve the equation for x : Cosh x = Sinh x + 1 / 2 . 1 1 Sol. - Coshx  Sinhx   e  x    x  ln 1  ln 2  x  ln 2 2 2 EX-19 – Verify the following identity : a) Sinh(u+v)=Sinh u. Cosh v + Cosh u.Sinh v b) then verify Sinh(u-v)=Sinh u. Cosh v - Cosh u.Sinh v

Sol.٢٠

a ) R .H .S .  Sinhu .Coshv  Coshu .Sinhv e u  e u e v  e v e u  e u e v  e v .   2 2 2 2 u v ( u v ) e e   Sinh( u  v )  L .H .S . 2 b ) L .H .S .  Sinh( u  (  v ))  Sinhu .Cosh(  v )  Coshu .Sinh(  v )  Sinhu .Coshv  Coshu .Sinhv  R .H .S .

EX-20 – Verify the following identities : 1 a) Sinhu .Coshv  Sinh( u  v )  Sinh( u  v ) 2 1 b ) Coshu .Coshv  Cosh( u  v )  Cosh( u  v ) 2 3 c ) Sinh 3 u  Sinh u  3Cosh 2 u .Sinhu  3 Sinhu  4 Sinh 3 u d ) Sinh 2 u  Sinh 2 v  Cosh 2 u  Cosh 2 v Sol. – 1 a ) R .H .S .  Sinh( u  v )  Sinh( u  v ) 2 1  Sinhu .Coshv  Coshu .Sinhv  Sinhu .Coshv  Coshu .Sinhv  2  Sinhu .Coshv  L .H .S . 1 b ) R .H .S .  Cosh( u  v )  Cosh( u  v ) 2 1  Coshu .Coshv  Sinhu .Sinhv  Coshu .Coshv  Sinhu .Sinhv  2  Coshu .Coshv  L .H .S . c ) L .H .S .  Sinh( 2 u  u )  Sinh 2 u .Coshu  Cosh2 u .Sinhu  2 Sinhu .Coshu .Coshu  ( Cosh 2 u  Sinh 2 u ).Sinhu  3 Sinhu .Cosh 2 u  Sinh 3 u  R .H .S .( I )  3 Sinhu .( 1  Sinh 2 u )  Sinh 3 u  3 Sinhu  4 Sinh 3 u  R .H .S .( II ) d ) L .H .S .  Sinh 2 u  Sinh 2 v  Cosh 2 u  1  ( Cosh 2 v  1 )  Cosh 2 u  Cosh 2 v  R .H .S .

2-5- Inverse hyperbolic functions : All hyperbolic functions have inverses that are useful in integration and interesting as differentiable functions in their own right .

٢١

0

y  Sinh 1 x

1

y  Cosh 1 x

D x : x R y : y

-١

-1

١

y  tanh1 x Dx : 1  x  1 R y : y

D x : x  1 R y : y  0

١

y  Coth1 x Dx : x  1 or x  1 Ry : y  0

١

y  Sech 1 x

y  Csch1 x Dx : x  0 R y : y  0

Dx : 0  x  1 R y : y  0

٢٢

Some useful identities : Sinh 1 x  ln( x  x 2  1 ) Cosh 1 x  ln( x  x 2  1 ) 1 1 x 3. tanh 1 x  . ln  2  1 x  1  x  1 1 1 4. Coth 1 x  . ln   tanh 2  x  1 x 2 1 1 x    Cosh 1 1 5. Sech1 x  ln   x x   1. 2.

1 6. Csch x  ln  x  1

x 2  1  1  Sinh 1  x x 

EX-21 - Derive the formula : Sinh  1 x  ln( x  x 2  1 )

Sol.e y  e y e2y  1 Let y  Sinh x  x  Sinhy   x 2 2e y  e 2 y  2 x .e y  1  0 2x  4 x2  4 y e   e y  x  x2  1 2 2 either y  ln( x  x  1 ) neglected sin ce x  x 2  1  0 1

or

y  ln( x 

x2  1 )

٢٣

Problems – 2 1. A body of unknown temperature was placed in a room that was held at 30o F . After 10 minutes , the body's temperature was 0oF , and 20 minutes after the body was placed in the room the body's temperature 15oF . Use Newton's law of cooling to estimate the body's initial (ans.:-30oF) temperature . 2. A pan of warm water 46oC was put in a refrigerator . Ten minutes later , the water's temperature was 39oC , 10 minutes after that , it was 33oC . Use Newton's law of cooling to estimate how cold the refrigerator was ? (ans.:-3oC) 3. Solve the following equations for values of θ from -180o to 180o inclusive: i) tan2θ + tan θ = 0 ii) Cot θ= 5 Cos θ iii) 3 Cos θ + 2 Sec θ + 7 = 0 iv) Cos2θ + Sin θ + 1 = 0 (ans.:i)-180,-45,0,135,180; ii)-90,11.5,90,168.5; iii)-109.5,109.5; iv)-90) 4. Solve the following equations for values of θ from 0o to 360o inclusive: i) 3 Cos 2θ – Sin θ + 2 = 0 ii) 3 tan θ = tan 2θ 2 iv) 3 Cot 2θ + Cot θ = 1 iii) Sin 2θ. Cos θ + Sin θ = 1 (ans.:i)56.4,123.6,270; ii)0,30,150,180,210,330,360; iii)30,90,150,270; iv)45,121,225,301) 5. If Sin θ = 3/ 5 , find without using tables the values of : i) Cos θ ii) tan θ (ans.: i) 4/5 ; ii) 3/4 ) 6. Find, without using tables, the values of Cos x and Sin x , when Cos 2x is : a) 1/8 , b) 7/25 , c) -119/169 3 7 4 3 5 12 ( ans . : a )  , ; b )  , ; c )  , ) 4 4 5 5 13 13 7. If Sin A = 3/5 and Sin B = 5/13 , where A and B are acute angles , find without using tables , the values of : a) Sin(A+B) , b) Cos(A+B) , c) Cot(A+B) (ans.: 56/65; 33/65; 33/56) 8. If tan A = -1/7 and tan B = 3/4 , where A is obtuse and B is acute , find (ans.: 135 ) without using tables the value of A – B .

