Itb - Engineering Mathematics 1 (feg1ma1) 84y4r

INSTITUTE TECHNOLOGY BRUNEI FACULTY OF ENGINEERING COURSE NOTE COURSE NUMBER: FEG1MA1 COURSE TITLE: ENGINEERING MATHEMATICS 1

DR. MD. FAZLUL KARIM

1

Differentiation Function: Let us consider the following question: To fence a rectangular region with a wire of length 100 m , what is the relation between the length and the width of the rectangle?

y

x Figure: A rectangle Refer to Figure, Let the rectangular region has length x and width y . Then

2 x + 2 y = 100 And so

y = 50 − x. y changes according to x , and a value of x will determine a value of y . For example, if x = 1 , then y = 49 ; if x = 19.4 , then y = 30.6 . We note that since x is the length of a rectangle, x ≥ 0 . It is also clear that x ≤ 50 . Hence, for this particular case, the value of x is restricted to the interval [ 0,50] . Similarly, y is also restricted to the interval [ 0,50] . The relation y = 50 − x exhibits the concept of a function. Definition: If a variable y depends on a variable x in such a way that each value of x determines exactly one value of y, then we say that y is a function of x. Definition: A function f from set A to set B , written as f : A → B , is a rule that assigns, to each element in A , a unique element in B . A is called the domain of f, and B the co-domain of f.

A

f

2

B

Figure: The function f In the above example, the function involved is

f : [ 0,50] → [ 0,50] f ( x ) = 50 − x

Concept of limit: Many are puzzled by the phrase “ h approaches 0 but h ≠ 0 ”. The following example gives you an idea of what the phrase means. Suppose one wishes to travel from A to B , where B is 100m on the right of A .

A

B

First he reaches the midpoint C of AB .

C A

B

Then he reaches the midpoint D of CB .

A

D

C

B

He travels in this way that he always reaches the midpoint of where he is and B . In other words, if h represents the distance between him and B , then h approaches 0 but h ≠ 0 . Can u think of another way?

f ( x ) = b means that when x approaches a but x ≠ a , then the function value The symbol lim x →a f ( x ) approaches b . Definition: If the values of f(x) can be made as close as we like to L by taking values of x sufficiently close to a (but not equal to a), then we write lim x→a f ( x ) = L , which is read “the limit of f(x) as x approaches a is L.

3

(

)

x2 + x − 1 = 1 Example. lim x →1 Intuitively, when x is close to 1, x 2 approaches 1 (say x = 1.01 , then x 2 = 1.0201 ; if

x = 1.001 , then x 2 = 1.002 ), so x 2 + x − 1 approaches 1. 2 Example: Find lim x→5 ( x − 4 x + 3)

2 2 Solution: lim x →5 ( x − 4 x + 3) =l imx →5 x − lim x →5 4 x + lim x →5 3

= ( lim x→5 x ) − 4 lim x→5 5 + lim x→5 3 = 5 2 − 4(5) + 3 = 8 2

Example: Find lim x →2

5x 3 + 4 x −3

Solution: lim x →2

(

)

5.2 + 4 5 x 3 + 4 lim x→2 5 x + 4 = = −44 = x −3 lim x→ 2 ( x − 3) 2−3 3

3

Continuity: A moving object cannot vanish at some point and reappear someplace else to continue its motion. Thus, we perceive the path of a moving object as an unbroken curve, without gaps, breaks or holes. Definition: A function f is said to be continuous at x = a provided the following conditions are satisfied (i)

f(a) is defined

(ii)

lim x→a f ( x) exists

(iii)

lim x →a f ( x ) = f ( a )

Example: Determine whether the following functions are continuous at x = 2.

 x2 − 4  x2 − 4 x2 − 4   , x ≠ 2 f ( x) = , g ( x) =  x − 2 , h( x ) =  x − 2 , x ≠ 2 x−2  3, x = 2  4, x = 2 Solution: In all the cases the functions are identical, except at x = 2, and hence all three have the same limit at x = 2, namely lim x →2 f ( x) = lim x→2 g ( x) = lim x→2 h( x) = lim x→2

x2 − 4 = lim x→2 ( x + 2 ) = 4 x −2

The function f is undefined at x = 2, and hence is not continuous at x = 2. The function g is defined at x = 2, but its value there is g(2) = 3 which is not same as the limit as x approaches 2, hence g is also not continuous at x = 2. The value of the function h at x = 2 is h(2) = 4, which is same as the limit as x approaches 2, hence h is continuous at x = 2.

4

Derivative: 3 2 Suppose a particle travels along the y -axis. Its position as time t is given by y = f ( t ) = t + 3t

. When t = 0 , y = 0 ; when t = 1 , y = 4 ; when t = 2 , y = 20 ; etc.

0

4

20

y

position of particle when t = 1 position of particle when t = 0

position of particle when t = 2

The average velocity of the particle between t = 0 and t = 1 is 4 while the average velocity of the particle between t = 0 and t = 2 is 10. We see that the average velocity is increasing when t increases. This shows that the velocity of the particle changes from time to time. Suppose t and t + h are two different times, then the average velocity between time t and t + h is

f ( t + h) − f ( t ) . h

If we let h approaches 0, it is reasonable that

lim h →0

f ( t + h) − f ( t ) h

is the “instantaneous velocity” of the particle at time t .

5

The gradient (slope) of a curve is the gradient of the tangent:

Gradient of chord: =

y2 − y1 ∆y = x2 − x1 ∆x

=

f ( x2 ) − f ( x1 ) x2 − x1

Gradient of tangent:

=

 ∆y  dy lim  ÷ = ∆x →0 ∆x   dx

Differentiation from first principles:

6

δy δx

=

f ( x + δ x) − f ( x) δx dy δ y   f ( x + δ x) − f ) x)  ' = f ( x) = lim  ÷ = δlim  ÷ δ x →0 δ x dx δx   x →0  

Definition: . Let f : (a, b) → R, C ∈ (a, b) . The derivative of f at x = c is the limit

lim h →0

f ( c + h) − f ( c) . h

If this limit exists, f is said to be differentiable at x = c . Remarks 1. We use the symbol f ' ( c ) to denote the derivative of f at x = c , i.e.

f ' ( c ) = lim h →0

f ( c + h) − f ( c) . h

7

Techniques of differentiation d (c ) = 0 ** dx d (5) = 0 Example1. Let f(x)=5, then dx n n −1 ** Let n be a positive integer and f ( x ) = x . Then f ' ( x ) = nx . Example: **

d d ( x 5 ) = 5 x 4 , ( x ) = 1x 0 = 1 dx dx

d (cf ( x)) = cf ′ dx

Example(i): (ii) **

d d (4 x 5 ) = 4 ( x 5 ) = 4.5 x 4 = 20 x 4 dx dx

d x 1 d 1 ( x) =  = dx  π  π dx π d [ f ( x) ± g ( x)] = d f ( x) ± d g ( x) dx dx dx

Example:

[

]

d 4 d d x + x 2 = ( x 4 ) + ( x 2 ) = 4x3 + 2x dx dx dx

Parametric differentiation:

dy dy dθ = dx dx dθ Implicit differentiation:

d  f ( y)  = d  f ( y)  x dy  dy   dx dx 

Logarithmic differentiation:

d ln( y) =  1  dy ( )  y ÷÷ dx dx  

The product rule:

8

y ( x) = u ( x)v( x) = uv dy dv du =u +v dx dx dx Example: Find

)(

[(

)(

dy d = 4x 2 −1 7 x3 + x dx dx

Solution: 2 = ( 4 x −1)

(

(

dy , if y = 4 x 2 −1 7 x 3 + x dx

)

)]

d d ( 7 x 3 + x ) + (7 x 3 + x ) ( 4 x 2 −1) dx dx

)(

) (

)

= 4 x 2 −1 21x 2 + 1 + 7 x 3 + x ( 8 x ) = 140 x 4 − 9 x 2 − 1 The quotient rule:

y ( x) =

u ( x) v( x)

du dv −u dy = dx 2 dx dx v v

Example: Differentiate: y =

Solution: dy =

2 cos t

dx

te 2t 2 cos t

d d (te 2t ) − te 2t (2 cos t ) dt dt 2 ( 2 cos t )

=

2 cos t ( 2te 2t + e 2t ) − te 2t ( −2 sin t ) 4 cos 2 t

=

e 2t (2t cos t + cos t + t sin t ) 2 cos 2 t

The chain rule: (function of a function) y is a function of u and u is a function of x

dy dy du = dx du dx

(

Example: Differentiate (i) y = ( 3 x −1) 9 ,(ii) y = 3 cos 5 x 2 + 2

9

)

Solution (i) Let u = 3 x −1, y = u 9 ∴

dy du = 9u 8 , =3 du dx

∴

dy dy du = . = (9u 8 ) .3= 27u 8 = 27( 3 x −1) 8 dx du dx

(ii) Let u = 5 x 2 + 2, y = 3 cos u ∴ ∴

dy du = −3 sin u , = 10 x du dx

(

dy dy du = . = −30 x sin u = −30 x sin 5 x 2 + 2 dx du dx

)

Chain Rule: If g is differentiable at x and f is differentiable at g(x), then the composition f og is differentiable at x. Moreover,

(f

′  g ) ( x ) = f ′( g ( x )) g ′( x )

Alternatively, if y = f ( g ( x ) ) , u = g ( x ) then, y = f (u) and dy dy du = . dx du dx Example :

( )

If h( x) = 4 cos x 3 , find h ' ( x ) .

Solution. We first find f and g such that h = f og . If g ( x) = x 3 , f (u ) = 4 cos u then

(f

( )

 g )( x) = f ( g ( x )) = 4 cos x 3 = h( x )

f ′(u ) = −4 sin u , g ′( x) = 3 x 2

Using Chain rule, h′ ( x ) = f ′( g ( x)) g ′( x) = ( − 4 sin g ( x) )3 x 2 = −12 x 2 sin ( x 3 ) Alternatively, let y = h(x) and let u = x 3 ,∴y = 4 cos u By the form of the chain rule h′( x) = Example:

(

(

)

If h ( x ) = x 2 + 3sin x + 1

100

, find h ' ( x ) .

Solution:

h ' ( x ) = differentiate

( )

100

= 100 ( x 2 + 3sin x + 1)

.differentiate ( x 2 + 3sin x + 1) 99

)

( )

dy dy du = . = ( − 4 sin u ) 3 x 2 = 12 x 2 sin x 3 dx du dx

( 2 x + 3cos x )

10

1

Example:

2 If f ( x ) =  x 2 sin x + 1  , find f ' ( x ) .  ÷ x 

Solution: −

1

1 1 2 1  f ' ( x ) =  x 2 sin x + ÷  2 x sin x + x 2 cos x − 2 ÷ 2 x  x  Example: Evaluate the derivatives of the following functions: (a ) sin (πx ) + cos( 3 x ), (b) x 2 sin Solution: (a )

x , (c )

cos x 1 − sin x

d (sin (πx ) + cos( 3 x ) ) = cos(πx )π − sin ( 3 x )3 = π cos(πx ) − 3 sin ( 3x ) dx

(b) By the product and chain rules: 3

d 2 1 1 ( x sin x ) = 2 x sin x + x 2 cos x . = 2 x sin x + x 2 cos x dx 2 2 x (c) By the Quotient rule: d  cos x  (1 − sin x ) ( −sin x ) − (cos x)(0 − sin x) − sin x + sin 2 x + cos 2 x =  = dx  1 − sin x  (1 − sin x ) 2 (1 − sin x ) 2 =

− sin x +1 1 = 2 (1 − sin x ) 1 − sin x

Implicit differentiation: dy , if xy = 1 Example: Find dx Solution: Differentiating implicitly yields 1 y = , from which it follows that x d ( xy ) d = (1) dx dx dy d ( x) ∴x +y =0 dx dx . dy ∴x + y =0 dx dy y 1 ∴ =− =− 2 dx x x Example: Find the slope of the curve y 2 − x +1 = 0 at the points (2,-1) and (2,1). Solution: Differentiating implicitly yields

11

[

]

d d y 2 − x +1 = ( 0) dx dx dy ∴2 y −1 = 0 dx dy  1 ∴ = dx  2 y dy dx

( 2 , −1)

1 dy =− , 2 dx

= ( 2 ,1)

1 2

We have accumulated the following formulae:

f ( x)

f '( x)

k (a constant)

( r ≠ 0)

xr sin x cos x

0

rx r −1 cos x − sin x

tan x

sec 2 x

cot x sec x csc x

− csc 2 x sec x tan x − csc x cot x 1

−1

sin x

1 − x2 1

−1

cos x

1 − x2

tan −1 x

1 1 + x2

cot −1 x

−1 1 + x2

sec −1 x

1 x x2 −1

csc −1 x

1 x x2 −1 12

d (log x) dx

1 x

d (sinh x ) dx

Coshx

d (cosh x ) dx d (tanh x) dx

Sinhx

sec h 2 x Derivative of Logarithmic and exponential function: d 1 (log x ) = dx x d (log( x 2 +1)) Example: Find dx 1 d 2 2x d (log( x 2 +1)) = 2 . x +1 = 2 Solution: dx x + 1 dx x +1

