Page 37 - Ranger SPM 2022 - Additional Mathematics

Additional Mathematics SPM Chapter 2 Differentiation
1 (b) (i) The equation of the tangent:
(b) When x = 2, y =
2 3 y = –5x + 2
= 1 Then, the gradient = –5
δx = 1.98 – 2 8 kx + 2 = –5
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= –0.02 At x = –2, (–2)k + 2 = –5
1 dy –2k = –7
1.98 3  y + dx δx k = 7
1

= + – 3  (–0.02) 2
8 16 (ii) Equation of the normal
1
= 0.12875 y – 10 = (x + 2)
3 = 3(0.12875) 5 52
1
1.98 3 = 0.38625 y = x + 5
5
5. (a) Given y = 2x – x + 3, solve the 6. A cylindrical container is open at
2
d y dy 2
2
+
equation 2   – 2y = 5. the top has a base radius of j cm. The
dx dx total surface area of the container is
(b) Given (kx + 2) is the gradient 300π cm .
2
function of a curve where k is (a) Show that the volume, I cm , of the
3
a constant. y + 5x – 2 = 0 is the πj 2
equation of a tangent to the curve at container is I = 2 (300 – j ).
point (–2, 10). Find (b) Find the value of j when dI = 0.
(i) the value of k, dj
(ii) the equation of the normal at the (c) Show that the value of j obtained
point (–2, 10). in (b) will make the volume of
the container maximised. Hence,
Solution calculate the maximum volume of
(a) y = 2x – x + 3 the container.
2
dy
= 4x – 1 Solution
dx
d y (a) Let h = height of the container
2
= 4
dx 2 Surface area of the container
2
d y dy 2 L = πj + 2πjh
2
dx 2   – 2y = 5 300π = π( j + 2jh)
+
2
dx
4 + (4x – 1) – 2(2x – x + 3) = 5 2jh = 300 – j 2
2
2
4 + 16x – 8x + 1 – 4x + 2x – 6 = 5 h = 300 – j 2
2
2
2j
12x – 6x – 6 = 0
2
Form 5 (2x + 1)(x – 1) = 0 I = πj h 300 – j 2
2x – x – 1 = 0
2
2
1
x = – , 1
2 = πj 2  2j 
πj
= 2 (300 – j )
2
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