9.Prove the following identities :

٢٤

i)

Sec 2  Csc 2  Sec 2 .Csc 2

Sin 2 ( 1  Sec 2 )  Sec 2  Cos 2 1  Sin  ( Sec  tan )2 iii ) 1  Sin tan 2   Cos 2 iv ) Sec  Sin  Sec  Sin Cos( A  B )  Cos( A  B ) v)  tan B Sin( A  B )  Sin( A  B ) vi ) CosB  CosA .Cos( A  B )  SinA .Sin( A  B ) tan A  tan B  tan C  tan A .tan B .tan C vii ) tan( A  B  C )  1  tan B .tan C  tan C .tan A  tan A .tan B If A , B ,C are angles of a triangle , show that : ii )

viii ) ix ) x) xi ) xii ) xiii )

xiv ) xv ) xvi )

xvii ) xviii ) xix ) xx )

tan A  tan B  tan C  tan A .tan B .tan C 1 tan x .Sin 2 h tan( x  h )  tan( x  h )  tan x  2 2 Cos x  Sin 2 h 1  Cos 2 x tan x  1  Cos 2 x Sin 4 A  Sin 2 A  tan 2 A Cos 4 A  Cos 2 A  1 1 Sin 4  Cos 4  ( Cos 4  3 ) 4 4 Sin 3 A .Cos 3 A  4 Cos 3 A .Sin 3 A  3 Sin 4 A 3 tan A  tan 3 A tan 3 A  1  3 tan 2 A Cos 1 (  x )    Cos 1 x  Cot 1 x   tan  1 x 2 Cosh( u  v )  Coshu .Coshv  Sinhu .Sinhv and then verify Cosh( u  v )  Coshu .Coshv  Sinhu .Sinhv 1 Coshu .Sinhv  Sinh( u  v )  Sinh( u  v ) 2 1 Sinhu .Sinhv  Cosh( u  v )  Cosh( u  v ) 2 Cosh 3 u  Coshu  4 Sinh 2 u .Coshu  4 Cosh 3 u  3Coshu ( Coshx  Sinhx )n  Coshnx  Sinhnx

1  Sin 1 1  Sin , prove that  and deduce formula for Sinθ , Cos u Cos Cosθ , tanθ in terms of u. (ans.:(u2-1)/(u2+1); 2u/(u2+1);(u2-1)/(u2+1))

10. If u 

٢٥

11. If Sin( x   )  2Cos( x   ) ; prove that : tan x 

2  tan . 1  2 tan

12. If Sin( x   )  Cos( x   ) ; prove that : tan x  1 . 13. If x  Cos  Cos 2 and y  Sin  Sin 2 . Show that : i ) x 2  y 2  Cos 2  2Cos 3  Cos 4 ii ) 2 xy  Sin 2  2 Sin 3  Sin 4 14. If Cos 2 A .Cos 2 B  Cos 2 , prove that : Sin 2 A .Cos 2 B  Cos 2 A .Sin 2 B  Sin 2 15. If S = Sin θ and C = Cos θ , simplify : S .C S. 1  S2 C S i) , ii ) , iii )  2 2 S C 1 S C. 1  C (ans.:i) Sinθ; ii)1; iii) Secθ.Cscθ) 16. Eliminate θ from the following equations : i) x  a .Csc and y  b .Sec ii ) x  Sin  Cos and y  Sin  Cos iii ) x  Sin  tan and y  Sin  tan  iv ) x  tan and y  tan 2 2

(ans. : i )

a

2

x

2



b

2

2

2

 1; ii)x  y  2; iii)

y

4 2

( x  y)



4 2

( x  y)

 1; iv)y 

2x 2

1 x

)

17. In the acute – angled triangle OPQ , the altitude OR makes angles A and B with OP and OQ . Show by means of areas that if OP=q , OQ=p , OR=r : p.q.Sin(A+B) = q.r.SinA + p.r. SinB. 18. Given that   Sin 1

1 , find Cosα , tanα , Secα , and Cscα. 2 3 1 2 ( ans . : ; ; ;2 ) 2 3 3

19. Evaluate the following expressions :

٢٦

1 ) 2 c ) Cot ( Cos  1 0 )

a ) Sin( Cos 1

b ) Csc ( Sec  1 2 ) d ) Sin  1 1  Sin  1 ( 1 )  f ) Cos  1 (  Sin ) 6 ( ans . : 1 / 2 ;2 / 3 ;0 ; ;0.6 ;2 / 3 )

e ) Cos( Sin 1 0.8 )

20. Find the angle α in the below graph ( Hint : α+β = 65o ) : 65o

٢١

β ٥٠

α (ans.: 42.2) 21. Let Sech u = 3/5 , determine the values of the remaining five hyperbolic functions . ( ans . : Coshu  5 / 3 ; Sinhu  4 / 3 ; tanh u  4 / 5 ; Cothu  5 / 4 ; Cschu  3 / 4 )

22. Rewrite the following expressions in terms of exponentials , write the final result as simply as you can : 1 a ) Sinh( 2.ln x ) b) Coshx  Sinhx c ) Cosh3 x  Sinh 3 x d ) ln( Coshx  Sinhx )  ln( Coshx  Sinhx ) (ans.:(x4-1)/(2x2); ex ; e--3x ; 0 ) 23. Solve the equation for x ; tanh x = 3/5 .

(ans.: ln 2 )

24. Show that the distance r from the origin O to the point P(Coshu,Sinhu) on the hyperbola x2 – y2 = 1 is r  Cosh2 u .

     and Sinh x = tan θ . Show that : 2 2 Cosh x = Sec θ , tanh x =Sin θ , Coth x = Csc θ , Csch x = Cot θ , and Sech x = Cos θ . 1 1 x 26. Derive the formula : tanh  1 x  ln ; x 1 2 1 x 27. Find : lim Cosh 1 x  ln x  . (ans.: ln 2 )

25. If θ lies in the interval 

x

٢٧

Chapter three Derivatives Let y = f ( x ) be a function of x . If the limit :

dy f ( x  x )  f ( x ) y  f ' ( x )  lim  lim x  0 x  o  x dx x

exists and is finite , we call this limit the derivative of f at x and say that f is differentiable at x . EX-1 – Find the derivative of the function : f ( x ) 

1 2x  3

Sol.: 1 f ' ( x )  lim

x 0

 lim

x 0

 lim

x 0



2( x  x )  3 f ( x  x )  f ( x )  lim x 0 x x 2 x  3  2( x  x )  3

x . 2( x  x )  3 2 x  3

.



