ADDMATH FORM 4
April 25, 2017 | Author: Nor Hanina | Category: N/A
Short Description
DIFFERENTIATION...
Description
10. DIFFERENTIATION Unit A : Determine the first derivative of the function y = axn using formula. Example Exercise 3 1. y= x a. y = x 4 b. y= x5
dy = 3x 3 – 1 dx
c.
y= x7
dy = dx
= 3x2 [5x4] [4x3]
2.
y = 2x
3
a.
dy = 2(3x2) dx
y = 3x
4
[ 7x6] 3
b.
y = 5x
b.
y= –8x
2
c.
y = 10x
c.
y = – 12 x
dy = dx
= 6x2
[12x3] 3.
y = – 2x
3
a.
dy = –2(3x2) dx
y= –5x
4
[15x2] 5
[20x] 2
dy = dx
= – 6x2 [ -20x3]
4.
f(x)
= x
-–2
a.
f ' ( x) = -2 x –2 – 1 = -2x
f(x) = x
-1
[-40x4 ]
b.
f’(x) =
f(x) = x
-5
f(x)
= 3 x -2
a.
f’(x) =
f(x) = 4x -1
b.
[- x62 ] c.
f’(x) =
f(x) = 6x -1 f’(x) =
-3
1 4 x 2 dy 1 = (4 x 4 – 1 ) dx 2 y=
f’(x) =
f(x) = 2x -4
[- x85 ]
[- x42 ] 6.
f(x) = x
[- x56 ]
f’(x) =
f ' ( x) = 3(-2 x -2-1) = -6x
c.
–3
[- x12 ] 5.
[-24x] -3
a.
y=
b.
3 4 x 2
y=
[- x62 ] c.
1 6 x 3
y=
1 –3 x 6
dy = dx
= 2 x3 [ 6x3]
127
[2x5 ]
[ 2x1 4 ]
7.
2 3 x 3 2 dy = (3 x3 – 1 ) dx 3 y=
a.
y=
2 x 3
b.
–6
dy = dx
f(x) =
c.
5 2x
f ’(x) =
4 3x 6
f(x) = f ’(x) =
= –2x
2
[ x47 ]
[- 2 5x 2 ]
[- x87 ]
Topic : DIFFERENTIATION Unit B : Determine first derivative of a function involving : (a) addition, or (b) subtraction of algebraic terms. 1.
Example
Exercise
y = x2 + 3x +4
a.
y = x2 +4x +3
a.
y = x2 -4x +3
b.
y = x2 + 5x +6
b.
y = x2 - 5x +6
dy = 2x + 3 dx [2x+4]
2.
y = x2 - 3x +4
[2x+5]
dy = 2x - 3 dx [2x-4]
3.
3
2
y = x + 4x + 5
a.
3
2
y = x +5x +7
[2x-5] 3
2
b.
y = x + 6x + 8
b.
y = x - 6x – 8
dy = 3x 2 + 8x dx [3x2 +10x]
4.
3
2
y = x - 3x -6
a.
3
2
y = x -5x -7
[3x2 + 12x] 3
2
dy = 3x 2 - 6x dx [3x2-10x]
5.
y = x(x + 5) y = x2 + 5x
a.
y = x(x - 6)
[3x2 -12x] 2
b.
y = x (x + 5)
b.
2
dy = 2x + 5 dx
[ 3x2 + 10x]
[2x-6]
6.
y = (x+1)(x + 5) y = x2 + 6x + 5
a.
y = (x+1)(x – 6)
y = (x +1)(x - 4)
dy = 2x + 6 dx
[3x2 -8x+1]
[2x-5]
7.
2
y = (x+3) y = x2 + 6x + 9
a.
y = (x+4)
2
b.
128
y = (3x+1)
2
dy = 2x + 6 dx [2(x+4)]
8.
y = x(x+3)2 y = x(x2 + 6x + 9) y = x3 + 6x2 + 9x
a.
y = x(x+4)2
dy = 3x 2 + 12x + 9 dx
y = x(3x+1)2
[27x2 +12x+1] [3x2 + 16x+16]
Example 9.
[6(3x+1)]
b.
Exercise
3 y = x2 + x y = x2 + 3x-1
a .
y = x2 +
b.
