ADDMATH FORM 4

127

10.

DIFFERENTIATION

Unit A : Determine the first derivative of the function y = axn using formula.

Example Exercise 1. y = x 3 dx dy = 3x 3 – 1 = 3x2 a. y = x 4 dx dy = [4x3] b. y = x 5 [5x4] c. y = x 7 [ 7x6] 2. y = 2x3 dx dy = 2(3x2) = 6x2 a. y = 3x 4 dx dy = [12x3] b. y = 5x 3 [15x2] c. y = 10x 2 [20x] 3. y = – 2x3 dx dy = –2(3x2) = – 6x2 a. y = – 5 x 4 dx dy = [ -20x3] b. y = – 8 x 5 [-40x4 _] c. y = – 12 x 2 [-24x] 4. f(x) = x-–2 ) ( ' x f = -2x –2 – 1 = - 2 x –3 a. f(x) = x -1 f’(x) = 2 1 x [- ] b. f(x) = x -5 f’(x) = 6 5 x [- ] c. f(x) = x -3 f’(x) = 2 6 x [- ] 5. f(x) = 3 x -2 ) ( ' x f = 3(-2 x -2-1) = - 6 x -3 a. f(x) = 4x -1 f’(x) = 2 4 x [- ] b. f(x) = 2x -4 f’(x) = 5 8 x [- ] c. f(x) = 6x -1 f’(x) = 2 6 x [- ] 6. y = 2 1 x4 dx dy = 2 1 (4 x 4 – 1 ) = 2 x3 a. y = 2 3 x4 dx dy = [ 6x3] b. y = 3 1 x 6 [2x5 _] c. y = 6 1  x – 3 4 1 2 x [ ]

128

7. y = 3 2  x3 dx dy = 3 2  (3 x3 – 1 ) = –2x2 a. y = 3 2  x – 6 dx dy = 7 4 x [ ] b. f(x) = x 2 5 f ’(x) = 2 5 2 x [- ] c. f(x) = ₆ 3 4 x f ’(x) = 7 8 x [- ] Topic : DIFFERENTIATION

Unit B : Determine first derivative of a function involving : (a) addition, or (b) subtraction of algebraic terms.

Example

Exercise

1. y = x2 + 3x +4 3 + x 2 = dx dy a. y = x2 +4x +3 [2x+4] b. y = x2 + 5x +6 [2x+5] 2. y = x2 - 3x +4 3 -x 2 = dx dy a. y = x2 -4x +3 [2x-4] b. y = x2 - 5x +6 [2x-5] 3. y = x3 + 4x2 + 5 x 8 + x 3 = dx dy 2 a. y = x3 +5x2 +7 [3x2 +10x] b. y = x3 + 6x2 + 8 [3x2 + 12x] 4. y = x3 - 3x2 -6 x 6 -x 3 = dx dy ₂ a. y = x3 -5x2 -7 [3x2-10x] b. y = x3 - 6x2 – 8 [3x2 -12x] 5. y = x(x + 5) y = x2 + 5x 5 + x 2 = dx dy a. y = x(x - 6) [2x-6] b. y = x2(x + 5) [ 3x2 + 10x] 6. y = (x+1)(x + 5) y = x2 + 6x + 5 6 + x 2 = dx dy a. y = (x+1)(x – 6) [2x-5] b. y = (x2 +1)(x - 4) [3x2 -8x+1] 7. y = (x+3)2 y = x2 + 6x + 9 a. y = (x+4)2 b. y = (3x+1)2

129

6 + x 2 = dx dy [2(x+4)] [6(3x+1)] 8. y = x(x+3)2 y = x(x2 + 6x + 9) y = x3 + 6x2 + 9x 9 + x 12 + x 3 = dx dy 2 a. y = x(x+4)2 [3x2 + 16x+16] b. y = x(3x+1)2 [27x2 +12x+1]

Example

Exercise

9. x 3 + x = y 2

y = x

2

+ 3x

-1 2 -x 3 -x 2 = dx dy 2 x 3 -x 2 = dx dy a . _x 5 + x = y 2

2 5 x [2x - ]

b.

x 4 -x = y 3

2 2 4 x [3x + ] 10. 2 2 x 4 + x 3 + x = y

y = x

2

+ 3x

-1

+ 4x

-2

3 -2 -x 8 -x 3 -x 2 = dx dy 3 2 x 8 -x 3 -2 = dx dy a 2 2 x 6 + x 5 + x = y

2 3 5 12 x x [2x - - ]

b.

