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2.10.4 Differentiation Short Questions (Question 15 – 18)


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

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