A-Level — mathsAdvanced1 min read

Parametric Differentiation

Differentiate parametric equations at A-Level using the chain rule.

Tutor AI Team2026-02-25T01:57:10.944Z

For parametric curves $x = f(t)$ , $y = g(t)$ , we find $\frac{dy}{dx}$ using the chain rule.

The Formula

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

Problem

$x = t^2$ , $y = 2t$ .

$\frac{dx}{dt} = 2t$ , $\frac{dy}{dt} = 2$ . $\frac{dy}{dx} = \frac{2}{2t} = \frac{1}{t}$ .

At $t = 3$ : gradient = $\frac{1}{3}$ . Point: $(9, 6)$ . Tangent: $y - 6 = \frac{1}{3}(x - 9)$ .

Solution

Problem

$x = 3\cos\theta$ , $y = 3\sin\theta$ .

$\frac{dy}{dx} = \frac{3\cos\theta}{-3\sin\theta} = -\cot\theta$ .

Solution

1. $x = t^3$ , $y = t^2$ . Find $\frac{dy}{dx}$ at $t = 2$ .
1. $x = e^t$ , $y = e^{2t}$ . Find $\frac{dy}{dx}$ .
1. Find the equation of the tangent to $x = 4\cos\theta$ , $y = 3\sin\theta$ at $\theta = \frac{\pi}{4}$ .

Want to check your answers and get step-by-step solutions?

✓
$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ .
✓
Find tangents by evaluating at a specific $t$ .
✓
For second derivative: differentiate $\frac{dy}{dx}$ with respect to $t$ , then divide by $\frac{dx}{dt}$ .

Tags:

#a-level#maths#pure#calculus#parametric

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