2 3 x 5 -x 4 -x = y

2 3 2 4 10 x x [3x + + ] 11. x 5 + x 4 + x = y 3

y = x

2

+ 4 + 5x

-1 2 -x 5 -x 2 = dx dy 2 x 5 -x 2 = dx dy a . x 6 -x 5 + x = y 3

2 6 x [2x + ]

b.

x 7 -x 2 -x = y 3

2 7 x [2x + ]

130

12. 2 3 x 5 + x 4 + x = y

y = x + 4x

-1

+ 5x

-2 3 -2 -x 10 -x 4 -1 = dx dy 3 2 x 10 -x 4 -1 = dx dy a . 2 3 x 6 -x 5 + x = y

2 3 5 12 x x [1- + ]

b.

2 3 x 7 -x 2 -x = y

2 3 2 14 x x [1+ + ] Topic : DIFFERENTIATION

Unit C : To determine the first derivative of a product of two polynomials.

Example Exercise 1 y= x ( x3+1) u=x , v =x3+1 1, 3x2 dx dv dx du   ) (uv dx d dx dy_ = dx du v dx dv u  = x(3x2)+(x3+1)(1) = 3x3+x3+1 = 4x3 +1 a. y= x ( x4+2) . [ 5x4+2] b. y= ( x 5+1)x [ 6x5+1] c. y= ( x3-1)x [4x3 -1] 2 y= 2x2 ( x3+1) u=2x2 , v =x3+1 4 3x2 dx dv x dx du_{ } ) (uv dx d dx dy_ = dx du v dx dv u  = 2x2(3x2)+(x3+1)(4x) = 6 x4+4x4 +4x = 10x4 +4x a. y= 3x2 ( x3-1) b. y= ( x3-1)(5x2) c. y= ( x3-1)(-4x2)

131

[15x4-6x] [25x4-10x] [-20x4+8x] 3 f(x)= (x+1) ( x3+1) u=x+1 , v =x3+1 2 3 ; 1 x dx dv dx du _{ } ) ( ) ( ' uv dx d x f  = dx du v dx dv u  = (x+1)(3x2)+(x3+1)(1) = 3x3+3x2 +x3+1 = 4x3+3x2+1 a. f(x)= (x-1) ( 1+x3) [4 x3-3x2+1] b. f(x)= (1-x) ( x3+2) [-4x3+3x2-2 ] c. f(x)= (2-x) ( x3+3) [-4x3+6x2-3] 4 y= (x+1) ( 2 x +1) u=x+1 , v = 2 x +1 2 1 1   dx dv dx du ) (uv dx d dx dy  = dx du v dx dv u  = (x+1)( 2 1 )+ ( 1 2 x )(1) = 1 2 1 2 1 2 1 _{  } x x = 2 1 1  x a. y= (x-1) ( 3 x +1) b. y= (2x+1) ( 2 x +1) c. y= (3-2x) ( 2-2 x )

132

[ 3 2 3 2 _ x ] [ 2 1 2 2x ] [ 2 1 5 2x ] 5 y= (x2+1) ( 2 x +1) u=x2+1 , v = 2 x +1 2 1 2   dx dv x dx du ) (uv dx d dx dy_ = dx du v dx dv u  = (x2+1)( 2 1 )+ ( 1 2 x )(2x) = 2 2 1 x + x x 2 2 1_ 2 _ = 2 1 2 2 3 2 _{ } x x a. y= (x4+1) ( 3 x +1) [ 3 1 4 3 5 4_ 3_ x x ] b. y= (2+x2) ( 4 x -1) [ _] 2 1 2 4 3 2_{ } x x c. y= (3-x3) ( 3 x +1) [ - _{3 1} 3 4 3_ 2_ x x ]

133

Topic : DIFFERENTIATION

Unit D : To determine the first derivative of a quotient of two polynomials using formula.

