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2.10.2 Differentiation SPM Practice (Paper 1 Question 11 – 20)


Question 11:
Given that the graph of function f ( x ) = h x 3 + k x 2  has a gradient function f ‘ ( x ) = 12 x 2 − 258 x 3 such that h and k are constants. Find the values of h and k.

Solution:
f ( x ) = h x 3 + k x 2 = h x 3 + k x − 2 f ‘ ( x ) = 3 h x 2 − 2 k x − 3 f ‘ ( x ) = 3 h x 2 − 2 k x 3 But it is given that  f ‘ ( x ) = 12 x 2 − 258 x 3 Hence, by comparison,  3 h = 12  and   2 k = 258 h = 4    k = 129


Question 12:
Given that  y = x 2 x + 3 , show that  d y d x = x 2 + 6 x ( x + 3 ) 2 Find  d 2 y d x 2  in the simplest form .

Solution:
y = x 2 x + 3 d y d x = ( x + 3 ) ( 2 x ) − x 2 .1 ( x + 3 ) 2 = 2 x 2 + 6 x − x 2 ( x + 3 ) 2 d y d x = x 2 + 6 x ( x + 3 ) 2  (shown) d 2 y d x 2 = ( x + 3 ) 2 ( 2 x + 6 ) − ( x 2 + 6 x ) .2 ( x + 3 ) ( x + 3 ) 4 d 2 y d x 2 = ( x + 3 ) [ ( x + 3 ) ( 2 x + 6 ) − 2 ( x 2 + 6 x ) ] ( x + 3 ) 4 d 2 y d x 2 = [ 2 x 2 + 6 x + 6 x + 18 − 2 x 2 − 12 x ] ( x + 3 ) 3 d 2 y d x 2 = 18 ( x + 3 ) 3

Question 13:
If  y = x 2 + 4 x , show that  x 2 d 2 y d x 2 − 2 x d y d x + 2 y = 0.

Solution:
y = x 2 + 4 x d y d x = 2 x + 4 d 2 y d x 2 = 2 x 2 d 2 y d x 2 − 2 x d y d x + 2 y = x 2 ( 2 ) − 2 x ( 2 x + 4 ) + 2 ( x 2 + 4 x ) = 2 x 2 − 4 x 2 − 8 x + 2 x 2 + 8 x = 0  (Shown)


Question 14:
Given y = x (6 – x), express y d 2 y d x 2 + x d y d x + 18 in terms of x in the simplest form.

Hence, find the value of x which satisfies the equation  y d 2 y d x 2 + x d y d x + 18 = 0

Solution:
y = x ( 6 − x ) = 6 x − x 2 d y d x = 6 − 2 x d 2 y d x 2 = − 2 ∴ y d 2 y d x 2 + x d y d x + 18 = ( 6 x − x 2 ) ( − 2 ) + x ( 6 − 2 x ) + 18     = − 12 x + 2 x 2 + 6 x − 2 x 2 + 18     = − 6 x + 18 y d 2 y d x 2 + x d y d x + 18 = 0    − 6 x + 18 = 0   x = 3


Question 15:
Find the coordinates of the point on the curve, y = (4x – 5)2 such that the gradient of the normal to the curve is 1 8 .

Solution:
y = (4x – 5)2
d y d x = 2(4x – 5).4 = 32x – 40

Given the gradient of the normal is 1/8, therefore the gradient of the tangent is –8.
d y d x = –8
32x – 40 = –8
32x = 32
x = 1
y = (4(1) – 5)2= 1

Hence, the coordinates of the point on the curve, y = (4x – 5)2 is (1, 1).



Question 16:
A curve has a gradient function of kx2 – 7x, where k is a constant. The tangent to the curve at the point (1, 4) is parallel to the straight line y + 2x–1 = 0. Find the value of k.

Solution:
Gradient function of kx2– 7x is parallel to the straight line y + 2x–1 = 0
d y d x = kx2– 7x

y + 2x –1 = 0, y = –2x + 1, gradient of the straight line = –2
Therefore kx2– 7x = –2

At the point (1, 4),
k(1)2 – 7(1) = –2
k – 7 = –2
k = 5


Question 17:

In the diagram above, the straight line PR is normal to the curve   y = x 2 2 + 1 at Q. Find the value of k.

Solution:
y = x 2 2 + 1 d y d x = x At point  Q ,   x -coordinate = 2 , Gradient of the curve,  d y d x = 2 Hence, gradient of normal to the curve,  P R = − 1 2 3 − 0 2 − k = − 1 2 6 = − 2 + k k = 8


Question 18:
The normal to the curve y = x2 + 3x at the point P is parallel to the straight line 
y = –x + 12. Find the equation of the normal to the curve at the point P.

Solution:
Given normal to the curve at point P is parallel to the straight line y = –x + 12.
Hence, gradient of normal to the curve = –1.
As a result, gradient of tangent to the curve = 1

y = x2 + 3x
d y d x = 2x + 3
2x + 3 = 1
2x = –2
x = –1
y = (–1)2+ 3(–1)
y = –2
Point P = (–1, –2).

Equation of the normal to the curve at point P is,
y – (–2) = –1 (x – (–1))
y + 2 = – x – 1
y = – x– 3


Question 19:
The volume of water V cm3, in a container is given by   V = 1 5 h 3 + 7 h , where h cm is the height of the water in the container. Water is poured into the container at the rate of 15cm3s-1. Find the rate of change of the height of water in cms-1, at the instant when its height is 3cm. 

Solution:
V = 1 5 h 3 + 7 h d V d h = 3 5 h 2 + 7 = 3 h 2 + 35 5 Given  d V d t = 15 ,  h = 3 Rate of change of the height of water = d h d t d h d t = d h d V × d V d t ← Chain rule d h d t = 5 3 h 2 + 35 × 15 d h d t = 75 62  cms − 1



Question 20:
A wire of length 88 cm is bent to form a circle. When the wire is heated, the length increases at the rate of 0.3 cms-1.
(a) Calculate the rate of change in the radius of the circle.
(b) Hence, calculate the radius of the circle after 5s.

Solution:
Length of circumference of a circle,  L = 2 π r d L d r = 2 π

(a)
Given  d L d t = 0.3 Rate of change in the radius of the circle = d r d t d r d t = d r d L × d L d t d r d t = 1 2 π × 0.3 d r d t = 0.0477  cms − 1

(b)
2 π r = 88 r = 88 2 π = 44 π Hence, the radius of the circle after  5 s = 44 π + 5 ( 0.0477 ) = 14.24  cm


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