A curve C has the equation $y^3 - 3y = x^3 + 8.$ (a) Find \( \frac{dy}{dx} \) in terms of x and y - Edexcel - A-Level Maths Pure - Question 3 - 2009 - Paper 3

To find the gradient of C when ( y = 3 ), we substitute ( y = 3 ) into the equation we derived:

Using:

$\frac{dy}{dx} = \frac{x^2}{y^2 - 1},$

we first find ( \frac{dy}{dx} ):

Substituting ( y = 3 ):

$\frac{dy}{dx} = \frac{x^2}{3^2 - 1} = \frac{x^2}{9 - 1} = \frac{x^2}{8}.$

Next, we need to find the corresponding x value when ( y = 3 ). Substituting ( y = 3 ) back into the original equation:

$3^3 - 3(3) = x^3 + 8$

This simplifies to:

$27 - 9 = x^3 + 8 \Rightarrow 18 = x^3 + 8 \Rightarrow x^3 = 10 \Rightarrow x = \sqrt[3]{10}.$

Now substituting this value into the gradient equation:

$\frac{dy}{dx} = \frac{(\sqrt[3]{10})^2}{8} = \frac{10^{2/3}}{8}.$

So the gradient of C at the point where ( y = 3 ) is:

$$\frac{10^{2/3}}{8}.$