Example : 1   x x y 1   dx dy x u 1 1    dx dv x v





 













2 2 2 2 1 1 1 1 1 1 1 1             x x x x x x x v dx dv u dx du v dx dy 1. 3 2 5   x x y 2 15 (2x+3) 2. y = 5 4 3  x x 3. y = 7 2 6  x x

134

2 15 (4x+5) 2 42 (2x-7)

-4. y = 2 3 4 5   x x 2 22 (3x-2)

-5. y = x x   1 4 1 2 5 (1+x)

-6. x x y 2 1 1    7. y = 3 2  x x

135

2 1 (1-2x ) x-3x 8. y = x x 5 1 3 2  22 6x-15x (1-5x ) 9. y =

10

4

2 3



x

x

42 22 4x +120x (x +10) 10. 2 3

5x - 3x

y =

8 + x

11. y =

5

4

1

3 2





x

x

136

4 33 2 -5x +6x +80x-24 (8+x ) 4 2 3 2 -5x +15x -8x (5x -4) Topic : DIFFERENTIATION

Unit E : Determine the first derivative of composite function using chain rule.

Example Exercise 1. 2 ) 2 (   x y ) 2 ( 2 1 ) 2 ( 2 1      x x dx dy a. 4 ) 3 (   x y [4(x+3)3_] b. 5 ) 2 (   x y [5(x+2)4] c. 3 ) 8 (   x y [3(x+8)2] 2. 2 ) 2 3 (   x y ) 2 3 ( 6 3 ) 2 3 ( 2 1       x x dx dy a. 4 ) 3 2 (   x y [8(2x+3)3] b. 5 ) 2 4 (   x y [20(4x+2)4] c. 3 ) 8 5 (   x y [15(5x+8)2] 3. 2 ) 2 ( 3   x y ) 2 ( 6 1 ) 2 ( 2 3 1       x x dx dy a. 4 ) 2 ( 5   x y [20(x+2)3] b. 5 ) 2 4 ( 3   x y [60(4x+2)4] c. 3 ) 8 2 ( 2   x y [12(2x+8)3] 4. 2 ) 2 ( 2   x y 3 3 3 2

)

2

(

4

)

2

(

4

1

)

2

(

2

2

)

2

(

2



























  

x

x

x

dx

dy

x

y

a. 4 ) 2 ( 5   x y 5 20 (x+2)

[-

]

b. 5 ) 2 ( 3   x y 6 15 (x+2)

[-

]

c. 3 ) 8 ( 2   x y 4 6 (x+8)

[-

]

137

5. 2 ) 2 ( 5 2   x y 2 ) 2 ( 5 2 _   x y 1 ) 2 ( 2 5 2  3   x y 3 ) 2 ( 5 4    x y a. 3 ) 5 ( 4 3   x y 4 9 4(x+5)

[-

]

b. 6 ) 3 2 ( 5 4   x y 7 24 5(2x-3)

[-

]

c. 4 ) 4 3 ( 2 5    x y 5 10 (3x-4)

[

]

Topic : DIFFERENTIATION

Unit F : Determine the Gradient of a Tangent and a Normal at a point on a Curve.

Example 1 : Find the gradient of the tangent to

the curve y 2x33x2 7x5 at the point (-2,5) Solution: y2x3 3x2 7x5 6x2 6x7 dx dy At point (-2,5), x=-2

Hence, the gradient of tangent at the point (-2,5)

5 7 ) 2 ( 6 ) 2 ( 6 2 w hen 7 6 6 2 2 2               x x x x when dx dy mT

Example:

Given

f(x)x(2x3)

and the gradient

of tangent at point P on the curve

y = f(x) is 29, find the coordinates of the

point P.

Solution: y f(x) y = 2x2 – 3x since f(x) = x(2x-3) 4x3 dx dy At point P, 29 dx dy 4x – 3 = 29 x = 8 y = 104 The coordinates of P is (8 , 104)

(1)

Given that the equation of a parabola is

2 2 4 1 x x

y    , find the gradient of the tangent

to the curve at the point (-1,-3)

(2) Find the gradient of the tangent to the curve





2





3





x

x

138

8



T

m m_T 7

(3) Given that the gradient of the tangent at point P on the curve y 



2x5



2 is – 4, find the coordinates the point P. P(2 , 1) (4) Given ( ) 4₂ x x x

f   and the gradient of tangent

is 28. Find the value of x.