(

)

d   x 2 sin x   Example: Find ln dx   1 + x  Solution: =

d   x 2 sin x  d  1   = 2 ln x + ln(sin x) − ln (1 + x )  ln  2 dx   1 + x  dx  

2 cos x 1 2 1 + − = + cot x − x sin x 2(1 + x) x 2(1 + x)

Example: Differentiate (sec x) tan x Solution: Let y = (sec x ) tan x ∴log y = tan x log sec x

Differentiating both sides w. r. to x, 1 dy 1 . = tan x. sec x tan x + sec 2 x log sec x y dx sec x dy tan x ∴ = y tan 2 x + sec 2 x log sec x = ( sec x ) tan 2 x + sec 2 x log sec x dx

(

)

(

dy 1 + sin x − 1 − sin x , if y = tan −1 dx 1 + sin x + 1 − sin x Solution: On rationalizing the denominator, Example: Find

13

)

1 − cos x y = tan −1 = tan −1 sin x ∴

2 sin 2 2 sin

x 2

x x cos 2 2

= tan −1 tan

x x = 2 2

dy 1 = dx 2

Example: If y = tan −1

dy 1 + x 2 −1 , , find dx x

Solution: Putting

x = tan θ ,

1 + x − 1 sec θ − 1 1 − cos θ = = = x tan θ sin θ

∴ y = tan −1 tan ∴

2 sin 2

2

2 sin

θ 2

θ θ cos 2 2

= tan

θ 2

θ θ 1 = = tan −1 x 2 2 2

dy 1 1 = . dx 2 1 + x 2

Differentiations of parametric equations: When x and y are given in of a parameter, say θ , then by the function of a function rule of differentiation d  dy    2 dy dy dx d y dθ  dx  = / , = dx dx dθ dθ dx 2 dθ dy , if x = a(θ − sin θ ), y = a((1 + cos θ ) Example: Find dx Solution:

dy dy dx a sin θ = / =− =− dx dθ dθ a(1 − cos θ )

θ θ cos 2 2 = − cot θ θ 2 2 sin 2 2

2 sin

Higher (successive) derivatives: dy y = x3 , = 3x 2 dx d  dy  d (3x 2 ) = 6 x  = dx  dx  dx d  dy  d2y   is denoted by Now, (i.e. 2nd derivative of y with respect to x) 2 dx  dx  dx Similarly,

d3y ,(i.e. 3nd derivative of y with respect to x) is here 6. 3 dx

If y = f (x), the successive derivatives are also denoted by

14

y1

y2

−

−

y′

y ′′

−

−

f ′( x)

f ′′( x )

yn yn −

−

f n ( x)

− − d D stand for the symbol . dx Df ( x)

D 2 f ( x)

D n f ( x)

##The n th derivatives of some special functions f n (x ) f ( x)

k (a constant)

0

xn

n!

e ax

a n e ax

1 x +a

( − 1) n n! ( x + a ) n +1

log( x + a )

( −1) n −1 (n −1)! ( x + a ) n +1

Sin(ax+b) Cos(ax+b)

 nπ  a n sin  + ax + b   2   nπ  a n sin  + ax + b   2 

Example: If y = a cos(log x) + b sin(log x), show that x 2 y 2 + xy1 + y = 0 Solution : Given y = a cos(log x) + b sin(log x), Differentiating 1 1 y1 = −a sin(log x ). + b cos(log x ). x x ∴xy1 = −a sin(log x ) + b cos(log x ) Differentiating again, 1 1 xy 2 + y1 = −a cos(log x). − b sin(log x ). x x 2 ∴ x y 2 + xy1 = −( a cos(log x ) + b sin(log x ) ) = − y ∴ x 2 y 2 + xy1 + y = 0 Example: If

y=

x2 + x −1 , find y n . x3 + x 2 − 6x

Solution: x 3 + x 2 − 6 x = x( x 2 + x − 6) = x( x + 3) ( x − 2)

15

Let

x2 + x −1 A B C = + + 3 2 x + x − 6x x x + 3 x − 2

Multiplying both sides by x( x +3)( x − 2) , we get, x 2 + x −1 = A( x + 3) ( x − 2) + Bx( x − 2) + Cx ( x + 3) Putting x = 0,-3,2 successively on both sides, we get 1 1 1 A = , B = ,C = 3 3 2 1 1 1 1 1 1 ∴y = . + . + . 6 x 3 x +3 2 x +2 1 1  1 1 1 1 n ∴ y n = ( −1) n!  . n +1 + . + . n +1 n +1  3 ( x + 3) 2 ( x + 2)  6 x Applications of differentiations: Example: A particle P moves along the x-axis in such a way that its position at time t is given by x = 2t 3 − 15t 2 + 24t ft. (a) Find the velocity and acceleration of P at time t. (b) In which direction and how fast is P moving at 2 s? Is it speeding up or slowing down at that time? (c) When is P instantaneously at rest? When is its speed instantaneously not changing? Solution: (a) The velocity and acceleration of P at time t are ds v= = 6t 2 − 30t + 24 = 6(t −1)(t − 4) ft/s dt dv a= = 12t − 30 = 6( 2t − 5) ft/s2 dt (b) At t = 2, we have v = -12 and a = -6. Thus, P is moving to the lef with speed 12 ft/s, and, since the velocity and acceleration are both negative, its speed is increasing. (c) P is at rest when v = 0, that is, when t = 1 or t = 4 s. Its speed is unchanging when a =0, that is, at t = 5/2 s. Example: Determine the rate of change of voltage, given v = 5t sin 2t volts when t = 0.2s. dv = ( 5t ) 2 cos 2t + 5 sin 2t = 10t cos 2t + 5 sin 2t Solution: Rate of change of voltage is dt dv = 10t cos 2t + 5 sin 2t = 10(0.2) cos 2(0.2) + 5 sin 2(0.2) = 3.7892 When t = 0.2, dt Example: The luminous intensity I candelas of a lamp at varying voltage V is given by l = 4 ×10 −4 V 2 . Determine the voltage at which the length is increasing at a rate of 0.6 candelas per volt. dI Solution: The rate of change of light with respect to voltage is given by . dV Since l = 4 ×10 −4 V 2 dI ∴ = 4 ×10 −4 ( 2 )V = 8 ×10 −4 V dV When the light is increasing at 0.6 candelas per volt then

16

0.6 =8 ×10 −4 V , ⇒∴V = 750volts −kt Example: Newton’s law of cooling is given by θ = θ0 e , where the excess of temperature at

zero time is θ0 C. Determine the rate of change of temperature after 40 s, given that θ0 =16  C , k = −0.03. Solution: The rate of change of temperature is dθ = θ0 ( − k ) e −kt dt When θ0 =16  C , k = −0.03, t = 40, dθ 0.03 ( 40 )  

dt

=16( 0.03)e

=1.594 C / s

Example: The distance x metres moved by a car in a time t seconds is given by x = 3t 3 − 2t 2 + 4t − 1. Determine the velocity and acceleration when (a) t = 0 and (b) t = 1.5 s. dx = 9t 2 − 4t + 4m / s Solution: Velocity v = dt Acceleration a =

d 2x = 18t − 4m / s 2 dt 2

(a) When time t = 0, v = 9( 0 ) 2 − 4( 0 ) + 4 = 4m / s, a = 18( 0 ) − 4 = −4m / s 2 (b) When time t = 1.5, v = 9(1.5) 2 − 4(1.5) + 4 =18.25m / s, a =18( 01.5) − 4 = 23m / s 2 Partial Differentiation Area of a rectangle depends upon its length and breadth, hence we can say that area is the function of two variables, i.e., its length and breadth. z is called a function of two variables x and y if z has one definite value for every pair of x and y. Symbolically, it is written as z = f (x,y). The variables x and y are called independent variables while z is called the dependent variable. Similarly, we can define z as a function of more than two variables. Partial derivatives: Let z = f(x,y) be function of two independent variables x and y. If we keep y constant and x varies then z becomes a function of x only. The derivative of z with respect of x, keeping, y as constant is called partial derivatives of z, w.r.to x and is denoted by the symbols ∂z f ( x + δx, y ) − f ( x, y ) ∂z ∂f , , f x ( x, y ) etc. Then = lim ∂x →0 ∂x ∂x ∂x δx ∂z f ( x, y + δy ) − f ( x, y ) = lim ∂y →0 Similarly, ∂y δy 2 ∂z ∂z ∂ z ∂2 z ∂2 z = p, = q, = s, = r , =t Notation: ∂x ∂y ∂x∂y ∂x 2 ∂y 2 Example: If z = 5 x 4 + 2 x 3 y 2 − 3 y, find ( a ) Solution: (a)To find

∂z ∂z , (b) ∂x ∂y

∂z , y is kept constant ∂x

17

∂z d d 3 d = 5x 4 + 2 y 2 x −3y (1) = 20 x 3 + 6 x 2 y 2 − 0 = 20 x 3 + 6 x 2 y 2 ∂x dx dx dx

(

)

( )

∂ z , x is kept constant ∂ y ∂z d d d = 5x 4 (1) + 2 x 3 y 2 −3 ( y ) = 5 x 4 (0) + 2 x 3 .2 y − 3.1 = 4 x 3 y − 3 ∂y dy dy dy 1 ∂z 1 ∂z = Example: If z = sin xy show that y ∂x x ∂y ∂z 1 ∂z = y cos xy,∴ = cos xy , since y is kept constant Solution: ∂x y ∂x ∂z 1 ∂z = x cos xy,∴ = cos xy , since x is kept constant ∂y x ∂y 1 ∂z 1 ∂z = Hence . y ∂x x ∂y

(b)To find

(

)

(

)

( )

Application: Error determination ∂y dy ∂y dy dy lim ∂x →0 = , = ( approx), δy = .δx (approx ) ∂x dx ∂x dx dx δx is known as absolute error in x δx is known as the relative error in x x δx ×100 is known as percentage error in x. x Example: Pressure p of a mass of a gas is given by pV = mRT , where m and R are constants, V ∂p ∂p , is the volume and T the temperature. Find expressions for . ∂T ∂V mRT Solution: Since pV = mRT then p = V ∂p mR d mR = (T ) = , V is kept constant. Hence ∂T V dT V

∂p d 1 mRT = mRT   = − 2 , T is kept constant. ∂V dV  V  V E2 . Using calculus, find the R approximate percentage change in P when E is increased by 3% and R is decreased by 2%. E 2 ⇒ log P = 2 log E − log R Solution: P = R On differentiating we get,

Example: The power dissipated in a resistor is given by P =

18

∂P ∂E ∂R 100∂P 100∂E 100∂R =2 − ⇒ =2× − P E R P E R 100∂P ∴ = 2 ×3 − ( −2) = 8 P ∴ Percentage change in P = 8.

Partial derivatives of higher order: Let z = f (x,y), then

∂z ∂z , being the functions of x and y can be further differentiated ∂x ∂y

partially w.r. to x and y. Symbolically ∂  ∂z  ∂2 z ∂2 f or , or , f xx  = ∂x  ∂x  ∂x 2 ∂x 2  ∂z  ∂2 z ∂2 f   = or , or , f yy  ∂y  ∂y 2 ∂y 2   ∂  ∂z  ∂2 z ∂2 f   = or , or , f xy , f xy = f yx   ∂x  ∂y  ∂x∂y ∂x∂y Example: Prove that y = f ( x + at ) + g ( x − at ) satisfies ∂ ∂y

2 ∂2 y 2∂ y   = a  ∂x 2  , where f and g are assumed to be at least twice differentiable and a is any ∂t 2   constant. Solution: y = f ( x + at ) + g ( x − at ) --------------(1) Differentiating (1) w.r. to x partially we get ∂y = f ′( x + at ) + g ′( x − at ) ∂x ∂2 y ∴ 2 = f ′′( x + at ) + g ′′( x − at ) ∂x Differentiating (1) w.r. to t partially we get

19

∂y = af ′( x + at ) − ag ′( x − at ) ∂t ∂2 y ∂2 y ∴ 2 = a 2 f ′′( x + at ) + a 2 g ′′( x − at ) = a 2 [ f ′′( x + at ) + g ′′( x − at )] = a 2 ∂t ∂x 2 Example: If u = e xyz , find the value of

∂3 u ∂ x∂ y∂ z

Solution: u = e xyz ∂u ∴ = e xyz ( xy ) ∂z ∂2 u or , = e xyz ( x ) + e xyz ( xz )( xy ) = e xyz ( x + x 2 yz ) ∂y∂z or ,

∂3 u = e xyz 1 + 2 xyz ) + e xyz ( yz ) .( x + x 2 yz ) ∂x∂y∂z

(

)

[

= e xyz 1 + 3 xyz + x 2 y 2 z 2

]

Extremum problem (Maxima and Minima): One application of the derivative is to determine the maximum and minimum of a differentiable function. We shall give examples of this application. Local maximum or local minimum is generally known as local extremum . Look at the graph of f : [ a, b ] → ¡ in the figure and note that (i) (ii) (iii)

f ( c1 ) is a local maximum,

f ( c2 ) is a local minimum,

f ( a ) is a minimum on [ a, b ]

f ( c3 ) is a local maximum on [ ,b ] and it is also maximum on [ a, b ] , (iv) (v) f(a) is not a local minimum on [a,b] f (c1 ) is a maximum on [α, β] (vi) f (c 2 ) is a minimum on [α, β] (vii) Theorem: If f(c) is a local extremum and f if differentiable at c, then f ′(c) = 0.

y .

f

a

α

c1

c2

β

Turning points: (Stationary points)

20

c3

b

x

P is a maximum

Q is a minimum

dy = 0; dx 2 2 d y d y Max if : Min if : < 0 >0 2 2 dx dx 2 d y Inflexion point if: =0 dx 2 At a turning point:

Definition:

A number x0 is called a critical number of f if f ' ( x0 ) = 0 or f is not

differentiable at x = x0 .

