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4. (a) 2πr + 2πrh = 500 dx = dx × dy
2
2πrh = 500 – 2πr dt dy dt
2
250 – πr 2 2
h = = × (–21)
πr 3(4) 2
[1]
= – 7
(b) V = πr h 8 [3]
2
= πr × 250 – πr 2
2
πr 9. (a) Surface area, A = 4πr
2
= 250r – πr 3 dA
dr = 8πr
dV = 250 – 3πr dA
2
dr dA = dr × dr
250 – 3πr = 0 = 8π(1.5) × 0.25
2
r = 5.15 cm
= 3π
250 – π(5.15) 2 [3]
h = 4
π(5.15) (b) Volume, V = πr
3
= 10.3 cm dV 3
= 4πr
2
2
V = π(5.15) (10.3) dr dV
2
= 1 = 858.23 cm [5] dV = dr × dr
2 = 4π(1.5) × 0.25
2
= √3 5. Let the side of cube = s = 2.25π
3
V = s [3]
= √(7) – dV dy
2
2
ds = 3s 10. (a) dx = 6x
2
2
= 1 8 + 3 dV = dV × ds When x = 1.2,
2
1 dt ds dt dy = 6(1.2)
= 2 2 = 3(30) × 0.05 dx = 7.2
2
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= 135 cm s
3 –1
y = 3(1.2) + 5
2

[3]
= 9.32
6. dy = 3x + 6x Equation of the tangent at point A,
2
–3
dx
dy = dy × dx y – 9.32 = 7.2(x – 1.2)
dx y = 7.2x + 0.68
= [3(2) + 6(2) ] × 0.3 [4]
2
–3
= 3.825 (b) dy = dy × dx
[3] dx
= 6(0.95) × (–0.25)
7. A = 4πx + 24πx = –1.425
3
–1
dA = 12πx – 24πx [2]
2
–2
dx
2
12πx = 24πx –2 11. xy + 2y = –2x
2
2
x = 2 y(x + 2) = –2x
4
1 y = –2x 2
x = 2 4 x + 2
1 1 –1 dy –4x(x + 2) + 2x 2
3
A = 4π(2 ) + 24π(2 ) dx = (x + 2) 2
4
4
= 84.54 m 2
2
[4] = –2x – 8x
(x + 2) 2
3
8. dy = x dy
2
dx 2 When x = –3, dx = 6
When y = 8, x = 4 When x = –3, y = 18
y – 18 = 6(x + 3)
y = 6x + 36
[5]
Cambridge IGCSE
TM
202 Ace Your Additional Mathematics
Answers Add Math.indd 202 14/03/2022 12:29 PM