3 2 

x Topic : DIFFERENTIATION

Unit G : Determine the Equation of a Tangent and a normal at a Point on a Curve. Example 1 :

Find the equation of the tangent at the point (2,7) on the curve 3 2 5 x y Solution: 12 6(2) dx dy 2, w hen x 6 5 3 2       x dx dy x y Gradient of tangent, m_T=12 Equation of tangent is

y



y

₁



m

_T



x



x

₁







0

17

12x

y

24

12

7

2

12

7



















x

y

x

y

Example 2 :

Find the equation of the normal at the point x = 1 on the curve y42x3x2 Solution: 4 6(1) 2 dx dy , 1 w hen x 6 2 3 2 4 2            x dx dy x x y Gradient of normal, 4 1   N m when x = 1 , y = 4 – 2(1) + 3(1)2 = 5 Equation of normal is yy1 m_N



xx1







0

21

-y

4

x

)

1

(

1

20

4

1

4

1

5





















x

y

x

y

(1)

Find the equation of the tangent at the point (1,9) on the curve y



2x5



2

(2) Find the equation of the tangent to the curve

y





2

x



1



x



1



at the point where its x-coordinate is -1.

139

21 12    x y y3x3

(3) Find the equation of the normal to the curve y 2x2 3x2 at the point where its x-coordinate is 2. 0 22 5    y x

(4) Find the gradient of the curve

3 2 4   x y at the

point (-2,-4) and hence determine the equation of the normal passing through that point.

0 30 8 ; 8      x y mT Topic : DIFFERENTIATION

Unit H : Problem of Second Derivatives

Example :

Given that

f

(

x

)





3



2

x

2



5, find f ”(x) . Hence, determine the value of f ”(1)

Solution:

































300

60(1)(5)

(1)

"

1)

-(6x

)

2x

-60(3

3

18

)

2x

-20(3

)

2

3

(

16

2x

-3

20

)

2

3

(

320

)

2

3

(

20

)]

4

(

2

3

4

)[

20

(

]

20

[

)

2

3

(

)

(

"

2x

-3

-20x

4

2

3

5

)

(

'

2

3

)

(

2 3 2 2 3 2 2 2 3 2 3 2 2 4 2 3 2 4 2 4 2 4 2 5 2





















































f

x

x

x

x

x

x

x

x

x

x

x

f

x

x

x

f

x

x

f

(1) Given that

y2x3 4x2 6x3

, find

₂

2 dx y d

(2) Given that f(x)2x2



403x



, find f”(x).

) (x f y ) ( ' x f dx dy  ) ( ' ' 2 2 x f dx y d _

First derivative

Second derivative

Differentiate the first time

140

12x + 8

160 – 36x

(3) Given that f(x)(4x1)5, find f ”(0)

–320

(4) Given that

s



3



t

2



1



2,calculate the value of

2 2 dt s d when t = 2 1 . –3 Topic : DIFFERENTIATION

Unit I : Determine the Types of Turning Points

(Minimum and Maximum Points)

Example :

Find the turning points of the curve

y2x3 12x2 18x3 and determine whether each of them is a maximum or a minimum point. Solution:

3

18

12

2

3 2











dx

dy

x

x

x

y

At turning points, 0 dx dy







3 or x 1 0 3 1 0 3 4 0 18 24 6 2 2            x x x x x x x

Substitute the values of x into

(1) Find the coordinates of two turning points on the curve y x



x2 3



141

y2x3 12x2 18x3

When x = 1 , y 2(1)3 12(1)2 18(1)311

When x = 3 , y 2(3)3 12(3)2 18(1)333 Thus the coordinates of the turning points are

and ₂ 12 24 2   x dx y d When x = 1 , 2 12(1) 24 12 0 2      dx y d ,

Thus (1 , 11) is the point

When x=3, ₂ 12(3) 24 12 0 2     dx y d

Thus , (3 , –33) is the point

(1 , –2) and (–1 , 2)

(2) Determine the coordinates of the minimum point of y x2 4x4. (2 , 0) (3) Given 2 5 3 2 3  2   x x y is an equation of a

curve, find the coordinates of the turning points of the curve and determine whether each of the turning point is a maximum or minimum point.

142

Topic : DIFFERENTIATION

Unit J : Problems of Rates of Change

Task 1 : Answer all the questions below.

(1) Given that y3x2 2x and x is increasing at a constant rate of 2 unit per second, find the rate of change of y when x = 4 unit.

