Page 50 - Focus SPM 2022 - Additional Mathematics

Additional Mathematics SPM Chapter 2 Differentiation
1
≈ 2 × (–0.02) dx = 63.98 – 64 I Solving problems involving small
3
3(64) = –0.02 changes and approximations of
3
≈ 1 × (–0.02) When y = √64 , certain quantities
48 x = 64 1. The formula dy ≈ dy or dy ≈ dy × dx is used in
≈ – 1 or –0.0004167 dx dx dx
2 400 solving problem which involves a small changes,
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either in x or y.
1
1
Therefore, √63.98 ≈ √64 + – 2 400 2 2. Recognising two quantities involved in the
3
3
≈ 4 – 1 problem and forming the function or relation
2 400 between the two quantities are the main things
≈ 3.9996 in solving the problems.
Try Questions 17 – 18 in ‘Try This! 2.4’
24
SPM Highlights Find the approximate change in the surface area of a
sphere when the radius of the sphere decreases from
5 3.0 cm to 2.97 cm.
It is given that the equation of a curve is y = x 2 . Solution
(a) Find the value of dy when x = 3. Area = 4π (radius)
2
dx
2
(b) Hence, estimate the value of 5 . y = 4πx
(2.98) 2
and dy = 8πx
Solution dx = 8π(3)
(a) y = 5x = 24π
–2
When x = 3, dy = –10x dy
–3
dx Therefore, dy ≈ × dx
= – 10 dx
x 3 ≈ 24π × (2.97 – 3.0)
= – 10 ≈ 24π × (–0.03)
(3) 3 ≈ –0.72π cm
2
2
= – 10 \ The surface area of the sphere decreases 0.72π cm .
27 (a negative value is obtained)
5 5
(b) Let = + dy
(2.98) 2 (3) 2 SPM Highlights
such that x = 3 and dx = 2.98 – 3
= –0.02 It is given that L = 4t – t and x = 3 + 6t.
2
Small changes in y, (a) Express dL in terms of t.
dy dx
dy ≈ × dx (b) Find the small change in x when L changes from Form 5
dx
10 3 to 3.4 when t = 1.
≈ – × (–0.02)
27 Solution
1 2
≈ or 0.007407 (a) L = 4t – t x = 3 + 6t
135 dL dx
5 5 1 dt = 4 – 2t dt = 6
Therefore, ≈ +
(2.98) 2 (3) 2 135 dL dL 1
5 1 dx = dt × dx
≈ +
9 135 dt
≈ 76 or 0.5630 = (4 – 2t) × 1
135 2 – t 6
=
3

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