21

Determination of Maxima and Minima: 3 2 Example: Find the extremum of f : [ 0, 6] → ¡ , f ( x ) = x − 2 x + 7 .

Solution: Since f is continuous on [ 0, 6] , the Extremum Theorem tells us that f has a maximum and a minimum. f ′( x) = 0

∴3 x 2 − 4 x = 0 ⇒ x (3 x − 4) = 0

Or, x = 0 or 4/3 So, the critical numbers of f are 0,

4 and 6 (observe that 0 and 6 are critical numbers of f 3

because f is not differentiable at 0 or 6).

f ( 0) = 7  4  31 f  ÷= 3 9 f ( 6 ) = 151 Thus, 151 is the maximum and

31 is the minimum. 9

Let us note the following result which is useful in determining whether a local extremum occurs at a critical point. Example:

Find the extremum of f ( x ) = sin x + cos x, x ∈ R

Solution:

Since sin and cos are periodic functions with period 2π , it is enough to look at the

problem for x ∈ [ 0, 2π ] .

f ' ( x ) = cos x − sin x and to find the critical number we let f ' ( x ) = 0 . Then we get tan x = 1 or x =

π 5 , π 4 4

π 5 , π and 2π . ( 0 and 2π are critical numbers because f is not 4 4 differentiable at 0 and 2π when we restrict the domain to [ 0, 2π ] ). f is continuous on Critical numbers of f are 0,

[ 0, 2π ] , thus

f has a maximum and a myinimum. They occur at the critical numbers.

22

π  5  π  ÷ = 2, f  π ÷ = − 2 and f ( 2π ) = 1 , we see that f  ÷ = 2 4 4  4 5  is the maximum and f  π ÷ = − 2 is the minimum. 4  Comparing f ( 0 ) = 1 , f 

Second derivative test: If c is a point the interval in which the function f (x) is defined and if f ′(c) = 0 and f ′′(c) ≠ 0 the f(c) is a maximum if f ′′(c) is negative and a minimum if f ′′(c) is positive. Example: Find for what values of x, the following expression is maximum and minimum respectively: 2 x 3 − 21x 2 + 36 x − 20 , find also the maximum and minimum values of the expression. Solution: Let f ( x) = 2 x 3 − 21x 2 + 36 x − 20 ∴ f ′( x) = 6 x 2 − 42 x + 36

Which exists for all values of x. Now, when f(x) is a maximum or a minimum, f ′( x ) = 0 ∴6 x 2 − 42 x + 36 = 0 ∴6( x −1)( x − 6 ) = 0 ∴x = 1,6

Now, when x =1, f ′′( x) = 12 x − 42 = −30 , which is negative When x = 6, f ′′( x) = 12 x − 42 = 30 , which is positive Hence the given expression is maximum for x = 1, and minimum for x = 6. The maximum and minimum values of the given expression are respectively f(1), i.e. -3 and f(6), i.e. -128. Example: Examine whether

1 x

x possesses a maximum or a minimum and determine the

same. Let 1 x

1 log x x 1 dy 1 1 1 ∴ . = 2 − 2 log x = 2 (1 − log x) − − − −(1) y dx x x x dy = 0, ⇒ 1 − log x,∴ x = e dx Differentiating (1) w. r . to x,  1 x 2 . −  − (1 − log x ) 2 x 2 2 1  dy  1 d y − 3 + 2 log x  x − 2  + . 2 = = 4 y dx y  dx  x x3 y = x , log y =

1

When x = e,

d2y −3+ 2 , which is negative = ee . 2 dx e3

Therefore, for x = e, the function is a maximum, and the maximum value is

23

1

ee .

Example: A liquid form of penicillin manufactured by a pharmaceutical firm is sold in bulk at a price of RM 200 per unit. If the total production cost for x units is C ( x) = 500,000 + 80 x + 0.003 x 2 And if the production capacity of the firm is at most 30,000 units in a specified time, how many units of penicillin must be manufactured and sold in that time to maximize the profit? Solution: Since the total revenue for selling x units is R ( x) = 200 x , the profit P(x) on x units will be P ( x ) = R ( x ) − C ( x ) = 200 x − (500,000 + 80 x + 0.003 x 2 ) ---(1) Since the production capacity is at most 30,000 units, x must lie in the interval [0,30,000]. dP = 200 − (80 + 0.006 x) = 120 − 0.006 x From (1) dx dP = 0, ⇒120 − 0.006 x = 0 Setting dx

∴x = 20,000

Since the critical number lies in the interval [0,30,000], the maximum profit must occur at one of the values x = 0, x = 20,000 or, x = 30,000. x = 0, ⇒ P ( x ) = −500,00

x = 20000, ⇒ P ( x ) = 700,000 x = 30000, ⇒ P ( x ) = 400,000

Therefore the maximum profit P= 700,000 occurs when x = 20,000 units manufactured and sold in the spefied times. Example: An object is hurled upward from the roof of a building 10 m high. It rises and then falls back; its height above ground t s after it is thrown is y = −4.9t 2 + 8t +10 m, until it strikes the ground. What is the maximum height above the ground that the object attains? With what speed does the object strike the ground? Solution: The vertical velocity at time t during flight is dy = −2(4.9)t + 8 = −9.8t + 8 m/s. dt The object is rising when v>0, that is, when 0 < t < 8 / 9.8, and is falling for t > 8/9.8. Thus, the object is at its maximum height at time t = 8/9.8 s, and this maximum height is v (t ) =

2

 8   8  y max = −4.9  +8  +10 ≈13.27 m.  9.8   9.8 

The time t at which the object strikes the ground is the positive root of the quadratic equation obtained by setting y = 0, Namely, t=

− 8 − 64 +196 ≈ 2.462 s. − 9.8

The velocity at this time is v = -(9.8)(2.462)+8 a speed of about 16.12 m/s.

≈ =16.12. Thus, the object strikes the ground with

24

Example: A boat sails 30miles to the east from a point P , then it changes direction and sails to the south. If this boat is sailing at a constant speed of 10miles/hr, at what rate is its distance from the point P increasing 2hours after it leaves the point P 7 hours after it leaves the point P ?

(i) (ii) Solution: (i) (ii)

Since the constant speed of the boat is 10miles/hr, so 2hrs after it leaves the point P , it has traveled 20 miles and it is still sailing east. Thus the rate of its distance from the point P is increasing at 10miles/hr. 7hrs after it leaves P , it has sailed east 30 miles in 3hrs and south 40 miles in 4hrs. Let its distance from P at time t after it starts sailing be s , where t ≥ 3 hrs, and a be the distance traveled along the south direction.

Then So

s 2 = 302 + a 2 30miles

ds da 2 s = 2a . dt dt When a = 40.s = 50. Then ds a da = dt s dt 40 = ×10 50 = 8miles/hr.

P a s

Thus the rate of its distance from the point P is increasing at 8miles/hr. Example: A d.c supply has e.m.f E = 12 V and internal resistance r = 1Ω . Prove using calculus that the power transferred to a load resistor R is a maximum when R = r = 1Ω. E2R P = [Hints: The power, P, in the load is given by ] ( R + r) 2 Solution: The power, P, in the load is given by P = Substituting E = 12 and r = 1 we find P =

E2R ( R + r) 2

144 R

( R + 1) 2

dP 144( R +1) −144( 2 ) R ( R +1) 144(1 − R ) = = dR ( R +1) 4 ( R +1) 3 For maximum power transfer to the load we require dP 144(1 − R ) ∴ = 0, ⇒ = 0, ⇒ R = 1 dR ( R + 1) 3 Now, 2

∴

25

(2)

(2)

d 2 P ( R + 1) (− 144) − 144(1 − R).3( R + 1) 288(2 − R) ∴ 2= = − dR ( R + 1) 6 ( R + 1) 4 3

R = 1, ⇒

2

d P − 288 = ⟨0 2 16 dR

(4)

2

Which show that he turning point, 1, is a maximum. Hence maximum power is transferred to the load when R = r. (2)

Integration Definition: If F ' ( x ) = f ( x ) , ∀x ∈ ( a, b ) , then f is the derivative of F and F is called

an antiderivative (integral or indefinite integral) of f on ( a, b ) . Again, If F (x) is an integral of f(x), and x = a and x = b be two given values of x, the quantity F(b) – F(a) is defined as the definite integral of f(x), denoted by the symbol b

∫ a

f ( x ) dx .

y = f (x )

a b Constant of integration: It may be noted that d d F ( x ) = f ( x), then we also have { F ( x) + c} = f ( x), where C is an arbitrary dx dx constant. Thus, a general value of the indefinite integral ∫ f ( x) dx = F ( x) +C .

Integration is differentiation in reverse:

26

dy = 2x dx 2x reverses this and we get back x 2 + c

differentiating. integrating

y = x 2 gives

2 2xdx = x +c ∫

c can be any number – c is an arbitrary constant

∫ 2xdx

is called an Indefinite Integral.

Example:

Find

∫ cos xdx.

Solution: Let F ( x ) = sin x . Then F ' ( x ) = cos x . Thus F ( x ) = sin x is an antiderivative of cos x . Therefore

∫cos xdx = {sin x + c c ∈R}

(You will see later that

e .)

1

∫ x dx = ln x + k , where ln x

is the logarithmic function with base

Example: Determine: ∫cos(3 x +7) dx Solution: Let 3x+7=u du 1 = 3, ⇒ dx = du Then dx 3 du 1 1 = sin u + c = sin(3x + 7) + c Therefore ∫ cos(3 x + 7) dx = ∫ cos u 3 3 3

∫ sin

r

x cos xdx where r is a rational number and r ≠ −1 .

Example:

Find

Solution:

Let u = sin x , then

∫ sin

r

du = cos x and du = cos xdx . dx

x cos xdx = ∫ u r du u r +1 +C r +1 sin r +1 x = + C, C ∈ ¡ r +1 =

27

Fundamental Integrals: x n +1 1 e mx + c, (ii ) ∫ dx = log x + c, (iii ) ∫ e mx dx = + c, (iv) ∫ e x dx = e x + c n +1 x m − cos mx sin mx (v) ∫ sin mxdx = + c, ∫ sin xdx = −cos x + c, (vi ) ∫ cos mxdx = + c, ∫ cos xdx = sin x + c m m tan mx (vii) ∫ sec 2 mxdx = + c, ∫ sec 2 xdx = tan x + c,(viii ) ∫ sec x tan xdax = sec x + c m (i ) ∫ x n dx =

(ix) ∫ sec 2 xdx = tan x + c, ( x) ∫ cos ec 2 xdx = − cot x + c,

Techniques and applications of Integration: Example:

Find

∫ tan

n

x sec 2 xdx, n ∈N

Let u = tan x , then du = sec 2 xdx. u n +1 tan n +1 x n 2 n tan x sec xdx = u dx = + c = + c, c ∈ R ∫ ∫ n +1 n +1

Solution:

Example: Find

∫ sin ax dx, a ∈ ¡ , a ≠ 0.

Solution: Let u = ax , then du = a dx and

∫ sin ax dx = ∫ sin u

du a

1 sin udu a∫ 1 = ( − cos u ) + C a 1 = − cos ax + C , C ∈ ¡ a =

Example: Determine Solution: Hence =4x+

3

∫ (4 + 7 x − 6 x

∫ (4 +

3

∫ (4 + 7 x − 6 x 2

2

) dx

) dx can be written as

3

∫ 4dx + ∫ 7 xdx − ∫ 6 x

2

dx

3 3 x 1+1 x 2+1 x − 6 x 2 )dx =4x+ . −6 +c 7 7 1 +1 2 +1

3 2 x − 2x 3 + c 14

2 x 3 − 3x 2 ∫ 4 x dx, (b) 1 − t dt Solution: Rearranging into standard integral form gives: 2 x 3 − 3x 2x3 3x 1 2 3 1 3 3 dx = ∫ 4x ∫ 4 x dx − ∫ 4 x dx = 2 ∫ x dx − 4 ∫ dx = 6 x − 4 x + c

Example: Determine (a)

(

28

)

Example: Find the indefinite integrals:

(a) ∫

x sin(3 ln x) dx, (b) ∫ dx, (c) ∫ e x 1 + e x dx x x +1 2

Solution: (a)

∫x =

2

x 1 dx , Let u = x 2 +1, du = 2 xdx, xdx = du 2 +1

(

)

1 du 1 1 = ln u + c = ln x 2 + 1 + c = ln x 2 + 1 + c ∫ 2 u 2 2

(b)

3 sin(3 ln x ) dx , Let u = 3 ln x, du = dx x x 1 1 1 = ∫ sin udu = − cos u + c = − cos(3 ln x) + c 3 3 3

∫

(c)

∫e

1 +e x dx , Let u =1 + e x , du = e x dx

x

1 2

3

(

2 2 = ∫ u du = u 2 + c = 1 + e x 3 3 Example: Find Solution:

)

3 2

+c

∫ tan x dx.

sin x

∫ tan x dx = ∫ cos x dx 1 = − ∫ du u = − ln u + C

( Let u = cos x, then du = − sin x dx )

= − ln cos x + C = ln cos x

−1

+C

= ln sec x + C , C is an arbitrary constant

Example: Find Solution:

∫x e

x2

dx.