1 2 44 ) 2 )( 22 ( 22 2 ) 4 ( 6 4 2 6 2 3 2               s unit dt dx dx dy dt dy dx dy x When x dx dy x x y dt dx

(2) Given that y 4x2 x and x is increasing at a constant rate of 4 unit per second, find the rate of change of y when x = 0.5 unit.

12 unit s –1

(3) Given that

x x

v9 1 and x is increasing at a constant rate of 3 unit per second, find the rate of change of v when x = 1 unit.

(4) SPM 2004 (Paper 1 – Question 21) [3 marks] Two variables, x and y, are related by the equation

3 2.

x x

y  Given that y increases at a constant

rate of 4 unit per second, find the rate of change of x when x =2.

If y = f (x) and x = g(t), then using the chain rule

dt dx dx dy dt dy  

, where

dt dy

is the rate of change of

y and

dt dx

143

30 unit s –1 5

8 _{unit s} –1

Task 2 : Answer all the questions below.

(1) The area of a circle of radius r cm increases at a constant rate of 10 cm2 per second. Find the rate of change of r when r = 2 cm. ( Use п = 3.142 ) Answer : 1 2

7957

.

0

4

10

4

)

2

(

2

2

2

10

























s

cm

dt

dr

dt

dr

dt

dr

dr

dA

dt

dA

dr

dA

cm

r

When

r

dr

dA

r

A

dt

dA











(2) The area of a circle of radius r cm increases at a constant rate of 16 cm2 per second. Find the rate of change of r when r = 3 cm. ( Use п = 3.142 )

0.8487 cm s –1

(3) The volume of a sphere of radius r cm increases at a constant rate of 20 cm3 per second. Find the rate of change of r when r = 1 cm. ( Use п = 3.142 )

(4) The volume of water , V cm³, in a container is given

by 8 ,

3 1 3

h h

V   where h cm is the height of the

water in the container. Water is poured into the container at the rate of 10cm3s-1.

Find the rate of change of the height of water, in cm3s-1, at the instant when its height is 2 cm. [3 marks]

144

1.591 cm s –1 6

5 cm s –1

Task 3 : Answer all the questions below.

Example :

The above figure shows a cube of volume 729 cm³.

If the water level in the cube, h cm, is increasing at

the rate of 0.8 cm s

1

, find the rate of increase of

the volume of water.

Solution :

Let each side of the cube be x cm.

Volume of the cube = 729 cm³

x³ = 729

x = 9

(1) A spherical air bubble is formed at the base of a pond. When the bubble moves to the surface of the water, it expands. If the radius of the bubble is expanding at the rate of 0.05 cm s1, find the rate at which the volume of the bubble is increasing when its radius is 2 cm.

 8 . 0 cm3 s –1

h cm

9 cm

9 cm

9 cm

h cm

Chain rule V = 9 x 9 x h = 81h dh dV ₌₈₁ dt dh =rate of increase of the water level = 0.8 cm s1

145

Rate of change of the volume of water,

1 3

64.8

0.8

81













s

cm

dt

dh

dh

dV

dt

dV

Hence, the rate of increase of the volume of water

is 64.8 cm³ s

1

.

(2)

If the radius of a circle is decreasing at the rate

of 0.2 cm s

1

, find the rate of decrease of the

area of the circle when its radius is 3 cm.

 2 .

1 cm2 s –1

(3) The radius of a spherical balloon increases at the rate of 0.5 cm s –1. Find the rate of change in the volume when the radius is 15 cm.



450 cm3 s –1

(4) The edge of a cube is decreasing at the rate of 3 cm s –1. Find the rate of change in the volume when the volume is 64 cm3.

146

(5) Diagram 1 shows a conical container with a diameter of 60 cm and height of 40 cm. Water is poured into the container at a constant rate of 1 000 cm3s-1.

Calculate the rate of change of the radius of the water level at the instant when the radius of the water is 6 cm.

(Use π = 3.142; volume of cone r2h

3 1

_

 )

6.631cm3 s –1

(6) Oil is poured into an inverted right circular cone of base radius 6 cm and height 18 cm at the rate of 2 cm3s-1. Find the rate of increase of the height of water level when the water level is 6 cm high. ( Use п = 3.142 )

0.1591 cm s –1

Topic : DIFFERENTIATION

Unit K : Problems of Small Changes and Approximations

Task 1 : Answer all the questions below.

