Page 48 - Focus SPM 2022 - Additional Mathematics

Additional Mathematics SPM Chapter 2 Differentiation

From dy = dy × dx G Solving problems involving rates of
dt dx dt change for related quantities and
3
8.5 = – × dx interpreting the solutions
2 dt
\ The rate of change of x, dx = – 17 units per second. 1. The rate of change of variable (or quantity y)
dt 3 can be determined by using the formula
x experiences a rate of decreasing at the moment when dy dy dx
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x = 2 because dx , 0. dt = dx × dt if y = f(x) and x = f(t).
dt
2. Observe that dy is the rate of change of y and
20 dx dt

The relationship between two variables p and q is dt is the rate of change of x.
p q = 4. Given that p decrease with a rate of 3 unit s ,
2
–1
find the rate of change of q at the moment when q = 64. 3. In the process of solving problems involving
rate of change of related quantities, recognising
Solution the quantities of x and y and the relationship
Given dp = –3 (Negative because the rate decreased) between these quantities are the main focus.
dt
From p q = 4,
2
q = 4 21
p 2 The area of the wave of water on the surface of a water
–2
q = 4p tank expands with a rate of 5 cm s . Find the rate of
2 –1
dq = –8p change of its cicular radius when the area of the circle
–3
dp is 6.25π cm .
2
= – 8
p 3 Solution
Given q = 64, therefore p q = 4 dA 2 –1
2
p (64) = 4 Given dt = 5 cm s
2
2
p = 1 Also, A = πr
2
16 dA
p = ± 1 dr = 2πr
4 When A = 6.25π
1
When p = ± , πr = 6.25π
2
4
dq = – 8 or dq = – 8 r = 2.5 cm
1
dp 1 2 3 dp 1 – 1 2 3 From dA = dA × dr
4
4
= –512 = 512 When r = 2.5, dt dr dt Form 5
Therefore, dq = dq × dp dr
dt dp dt 5 = 2π(2.5) × dt
dq dq dr 5
When = –512, = –512 × (–3) dt = 5π
dp dt = 1 536 unit s 1 –1
–1
π
When dq = 512, dq = 512 × (–3) = cm s
dp dt = –1 536 unit s \ The radius of the circular wave increases with a rate
–1
(q may increase or decrease at the rate of 1 536 units 1
–1
2
per second.) of cm s when the area of the circle is 6.25π cm .
π
Try Questions 13 – 14 in ‘Try This! 2.4’

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