1 2x  3

2 x  3  2( x  x )  3 2 x  3  2( x  x )  3

( 2 x  3 )  ( 2( x  x )  3 )

x . 2( x  x )  3 2 x  3 ( 2 x  3  2( x  x )  3 ) 2

( 2 x  3 )( 2 x  3  2 x  3 )



1 ( 2 x  3 )3

Rules of derivatives : Let c and n are constants, u , v and w are differentiable functions of x : 1. 2. 3. 4. 5.

d c0 dx d u n  nu n1 du  d  1    1 du dx dx dx  u  u 2 dx d cu  c du dx dx d ( u  v )  du  dv ; d ( u  v  w )  du  dv  dw dx dx dx dx dx dx dx d ( u .v )  u . dv  v du dx dx dx

1

d ( u .v .w )  u .v dw  u .w dv  v .w du dx dx dx dx v du  u dv d  u   dx dx where v  0 2 dx  v  v and

6.

dy for the following functions : dx 2 a ) y  ( x 2  1 )5 b ) y  ( 5  x )( 4  2 x ) 12 4 3 c ) y  ( 2 x 3  3 x 2  6 x ) 5 d) y  3 4 x x x ( x 2  x )( x 2  x  1 ) x2  1 e) y f ) y 2 x3 x  x2

EX-2- Find

Sol.dy  5 ( x 2  1 )4 .2 x  10 x ( x 2  1 )4 dx b ) dy  2( 5  x )( 4  2 x ) 2( 5  x )  ( 4  2 x ) dx  8( 5  x )( 2  x )( 2 x  7 )

a)

c)

d)

e)

dy   5 ( 2 x 3  3 x 2  6 x ) 6 ( 6 x 2  6 x  6 ) dx  30 ( 2 x 3  3 x 2  6 x )6 ( x 2  x  1 ) dy y  12 x  1  4 x  3  3 x  4   12 x  2  12 x  4  12 x  5 dx dy    122  124  125 dx x x x ( x  1 )( x 2  x  1 )  x3 dy x 3 ( x 2  x  1 )  ( x  1 )( 2 x  1 )  3 x 2 ( x  1 )( x 2  x  1 ) 3   4 6 dx x x y



f)



dy 2 x( x 2  x  2 )  ( x 2  1 )( 2 x  1 ) x2  2x  1   dx ( x 2  x  2 )2 ( x 2  x  2 )2

2

The Chain Rule: 1. Suppose that h = go f is the composite of the differentiable functions y = g( t ) and x = f( t ) , then h is a differentiable functions of x whose derivative at each value of x is :

dy dy dx   dx dt dt 2. If y is a differentiable function of t and t is differentiable function of x , then y is a differentiable function of x :

y  g( t ) and t  f ( x ) 

dy dy dt  * dx dt dx

EX-3 – Use the chain rule to express dy / dx in terms of x and y : a) b) c) d)

t2 t2  1 1 y 2 t 1 2  t 1 y   t  1 1 y  1 t y

and

t

2x  1

and

x

4t  1

and

x

and

1 1 t2 1 t 1 x

at at

t2 x2

Sol.t2 dy 2t( t 2  1 )  2t .t 2 2t a) y  2    2 2 2 t  1 dt ( t  1) ( t  1 )2 1 1  dt 1 1 2 2 t  ( 2 x  1 )   .( 2 x  1 ) .2  dx 2 2x  1 dy dy dt 2t 1 2 2x  1 1 1 . .    .  2 dx dt dx ( t  1 )2 2 x  1 (( 2 x  1 )  1 )2 2 x  1 2( x  1 )2

3

b)

dy 2t   2 t ( t 2  1 ) 2   2 dx ( t  1 )2 1 1  dx 1 2 2 2 x  ( 4t  1 )   ( 4 t  1 ) .4  dt 2 4t  1 dy dy dx 2t 2 t 4t  1    2   2 2 dx dt dt (t  1) ( t  1 )2 4t  1 y  ( t 2  1 )1 

x2  1 1 xy 2 ( x 2  1 ) .x  2    4 4 y 2 x 1 where x  4 t  1  t  4 1 1 2 where y  2  t 1 y t 1 2

c)

dy  t 1  t  1  t  1  ( t  1 ) 4( t  1 )  2  y    2 dt ( t  1 )3  t  1  t  1 (t 1) 4( 2  1 ) 4  dy      3 27  dt  t  2 ( 2  1 ) 1 dx 2 2 1  dx   3    3  x  2 1 dt 4 t t 2  dt  t  2 4  1 16  dy   dy dx   dx    dt  dt   27    4    27   t2   t2  

1  1  1 at x  2 1 x 1 2 dy  dy   12     1 2 1 y1 1  t dt t  dt  t  1 (  1 ) 1 t  ( 1  x ) 1  dt   ( 1  x ) 2 (  1 )  dx ( 1  x )2 1   dt   1  dx  x  2 ( 1  2 ) 2  dy   dy     . dt   1* 1  1  dx   dt  x  2  dx  x  2 x2

d) t

Higher derivatives : If a function y = f( x ) possesses a derivative at every point of some interval , we may form the function f '(x) and talk