Let u = x 2 , then du = 2 x dx .

29

∫ xe

x2

dx =

1 u e du 2∫

1 u e +C 2 1 2 = e x + C , C is an arbitrary constant 2 =

Example: Find

sin x

∫ 2 + cos x dx .

Solution: Let u = 2 + cos x, du = − sin x dx .

sin x

1

∫ 2 + cos x dx = −∫ u du = − ln u + C , C ∈ ¡ = − ln 2 + cos x + C Example: Find

∫

sin x dx. x

Solution: Let u =

∫

x , du =

1 2 x

dx

sin x dx = ∫ sin u 2 du x = −2 cos u + C , C ∈ ¡ = −2 cos x + C


x ∫ x2 dx. 2

Let u = x 2 , then du = 2 x dx.

x u ∫ x2 dx = ∫ 2 2

1 du 2

1 u 2 du 2∫ 1 2u = + C, C ∈ ¡ 2 ln 2 =

2

2x = +C 2 ln 2 Integration by parts: ( fg )′ = f

′ + fg ′, ⇒fg =∫ f g ′ +∫ fg ′,∴ ′ = fg −∫ fg ′ g ∫f g

Sometimes it is written as ∫udv =uv −∫vdu This is known as the formula of integration by parts.

30


∫ ln x dx.

∫ ln x dx = ∫ 1.ln x dx = ∫ f ' ( x ) g ( x ) dx, = f ( x ) g ( x ) − ∫ f ( x ) g ' ( x ) dx, = ( x + C ) ln x − ∫ ( x + C )

1 dx, C ∈ ¡ x

 C = ( x + C ) ln x − ∫ 1 + ÷dx x  = ( x + C ) ln x − x − C ln x + K , K ∈ ¡ = x ln x − x + K Example: Find

∫ sin

−1

xdx.

−1 Solution: We let f ' ( x ) = 1 and g ( x ) = sin x , then f ( x ) = x and g ' ( x ) =

∫ sin xdx = ∫ 1.sin xdx = ∫ f ' ( x ) g ( x ) dx = f ( x) g ( x) − ∫ f ( x) g '( x) −1

−1

= x sin −1 x − ∫ To find

∫

∫

x 1 − x2

=−

x 1 − x2

x 1 − x2

dx

dx , let u = 1 − x 2 , then du = −2 xdx.

dx = − ∫

1 du u 2

1 − 12 u du 2∫ 1

1 u2 =− +C 2 1 2 = − 1 − x2 + C, C ∈ ¡ Putting this into (1), we get

31

1 1 − x2

.

∫ sin

−1

xdx = x sin −1 x + 1 − x 2 + C .

Example:

Find

Solution:

∫e

x

∫e

x

sin xdx.

sin xdx = ∫ f ' ( x ) g ( x ) dx

= f ( x ) g ( x ) − ∫ f ( x ) g ' ( x ) dx

f ' ( x ) = e x , g ( x ) = sin x f ( x ) = e x , g ' ( x ) = cos x

= e x sin x − ∫ e x cos xdx

∫e

x

s cos xdx = ∫ f ' ( x ) g ( x ) dx

= f ( x ) g ( x ) − ∫ f ( x ) g ' ( x ) dx

f ' ( x ) = e x , g ( x ) = cos x

= e x cos x − ∫ e x ( − sin x ) dx

f ( x ) = e x , g ' ( x ) = sin x

= e x cos x + ∫ e x sin xdx

Put this into (1), then

∫e

x

(

sin xdx = e x sin x − e x cos x + ∫ e x sin xdx

)

= e x sin x − e x cos x − ∫ e x sin xdx

∫

x x x So 2 e sin xdx = e sin x − e cos x, thus

∫e

x

sin xdx =

1 x ( e sin x − e x cos x ) + C , C ∈ ¡ 2

Example: Evaluate: Solution:

π

∫

2 0

2θ sin θdθ

∫ 2θ sin θdθ =

2θ( −cos θ) − ∫ (−cos θ )2dθ = −2θ cos θ + 2 sin θ + c

π

π

∴ ∫ 2 2θ sin θdθ = [ − 2θ cosθ + 2 sin θ ] 02 0

Example: Solution:

∫

1 x −1 2

∫

=2

1

dx . x2 −1 Let x = sec u , then dx = sec u tan udu and Find

dx = ∫

1 sec 2 u − 1

sec u tan udu

= ∫ sec udu = ln sec u + tan u + C , C ∈ ¡

x 32

x2 −1

Refer to the figure, since secu = x ,

tan u = x 2 − 1 . 1

∫

Thus

x −1 2

1

dx = ln x + x 2 − 1 + C. 1

Example: Find

∫

Solution:

Let x = tan u , then dx = sec 2 udu .

∫

1 1+ x

dx.

1 + x2

1

dx = ∫

du 1 + tan 2 u 1 =∫ sec 2 udu sec u

2

1 + x2

x

= ∫ sec udu

u

= ln sec u + tan u + C , C ∈ ¡

1

Since tan u = x , we have sec u = 1 + x 2 ,

1

∫

thus

1+ x

2

dx = ln 1 + x 2 + x + C.

∫

1

dx.

Example:

Find

Solution:

x 2 + 2 x + 2 = ( x + 1) + 1. Let x + 1 = tan u , then dx = sec 2 udu .

∫

2

1 x2 + 2x + 2

=∫

x2 + 2x + 2

1 tan u + 1 2

dx = ∫

1

( x + 1) + 1

dx

sec 2 udu

= ∫ sec udu = ln sec u + tan u + C , C ∈ ¡ = ln

x2 + 2x + 2 + x + 1 + C

Partial fraction: Find

∫x

2

x dx + 2x − 3

33

Solution:

x x = , writing it in the partial fraction, we have x + 2 x − 3 ( x + 3)( x − 1) 2

x 3 1 1 1 = . + . x + 2x − 3 4 x + 3 4 x −1 2

x 1 1  3 1 1 1 3 1 dx = ∫  . + . dx + ∫ dx dx = ∫ 4 x +3 4 x −1 x + 2x − 3  4 x + 3 4 x −1  3 1 = ln x + 3 + ln x −1 + c, c ∈ R 4 4

∴∫

2

Example: Find

1

∫ ( x − 1) ( x + 2) dx.

Solution: First we shall express

1

( x − 1) ( x + 2 )

as a partial fraction.

1 1  1 1  =  − ( x − 1) ( x + 2 ) 3  x − 1 ÷ 3  x + 2 ÷ 1 1 1 1 1 Then ∫ ( x + 1) ( x + 2 ) dx = 3 ∫ x − 1 dx − 3 ∫ x + 2dx 1 To find ∫ x − 1 dx , we let u = x − 1 , then du = dx. 1 1 ∫ x − 1 dx = ∫ u du = ln u + C1 , C1 ∈ ¡ 1

= ln x − 1 + C1 Similarly, ∴∫

1

∫ x + 2 dx = ln x + 2 + C

2

,C 2 ∈ R

1

1 1 dx = ln x −1 − ln x + 2 + C , C ∈ R ( x −1)( x + 2) 3 3

THEOREM (Fundamental Theorem of Calculus) Suppose (i) f is integrable on [ a, b ]

(ii) F is continuous on [ a, b ] and F ' = f on ( a, b ) Then

∫

b

a

f = F ( b) − F ( a)

34

Definite integral: Definite integrals are those in which limits are applied. a

Example: Evaluate (a)

2 ∫x dx and (b) 0

2

∫( x

2

)

− 3x + 2 dx

−1

Solution: (a) a

2 ∫ x dx = 0

1 3 x 3

a

= 0

a3 3

(b) 2

∫(x

2

2

)

1 3 3 2 1 3 3 1  9 x − x + 2 x = (8) − (4) + 4 −  ( −1) − (1) + (−2)  = 3 2 3 2 2 3  2 −1

− 3 x + 2 dx =

−1

3

Example: Evaluate

∫ (4 − x )dx 2

−2

 x  ∫ (4 − x )dx = 4 x − 3  3

Solution:

3

2



−2

Example: Evaluate: Solution:

3

 −2

= 4.3 −

( − 2) 3 33  − 4 .( − 2 ) − 3  3 

 25 =  3 

π

∫

2 0

3 sin 2 xdx π

π

  1  2  3  π  2 3 sin 2 xdx = 3 − cos 2 x   = − cos 2  − cos 2(0)   ∫0  0 2   2   2 3 3 = − (cos π − cos 0) = − (−1 −1) = 3 2 2

Example: Find

1

∫ sin x dx. 0

Solution: We know that F ( x ) = − cos x is an antiderivative of f ( x ) = sin x. Now F is continuous on [ 0,1] and F ' = f on ( 0,1) ,

f is continuous on [ 0,1] and f is integrable on [ 0,1] .

By the Fundamental Theorem of Calculus,

∫

1

0

1

sin x dx = ∫ f 0

= F |10

= F ( 1) − F ( 0 )

Remark

∫

b

a

= − cos1 + cos 0 = 1 − cos1.

f is often written as

b

∫ f ( x ) dx to emphasize on the variable x . a

35

Example: Find

1

3

∫1 ( 1 + x ) 2 dx.

f ( x ) = ( 1 + x ) 2 is continuous on [ 1,3] , so it is integrable on [ 1,3] . We need to find one antiderivative of f . 1

Solution:

1

1

∫ ( 1 + x ) 2 dx = ∫ u 2 du

( let u = 1 + x, then du = dx )

2 1 = u2 + c 3 1 2 = ( 1 + x ) 2 + c, c is a constant 3 We need only one antiderivative of f , we choose

F ( x) =

3 2 ( 1+ x) 2 3

Thus

∫

3

1

1

( 1 + x ) 2 dx = F ( x ) |13 = F ( 3) − F ( 1) 3 3 2 2 ( 1 + 3) 2 − ( 1 + 1) 2 3 3 2 = 8−2 2 . 3

=

(

)

Example: Evaluate:

2

∫

0

3 xdx 2 x 2 +1

Solution: Let u = 2 x 2 + 1 du du = 4 x, dx = dx 4x whenx = 2, u = 9, x = 0, u =1

∴

Thus

Example: Solution:

2

∫

3 xdx 2 x +1

0

2

π

∫

2 0

=

3 9 du 3 9 = u 4 ∫1 u 4 ∫1

1 − 2

du =

cos 2 xdx

36

1 2

3u 4 1 2

9

= 1

3 2

[

]

9 − 1 =3

π

π

π

π

1 1 x 1 2 2 ∫02 cos xdx = 2 ∫02 2 cos xdx = 2 ∫02 (1 + cos 2 x )dx =  2 + 4 sin 2 x0 π 1 π = + sin π = 4 4 4 −1 1 sin x dx Example: ∫0 1−x2 1 −1 dx Solution: Putting sin x = θ ,∴dθ = 1− x2 2

When x =0, θ = 0; x = 1, θ = 1

∴I = ∫

Example:

π

π

θ 2  2 π2 dx = ∫ 2 θdθ =   = 0 8 1− x2  2 0

sin −1 x

0

π 2

a

∫

0

a 2 − x 2 dx

Solution:Put x = a sin θ, dx = a cos θdθ When x =0, θ = 0; x = a, θ = π

π 2

π

π

1 a2  sin 2θ  2 πa 2 ∴I = ∫ 2 a cos θdθ = a . ∫ 2 (1 + cos 2θ ) dθ = θ+ =   0 2 0 2  2 0 4 2

2

2

General properties of definite Integral: b

b

a

c

a

i.∫ f ( x ) dx = −∫ f ( x ) dx, ii.∫ f ( x ) dx = ∫ f ( x ) dx + ∫ f ( x ) dx, a < c < b b

a

a

a

c

a

a

na

a

0

0

0

0

iii.∫ f ( x ) dx = ∫ f ( a − x ) dx, iv. ∫ f ( x ) dx = n ∫ f ( x ) dx, if , f ( x ) = f ( a + x ) a

v.∫

−a

a

a

f ( x ) dx = 0, if f(x) is odd function, vi.∫ f ( x ) dx = 2 ∫ f ( x ) dx , if f(x) is even −a

0

function Cor: A function f(x) is said to be an odd function of x, if f(-x)= - f(x). e.g x, sinx A function f(x) is said to be an even function of x, if f(-x)= f(x). e.g f ( x ) = x 2 , cos x

Example: Show that Solution: π

π 2 0

∫

log tan xdx = 0 π

π

π

π  I = ∫ 2 log tan xdx = ∫ 2 log tan  − x dx = ∫ 2 log cot xdx = −∫ 2 log tan xdx = −I 0 0 0 0 2  ∴ 2 I = 0,∴ I = 0

37

Example: Show that

π 2 0

∫

sin x sin x + cos x

dx =

π 4

Solution: π 2 0

I =∫

sin x sin x + cos x π

∴ 2I = ∫ 2

sin x

π

0

dx =∫

π  sin  − x  2  π  π  sin  − x  + cos − x  2 2     π

dx +∫ 2

π

dx =∫ 2 0

cos x cos x + sin x

dx

cos x

dx 0 sin x + cos x cos x + sin x π π sin x + cos x π π dx =∫ 2 dx = [ x ] 02 = ,∴ I = 0 2 4 sin x + cos x 0

= ∫2

π 2 0

Example: A car is travelling at 72 km/h. At a certain instant its brakes are applied to produce a constant deceleration of 0.8 m/s2. How far does the car travel before coming to a stop? Solution: Let s(t) be the distance the car travels in the t seconds after the breaks are applied. Then s ′′(t ) = −0.8( m / s 2 ), so the velocity at time t is given by s ′(t ) = ∫−0.8dt = −0.8t +c1 m/s.