5 x
y = x3 -
4 x
dy = 2x - 3x - 2 dx dy 3 = 2x - 2 dx x [3x 2 + x42 ]
[2x - x52 ] 10.
y = x2 +
3 4 + x x2
a
y = x2 +
b.
5 6 + x x2
y = x3 -
4 5 x x2
y = x2 + 3x-1 + 4x-2 dy = 2x - 3x - 2 - 8x -3 dx dy 3 8 = 2- 2 - 3 dx x x 11.
x 3 + 4x + 5 y= x
[3x 2 + x42 + 10 ] x3
[2x - x52 - 12 ] x3 a .
b.
x 3 + 5x - 6 y= x
x 3 - 2x - 7 y= x
y = x2 + 4 + 5x -1 dy = 2 x - 5 x -2 dx
dy 5 = 2x - 2 dx x
[2x +
129
6 x2
]
[2x +
7 x2
]
12.
a .
x 3 + 4x + 5 y= x2
b.
x 3 + 5x - 6 y= x2
x 3 - 2x - 7 y= x2
y = x + 4x -1 + 5x -2 dy = 1 - 4x - 2 - 10x -3 dx dy 4 10 = 1- 2 - 3 dx x x
[1- x52 + 12 ] x3
[1+ x22 + 14 ] x3
Topic : DIFFERENTIATION Unit C : To determine the first derivative of a product of two polynomials. Example Exercise 1 y= x ( x3+1) a. y= x ( x4+2) b. y= ( x 5+1)x 3 u=x , v =x +1 du dv 1, 3x 2 dx dx dy d (uv) dx dx dv du =u v dx dx =
y= ( x3-1)x
x(3x2)+(x3+1)(1) .
=
3x3+x3+1 = 4x3 +1
[ 5x4+2]
2
c.
2
3
y= 2x ( x +1) u=2x2 , v =x3+1 du dv 4x 3x 2 dx dx dy d (uv) dx dx dv du =u v dx dx
a.
2
3
y= 3x ( x -1)
[ 6x5+1]
b.
= 2
2x (3x2)+(x3+1)(4x) =
6 x4+4x4 +4x = 10x4 +4x
130
3
2
y= ( x -1)(5x )
[4x3 -1]
c.
3
y= ( x -1)(-4x2)
[15x4-6x]
3
3
f(x)= (x+1) ( x +1) u=x+1 , v =x3+1
a.
[25x4-10x]
3
f(x)= (x-1) ( 1+x )
b.
3
f(x)= (1-x) ( x +2)
[-20x4+8x]
c.
f(x)= (2-x) ( x3+3)
du dv 1; 3x 2 dx dx d (uv) dx dv du =u v dx dx
f ' ( x)
=
(x+1)(3x2)+(x3+1)(1)
=
3x3+3x2 +x3+1
= 4x3+3x2+1
[4 x3-3x2+1]
4
x +1) 2 x u=x+1 , v = +1 2 du dv 1 1 dx dx 2
y= (x+1) (
a.
y= (x-1) (
b.
x +1) 3
dy d (uv) dx dx dv du =u v dx dx =
(x+1)(
1 )+ 2
x 1 )(1) 2 1 1 1 = x x 1 2 2 2 (
=
x 1
[-4x3+3x2-2 ]
1 2
131
y= (2x+1) (
x +1) 2
[-4x3+6x2-3]
c.
y= (3-2x) ( 2-
x ) 2
[ 2x 2 1 ]
[2x 2] 3
5
x +1) 2 x u=x2+1 , v = +1 2 du dv 1 2x dx dx 2 dy d (uv) dx dx dv du =u v dx dx
y= (x2+1) (
=
(x2+1)( (
=
a.