4

about its derivate , if it has one . The procedure is formally identical with that used before , that is : d 2 y d  dy  d f  ( x   x )  f ( x )    f ( x )  lim x0 x dx 2 dx  dx  dx if the limit exists . This derivative is called the second derivative of y with respect to x . It is written in a number of ways , for example, d2 f( x) y'' , f ''(x) , or . dx 2 In the same manner we may define third and higher derivatives , using similar notations . The nth derivative may be written : dny (n) (n) y ,f (x), n . dx EX-4- Find all derivatives of the following function : y = 3x3 - 4x2 + 7x + 10 Sol.dy d2y ,  9x2  8x  7  18 x  8 dx dx 2 d3y d4 y d5 y ,  18 0  .... dx 3 dx 4 dx 5 Ex-5 – Find the third derivative of the following function : 1 y   x3 x Sol.1 dy   12  3 x 2 dx 2 x 2 1 d y 2 3  3  x 2 4 dx 2 x 3 3 d y d3y 6  3 x 2      64  3 3 3 4 3 8 dx x dx x 8 x

5

Implicit Differentiation: If the formula for f is an algebraic combination of powers of x and y . To calculate the derivatives of these implicitly defined functions , we simply differentiate both sides of the defining equation with respect to x . EX-6- Find 2

2

dy for the following functions: dx 2

2

a ) x .y  x  y x y c)  2 at P(3,1) x  2y

b ) (x  y)

3



(x y)

3



4

x y

4

d) xy  2x - 5y  2 at P(3,2)

Sol. dy dy dy x  xy 2 )  y2( 2x )  2x  2 y   dx dx dx x 2 y  y dy dy dy b ) 3(x  y)2 ( 1  )  3( x  y )2 ( 1  )  4 x3  4 y3 dx dx dx 3 2 2 dy 4 x  3( x  y )  3( x  y ) dy 2 x 3  3 x 2  3 y 2     dx 3( x  y )2  3( x  y )2  4 y 3 dx 6 xy  2 y 3

a ) x2( 2 y

(x - 2y)(1 c) d) x

dy dy )  ( x  y )( 1  2 ) 1 dx dx  0  dy  y   dy   2  dx  ( x  2y) dx x ( 3 ,1 ) 3

dy dy dy y  2 22  dy   y25 0     2 dx dx dx 5  x  dx  ( 3 ,2 ) 5  3

Exponential functions : If u is any differentiable function of x , then :

7)

d u du a  a u .ln a . dx dx

d u du e  eu . dx dx

and

6

EX-7 –Find

dy for the following functions : dx

a ) y  2 3x

b) y  2 x .3 x

c) y  ( 2 x ) 2

d ) y  x.2 x

5x e) y  e (x  e )

f ) ye

2

1 5x 2

Sol.dy  2 3 x * 3 ln 2 dx dy b ) y  2 x .3 x  y  6 x   6 x . ln 6 dx dy c ) y  (2 x )2  y  2 2 x   2 2 x ln 2.2  2 2 x  1 ln 2 dx 2 2 2 2 dy d ) y  x.2 x   x.2 x ln2.2x  2 x  2 x (2x 2 ln2  1) dx 5x dy (x  e 5x ) e) ye   e (x  e ) ( 1  5 e 5 x ) dx 1 1 1  2 2 2 2 1 2 dy f ) y  e (1 5x )   e (1 5x ) ( 1  5 x 2 ) 2 .10 x  e 1 5 x dx 2 a ) y  2 3x 

5x 1  5x2

Logarithm functions : If u is any differentiable function of x , then :

8)

d 1 du loga u  . dx u .ln a dx EX-8 – Find

d 1 du ln u  . dx u dx

and

dy for the following functions : dx

a ) y  log10 e x

b ) y  log 5 ( x  1 )2

c ) y  log 2 ( 3 x 2  1 )3

d ) y  ln(x 2  2 )2

e ) y  ln(xy)  1

(2x 3  4 ) 3 .( 2 x 2  3 ) 2 f) y  ( 7 x 3  4 x  3 )2



2

Sol. –

7



3 5

dy  log 10 e  ln e  1 dx ln 10 ln 10 dy 2 b ) y  log 5 ( x  1 )2  2 log 5 ( x  1 )   dx ( x  1 ) ln 5 dy 3 18 x c ) y  3 log 2 ( 3 x 2  1 )  . 6x   dx 3 x 2  1 ln 2 ( 3 x 2  1 ) ln 2 2 2 48 x ln( x 2  2 ) dy 2 2 .2 x  d)  3 2 ln( x  2 ) dx x2  2 x2  2

a ) y  log 10 e x  y  x log 10 e 









dy 1 1 dy dy y   . 0  dx x y dx dx x( y  1 ) 2 5 f ) lny  ln( 2 x 3  4 )  ln( 2 x 2  3 )  2 ln( 7 x 3  4 x  3 ) 3 2 2 1 dy 2 6 x 5 4x 21 x 2  4  .  . 3  . 2  2. 3 y dx 3 2 x  4 2 2 x  3 7 x  4x  3 2  dy 2x 5x 21 x 2  4    2 y 3    2 3 dx 2x  4 2x  3 7 x  4x  3

e ) y  lnx  lny  1 

Trigonometric functions : If u is any differentiable function of x , then :

d s inu  cosu. du dx dx d 10 ) cosu   sin u. du dx dx 11 ) d tanu  sec 2 u. du dx dx 12 ) d cotu   csc 2 u. du dx dx 13 ) d secu  secu.tanu. du dx dx d 14 ) cscu   c scu.cotu. du dx dx 9)

EX-9- Find

dy for the following functions : dx

8

a ) y  tan(3x 2 ) c) y  2sin x  xCos x 2 2 e ) x  tan(xy)  0

b) y  (cscx  cotx)2 d ) y  tan 2 (cos x ) f) y  sec 4 x  tan 4 x

Sol.dy  sec 2 ( 3 x 2 ).6 x  6 x . sec 2 ( 3 x 2 ) dx dy b)  2(csc x  cot x )(  csc x .cot x  csc 2 x )  2 csc x .(csc x  cot x )2 dx dy c)  2 cos x . 1   x(  sin x ). 1  cos x   x . sin x dx 2 2  2 2 2  2 2 dy d)  2.tan(cosx ).sec2 (cosx ).(  sin x )  2.sin x .tan(cosx ).sec2 (cosx ) dx dy dy 1  y .sec2 ( xy ) cos2 ( xy )  y 2 e ) 1  sec ( xy ).( x  y )  0    dx dx x x.sec2 ( xy ) dy  4 sec3 x .sec x .tan x  4.tan3 x .sec2 x  4 tan x.sec2 x f) dx a)