Since s ′(0) = 72 km/h = 20 m/s, we have c1 = 20 . Thus,

s ′(t ) = 20 − 0.8t

38

And

s (t ) = ∫( 20 −0.8t ) dt = 20t −0.4t 2 +c 2 .

Since s(0) = 0, we have c 2 = 0 and s (t ) = 20t −0.4t 2 . When the car has stopped, its velocity will be 0. Hence, the stopping time is the solution t of the equation

0 = s ′(t ) = 20 − 0.8t

That is, t = 25 s. The distance travelled during deceleration is s(25) = 250 m. Applications of integration: Let us look at a few applications of integration. We shall see how the integral can be used to find the length of a curve, the area of a surface of revolution and the volume of a solid of a revolution. Length of a curve: The length of the curve y = f ( x ) , x ∈ [ a, b ] is defined as lim

P →0

n

∑T k =1

Tk −1

T1

Tn

Tk

y = f ( x)

T .

k −1 k

T0 x a = x0

a = x1

xk −1

b = xn

xk

The length lk of the line segment ing Tk −1 and Tk is given by

( xk − xk −1 )

lk =

2

+ ( f ( xk ) − f ( xk −1 ) ) . 2

If f is a function with continuous first order derivative on the interval [ a, b ] , then the length of the curve y = f ( x ) , x ∈ [ a, b ] is defined by the integral

∫

b

a

1 + ( f ' ( x ) ) dx. 2

(

3

)

Example: Find the length of the curve y = x 2 from (1,1) to 2,2 2 . 1

dy 3 2 = x Solution: dx 2 The required arc length 2

L = ∫ 1+

9 9 9 x dx , Let u = 1 + x, du = dx 4 4 4

x =1⇒ u =

13 22 ,x = 2⇒u = 4 4

1

39

22 / 4

3 22 / 4

1

4 8 2 L= u 2 du = u 9 13∫/ 4 27

13 / 4

3

3

8  22 2 13 2 = [  −  ] ≈ 2.09 27  4  4  3

Example:

Find the length of the curve y = ( x + 1) 2 from x = 0 to x = 4.

Solution:

Let f ( x ) = ( x + 1) 2 , then

f '( x) =

3

1 2 3 9 ( x + 1) 2 and 1 + ( f ' ( x ) ) = 1 + ( x + 1) . 2 4

=∫

1 + ( f ' ( x ) ) 2dx

4

0

1 9 x + 13dx 0 2 49 1 du =∫ u ( let u = 9 x + 13) 13 2 9 The length of the curve =∫

4

49

1 3 = u2 27 13 =

3  1  2 343 − 13  ÷ 27  

Example: Determine the length of an arc of the cycloid x = a (θ + sin θ ), y = a (1 − cos θ ) , measured from the vertex (i.e., the origin) 2

Solution: Here

2

ds θ  dx   dy  2 2 =   +  = a (1 + cos θ ) + sin θ = 2a cos dθ d θ d θ 2    

Also at the origin θ = 0 . Hence the required length, from θ = 0 to any point θ is

θ θ dθ = 4a sin . 2 2 Areas of a plane curves: θ

s = ∫ 2a cos 0

The definite integral

b

∫ a

b

f ( x ) dx, i.e, ∫ ydx represents the area bounded by the curves y = a

f(x), the x-axis and the two fixed ordinates x = a and x =b. Example: Find the area of the quadrant of the ellipse

x2 y2 + = 1 between the major and minor axes. a2 b2 Solution: The required area is a

a

0

0

= ∫ ydx = ∫ π

y=

b x2 y2 a 2 − x 2 dx, 2 + 2 = 1 , for the curve a a b

b 2 a cos θ .a cos θdθ , putting x = a sin θ a ∫0

40

π

π

2 = ab ∫ 2 (1 + cos 2θ ) dθ = ab θ + sin 2θ  = 1 πab 0 2 2  2 0 4

Example: Find the area A of the plane region lying above the x-axis and under the curve y = 3x − x 2 .

Solution: We need to find the points where the curve y = 3 x − x 2 meets the x-axis. These are solutions of the equation 0 = 3 x − x 2 = x (3 − x ). The only roots are x = 0 and 3. Hence, the area of the region is given by 3

A=∫

(

0

)

3 1 3 x − x dx = x 2 − x 3 2 3 2

3

= 0

27 27 9 − − (0 − 0) = square units. 2 3 2

Example: Find the area under the curve y =sin x, above y = 0 from x = 0 to Solution: The required area is π

x = π.

π

A = ∫ sin xdx = − cos x 0 = −( −1 −1) = 2 square units. 0

**Area between two curves If f and g are continuous functions on the interval [a,b] and if f ( x ) ≥ g ( x) for all x in [a,b], then the area of the region bounded above by y = f(x), below by y = g(x), on the left by x = a and on the right by the line x = b is b

A = ∫[ f ( x ) − g ( x)]dx a

Example: Find the area of the bounded, plane region R lying between the curves y = x 2 − 2 x and y = 4 − x 2 .

Solution: First, we must find the intersections of the curves, so we solve the equations simultaneously: y = x 2 −2x = y = 4 − x 2 2x 2 −2x −4 = 0 2( x − 2)( x +1) = 0

So x = 2 or x = -1 Since 4 − x 2 ≥ x 2 − 2 x,−1 ≤ x ≤ 2, The area A of R is given by

41

2

(

) (

2

)

A = ∫[ 4 − x 2 − x 2 − 2 x ]dx = ∫ (4 − 2 x 2 + 2 x ) dx = 4 x − −1

−1

2 3 x + x2 3

2

−1

2 2 = 4(2) − (8) + 4 − (−4 + +1) 3 3

= 9 square units. Example: Find the area of the region bounded above by y = x+6, bounded below by y = x 2 , and bounded on the sides by the lines x = 0 and x = 2. Solution: The required area 2

 x2 x3  34 A = ∫[( x + 6) − x ]dx =  + 6 x −  = 3 0 3 2 0 2

2

Example . Find the area of the region bounded by the curves y = x 2 and y = x 3 .

2 Solution: The two curves y = x 2 and y = x 3 meet when x 2 = x3 , i.e. x ( 1 − x ) = 0 which

gives x = 0 or x = 1 . Note that on the interval [ 0,1] , the graph of y = x 2 lies above that of

y = x 3 . Thus the area of the region between y = x 2 and y = x 3 is

1

1

 x3 x 4  x − x dx = ( ) 3 − 4 ∫0  0 1 = 12 1

2

y

3

y = x2 y = x3 x 1

Surface of Revolution: If f is a continuously differentiable non-negative function on [ a, b ] ,then the area of a surface generated by revolving the curve y = f ( x ) , x ∈ [ a, b ] , about the x -axis is

42

∫

b

a

2π f ( x ) 1 + ( f ' ( x ) ) dx 2

Suppose that a parametric curve in the upper half of the xy -plane defined by a pair of continuously differentiable functions

x = x ( t ) , y = y ( t ) , t ∈ [ a, b ]

∫

b

a

( x '( t ) ) + ( y '( t ) )

2π y ( t )

2

2

dt.

Example:

Find the surface area of a sphere with radius r.

Solution:

The surface of a sphere can be generated by revolving a semicircle y = r 2 − x 2

, x ∈ [ − r , r ] about the x -axis.

Let f ( x ) = r 2 − x 2 , x ∈ [ − r , r ] . Then f ' ( x ) =

−x r − x2 2

.

The surface area of a sphere = ∫ 2π f ( x ) 1 + ( f ' ( x ) ) dx r

2

−r

2

r

= ∫ 2π r − x 2

2

−r

 −x  1+  ÷ dx 2 2 r − x  

r

= ∫ 2π rdx −r

= 4π r 2

Volume of Revolution: The Disc Method: The volume of the solid of revolution generated by revolving about the x -axis the region bounded by the curve y = f ( x ) , x ∈ [ a, b ] , where f is continuous, and the x -axis from x = a to x = b is given by

∫ π ( f ( x) ) b

2

a

dx .

Example: Find the volume of a sold ball having radius a. Solution: The ball can be generated by rotating the half-disk, 0 ≤ y ≤ a 2 − x 2 ,−a ≤ x ≤ a about the x-axis. Therefore its volume is a

V =π ∫

−a

(

a2 − x2

) dx = 2π ∫ (a 2

a

0

2

)

1 − x 2 dx =2π a 2 x − x 3 3

a

0

4 = πa 3 cubic units. 3

Example: Find the volume of the solid of revolution when the region which is bounded by the curves y = 2 x and y = x 2 is revolved about (i) The x -axis, (ii) The y -axis

43

Solution: Let us first find the points of intersection between the two given curves. Solving the two equations y = 2 x and y = x 2 , we have

x 2 = 2 x, i.e. x ( x − 2 ) = 0,

So, x = 0 or x = 2 , and the curves meet at ( 0, 0 ) and ( 2, 4 ) .

y

( 2, 4 ) y = 2x

y = x2 T

4

(i)

2

x

The region T bounded by the two curves (refer to the figure above) is revolved about the x -axis. Note that a small vertical slice of the solid which is perpendicular to the axis of revolution is in the form of a washer with cross sectional area 2 2 π ( 2 x ) − ( x 2 )  .  

The volume obtained by revolving T about the x -axis is 2

 4 x3 x5  2 2 2  ∫0 π ( 2 x ) − ( x ) dx = π  3 − 5  0 64π = 15 2

(ii)

The region T is revolved about the y -axis. A small horizontal slice of the solid which is perpendicular to the axis of revolution is in the form of a washer with cross sectional area

 π  

( ) y

2

2

 y − ÷ 2

 . ÷ ÷ 

The volume obtained by revolving T about the y -axis is

 π ∫0   4

( y)

2

2

 y − ÷ 2

4

  y 2 y 3  8π dy = π  −  = ÷ ÷  2 12  0 3  44

The Shell Method: The volume of the solid of revolution is

∫

b

a

2π xf ( x ) dx.