60 cm

Water

40 cm

Diagram 1

x dx dy y dx dy x y

_

_





_{   }

x x y y in change small in change small where  





Small Changes

x

dx

dy

y

y

y

y

original original















new

Approximate Value

147

(1) Given that yx2 4x, find the small change in y when x increases from 2 to 2.01.

08

.

0

)

01

.

0

)(

8

(

8

2

4

)

2

(

2

01

.

0

2

01

.

2

01

.

2

2

4

2

4

2



































y

y

x

dx

dy

y

x

when

dx

dy

x

x

x

dx

dy

x

x

y











(2) Given that y x2 3x, find the small change in y when x increases from 6 to 6.01.

0.15

(3) Given that y  x2 x

2 , find the small

change in y when x decreases from 8 to 7.98.

–0.62

(4) Given that y 4 x, find

dx dy

.

Hence, find the small change in y when x increases from 4 to 4.02.

02

.

0

;

2





y

x dx dy



148

(1) Given the area of a rectangle , A3x2 2x, where x is the width, find the small change in the area when the width decreases from 3 cm to 2.98 cm. Answer : 1 2 2

s

cm

4

.

0

)

02

.

0

)(

20

(

20

3

2

)

3

(

6

02

.

0

98

.

2

3

98

.

2

3

2

6

2

3











































A

A

x

dx

dA

A

x

when

dx

dA

x

x

x

dx

dA

x

x

A











(2) A cuboid with square base has a total surface area, A3x2 4x, where x is the length of the side of the base. Find the small change in the total surface area when the length of the side of the base decreases from 5 cm to 4.99 cm.

–0.26 cm2

(3) The volume, V cm3 , of a cuboid with rectangular base is given by V  x3 2x2 3x, where x cm

is the width of the base. Find the small change in

the volume when the width increases from 4 cm to 4.05 cm.

(4) In a pendulum of length x meters, the period T seconds is given as 10 2 x T 



. Find dx dT . Hence, find the small change in T when x increases from 2.5 m to 2.6 m.

149

1.75 cm3 50 10

;

 

_

_



_x

T

dx dT second

Task 3 : Answer all the questions below.

Example :

The height of a cylinder is three times its radius.

Calculate the approximate increase in the total

surface area of the cylinder if its radius increases

from 7 cm to 7.05 cm.

Solution :

Let the total surface area of the cylinder be A cm².

A = Sum of areas of the top and bottom circular

surface + Area of the curved surface.

(1)

A cube has side of 6 cm. If each of the side of

the cube decreases by 0.1 cm, find the

approximate decrease in the total surface area

of the cube.

150

 

2 2 2

8

3

2

2

2

2

r

r

r

r

rh

r

A





















Approximate change in the total surface area is



A





 

2

cm

6

.

5

05

.

0

7

16

7

05

.

7

16

































r

r

dr

dA

A

dr

dA

r

A

Hence, the approximate increase in the total surface

area of the cylinder is 5.6π cm² .

(2)

The volume of a sphere increases from

288



cm3 to 290



cm3.

Calculate the

approximate increase in its radius.

72 1

cm

Task 4 : Answer all the questions below.

It is given that h=3r r dr dA r A   16 8 2   New r (7.05) Minus old r(7) Substitute r with the old value of r, i.e. 7

151

(1) Given that

5

4

x

y , calculate the value of

dx dy

if x = 2.

Hence, estimate the values of

_{5 5} 98 . 1 4 (b) 03 . 2 4 ) a ( Solution : 16 5 20 , 2 When 20 20 4 4 6 6 6 5 5              x dx dy x x x dx dy x x y

x

dx

dy

y

y

y

y

a

original original















)

(

_new (2) Given 27₃ x

y , find the value of

dx dy

when x = 3.

Hence, estimate the value of .

03 . 3 27 3

1





dx dy ; 0.97

152





.116525 0 320 3 32 4 2 03 . 2 16 5 2 4 2 4 03 . 2 2 2 4 , 2 4 03 . 2 4 5 5 5 5 5                     x dx dy x and y where y  





0.13125 160 1 32 4 2 98 . 1 16 5 2 4 1.98 4 (b) 5 5                  x dx dy y y y y y original new original new





(3) Given 32₄ x y , find dx dy .

Hence, estimate the value of .

99 . 1 32 4 5 128 x dx dy





; 2.04

Task 5 : Answer all the questions below.