EX-10- Prove that : a ) d tan u  sec 2 u . du dx dx

b ) d sec u  sec u .tan u . du dx dx

Proof :

cosu.cosu. du  sinu.(  sinu ) du dx dx a ) L.H .S .  d tanu  d sinu  2 dx dx cosu cos u 2 2  cos u 2 sin u . du  12 . du  sec2 u. du  R.H .S . dx cos u dx dx cos u b)

L.H .S .  d secu  d 1   12 (  sinu ) du dx dx cosu dx cos u  1 . sinu . du  secu.tanu. du  R.H .S . cosu cosu dx dx

The inverse trigonometric functions : If u is any differentiable function

9

of x , then :

15 ) 16 ) 17 ) 18 ) 19 ) 20 )

d sin  1 u  1 du 2 dx 1  u dx 1 du d cos  1 u   2 dx dx 1 u d tan  1 u  1 du 2 dx 1  u dx d cot  1 u   1 du 2 dx 1  u dx d sec  1 u  1 du 2 dx u u  1 dx d csc  1 u   1 du dx u u 2  1 dx

EX-11- Find

1 u1 1 u1

u 1 u 1

dy in each of the following functions : dx

b ) y  sin -1 x  1 x1 d ) y  sec -1 5 x

a ) y  cot -1 2  tan  1 x x 2 c ) y  x.cos -1 2 x  1 1  4 x 2 2 -1 e ) y  x.ln(sec x )

f ) y  3 sin

-1

2x

Sol. – a)

b)

1 1 4  1  2.  2   .  2 2 2  x   x 2 4 x 1  1   x 2 dy 1 ( x  1 ).1  ( x  1 ).1 1  .  2 2 dx ( x  1) ( x  1) x  x  1 1   x  1

dy  dx

1

2

 8x 2 1 dy  cos 1 2 x  cos 1 2 x  . x 2 2 4 dx 1 4x 1 4x dy 5 1   d) dx 5 x 25 x 2  1 x 25 x 2  1 c)

10

dy x 1   ln(sec  1 x )  dx sec  1 x x x 2  1 dy 2  3 sin 2 x .ln 3. dx 1  4x2

e)

1  ln(sec  1 x ) 1 x  1 . sec x 2

1

f)

EX-12- Prove that : a ) d sin  1 u  dx Proof : a)

1 du 2 dx 1 u

b ) d tan  1 u  1 2 du dx 1  u dx

1 u y

1 - u2 dy dy Let y  sin -1 u  u  sin y  du  cos y .  1  u2 dx dx dx dy 1 du d 1 du 1     sin u dx dx 1  u 2 dx 1  u 2 dx

b) 1  u2

u y 1





2 dy Let y  tan -1 u  u  tany  du  sec 2 y . du  1  u 2 dx dx dx dy   1 2 . du  d tan  1 u  1 2 . du dx dx 1  u dx 1  u dx

11

Hyperbolic functions : If u is any differentiable function of x , then :

21 ) 22 ) 23 ) 24 ) 25 ) 26 )

d sinh u  cosh u . du dx dx d cosh u  sinh u . du dx dx d tanh u  sec h 2 u . du dx dx d coth u   csc h 2 u . du dx dx d sec hu   sec h u .tanh u . du dx dx d csc hu   csc h u .coth u . du dx dx

EX-13 - Find

dy for the following functions : dx

a ) y  coth(tanx) c ) y  ln tanh x 2 3 e ) y  sech x

b ) y  sin -1 (tanh x ) d ) y  x.sinh2x - 1 .cosh 2 x 2 2 f ) y  csch x

Sol. dy   csc h 2 (tan x ). sec 2 x dx dy sec h 2 x sec h 2 x  sec h x   b) dx 1  tanh 2 x sec h 2 x 1 cosh2 x dy 1 sec h 2 x . 1  2  c) 2 2 dx tanh x sinh x 2 2 2. cosh x 2 1 1    csc h x 2 sinh x .cosh x sinh x 2 2 a)

12

dy  x cosh 2 x .2  sinh 2 x  1 sinh 2 x .2  2 x cosh 2 x dx 2 dy  3 sec h 2 x(  sec h x .tanh x )  3 sec h 3 x .tanh x e) dx dy  2 csc h x(  csc h x .coth x )  2 csc h 2 x .coth x f) dx

d)

EX-14- Show that the functions : x   2 sinh t and y  1 sinh t  cosh t 3 3 3 3 3 Taken together , satisfy the differential equations : dy dy i ) dx  2  x  0 and ii ) dx   y0 dt dt dt dt Proof x   2 sinh t  dx   2 cosh t dt 3 3 3 3 dy 1  cosh t  1 sinh t y  1 sinh t  cosh t  dt 3 3 3 3 3 3 3

dy i ) dx  2  x   2 cosh t  2 cosh t  2 sinh dt 3 dt 3 3 3 3 dy ii ) dx   y   2 cosh t  1 cosh t  1 sinh t  3 dt dt 3 3 3 3 3

t  2 sinh t  0 3 3 3 1 t t sinh  cosh 0 3 3 3

EX-15 - Prove that : a)

d du tanh u  sec h 2 u . dx dx

and

b)

du d sec h u   sec h u . tanh u . dx dx

Proofa)

b)

cosh u .cosh u . du  sinh u . sinh u . du d tanh u  d  sinh u   dx dx 2 dx dx  cosh u  cosh u du 2 2 (cosh u  sinh u ) 1 dx  . du  sec h 2 u . du  2 2 dx cosh u cosh u dx d 1  1 . sinh u . du   sec h u .tanh u . du 2 dx cosh u dx dx cosh u

13

The inverse hyperbolic functions : If u is any differentiable function of x , then :

