Short Questions (Question 14 – 16)

Given y = x (6 – x), express $y\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}+18$ in terms of x in the simplest form.

Hence, find the value of x which satisfies the equation  $y\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}+18=0$

Find the coordinates of the point on the curve, y = (4x – 5)² such that the gradient of the normal to the curve is $\frac{1}{8}$ .

y = (4x – 5)²

$\frac{dy}{dx}$ = 2(4x – 5).4 = 32x – 40

Given the gradient of the normal is 1/8, therefore the gradient of the tangent is –8.

$\frac{dy}{dx}$ = –8

32x – 40 = –8

32x = 32

x = 1

y = (4(1) – 5)²= 1

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.

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

$\frac{dy}{dx}$ = 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