Page 36 - Ranger SPM 2022 - Additional Mathematics

Additional Mathematics SPM Chapter 2 Differentiation
For the perimeter to be minimum, (b) Equation of the tangent
dP = 0 4 4

dx y – = 4 x – 3 
3
2 – 200π = 0 y = 4x – 16 + 4
(a) the coordinates of point A,Sdn Bhd. All Rights Reserved.
x
2
3
3
2
2x = 200π y = 4x – 4
x = 10π
3. The equation of a curve is y = x + 2x – 3.
2
200π
∴ P = 210π  + dy
10π (a) Find the value of dx when x = –2.
= 20 + 20 (b) Calculate the approximate change in
π
π
= 40 cm y in terms of h when x changes from
π
–2 to (–2 + h) where h is a small
CAUTION! value.
Solution
For the perimeter to be minimum,
2
d P 400π (a) y = x + 2x – 3
2
dx 2 = x 3 dy
When x = 10, dx = 2x + 2
π
d P 400π  0 dy
2
dx 2 = 10π  3 When x = –2, dx = 2(–2) + 2
Hence, the perimeter is minimum = –2
(b) When x = –2
and δx = –2 + h – (–2)
2. The point A lies on the curve = h
1
y = (3x – 2) . Given that the gradient δy  dy × δx
2
3
dx
1
of the normal at A is – . Find δy = –2h
4
Penerbitan Pelangi
1
(b) the equation of the tangent at point 4. Given that y = x 3 .
A. dy
(a) Find the value of when x = 2.
Solution dx 3
1 (b) Hence, estimate the value of 1.98 3 .
(a) y = (3x – 2) 2
3
dy 2 Solution
dx = (3x – 2)(3) 1
3
–3
= 2(3x – 2) (a) y = x 3 = x
Gradient of tangent = 4 dy = –3x –4 Form 5
2(3x – 2) = 4 dx
4
x = When x = 2,
3 dy 3
1 4 2 = – 4
– 2
Therefore, y =     dx 2
3
3 3 = – 3
= 4 16
3
,

Then, A  4 4 
3 3
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02 Ranger Mate Tambahan Tg5.indd 189 25/02/2022 9:23 AM