27 ) 28 ) 29 ) 30 ) 31 ) 32 )

d sinh  1 u  1 du dx 1  u 2 dx d cosh1 u  1 du dx u 2  1 dx d tanh 1 u  1 du dx 1  u 2 dx d coth1 u  1 du dx 1  u 2 dx d sec h 1 u   1 du dx u 1  u 2 dx d csc h 1 u   1 du 2 dx u 1  u dx

EX-16 - Find

u 1 u 1

dy for the following functions : dx

a ) y  cosh -1 (sec x )

b ) y  tanh -1 (cos x )

c ) y  coth -1 (sec x )

d ) y  sech -1 (sin 2 x )

Sol.a) b) c)

d)

dy x  sec x  sec x2.tan x  sec x .tan 2 dx sec x  1 tan x dy x   csc x   sin x2   sin 2 dx 1  cos x sin x dy sec x .tan x sec x .tan x     csc x dx 1  sec 2 x  tan 2 x dy 2.cos 2 x   2 csc 2 x dx sin 2 x . 1  sin 2 2 x

EX-17 – Verify the following formulas : 1 a ) d cosh  1 u  . du 2 dx u  1 dx u 1 b ) d tanh  1 u  1 2 . du dx 1  u dx

14

where tan x  0

where cos 2 x  0

Proof Let y  cosh -1 u  u  cosh y du  sinh y . dy  dy  1 . du dx dx dx sinh y dx 2 2 cosh y  sinh y  1  u 2  sinh 2 y  1  sinh y  u 2  1 dy 1 1  . du  d cosh 1 u  . du 2 2 dx dx u  1 dx u  1 dx b ) Let y  tanh 1 u  u  tanh y du  sec h 2 y . dy  dy  1 . du 2 dx dx dx sec h y dx 2 2 sec h y  tanh y  1  sec h 2 y  u 2  1  sec h 2 y  1  u 2 dy  1 2 . du  d tanh 1 u  1 2 . du dx 1  u dx dx 1  u dx a)

The derivatives of functions like uv : Where u and v are differentiable functions of x , are found by logarithmic differentiation : Let y  u v  ln y  v .ln u 1 . dy  v . du  ln u . dv dx y dx u dx dy  y  v . du  ln u . dv   u dx dx dx  33 ) d uv  uv . v . du  ln u . dv  dx dx   u dx dy for : dx a ) y  x cosx

EX-18- Find

b) y  (lnx  x)tanx

Sol. a)

dy cos x   ln x .(  sin x ) y  x cos x  ln y  cos x .ln x  1 . y dx x dy   y  cos x  sin x .ln x   x  dx or by formula , where u  x and v  cosx dy  y  cos x  sin x .ln x   x  dx

15

b)

y  (ln x  x )tan x  ln y  tan x .ln(ln x  x ) dy  1.  tan x .( 1  1 )  ln(ln x  x ). sec 2 x y dx ln x  x x dy  ( x  1 ).tan x    y  ln(ln x  x ). sec 2 x  dx  x (ln x  x )  or by formula , where u  lnx  x and v  tanx dy  y . tan x ( 1  1 )  ln(ln x  x ). sec 2 x   ln x  x x  dx  ( x  1 ).tan x   y .  ln(ln x  x ). sec 2 x   x (ln x  x ) 

16

Problems -3 dy for the following functions : dx y  ( x  3 )( 1  x ) y  ax  b x 4 3 x y 2x  3 y  3 x 3  2 x  52 x 2   3 1  y   x  x3  

1. Find 1) 2) 3) 4) 5)

6)

y  ( 2 x  1 )2 ( 3 x  2 )3 

7)

y  ln(ln x )

8)

y  ln( Cosx )

9)

y  Sinx 3

10 ) 11 ) 12 ) 13 ) 14 ) 15 )

16 ) 17 ) 18 )

1 ( x  2 )2

( ans . : 4  2 x ) ( ans . :  b2 ) x 1 ( ans . : ) ( 2 x  3 )2 ( ans . : 9 x 2  1  103 ) x x 3( x 6  1 ) ( ans . : ) x4 2

2

( ans . : ( 2 x  1 )( 3 x  2 ) ( 30 x  1 ) 

(x  2)

1 ) x .ln x ( ans . :  tan x )

( ans. :

( ans . : 3 x 2 .Cosx3 ) 30 x .Sin( 5 x 2  4 ) y  Cos 3 ( 5 x 2  2 ) ( ans. : ) Cos4 ( 5 x 2  4 ) y  tan x . sin x ( ans . : Sinx  tan x .Secx ) y  tan( Secx ) ( ans . : Sec 2 ( Secx ).Secx .tan x ) 6 y  Cot 3  x  1  ( ans . : .Cot 2  x  1  .Csc 2  x  1  ) 2  x1  x1  x1 ( x  1) Cosx ) y  Cosx ( ans . :  x .Sinx  2 x x 2 Sec 2 x  7 y  tan 2 x  7 ( ans . : ) 2 2 x  7 tan 2 x  7 y  x 2 .Sinx ( ans . : x 2 .Cosx  2 x .Sinx ) 2  5 Cot 5 x ) y  Csc 3 5 x ( ans . : . 2 3 5x Csc 3 5 x y  xSin(ln x )  Cos(ln x ) ( ans . : 2.Cos(ln x ) )

17

3

)

19 )

y  Sin 1 ( 5 x 2 )

20 )

y  Cot  1  1  x   1 x  1

21 )

y  tan

22 )

y  Sec1 ( 3 x 2  1 )3

( ans . : 

4x  2

2 23 ) y  Sin1 x  x 2 .Sec1 x 2 x 2

y  Sin1 2 x .Cos1 2 x

25 )

y3

26 )

y  tan 1 (ln x )

27 )

y3 

28 )

sin x .cos x 1  2.ln x 1 5 x . tan x y 3 ( 3  2 x ). x

29 )

y  sec 1 e 2 x

30 )

y  (cos x )

31 )

y  (sin x )tan x

32 )

y  2 x 2  cosh 2 ( 5 x )

33 )

y  sinh(cos 2 x ) y  csc h 1 x 2. y  x .tanh 2 x

34 ) 35 )

x

4x  x

( ans . :

2

(2  x) (2  x)  x

( ans. : ( ans. :