3
y= (x4+1) (
[ 2x 5 1 ]
2
b.
x +1) 3
y= (2+x2) (
x -1) 4
2
c.
y= (3-x3) (
x +1) 3
1 )+ 2
x 1 )(2x) 2
1 2 x + 2
1 x 2 2x 2 =
3 2 1 x 2x 2 2
[ 5 x 4 4x 3 1 ] 3
3
132
[
3 2 1 x 2x ] 4 2
[ - 4 x 3 3x 2 1 ] 3
Topic : DIFFERENTIATION Unit D : To determine the first derivative of a quotient of two polynomials using formula. Example : y
ux dy 1 dx
x x 1
1. y
5x 2x 3
v x 1 dv 1 dx
du dv v u dy dx 2 dx dx v x 11 x1 x 12 x 1 x x 12 1 x 12
15 (2x+3)2
2.
y =
3x 4x 5
3. y =
133
6x 2x 7
15 (4x+5)2
4. y =
5x 4 3x 2
42 - (2x-7) 2
5. y =
1 4x 1 x
5 - (1+x) 2
22 - (3x-2) 2
6.
y
1 x 1 2x
7. y =
134
x2 x3
1 (1-2x)2
8. y =
3x 2 1 5x
x x-3
9. y =
4x3 x 2 10
6x-15x 2 (1-5x)2
5x 2 - 3x 10. y = 8 + x3
4x 4 +120x 2 (x 2 +10)2
x2 1 11. y = 5x 3 4
135
-5x 4 +6x 3 +80x-24 (8+x 3 )2
-5x 4 +15x 2 -8x (5x3 -4)2
Topic : DIFFERENTIATION Unit E : Determine the first derivative of composite function using chain rule. Example Exercise 2 1. a. b. c. y ( x 3) 4 y ( x 2) y ( x 2) 5
y ( x 8) 3
dy 2( x 2 ) 1 1 dx 2( x 2)
2.
y (3x 2) 2 dy 2 (3 x 2)1 3 dx 6(3 x 2)
3.
y 3( x 2) dy 3 2( x 2) 1 1 dx 6( x 2)
a.
y (2 x 3) 4
b.
y (4 x 2) 5
[8(2x+3)3] 2
a.
y 5( x 2)
y 3(4 x 2)
y
2 ( x 2) 2
a.
y
b.
5 ( x 2) 4
y (5 x 8) 3
5
[15(5x+8)2]
c.
y 2(2 x 8) 3
[60(4x+2)4]
[20(x+2)3]
4.
c.
[20(4x+2)4]
b.
4
[3(x+8)2]
[5(x+2)4]
[4(x+3)3]
y
3 ( x 2) 5
[12(2x+8)3]
c.
y
2 ( x 8) 3
y 2( x 2) 2 dy 2 2( x 2) 3 1 dx 4( x 2) 3 4 ( x 2) 3 20 [- (x+2) 5]
136
15 [- (x+2) 6]
6 [- (x+8) 4]
5.
a. 2 2 5( x 2) 2 y ( x 2) 2 5 2 y 2( x 2) 3 1 5
y
y
y
b.
3 4( x 5) 3
y
4 5(2 x 3) 6
c.
y
5 2(3 x 4) 4
4 5( x 2) 3
[-
9 4(x+5)4
24 [- 5(2x-3) 7 ]
]
10 [ (3x-4) 5]
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 2 x 3 3x 2 7 x 5 at the point (-2,5) Solution:
y 2 x 3 3x 2 7 x 5 dy 6x 2 6x 7 dx
Example: Given f ( x) x(2 x 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)
dy 4x 3 dx dy At point P, 29 dx
At point (-2,5), x=-2 Hence, the gradient of tangent at the point (-2,5)
dy when x 2 dx 6 x 2 6 x 7 when x 2
mT
4x – 3 = 29 x=8 y = 104
6( 2) 2 6( 2) 7 5
The coordinates of P is (8 , 104)
(2) Find the gradient of the tangent to the curve (1) Given that the equation of a parabola is 2 y x 2x 3 at the point (3,6). y 1 4 x 2 x , find the gradient of the tangent to the curve at the point (-1,-3)
137
mT 8 (3) Given that the gradient of the tangent at point P on 2 the curve y 2 x 5 is – 4, find the coordinates the point P.
mT 7
(4) Given f ( x) x
4 and the gradient of tangent x2
is 28. Find the value of x.