(1) Given that x y   2 20

, find the approximate change in x when y increases from 40 to 40.5.

(2) SPM 2003 (Paper 1 – Question 16)

Given that y x2 5x, use differentiation to find the small change in y when x increases from 3 to 3.01. [3 marks]

153

160 1 0.11 (3) Given 3 3 4 r

v



, use the differentiation method

to find the small change in v when r increases from 3 to 3.01.  27 . 0 (4) Given that 5₃ , x

y  find the value of

dx dy

when x = 4.

Hence, estimate the value of (a)





3 02 . 4 5 (b)





3 99 . 3 5 256 15





dx dy ; (a) 0.07930 ; (b) 0.07871 Topic : DIFFERENTIATION

Unit L : Problems of Maximum and Minimum Values of a Function.

Task 1 : Answer all questions below.

x = α that satisfies

0 dx dy

will minimize y if the value of

₂ 0 2  dx y d

at x = α

x = β that satisfies

0 dx dy

will maximize y if the value of

₂ 0 2  dx y d

at x = β

154

Example 1 :

Given Lx2(405x), find the value of x for which L is maximum.

Hence, determine the maximum value of L.

Solution: 27 10240 3 16 5 40 3 16 max 3 16 0 80 80 3 16 30 80 3 16 0 15 80 , 0 0 ) 15 80 ( 0 15 80 0 30 80 15 80 5 40 ) 5 40 ( 2 m ax 2 2 2 2 2 2 2 2 3 2 2                                                     L L imise will x dx L d dx L d x x x Knowing x x x x dx dL when x dx L d x x dx dL x x L x x L

(1) Given L x2(252x), find the value of x for which L is maximum.

Hence, determine the maximum value of L.

27 15625 m ax 3 25

_,





L

x

(2)

Given y36x4x2 7, find the value of x for which y is maximum.

Hence, determine the maximum value of y.

88 , _{m ax} 2 9   y x

155

Example 2 :

Given 3 ( 2 16)

x x

L



 , find the value of x for

which L is minimum.

Hence, determine the minimum value of L.

Solution:

 

                 36 2 16 2 3 min 2 0 18 ) 2 ( 96 6 2 8 , 0 48 6 48 6 0 48 6 0 96 6 48 6 48 3 ) 16 ( 3 2 m in 2 2 3 2 2 3 2 2 2 3 2 2 2 1 2 2                                           L L imise will x dx L d dx L d x x x Knowing x x x x x x dx dL when x dx L d x x dx dL x x L x x L (1) Given 4 ( 2 128) x x

L



 , find the value of x for

which L is minimum.

Hence, determine the minimum value of L.



192 , 4 _min   L x

(2)

Given (3 8 2) 2 1 2    x x

y , find the value of x

for which y is minimum.

Hence, determine the minimum value of y.

(3) SPM 2003(Paper 1, No 15) Given that y=14x(5 – x), calculate

(a)

the value of x when y is a maximum

156

3 5 m in 3 4

_,







y

x

(a) 2.5 ; (b) 87.5

Task 3 : Answer all the questions below.

(1) The diagram below shows the skeleton of a wire Box with a rectangular base of x by 3x and the heigh of h.

Given the total length of the wire is 348 cm,

(a)

express h in terms of x,

(b)

show that the volume of the box, V cm3 , is given by V = 3x2 (87 – 4x),

(c)

find the stationary value of V, stating whether it is maximum or minimum value.

(a) h = 87 – 4x (b) – (c)

V

m ax = 18291.75 cm3

(2) The diagram below shows a solid cylinder with circular cross-section of radius r and the height of h.

Given the total surface area of the cylinder is 300 cm2 ,

(a) express h in terms of r,

(b) show that the volume of the cylinder, V cm3 , is given by V = 150 r – п r 3 ,

(c) determine the maximum volume of the cylinder when r varies.

(a) h = _rr 2 150

(b) – (c) 100 50 or399



(3) A length of wire 160 m is bent to form a sector OPQ, of a circle of centre O and radius r as shown in the diagram below.

x 3x h r h r θ

157

(a)

Show that (i) 1602

r



(ii) the area, A, of the sector is given by A = 80r – r 2 (b) Find the value of θ and r when the area is a maximum.

(c) Determine the maximum area. _{(b) r = 40 m , θ = 2 rad (c) 1600}

m ax A