1 3 x y



2x

 2

x( x  1 )( x  2 ) ( x 2  1 )( 2 x  3 )

4

1 ) 1  x2

2 6 x ( ans. : ) ( 4 x3  1 ) 4 x3  2 18 x ( ans. : ) 3 x 2  1 ( 3 x 2  1 )6  1

3

24 )

10 x ) 1  25 x 4

( ans . :

 2 x .Sec

2

x1

1  4x 

1 x2



2

2x 2

x 1



18

)

) 2x  3   2

1 ) x ( 1  (ln x )2 ) 3 y cot x tan x 2   ( ans . : ( )) 4 2 2 x ( 1  2 ln x ) 1 2 ))  ( ans . : 2 y ( 14  1 2 3 x ( 1  x ).tan x 3  2 x 2 ( ans . : ) 4x e 1 y ( ans . : (ln cos x  2 x .tan x )) 2 x ( ans . : y ( 1  sec 2 x .ln sin x )) 2 x  5 cosh( 5 x ). sinh( 5 x ) ( ans . : ) 2 x 2  cosh 2 ( 5 x ) ( ans . : 2 sin 2 x .cosh(cos 2 x )) ( ans . : 12 .csc h 1 .coth 1 ) x x x 2 ( ans . : x .tanh x ( x sec h x  2 tanh x )) ( ans . :

x 2

x 4

4

2( Cos1 2 x  Sin1 2 x )

1

1

)

3  sin x . cos x tan x 36 ) y  ln x 37 ) y  log4 sin x

38 )

y  e ( x e

39 )

y  ex

40 )

y  7 csc

41 )

y  ln( x 2  2 )2 cos x

42 )

y  sinh  1 (tan x )

43 )

y  1  (ln x )2

44 )

x y e ln x

2

2

5x

cos x  sin x  3 tan x . sec x 1 ( ans. : )  3 2x sin x .cos x  tan x 2

2

2

( ans. : cot x ) ln 4 5x ( x e5 x ) ( ans. : ( 2 x  5e )e ) 2

)

( ans. : ( x 2 sec2 x  2 x tan x )e x

tan x 2 x 3



2



y  x 3 log 2 ( 3  2 x )

46 )

y  2 cosh  1 x  x 2 2

csc 2 x 3

tan x

ln7

x2  4

2. Verify the following derivatives :   a ) d 5 x  ( x  1 )2   6  12 dx  x x  b) d x ( ax 2  bx  c )  1 ( 5 ax 2  3bx  c ) dx 2 x



)

csc 2 x  3 .cot 2 x  3 ) 2x  3 ( ans . : 4 x2.cos x  2 ln( x 2  2 ) sin x ) x 2 ( ans . : sec x ) ln x ) ( ans . : x 1  (ln x )2 e x ( x ln x  1 ) ) ( ans . : x (ln x )2 2x3 ( ans . : 3 x 2 log 2 ( 3  2 x )  ) ( 3  2 x ) ln 2 x2 ) ( ans . : 2 x 4

( ans. :

45 )

7

2



3. Find the derivative of y with respect to x in the following functions : 2 18 x 2 y 2 3 u ) a) y 2 and u  3x  2 ( ans . : ( 3 x 3  2 )3 u 1 x  4x ) b ) y  u  2 u and u  x 2  3 ( ans . : 2 x 3

19

4. Find the second derivative for the following functions : a)

y  ( x  1 )3 x

b)

f ( x )  2 x  2 2 at x  2 x

c)

x 2  2 xy  y 2  16 x  0

( ans . : 6 x  63  125 ) x x (ans. : 1 ) 4 ( ans . :  x

3 2

)

5. Find the third derivative of the function : (ans. : - 3 ) 8y

y  x3

v ( vu''  uv'' )  2 v' ( vu'  uv' ) 6. Show for y  u that y''  . v v3

7. Show for y = u.v that y''' = uv''' + 3u' v'' + 3u'' v' + u''' v . 8. Show that y  35 x 4  30 x 2  3 satisfies ( 1  x 2 ) y'' 2 xy' 20 y  0 .

9. Find

dy for the following implicit functions : dx

20

a)

b)

3 x 2  5 y 2 x 2  4 y ) ( ans . : 10 x  1 y  2 x y y ( ans . : ) 2 xy  x

5 y2 3 x  4x y  x 3

xy  1  y 3 2

c)

3 xy  ( x  y )

d)

x 3  x . tan  1 y  y

e)

sin  1 ( xy )  cos  1 ( x  y )

3

3

f )

y 2 . sin( xy )  tan x

g)

sinh y  tan 2 x

( ans . :

3x2 x3  y3  2 y

) 2x  3 y2 x3  y3 ( 1  y 2 )( 3 x 2  tan  1 y ) ( ans . : ) 1  y2  x y 1  ( x  y ) 2  1  ( xy ) 2 ( ans . : ) 1  ( xy ) 2  x 1  ( x  y ) 2 sec 2 x  y 3 .cos( xy ) ( ans . : ) 2 y . sin( xy )  xy 2 .cos( xy ) 2 ( ans . : 2. tan x . sec x ) cosh y

10. Prove the following formulas : d cot u   csc 2 u . du a ) dx dx d b) csc u   csc u . cot u . du dx dx d 1 1 c) cos u   . du 2 dx dx 1 u d sec  1 u  1 d ) . du 2 dx u u  1 dx d sinh u  cosh u . du e) dx dx d f ) csc h u   csc h u . coth u . du dx dx d 1 du 1 g ) sinh u  . dx 1  u 2 dx d sec h  1 u   1 h) . du 2 dx dx u 1 u 11. Show that the tangent to the hyperbola x2- y2 = 1 at the point P(coshu, sinhu) , cuts the x-axis at the point ( sechu , 0 ) and except when vertical , cuts the y-axis at the point ( 0 , -cschu ) .

