Page 34 - Ranger SPM 2022 - Additional Mathematics

Additional Mathematics SPM Chapter 2 Differentiation

1
Example 14 Area = 6(6) – (6) 2
The diagram below shows a garden besides = 36 – 18 2
Joshua’s house. = 18 m 2
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B

Example 15
x m
Garden A boy blows a spherical balloon and its
volume increases with a rate of 10 cm /s.
3
Find the rate of change of its surface area
A y m C
when the radius, r is 4 cm.
BC is a wall from Joshua’s house and the
garden is fenced by 12 m of fencing. Solution
4
(a) Show that the area of the garden is Volume of a sphere, V = πr 3
1
A = 6x – x m . dV 3
2
2
2 = 4πr 2
(b) Find the length of AB and AC so that dr
the area of the garden is maximum. Given dV = 10 cm /s
3
Hence, find the maximum area of the dt
garden. dV = dV × dr

dt dr dt
Solution 2 dr
10 = 4πr × dt
(a) x + y = 12 dr = 10
y = 12 – x dt 4πr 2 ................a
1
Area of the garden, A = xy Surface area of a sphere, A = 4πr 2
2 dA
1
= x(12 – x) dr = 8πr
2
1
= 6x – x 2 dA = dA × dr
2 dt dr dt
dr
(b) For maximum area, = 8πr × dt ................
dA = 0
dx Substitute a into ,
6 – x = 0 dA 10
x = 6 dt = 8πr × 4πr 2
20
y = 12 – 6 = r Form 5
= 6 When r = 4 cm,
Hence, AB = AC = 6 m dA = 20
d A dt 4
2
= –1  0, the area is maximum. = 5 cm /s
2
dx 2

187

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