Example:

Find the volume of the solid generated by revolving the region bounded by the

1 curve y = 2 and the x -axis between x = 1 and x = 2 about x (i) The y -axis (ii)

The x -axis

(iii)

The line x = −1

Solution: (i)

The given region is revolved about the y -axis. Subdivide the region into small strips that are parallel to the axis of revolution, then each strip when revolved generates a cylindrical shell. The volume of the solid of revolution about the y -axis is

∫

2

1

2π x

1 2 dx = 2π ln x 1 2 x = 2π ln 2

y

y=

1 x2

x 1

2

(ii)When the given region is revolved about the x -axis, we subdivide the region into small strips that are perpendicular to the axis of revolution. Each strip when revolved generates a disc. The volume of the solid of revolution is 2

−1 7  1  ∫1 π  x 2 ÷dx = π 3x3 1 = 24 π 2

45

(iii)When the given region is revolved about the line x = −1 , a vertical strip at coordinate x is at distance x − ( −1) = x + 1 from the axis of revolution; and when revolved it generates a cylindrical shell. The volume of solid of revolution is

y=

1 x2 x

−1

∫

2

1

1

2

2

1 1 1   2π ( x + 1) 2 dx = 2π ln x −  = 2π  ln 2 + ÷ x x 1 2  

Multiple Integrals: Multiple integral is a natural extension of a definite integral to a function of two variables (double integral) or three variables (triple integral) or more variables. Double and triple integrals are useful in evaluating are, volume, mass, centroid and moments of inertia of plane and solid regions. Double integral over region A may be evaluated by two successive integrations. If A is described as f 1 ( x ) ≤ y ≤ f 2 ( x), [ y1 ≤ y ≤ y 2 , a ≤ x ≤ b ] Then

∫∫

b

y2

a

y1

f ( x, y ) dxdy = ∫[ ∫ f ( x, y )dy ]dx

c 2 : y = y 2 ( x) c1 : y = y1 ( x)

R

a

b

46

b

b

∫ f ( x, y )dx, ∫ f ( x, y )dy

Partial definite integrals:

a

1

Example:

2 2 ∫ xy dx = y ∫ xdx = 0

1

∫ xy 0

0

1

2

a

1

dy = x ∫ y 2 dy = 0

xy 3

3 1

=

0

2

y x 2

2 1

=

0

y2 2

x 3

This two-stage of integration process is called iterated integration. We introduce the following notation: d b   f ( x , y ) dxdy = ∫ f ( x, y ) dx dy ∫∫ ∫ c a c a  b d b d   f ( x, y )dydx = ∫ ∫ f ( x, y )dy dx ∫∫ a c a c  d b

These integrals are called iterated integrals. 3 4

4 3

1 2

2 1

Example: Evaluate ( a ) ∫∫ ( 40 − 2 xy )dydx, (b) ∫∫ ( 40 − 2 xy )dxdy

Solution: (a) 3

4



3

3

1

2



1

1

2 4 ∫∫ ( 40 − 2 xy )dydx = ∫ ∫ ( 40 − 2 xy ) dy dx = ∫ [40 y − xy ]2 dx = ∫ (80 −12 x)dx 3 4

1 2

[

]

2 3 1

= 80 x − 6 x = 112 Similarly,(b) 112 Example: Evaluate the double integral 2 ∫∫ y xdA R

over the rectangle R = { ( x, y ) : −3 ≤ x ≤ 2,0 ≤ y ≤ 1}. Solution: 1 1 2 2 2 1 2 1y x  5 2 5 5 3  2 2 y xdA = y xdxdy = dy = − y dy = − y  ∫∫ ∫ ∫ ∫ 6  = − 6 0 ∫−3 0 2 2  0   −3 R 0 Example: Evaluate

π

2 0

Solution: We have I = =

π 2 0

∫

π

∫ ∫

0

cos( x + y ) dxdy

π 2 0

∫

π

dy ∫ cos( x + y ) dx = ∫ 2 dy[ sin ( x + y ) ] 0 π

0

π

0

π

π

[sin(π + y) − sin(0 + y)]dy = ∫ 2 [ − sin y − sin y ]dy = 2[ cos y ] 02 = −2 0

47

Example: Evaluate

1

x

∫ dx ∫ e 0

Solution:

0

y x

dy

x

1

y  xy  x 1 1 1 x2  1 x dx e dy = dx ∫0 ∫0 ∫0 xe  = ∫0 ( xe − x)dx = ( e −1) ∫0 xdx = ( e −1)  2  = 2 ( e −1)  0 0 xydxdy Example: Evaluate ∫∫ over the region in the positive quadrant for which x + y ≤ 1 . 1

Solution: B(0,1) x+y=1 O

A(1,0)

x

Solution: x+y=1 represents a straight line AB in the figure. The limits for y are 1 – x and 0. 1

1−x

0

0

∫ xdx ∫

Required integral =

1

ydy = ∫

0

1−x

y2  xdx    2 0

=

1 1 2 ∫0

( xdx )(1 − x ) 2

1

1 1 1 x2 x3 x4  1 +  = = ∫0 x − 2 x 2 + x 3 dx =  − 2 2 2 2 3 4  0 24

(

)

Area in polar coordinate: Area = ∫∫rdrdθ 2π

Evaluate, I =∫0 2π

Solution: I = ∫0 =

a2 4

a

θ ∫ rdrd θ a sin

a

2 2π  r  a2 θ =∫   dθ = ∫a sinrdrd θ 0 2  2 a sin θ a

2π

∫

0

cos 2 θdθ =

a2 4

π ∫ (1 + cos 2θ ) dθ 2

0

2π

sin 2θ  πa 2  θ − =  2  0 2

Triple integral: Let V be a given three-dimensional domain in space, bounded by a closed surface S. Let f(x,y,z) be a continuous function in V of the rectangular coordinates x,y,z. Then a triple integral of f over the domain V is defined as lim ∆vi →0 ∑ f ( pi ) ∆vi = ∫∫∫ f ( p ) dv = ∫∫∫ f ( x, y , z ) dxdydz ------------(1) V

V

Volume: Volume of a solid contained in the domain V is given by the triple integral (1) with f(x,y,z) = 1, dxdydz i.e. volume = ∫∫∫ V

Evaluate: Solution:

∫∫∫ ( x + y + z ) dxdydz, R : 0 ≤ x ≤ 1,1 ≤ y ≤ 2,2 ≤ z ≤ 3 R

1

2

1

2

0

2

∫ dx ∫ dy ∫ ( x + y + z )dz = ∫ dx ∫ 0

1

( x + y + z ) 2  dy   2  2 3

3

1

48

1

=

2

[

]

1

2

1 1 2 2 dx dy ( x + y + 3) − ( x + y + 2 ) = ∫ dx ∫ ( 2 x + 2 y + 5).1dy 2 ∫0 ∫1 20 1

1  ( 2 x + 2 y + 5) 2  1 1 1 1 2 2 = ∫ dx   = ∫0 dx ( 2 x + 4 + 5) − ( 2 x + 2 + 5) = ∫0 ( x + 4 ) dx 20  4 1 8 2

[

]

1

 x2  9 =  + 4 x = 2 0 2 Example: Find the total mass of the region in the cube 0 ≤ x ≤ 1,0 ≤ y ≤ 1,0 ≤ z ≤ 1 with density at any point given by xyz.

Solution:

Mass

1

1

1

1

1

0

0

0

0

0

x

2

∫ ∫ ∫ xyzdxdydz = ∫ ∫  2

1

= 1

1

 1 1  zy  1 1 1 z  1 yz  dydz = ∫   dz = ∫0 zdz =   = 0 2  2 0 4 4  2 0 8 0 2

2

Complex numbers For any real number x, x2 ≥ 0 and therefore x2 + 1 ≥ 1>0 for x ∈ R. Thus the equation x2 + 1 = 0 has no solution in R. This equation has a solution in the complex number field. Note that i, -i are the solutions of the equation x2 + 1 = 0. i = −1 . Definition: A number of the form a + ib is called a complex number when a and b are real numbers and i = −1 . We denote the set of all complex numbers by C. Two complex numbers z1 = (a1,b1) and z2 = (a2,b2) are equal if a1=a2 and b1=b2. A pair of complex numbers a + ib and a – ib are said to be conjugate of each other. Fundamental operations with complex numbers: 1. Addition: z1 + z2 = ( a1 + ib1 ) + ( a 2 + ib2 ) = ( a1 + a 2 ) + i ( b1 + b2 ) = (a1+a2, b1+b2)

2. Subtraction: z1 - z2 = ( a1 + ib1 ) − ( a 2 + ib2 ) = ( a1 − a 2 ) + i ( b1 − b2 ) = (a1-a2, b1-b2) 3. Multiplication:z1z2 = a1b2+a2b1) 4. Division:

( a1 + ib1 )( a 2 + ib2 ) = ( a1a2 − b1b2 ) + i( a 2 b1 + a1b2 ) =

a1 + ib1 a1 + ib1 a 2 − ib2 a1 a 2 + b1b2 a b −ab = . = + i 2 21 12 2 2 2 a 2 + ib2 a 2 + ib2 a 2 − ib2 a 2 + b2 a 2 + b2

Graphical representation of a complex number: y

(a,b) x 49

(a1a2 – b1b2,

If z = (a,b), then a and b are the real and imaginary parts of the complex number z. The complex numbers can be represented as points in R2 and we call the x-axis and y-axis to be the real and imaginary axis respectively. The plane itself is called the Argand plane. Also one can think of a complex number as a vector from the origin to the point (a,b). Example: Let z1 = 3 + i and z2 = −1 + 2i . Then their sum z1 + z2 is given by

z1 + z2 = (3 + i ) + (−1 + 2i ) = 2 + 3i And their difference z1 − z2 is given by

z1 − z2 = (3 + i ) − (−1 + 2i ) = 4 − i Their product z1 z2 is given by

z1 z2 = (3 + i )(−1 + 2i ) = −3 + 6i − i + 2i 2 = −3 + 5i − 2 = −5 + 5i , Example: We find the real and imaginary part of 1/ z . Since

( x + iy )( x − iy ) = x 2 + y 2 , We have

1 1 1 x − iy = = . z x + iy x + iy x − iy x − iy x iy = 2 = 2 − 2 2 2 x +y x +y x + y2

And therefore

R

1 x Rz 1 −y Iz = 2 = 2 , I = 2 =− 2 2 2 2 z x +y x +y z x +y x + y2

From these equation, we conclude that

Rz > 0 ⇔ R

1 >0 z

And

Iz > 0 ⇔ I

1 <0 z

Modulus of a complex number (Absolute value)

50

For any complex number z = a + ib, a 2 + b 2 is a nonnegative real number and therefore there is a nonnegative square root for a 2 + b 2 . This nonnegative square root of a 2 + b 2 is the modulus of the complex number. Note also that z z = (a + ib)(a − ib) = a 2 + b 2 . Definition: Let z = a + ib be a complex number. The nonnegative square root of a 2 + b 2 is the modulus (or the absolute value) of the number z and is denoted by z ; thus we have

z := z z = a 2 + b 2 Argument of a complex number (Polar form): Definition: Let z = a + ib be any nonzero complex number. Let r = z . The argument of a nonzero complex number z , denoted by arg z , is the angle θ defined by the equations

y

(a,b) b

θ

x

a a = r cos θ , b = r sin θ The argument of a complex number z is the angle between the real axis and the line segment ing the origin to the point z . We can write any nonzero complex number z in the following polar form

z = r (cos θ + i sin θ )

We write

eiθ = cos θ + i sin θ

And thus z = reiθ = r∠θ R is called the modulus (or magnitude) of z and θ is called the argument (or amplitude) of z. Example: Express the complex number 2 +2 3i in polar form. Solution: Modulus or absolute value, r = 2 +2 3i = 4 +12 =4 ∴2 = 4 cos θ,2 3 = 4 sin θ

π 3 πi π π  Then 2 +2 3i = r ( cos θ + i sin θ ) = 4 cos + i sin  = 4e 3 3 3  −1 3= Amplitude or argument , θ = tan

51

Example: Let z1 = 1 + i . Then

z1 = 12 + 12 = 2 And the argument θ1 is given by

1 = 2 cos θ1 , 1 = 2 sin θ1 Clearly

θ1 = π / 4

Thus

1 + i = 2eiπ / 4 Let z2 = i . The modulus of z2 is given by z2 = 02 + 12 = 1 And the argument is given by

0 = cos θ 2 , 1 = sin θ 2

Therefore,

θ2 = π / 2

And

i = eiπ / 2 Let z3 = −1 . In this case, we have z3 = 1 and the argument θ 3 = π and therefore −1 = eiπ Let z4 = 1 − i . Then z4 = 12 + (−1) 2 = 2 And the argument θ 4 is given by

1 = 2 cos θ 4 ,

− 1 = 2 sin θ 4

Clearly,

θ 1 = −π / 4

Thus

1 − i = 2e − iπ / 4 Example: Convert (a) 4∠ 30  , (b) 7∠−145  into Cartesian form. Solution(a) Using trigonometric ratios, x = 4 cos 30  =3.464, y = 4 sin 30  = 2 Hence 4∠30  = 3.464 +i 2 (b) 7∠−145  lies in the 3rd quadrant. Angle θ = 180  −145 = 35 Hence x =7 cos 35 =5.734, y =7 sin 35  = 4.015 Hence 7∠−145  = −5.734 −i 4.015 Example: Solve the complex equation 2( x + iy ) = 6 − i3 Solution: 2( x + iy ) = 6 − i3 ⇒ 2 x + i 2 y = 6 − i3 , Equating the real and imaginary parts gives:

52

2 x = 6,2 y = −3 3 ∴x = 3, y = − 2

De Moivre’s Theorem: cosθ + i sin θ ) n = cos nθ + i sin nθ **Roots of a complex number:

cos θ + i sin θ = cos( 2mπ + θ ) + i sin ( 2mπ + θ ) , m ∈ I 1

1

( cosθ + i sin θ ) n = [ cos( 2mπ + θ ) + i sin ( 2mπ + θ ) ] n

= cos

2mπ + θ 2mπ + θ + i sin n n

Giving m the values of 0,1,2,…,n-1 successively, we get the required roots. 1

Example: Find the different values of (1 + i ) 3 Solution: 1 + i = r (cosθ + i sin θ ) 1 = r cosθ ,1 = r sin θ , ⇒ r = 2 , tan θ = 1, ⇒ θ =

π 4

π π  1 + i = 2  cos + i sin  4 4  1

∴(1 + i )

1 3

1

  π π  3   π π  3 =  2  cos + i sin  =  2  cos(2nπ + ) + i sin( 2nπ + )  4 4  4 4     

1

1 π 1 π   = 2 6 cos (2nπ + ) + i sin (2nπ + ) 3 4 3 4   Putting n = 0,1,2 we get 3 values 1 6