21

Chapter four Applications of derivatives 4-1- L'Hopital rule : Suppose that f ( xo ) = g ( xo ) = 0 and that the functions f and g are both differentiable on an open interval ( a , b ) that contains the point xo . Suppose also that g' ( x )  0 at every point in ( a , b ) except possibly xo . Then : f'( x ) f(x) provided the limit exists .  lim lim x x g( x ) x  x g' ( x ) Differentiate f and g as long as you still get the form 0 or  0  at x = xo . Stop differentiating as soon as you get something else . L'Hopital's rule does not apply when either the numerator or denominator has a finite non-zero limit . EX-1 – Evaluate the following limits : 2 x 5 3 sin x 1 ) lim 2 ) lim 2 xo x2 x x 4 x 3 ) lim x  sin 4 ) lim  ( x   ).tan x 3 x 0 2 x x  2 0

0

Sol. – 1 ) lim sin x  0 u sin g L' Hoptal' s rule  x 0 x 0  lim cos x  cos 0  1 x 0 1 2)

x2  5  3 0 lim u sin g L' Hoptal' s rule   2 x 2 x 4 0 x 2 1 1 1  lim x  5  lim   x 2 x 2 2x 2 x2  5 2 4  5 6

x  0 u sin g L' Hoptal' s rule  3 ) lim x  sin 3 x 0 0 x x  0 u sin g L' Hopital' s rule   lim 1  cos 2 x 0 0 3x 1 sin x 1  lim  6 x 0 x 6

١

4 ) lim  ( x   ) tan x  0. we can' t u sin g L' Hoptal' s rule  2 x  2 x  2 . lim sin x  0 u sin g L' Hopital' s rule   lim  cos x x   0 x  2 2 1 . lim sin x  1 . sin   1  lim    sin x x   2 x sin  2 2 2

4-2- The slope of the curve : Secant to the curve is a line through two points on a curve. Slopes and tangent lines : 1. we start with what we can calculate , namely the slope of secant through P and a point Q nearby on the curve . 2. we find the limiting value of the secant slope ( if it exists ) as Q approaches p along the curve . 3. we take this number to be the slope of the curve at P and define the tangent to the curve at P to be the line through p with this slope . The derivative of the function f is the slope of the curve : dy the slope  m  f ' ( x )  dx EX-2- Write an equation for the tangent line at x = 3 of the curve : 1 f( x) 2x  3 Sol.1 m  f'( x )    m x  3  f ' ( 3 )   1 3 27 ( 2x  3 ) 1  1 f(3) 2* 3  3 3 The equation of the tangent line is : y  1   1 ( x  3 )  27 y  x  12 3 27

٢

4-3- Velocity and acceleration and other rates of changes : - The average velocity of a body moving along a line is :

vav 

s f ( t  t )  f ( t ) displaceme nt   t t time travelled

The instantaneous velocity of a body moving along a line is the derivative of its position s = f ( t ) with respect to time t . i.e. v 

ds s  lim t 0 t dt

- The rate at which the particle’s velocity increase is called its acceleration a . If a particle has an initial velocity v and a constant acceleration a, then its velocity after time t is v + at . average acceleration  a  av

v t

The acceleration at an instant is the limit of the average acceleration for an interval following that instant , as the interval tends to zero . i.e.

a  lim t  0

v t

- The average rate of a change in a function y = f ( x ) over the interval from x to x + Δx is : average rate of change 

f ( x  x )  f ( x ) x

The instantaneous rate of change of f at x is the derivative.

f ' ( x )  lim x  0

f ( x  x )  f ( x ) provided the limit exists . x

EX-3- The position s ( in meters ) of a moving body as a function of time t ( in second ) is : s  2 t  5 t  3 ; find : a) The displacement and average velocity for the time interval from t = 0 to t = 2 seconds . b) The body’s velocity at t = 2 seconds . 2

٣

Sol.a ) 1)

s  s( t  t )  s( t )  2( t  t ) 2  5 ( t  t )  3  2 t 2  5 t  3   ( 4 t  5 )t  2( t )2 at t  0 and t  2  s  (4 * 0  5) * 2  2 * 2 2  18 ( 4 t  5 )t  2( t )2  s 2 ) v av    4 t  5  2.t t t at t  0 and t  2  v av  4 * 0  5  2 * 2  9

b)

d f ( t )  4t  5 dt v ( 2 )  4 * 2  5  13 v( t ) 

EX-4- A particle moves along a straight line so that after t (seconds) , its distance from O a fixed point on the line is s (meters) , where s  t  3 t  2 t : i) when is the particle at O ? ii) what is its velocity and acceleration at these times ? iii) what is its average velocity during the first second ? iv) what is its average acceleration between t = 0 and t = 2 ? Sol. – 3

2

i ) at s  0  t  3 t  2 t  0  t ( t  1 )( t  2 )  0 3

ii )

2

either t  0 or t  1 or t  2 sec . velocity  v ( t )  3t  6 t  2  v ( 0 )  2 m / s 2

 v ( 1 )  1m / s  v( 2 )  2m / s acceleration  a ( t )  6 t  6  a ( 0 )  6 m / s  a( 1 )  0 m / s

2

 a( 2 )  6 m / s

2

s( 1 )  s( 0 ) 1  3  2  0 v av  s    0m / s t 10 1 v( 2 )  v( 0 ) 2  2 iv ) a av  v    0m / s 2 t 20 2

iii )

٤

2

4-4- Maxima and Minima : Increasing and decreasing function : Let f be defined on an interval and x1 , x2 denoted a number on that interval : - If f(x1) < f(x2) when ever x1 < x2 then f is increasing on that interval . - If f(x1) > f(x2) when ever x1 < x2 then f is decreasing on that interval . - If f(x1) = f(x2) for all values of x1 , x2 then f is constant on that interval . The first derivative test for rise and fall : Suppose that a function f has a derivative at every point x of an interval I. Then : - f increases on I if f ' ( x )  o , x  I - f decreases on I if f ' ( x )  o , x  I If f ' changes from positive to negative values as x passes from left to right through a point c , then the value of f at c is a local maximum value of f , as shown in below figure . That is f(c) is the largest value the function takes in the immediate neighborhood at x = c . f' = 0

f increasing

f decreasing

f increasing

f'=0

a

+ + f'>0 c

- - - d

f'