x
P(2 , 1)
Topic : DIFFERENTIATION Unit G : Determine the Equation of a Tangent and a normal at a Point on a Curve. Example 1 : Example 2 : Find the equation of the tangent at the point (2,7) Find the equation of the normal at the point x = 1 2 on the curve y 3x 5 on the curve y 4 2 x 3x 2 Solution: Solution: 2 y 4 2 x 3x 2 y 3x 5 dy dy 2 6 x 6x dx dx dy dy when x 1, 2 6(1) 4 when x 2, 6(2) 12 dx dx Gradient of normal, m N
Gradient of tangent, m T =12 Equation of tangent is
when x = 1 , y = 4 – 2(1) + 3(1)2 = 5 Equation of normal is
y y1 mT x x1 y 7 12x 2
y y1 m N x x1
1 x 1 4 4 y 20 1( x 1) x 4 y - 21 0 y 5
y 7 12 x 24 y 12x 17 0 (1) Find the equation of the tangent at the point (1,9) 2 on the curve y 2 x 5
1 4
(2) Find the equation of the tangent to the curve y 2 x 1x 1 at the point where its x-coordinate is -1.
138
2 3
y 12x 21 (3) Find the equation of the normal to the curve y 2 x 2 3 x 2 at the point where its x-coordinate is 2.
y 3x 3
(4) Find the gradient of the curve y
4 at the 2x 3
point (-2,-4) and hence determine the equation of the normal passing through that point.
mT 8 ; x 8 y 30 0
x 5 y 22 0
Topic : DIFFERENTIATION Unit H : Problem of Second Derivatives Example :
y f (x) Differentiate the first time
dy f ' ( x) dx
First derivative
f ( x) 3 2 x 2
Second derivative
5
5
4 x -20x3 - 2x
f ' ( x) 5 3 2 x 2
Differentiate the second time
d2y f ' ' ( x) dx 2
Given that f ( x) 3 2 x 2 , find f ”(x) . Hence, determine the value of f ”(1) Solution:
4
2 4
3
f " ( x) (3 2 x 2 ) 4 [20] (20 x)[4 3 2 x 2 (4 x)] 20(3 2 x 2 ) 4 320x 2 (3 2 x 2 ) 3
20(3 - 2x ) 18x
3
20 3 - 2x 2 16 x 2 (3 2 x 2 ) 2 3
2
3
60(3 - 2x ) (6x - 1) 2 3
(1) Given that y 2 x 3 4 x 2 6 x 3 , find
2
f " (1) 60(1)(5) 300 (2) Given that f ( x) 2 x 2 40 3 x , find f”(x).
d2y dx 2
139
12x + 8
(3) Given that f ( x) (4 x 1) , find f ”(0) 5
160 – 36x
2
(4) Given that s 3 t 2 1 ,calculate the value of 2
d s 1 when t = . 2 2 dt
–3
–320
Topic : DIFFERENTIATION Unit I : Determine the Types of Turning Points Example : Find the turning points of the curve y 2 x 3 12 x 2 18 x 3 and determine whether each of them is a maximum or a minimum point. Solution:
(Minimum and Maximum Points) (1) Find the coordinates of two turning points on the curve y x x 2 3
y 2 x 3 12 x 18x 3 2
dy dx dy 0 dx 6 x 2 24 x 18 0
At turning points,
x 2 4x 3 0 x 1x 3 0 x 1 or x 3 Substitute the values of x into
140
(1 , –2) and (–1 , 2)
y 2 x 3 12 x 2 18 x 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
d y 12 x 24 dx 2 d2y When x = 1 , dx 2 12(1) 24 12 0 ,
Thus (1 , 11) is the d2y 12 (3) 24 12 0 dx 2 Thus , (3 , –33) is the
point
When x=3,
point (2) Determine the coordinates of the minimum point of y x 2 4x 4 .
(2 , 0)
(3) Given y
2 3 x 2 x 2 5 is an equation of a 3
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.
min. point = (0 , –5) ; max. point = (2 ,
141
23 3
)
Topic : DIFFERENTIATION Unit J : Problems of Rates of Change
If y = f (x) and x = g(t), then using the chain rule dy dy dx dy is the rate of change of , where dt dx dt dt dx y and is the rate of change of x. dt
Task 1 : Answer all the questions below. (1) Given that y 3x 2 2 x and x is increasing at a constant rate of 2 unit per second, find the rate of change of y when x = 4 unit.