1

1

π π  9π 9π  6  17π 17π   2 cos + i sin ,2 6 cos + i sin ,2 cos + i sin  12   12 12   12 12   12 Applications of complex numbers: The effect of multiplying a phasor by i is to rotate it in a positive direction (i.e. anticlockwise) on an Argand diagram through 90  without altering its length.. Similarly, multiplying a phasor by – i rotates the phasor through - 90  . These facts are used in a.c. theory since certain quantities in the phasor diagrams lie at 90  to each other. For example, in the R-L series circuit V L leads i by 90  and may be written as i V L . Thus V R + iV L = V ,V R = iR, V = IX L , where X L is the inductive reactance, 2πfLΩ and V = IZ (Z is the impedance) then R + iX L = Z Vc Similarly for the R – C circuit lags I by and 90  V R − iVc = V ,∴R − iX c = Z , X c =

1 2πfC

Example: Determine the resistance and series inductance (or capacitance) for each of the following impedances, assuming a frequency of 50 Hz: ( a )( 4.0 + i 7.0 )Ω, (b) −i 20Ω

53

Solution: (a) Impedance , Z = ( 4.0 + i 7.0 )Ω, hence Resistance = 4.0Ω and reactance = 7.0 Ω , since the imaginary part is positive, the reactance is inductive, i.e. X L = 7.0Ω, X L = 2πfL , then inductance, X 7.0 L= L = = 0.0223H or 22.3 mH 2πf 2π (50)

(b)Impedance , Z = ( 0 −i 20.0 )Ω, hence Resistance = 0 and reactance = 20 Ω , since the imaginary part isnegative, the reactance is capacitive, i.e. X c = 20Ω, X c = C=

1 , then capacitance, 2πfC

1 1 = µF = 159.2 µF 2πfX c 2π (50)( 20)

Exponential and circular functions of a complex number: If z = x + iy then we define: z2 + 2! z3 sin z = z − + 3! z2 cos z = 1 − + 2!

z3 z4 + + ..... 3! 4! z5 − ... 5! z4 −. 4!     z2 z4 z3 z5 ∴cos z + i sin z = 1 − + − ....  + i z − + − ...  2 ! 4 ! 3 ! 5 !     2 3 ( iz ) + ( iz ) + ... = e iz = 1 + iz + 2! 3! ∴cos z + i sin z = e iz ez =1+ z +

similarly , cos z − i sin z = e −iz cos z =

e iz + e −iz e iz − e −iz , sin z = 2 2i

54

Hyperbolic functions: We will study certain combinations of e x and e −x , called hyperbolic functions. The function e x can be expressed as the sum of an odd function and an even function: e x − e−x e x + e −x ex = (odd ) + (even) 2 2 The odd function is called the hyperbolic sine of x and the even function is called the hyperbolic cosine of x. They are denoted by e x − e −x e x + e −x sinh x = , cosh x = 2 2 x −x x e −e e + e −x tanh x = x , coth x = e + e −x e x − e −x sinh x + cosh x = e x , cosh x − sinh x = e −x

Relation between circular and hyperbolic functions: sin ix = i sinh x, sinh ix = i sin x cos ix = cosh x, cosh ix = cos x

tan ix = i tanh x, tanh ix = i tan x

Derivatives of Hyperbolic functions: d d  e x − e −x  (sinh x ) = dx dx  2

 e x + e −x  = = cosh x 2 

d d  e x + e −x  (cosh x ) = dx dx  2

 e x − e −x  = = sinh x 2 

d d  sinh x  cosh 2 x − sinh 2 x 1 (tanh x) = = = sec h 2 x  = 2 2 dx dx  cosh x  cosh x cosh x d d (coth x) = − cos ech 2 x, (cos echx) = − cos echx cot x dx dx

Example:

( )

( )

( )

( )

d d (cosh x 3 ) = sinh x 3 x 3 = 3 x 2 sinh x 3 dx dx

55

Example: Differentiate the following hyperbolic functions (a ) y = 4 sin 3tch 4t , (b) y = ln( sh3θ) − 4ch 2 3θ Solution: (a) y = 4 sin 3tch4t dy ∴ = ( 4 sin 3t )( 4 sh 4t ) + ( ch4t ) 4( 3 cos 3t ) = 4( 4 sin 3tsh 4t + 3 cos 3tch4t ) dt (b) y = ln ( sh3θ ) − 4ch 2 3θ dy 1 ∴ = 3ch3θ − 4.2ch3θ.3sh3θ = 3( coth 3θ −8ch3θsh3θ ) dθ sh3θ

56

Vectors and Scalars A vector is a quantity having both magnitude and direction, such as displacement, velocity, force and acceleration. Graphically a vector is represented by an arrow OP defining the direction, the magnitude of the vector being indicated by the length of the arrow. O  A P  Analytically a vector is represented by a letter with an arrow over it, as A and its magnitude is  denoted by A or A. A scalar is a quantity having magnitude but no direction, e.g. mass, length, time and any real number. Vector algebra: 1. Two vectors A and B are equal if they havethe same magnitude and direction regardless  of the position of their initial points. Thus A = B.  A

 B

 2. A vector having direction opposite to that of vector A but having the same magnitude is  denoted by - A .  A  -A

   3. The sum or resultant of vectors A and B is a vector C formed by placing the initial    point of B on the terminal point of A and then ing the initial point of A to the     terminal point of B , i.e. C = A + B .

 A

 B

   C = A+B

      = A + (- B ), the subtraction of B from A may be regarded as the addition A- B       of - B to A . If A = B , then A - B is defined as the null or zero vector and is represented by the symbol 0.   5. The product of a vector A by a scalar m is a vector m A with magnitude m times   the magnitude of A and with direction the same as or opposite to that of A , according as m is positive or negative. 4.

57

Laws of vector Algebra:

    1. A + B = B + A       2. A + ( B + C ) = ( A + B )+ C   3. m A = A m   4. m(n A )= (mn) A

commutative law for addition Associative law for addition commutative law for multiplication Associative law for multiplication

 Unit vector: A unit vector is a vector is a vector having unit magnitude, if A is a vector with   A magnitude A ≠ 0 , then is a unit vector having the same direction as A . A    The rectangular unit vectors i , j , k : An important set of unit vectors are those having the directions of the positive x, y and z axes of a three dimensional rectangular coordinate system,    and are denoted respectively by i , j and k .  Components of a vector: Any vector A in 3 dimensions can be represented with initial point at the origin O of a rectangular coordinate system. Let ( A1 , A2 , A3 ) be the rectangular coordinates     of the terminal point of vector A with initial point at O. The vectors A1i , A2 j and A3 k are  called the component vectors of A .       Vector A = A1i + A2 j + A3 k , the magnitude of A is A = A = A12 + A22 + A32  In particular, the position vector or radius vector r from O to the point (x,y,z) is written     r = xi + yj + zk and has magnitude r = x 2 + y 2 + z 2 .

z  A1 i

x

 A

 A3 k

y  A2 j             Example: Given r1 = 3i − 2 j + k , r2 = 2i − 4 j − 3k , r3 = −i + 2 j + 2k , find the magnitude        of (a )r3 , (b)r1 + r2 + r3 , (c )2r1 −3r2 −5r3     2 Solution: (a) r3 = − i + 2 j + 2k = ( −1) + 2 2 + 2 2 = 3 (b)               r1 + r2 + r3 = (3i − 2 j + k ) + (2i − 4 j − 3k ) + ( −i + 2 j + 2k ) = 4i − 4 j    2 r1 + r2 + r3 = 4 2 + ( − 4 ) = 4 2

Example: Find a unit vector        r1 = 2i + 4 j − 5k , r2 = i + 2 j + 3k 









parallel 

to

the

resultant

Solution: Resultant R = r1 + r2 = 3i + 6 j − 2k , R = 3 2 + 6 2 + ( − 2) 2 = 7         R 3i + 6 j − 2k 3i 6 j 2k Then a unit vector parallel to R is . = = + − R 7 7 7 7

58

of

vectors

Product of vectors: Two vectors are multiplied in two manners, i.e., scalarly and vectorly.

  The scalar or dot product: The dot product of vectors A and B is the scalar quantity obtained by multiplying the product of the magnitudes of the vectors and the cosine of the angle between them.    In symbols, A.B = AB cos θ, 0 ≤ θ ≤ π , where θ is the angle between A and B .  B

θ  A

Thefollowing laws are valid:   

1. A.B = B. A        2. A.( B +C ) = A.B + A.C        3.m( A.B ) =(mA).B = A.( mB ) =( A.B ) m       4.i .i = j . j = k .k =1, i . j = j .k = k .i =0

5.If         A = A1 i + A2 j + A3 k , B = B1i + B2 j + B3 k   A.B = A1 B1 + A2 B2 + A3 B3 , A. A = A 2 = A12 + A22 + A32       6.If A.B = 0 , and A and B are not null vectors, then A and B are perpendicular.

  The vector or cross product: The vector or cross product of two vectors A and B is a vector        C = A × B . The magnitude of A × B is defined as the product of the magnitudes of A and B    and the sine of the angle θ between them. The direction of the vector C = A × B is      perpendicular to the plane A and B such that A , B and C form a right-handed system.    In symbols, A × B = AB sin θη,0 ≤θ ≤ π , where ηis a unit vector indicating the direction of         A × B . If A = B , or if A is parallel to B then sin θ = 0, ⇒ A × B = 0 

η  B

θ

 A

The following laws are valid:    

1. A × B = −B × A        2 . A × ( B + C ) = A × B + A ×C        3.m( A × B ) = ( mA) × B = A ×(mB ) = ( A × B ) m                4.i ×i = j × j = k × k = 0, i × j = k , j × k = i , k ×i = j

5.If

59

        A = A1i + A2 j + A3 k , B = B1i + B2 j + B3 k    i j k   A × B = A1 A2 A3 B1 B2 B3

    6.The magnitude of A × B is the same as the area of a parallelogram with sides A and B .   Triple product: The vector product of two vectors B and C is a vector quantity. So this    product ( B ×C ) may be multiplied scalarly or vectorially with a third vector A to give two       triple products namely A. B ×C and A × B ×C . The former being a scalar quantity is termed as scalar triple product or box product and the latter being a vector quantity is called a vector triple product.

(

)

(

)

The following laws are valid:     

(

)

1.( A.B )C ≠ A B.C             2. A.( B ×C ) = B.(C × A) = C.( A ×B ) = volume of the parallelepiped having A, B and C

as edges. If             A = A1i + A2 j + A3 k , B = B1i + B2 j + B3 k , C = C1i + C 2 j + C 3 k A1    A.( B × C ) = B1

A2

A3

B2

B3

C1 C 2      3. A × B × C ≠ A × B      4. A × B ×C = A.C

(

(

) ( ) (

) )

C3  ×C     B − A.B C

(

)

















Example: Find the angle between A = 2i + 2 j − k and B = 6i −3 j + 2k Solution: 

A.B = AB cos θ  A.B = 2.6 + 2(−3) + ( −1) 2 = 4   A = 2 2 + 2 2 + ( −1) 2 = 3, B = 6 2 + ( −3) 2 + 2 2 = 7  A.B 4 ∴cos θ = = = 0.1905 AB 3.7 ∴θ = 79 

















Example: Determine the value of a so that A = 2i + aj + k and B = 4i − 2 j − 2k are perpendicular. Solution:     If A and B are perpendicular then A.B = 0  Then A.B = 2.4 + a (−2) +1(−2) = 8 − 2a − 2 = 0, ⇒ a = 3     Example: Find the work done in moving an object along a vector r = 3i + 2 j −5k if the     applied force is F = 2i − j − k Solution:

60

(

)(

)

       Work done = F .r = 2i − j − k . 3i + 2 j − 5k = 6 − 2 + 5 = 9         A = 2i − 3 j − k B = i + 4 j − 2k , Example: If and         ( a ) = A × B, (b) B × A, (c ) A + B × A − B

(

 i



Solution: (a) A ×B = 2

(

1

)

) (

 j

(

)

k

   −1 =10i +3 j +11k

−3 4

find

)

−2

          (c) A + B = 3i + j − 3k , A − B = i − 7 j + k  i

    ( A +B ) ×( A −B ) = 3 1

 j

1 −7

 k

   −3 =−20i −6 j −22k 1









Example: Determine a unit vector perpendicular to the plane of A = 2i −6 j −3k and     B = 4i + 3 j − k

    Solution: A × B is a vector perpendicular to the plane of A and B .  i  A ×B =2 4

 j −6 3

 k    −3 =15i − 10 j +30k − 1

  A unit vector perpendicular to A × B is   A ×B   = A ×B

   15i − 10 j +30k

(15)2

+(− 10 )

2

+(30 )

2

   3i 2j 6k = − + 7 7 7

Example: Find the volume of the parallelepiped whose three co-terminus edges are represented     by the vectors i +2 j ,4 j and j +3k Solution: Volume of the required parallelepiped is scalar triple product of the vectors     i +2 j ,4 j and j +3k 1 =0 0

2 4 1

0 0 =12 3

61

Matrices and Matrix Operations The following rectangular array with three rows and seven columns might describe the number of hours that a student spent studying three subjects during a certain week: Mon Tue Math Engl Che

Wed 2 0 4

Thu 3 3 1

Fri 2 1 3

Sat 4 4 1

Sun 1 4 0

4 2 0

2 2 2

If we suppress the headings, then we are left with the following rectangular array of numbers with 3 rows and 7 columns called a “matrix”. 2 0 4

3 3 1

2 1 3

4 4 1

1 4 0

4 2 0

2 2 2

Definition: A matrix is a rectangular array of numbers. 2 3 2 Examples: 4 2, [2 3], 1 ,     A general m × n matrix might be written as

62

 a11 a12 − − a1n     a21 a22 − − a2n  A=  −−−−−−−−−    a a − − a  mn   m1 m2 A matrix with only one row is called a row matrix and a matrix with only one column is called a column matrix. A matrix A with n rows and n columns is called a square matrix of order n and the entries a 11 , a 22

,…., a nn are said to be main diagonal of A .

 a11 a12 − − a1n     a21 a22 − − a2n  A=  −−−−−−−−−    a a − − a  nn   n1 n2 Operations on Matrices Equality: Two matrices are defined to be equal if they have the same order and their corresponding entries are equal. Consider the matrices

 1 2  1 2  1 2 0  A =   , B =   ,C =    3 x   3 5  3 4 0

If x = 5, then A = B. There is no value of x for which A=C since A & C have different orders. Definition: If A and B are the matrices of the same order, then the A+B is the matrix obtained by adding the entries of B to the corresponding entries of A, and the difference A-B is the matrix obtained by subtracting the entries of B from the corresponding entries of A. Matrix of different sizes cannot be added or subtracted.

