A curve is defined by the parametric equations $x = t^3 + 2 ,$ $y = t^2 - 1$ 3 (a) Find the gradient of the curve at the point where $t = -2$ 3 (b) Find a Cartesian equation of the curve. - AQA - A-Level Maths Pure - Question 3 - 2017 - Paper 2

To find the gradient of the curve defined by the parametric equations, we need to apply the formula for the gradient in parametric form:

$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

First, we differentiate $x$ and $y$ with respect to $t$:

1. For $x = t^3 + 2$, we get: $\frac{dx}{dt} = 3t^2$

2. For $y = t^2 - 1$, we find: $\frac{dy}{dt} = 2t$

Now, substituting these into the gradient formula:

$\frac{dy}{dx} = \frac{2t}{3t^2} = \frac{2}{3t}$

Next, we evaluate the gradient at $t = -2$:

$\frac{dy}{dx} = \frac{2}{3(-2)} = \frac{2}{-6} = -\frac{1}{3}$

Thus, the gradient of the curve at the point where $t = -2$ is $-\frac{1}{3}$.