(2) Given that y 4 x 2 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.
dx 2 dt y 3x 2 2 x dy 6x 2 dx When x 4 dy 6( 4) 2 dx 22 dy dy dx dt dx dt ( 22)(2) 44 unit s 1 12 unit s –1
(3) Given that v 9 x
1 and x is increasing at a x
(4) SPM 2004 (Paper 1 – Question 21) [3 marks] Two variables, x and y, are related by the equation
2 y 3x . Given that y increases at a constant x
constant rate of 3 unit per second, find the rate of change of v when x = 1 unit.
rate of 4 unit per second, find the rate of change of x when x =2.
142
30 unit s
8 5
–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 )
unit s –1
(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 )
Answer :
dA 10 dt A r2 dA 2 r dr When r 2 cm dA 2 (2) dr 4 dA dA dr dt dr dt dr 10 4 dt dr 0.7957 cm s 1 dt 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 V
1 3 h 8h, where h cm is the height of the 3
water in the container. Water is poured into the container at the rate of 10 cm3 s -1. Find the rate of change of the height of water, in cm 3 s -1 , at the instant when its height is 2 cm. [3 marks]
143
1.591 cm s
5 6
–1
Task 3 : Answer all the questions below. Example :
cm s –1
(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.
h cm
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
h cm 9 cm 9 cm 0.8 cm3 s –1
9 cm
Chain rule dh =rate of increase of dt the water level 144 V = 9 x 9 x h = 81h
dV =81 dh
= 0.8 cm 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.
Rate of change of the volume of water, dV dV dh dt dh dt
81 0.8
64.8 cm 3 s 1 Hence, the rate of increase of the volume of water is 64.8 cm³ s 1 .
1.2 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.
(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.
450 cm3 s –1
145
–144 cm3 s –1
(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 cm3 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 cm3 s -1 . Find the rate of increase of the height of water level when the water level is 6 cm high. ( Use п = 3.142 )
60 cm
40 cm Water
Diagram 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
1 2 r h ) 3
6.631cm3 s –1
0.1591 cm s –1
Topic : DIFFERENTIATION Unit K : Problems of Small Changes and Approximations Small Changes
y dy x dx
y
Approximate Value
dy x dx
y new y original y
where y small change in y x small change in x
y original
Task 1 : Answer all the questions below.
146
dy x dx
(1) Given that y x 2 4 x , find the small change in y when x increases from 2 to 2.01.
(2) Given that y x 2 3 x , find the small change in y when x increases from 6 to 6.01.
y x 2 4x dy 2x 4 dx x 2 2.01
x 2.01 2 0.01 dy 2(2) 4 when x 2 dx 8 dy x dx y (8)(0.01)
y
y 0.08 0.15
(3) Given that y 2 x x , find the small change in y when x decreases from 8 to 7.98. 2
(4) Given that y 4 x , find
dy . dx
Hence, find the small change in y when x increases from 4 to 4.02.
–0.62
Task 2 : Answer all the questions below.
147
dy dx
2 x
; y 0.02
(1) Given the area of a rectangle , A 3x 2 2 x , where x is the width, find the small change in the area when the width decreases from 3 cm to 2.98 cm. Answer :
(2) A cuboid with square base has a total surface area, A 3x 2 4 x , 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.
A 3x 2 2 x dA 6x 2 dx x 3 2.98
x 3 2.98 0.02 dA 6(3) 2 when x 3 dx 20 dA x dx A (20)(0.02)
A
A 0.4 cm2 s 1
–0.26 cm2
(3) The volume, V cm3 , of a cuboid with rectangular base is given by V x 3 2 x 2 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 T 2
x dT . Find . 10 dx
Hence, find the small change in T when x increases from 2.5 m to 2.6 m.
148
1.75 cm3
Task 3 : Answer all the questions below. Example :
dT dx
10 x
; T 50 second
(1) A cube has side of 6 cm. If each of the side of
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.
the cube decreases by 0.1 cm, find the approximate decrease in the total surface area of the cube.
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.
7.2 cm2
149
A 2r 2 2rh 2r 2r 3r 2
8r 2
(2) The volume of a sphere increases from
288 cm3 to 290 cm3 . Calculate the approximate increase in its radius.
It is given that h=3r
Approximate change in the total surface area is A A dA A 8 r 2 r dr dA 16r dA dr A r dr New r (7.05) 16 r 7.05 7
16 7 0.05 5.6 cm2
Minus old r(7)
Substitute r with the old value of r, i.e. 7
Hence, the approximate increase in the total surface area of the cylinder is 5.6π cm² .