63

Example:

Then

 1 2  1 2  1 2 0  A =   , B =   ,C =    3 4  3 5  3 4 0

2 4  0 0  , A − B =   A + B =  6 9   0 − 1 The expressions A+C, B+C, A-C and B-C are undefined. Definition: If A is any matrix and c is any scalar, then the product cA is the matrix obtained by multiplying each entry of the matrix A by c. For the matrix 1 2  A = 3 4     we have, 2 4  2A = 6 8     Definition:If A is an m × r matrix and B is an r × n matrix, then the product AB is the m × n matrix whose entries are determined as follows: Multiply the corresponding entries from the row and column together and then add up the resulting products. Consider

 A=  

 31  021     ,B=  23  043     11 

The entry in row 1 and column 1 of AB is computed as follows: (1.1) + (2.3) + (0.1) = 7 The computations for the remaining products are (1.3)+ (2.2) + (0.1) = 7 …………………………

64

7 7 AB =  15 17     A B = AB m× r r× n = m× n × Transpose of a matrix: If A is an m n matrix, then the transpose of A,denoted byA T is defined to be the n × m matrix that results from interchanging the rows and columns of A. If  7 7 A = 16 2  , then   7 At =  7 

16   2 

7 A = 16 

7 , then 2 

Trace: If A is a square matrix, then the trace of A, denoted by tr(A), is defined to be the sum of the entries on the main diagonal of A. If

tr ( A) = 7 + 2 = 9

Inverses; Rules of Matrix Arithmetic Properties of Matrix Operations: for real numbers a and b always have ab = ba, which is called commutative law for multiplication. Matrices, however, AB and BA need not be equal. It can happen that the product AB is defined but BA is undefined. Consider the matrices

−1 0  1 2   A = , B = 2  3 0   3    −1 −2   3  AB =  11 , BA =  −3 4   

6  0 

Thus AB ≠ BA.

Properties of Matrix Arithmetic: The following rules of matrix arithmetic are valid (i)A + B = B + A (commutative law for addition) (ii) A + (B + C) = (A + B) + C (Associative law for addition) (iii) A(BC) = (AB)C (Associative law for multiplication) (iv)A(B + C) = AB + AC (Left distributive law) (v)(B + C)A = BA + CA (Right distributive law) Zero matrix: A matrix, all of whose entries are zero, such as 0 0  A = 0 0     is called zero matrix. Identity Matrix: The n-square identity or matrix denoted by I is the n-square matrix with 1’s on the diagonal and 0’s elsewhere. Example: 1 0  I = 0 1    

65

A ≠0 , then A is said to be nonsingular matrix; Singular & Nonsingular matrices: If otherwise it is said to be singular. Thus 1 2  1 4  A = 4 6   and B =  2 8   are nonsingular and singular respectively.     The inverse of a matrix: Let A be a nonsingular matrix of order n. Let B be another square matrix of the same order such that AB = BA = I, where I is the unit matrix of order n. Then B is said to be the inverse of A which is written as A −1 . Thus A A −1 = A −1 A = I.

The matrix a b  A = c d     is invertible if ad – bc ≠ 0, in which case the inverse is given by the formula  d - b 1   A -1 = ad − bc  - c a  Example: Inverse of 2 3 1  5 - 3 -1 A = 4 5  is A = − 2  - 4 2      Example: Find the inverse of 3  A = 2 1 

2   −1 1  

1 −3 2

Here A = 8 Cofactors of A are a11 = -1 a12 = -3 a13 = 7 a 21 = 3 a22 = 1 a23 = -5 a31 = 5 a32 = 7 a33 = -11  −1  A = − 3 adt  7 

5   7  −11 

3 1 −5

5   −1 3 1  A =  −3 1 7  8   7 −5 −11 −1

Task: Find the inverse of the following matrices: 1  2 4 

2 −1 1

0  −11  3 ,  − 4  8  6

2 0 −1

2   1  −1 

66

Using Row Operations to find the inverse of 1  A = 2 1 

2 5 0

3  3 8 

Solution: We want to reduce A to the identity matrix by row operations and simultaneously apply these operations on I to transform it into A −1 . The computations are as follows:

67

 1 2 31 0   1 2 31 0   1 2 3 1 0       2 5 3 0 1 0 →  0 1 − 3 2 1 0  →  0 1 − 3 2 1 0  →     1 0 80 1 0 2 5−− 1 0 1 0 − 1 5 2 1     1 2 31 0  1 2 0− 14 6 3  1 0 − 40 6 91      0 1 − 3 2 1 0 → 0 1 03 −− 351  → 0 1 03 −− 351         

68

Thus A

−1

 − 40  =  13  5 

9   −3 −1  

16 −5 −2

Triangular Matrices: A square matrix in which all the entries below the main diagonal are zero is called upper triangular and a square matrix in which all the entries above the main diagonal are zero is called lower triangular. A matrix that is either upper triangular or lower triangular is called triangular. Examples:

1  0 0 

3  2 3 

2 4 0

2  1 3 

0 3 2

0  0 7 

upper triangular lower triangular Orthogonal Matrices: A square matrix A with the property At = A-1 is said to be orthogonal matrix. It follows from this definition that a square matrix A is orthogonal if and only if A At = AtA = I. Example:  cos θ − sin θ   is orthogonal for all choices of θ, since A =  cos θ   sin θ sin θ  cos θ − sin θ   1 0   cos θ  = =I A t A =  cos θ   0 1   − sin θ cos θ  sin θ Rank of a matrix: The rank of a matrix A is equal to the order of the highest ordered nonvanishing determinant in A. It follows, therefore, that for a nonsingular square matrix of order n, the rank is equal to n.

 0 0  , r (A) = 0 , since all elements are 0  0 0

Examples: A = 

2 4 2 A=  4

1  , r (A) = 2, since A ≠ 0 3 1  , r (A) = 1, since A = 0 2

1  4 3 

3 1   6 and 4 3 1  

A= 

Task: Find the rank of 2 5 2

2 5 2

3  8 1 

Example: Solve by Cramer’s rule: x + y + z = 11 2x −6 y − z = 0 3x + 4 y + 2 z = 0

69

1 D=2 3

Solution:

1 Dz = 2 3 x=

1 −6 4 1 −6 4

1 11 −1 =11, D x = 0 2

0

1 −6 4

1 1 −1 = −88, D y = 2 2

3

11 0 0

1 −1 = −77, 2

11 0 = 286 0

Dy Dx D = −8, y = = −7, z = z = 26 D D D

Matrix Inversion Method: Consider the system a11x1 + a12x2 + … + a1nxn = b1 a12x2 + a22x2 + … + a2nxn = b2 ……………………………… an1x1 + an2 x2 + … + annxn = bn The system can be written in the matrix form as AX = B where

 x1   b1   a11 a12 − − a1n         x2   b2   a21 a22 − − a2n      A=  , X = − ,B = −  −−−−−−−−−−−−       a a − − a   −   −   n1 n2 nn   x   b   n  n AX = B

⇒ A−1 AX = A−1B ⇒ IX = A−1 B ⇒ X = A−1 B If A is known, then the solution vector X can be found out from the above matrix relation. Example: Solve the equations 3x1 + x2 + 2x3 = 3 2x1 – 3x2 – x3 = -3 x1 + 2x2 + x3 = 4 The given system can be written as AX = B Where

2 3 1  x1   3        A =  2 − 3 − 1, X =  x 2 , B =  − 3  1 2 x   4  1    3   70

Here A = 8 Cofactors of A are a11 = -1 a12 = -3 a13 = 7 a 21 = 3 a22 = 1 a23 = -5 a31 = 5 a32 = 7 a33 = -11  −1  adt A =  − 3  7 

3 1 −5

5   7  −11 

5   −1 3 1  A =  −3 1 7  8   7 −5 −11 −1

It follows therefore

 x1  −1 3 5  3   1    1      7 − 3  =  2   x 2  = − 3 1     x  8  − 5 −11  7  4  −1  3 which gives x1 = 1, x2 = 2 and x3 = -1. Eigenvectors and Eigenvalues: Let 3 − 2   − 1 2 , u =  , v =   A =  1 0  1  1  The images of u and v under multiplication by A are shown in figure. In this section, we study equations such as Ax = 2x or Ax = -4x and we look for vectors that are transformed by A into a scalar multiple of themselves. Definition: An eigenvector of n × n matrix A is a nonzero vector x such that Ax = λx for some scalar λ. A scalar λ is called an eigenvalue of A if there is nontrivial solution x of Ax = λx; such an x is called an eigenvector corresponding to λ. Example: Let 1 6  − 6  3  , u =  , v =   A =  5 2   5  − 2 Are u and v eigenvectors of A ? Solution: 1 Au =  5  1 Av =  5  1 A =  5

6  − 6   24  − 6     = = −4       5   = −4u 2  5   − 20    6  3   − 9   3      = ≠ λ      − 2  2  − 2   11    6 − 6  3  , u =  , v =   2   5  − 2

Thus u is an eigenvector corresponding to an eigenvalue –4, but v is not an eigenvector of A , because Av is not a multiple of v.

71

To find the eigenvalues of an n × n matrix we rewrite Ax = λx as Ax = λIx Or equivalently (A - λI)x = 0 …..(i) For λ to be an eigenvalue, there must be a nonzero solution of this equation. However (i) has a nonzero solution if and only if det(A - λI) = 0 This is called the characteristic equation of A; the scalars satisfying this equation are the eigenvalues of A. When expanded, the determinant det(A - λI) is a polynomial p in λ called the characteristic polynomial of A. Example: Find the eigenvalues and eigen vectors of 8 − 4   A = 2 2    Solution: The eigen values2 are the roots of the characteristic equation 8 −λ −4 =0 2 2 −λ or , λ2 −10λ + 24 = 0 or , ( λ − 4 )( λ − 6 ) = 0 i.e.λ = 4,6 The two distinct eigenvalues are 4 and 6 Eigenvector corresponding to eigen value λ = 4 :

( A − λI ) X

=0

 8 − 4 − 4  x1   4 − 4  x1    = 0, ⇒    = 0 or ,  2 − 4  x 2   2  2 − 2  x 2  4 x1 − 4 x 2 = 0 2 x1 − 2 x 2 = 0,∴ x1 = x 2 1 X 1 = c1   1 Eigenvector corresponding to eigen value λ = 6 :

( A − λI ) X

=0

 8 − 6 − 4  x1   2 − 4  x1    = 0, ⇒    = 0 or ,  2 − 6  x 2   2  2 − 4  x 2  2 x1 − 4 x 2 = 0 2 x1 − 4 x 2 = 0,∴ x1 = 2 x 2 1 X 2 = c 2    2

Eigenvalues of 3

×

3 matrix: Find the eigenvalues of

72

0  A = 0 4 

1 0 −17

0  1 8 

Solution: The characteristic polynomial of A is det(A - λI) = −λ 1

1 0

4

−17 3

0 −λ 8 −λ

2

= -λ + 8λ - 17λ + 4 The eigenvalues of A must satisfy the cubic equation -λ3 + 8λ2 - 17λ + 4 = 0 or λ3 - 8λ2 + 17λ - 4 = 0 or (λ - 4) (λ2 - 4λ + 1) = 0 Therefore λ = 4 , 2 + 3, 2 − 3 Theorem: If A is an n n triangular matrix (upper , lower or diagonal), then the eigenvalues of A are the entries on the main diagonal of A. Example: The eigenvalues of the lower triangular matrix 2  1 −1 

0 4 3

0  0 2 

are λ = 2 , 4 and 2.

73

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