1 72
Task 4 : Answer all the questions below.
150
cm
(1) Given that y
dy 4 , calculate the value of if 5 dx x
27 dy , find the value of when x = 3. 3 dx x 27 Hence, estimate the value of . 3.033
(2) Given y
x = 2. Hence, estimate the values of 4 4 (a ) (b) 5 2.03 1.98 5 Solution : y
4 4 x 5 x5
dy 20 20 x 6 6 dx x dy 20 5 When x 2, 6 dx 16 x
(a)
y new y original y y original
dy x dx dy dx
151
1
; 0.97
4 4 4 5 y , where y 5 and x 2 2.03 5 2.03 2 2 dy 4 5 x dx 2 4 5 5 2.03 2 2 16 4 3 32 320 0.116525
(3) Given y
32 dy , find . 4 dx x
Hence, estimate the value of
32 . 1.99 4
(b) y new y original y y new y original
dy x dx
4 4 5 5 1.98 2 5 1.98 2 16
4 1 32 160
0.13125
Task 5 : Answer all the questions below. 20 (1) Given that y , find the approximate 2 x change in x when y increases from 40 to 40.5.
dy dx
128 x5
; 2.04
(2) SPM 2003 (Paper 1 – Question 16) Given that y x 2 5 x, use differentiation to find the small change in y when x increases from 3 to 3.01. [3 marks]
152
1 160
(3) Given v
4 r 3 , use the differentiation method 3
0.11
(4) Given that y
to find the small change in v when r increases from 3 to 3.01.
5 dy when , find the value of 3 dx x
x = 4. Hence, estimate the value of (a)
5
4.02
3
dy dx
0.27
(b)
15 256
dy 0 will minimize y if the value of dx dy x = β that satisfies 0 will maximize y if the value of dx Task 1 : Answer all questions below.
153
3.99 3
; (a) 0.07930 ; (b) 0.07871
Topic : DIFFERENTIATION Unit L : Problems of Maximum and Minimum Values of a Function.
x = α that satisfies
5
d2y 0 at x = α dx 2 d2y 0 at x = β dx 2
(1) Given L x 2 (25 2 x) , find the value of x for which L is maximum. Hence, determine the maximum value of L.
Example 1 : Given L x 2 (40 5 x) , find the value of x for which L is maximum. Hence, determine the maximum value of L. Solution: L x 2 ( 40 5 x ) L 40 x 2 5 x 3 dL 80 x 15 x 2 dx d 2L 80 30 x dx 2 dL when 0 dx 80 x 15 x 2 0 x(80 15 x ) 0 Knowing x 0, 80 15 x 0 16 x 3 2 d L 16 80 30 80 2 dx 3 d 2L 80 0 dx 2 16 x will max imise L 3 16 16 40 5 3 3 10240 27 2
Lm ax
x
25 3
, Lm ax 15625 27
(2) Given y 36 x 4 x 2 7 , find the value of x for which y is maximum. Hence, determine the maximum value of y.
x 92 , y m ax 88 Task 2 : Answer all questions below.
154
Example 2 :
16 Given L 3 ( x 2 ) , find the value of x for x
(1) Given L 4 ( x 2
128 ) , find the value of x for x
which L is minimum. Hence, determine the minimum value of L.
which L is minimum. Hence, determine the minimum value of L. Solution: 16 ) x L 3x 2 48 x 1 dL 6x 48x 2 dx d 2L 6 96x 3 dx 2 dL when 0 dx 6x 48x 2 0 L 3 ( x 2
6x 48x 2 48 x2 Knowing x 0, x 3 8 x2
6x
d 2L 6 96 ( 2) 3 2 dx d 2L 18 0 dx 2 x 2 will min imise L 2 16 Lm in 3 2 2 36
x 4 , Lmin 192
(2) Given y
1 (3x 2 8 x 2) , find the value of x 2
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 (b) the maximum value of y [3 marks]
155
x 43 , y m in 53
(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. h x 3x 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) Vm ax = 18291.75 cm3 (2) The diagram below shows a solid cylinder with circular cross-section of radius r and the height of h. r 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. 150r 2 (b) – (c) r
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. (a) h =
r θ
156
(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) Am ax 1600
157
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