A-level · Differentiation

Parametric differentiationExam-style questions

Finding the gradient of a curve whose x and y are both given in terms of a parameter. Why the gradient is dy/dt divided by dx/dt; finding the value of t at a point before substituting; tangents and normals through the point, not through t; horizontal and vertical tangents; and the second derivative, where the obvious shortcut fails.

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30 questions · A-level level · mark scheme and worked solution on each · no sign-up to try

DifficultyAll Level 2 Level 3 Level 4 Level 5

1
A curve has parametric equations $x = t^2,\ y = t^3$ . Find $\dfrac{dy}{dx}$ in terms of $t$ .
Level 2 · 2 marksTry it
2
A curve has parametric equations $x = t^2 + 1,\ y = t^3 - 3t$ . Find $\dfrac{dy}{dx}$ in terms of $t$ , and its value at $t = 2$ .
Level 2 · 3 marksTry it
3
A curve has parametric equations $x = 4t, \; y = t^{2} - 1$ .

Find $\dfrac{dy}{dx}$ in terms of $t$ . (2 marks)
Level 2 · 2 marksTry it
4
An ellipse has parametric equations $x = 2\cos t,\ y = 3\sin t$ . Find $\dfrac{dy}{dx}$ in terms of $t$ , and its exact value at $t = \dfrac{\pi}{6}$ .
Level 3 · 3 marksTry it
5
A curve has parametric equations $x = t + \dfrac{1}{t},\ y = t - \dfrac{1}{t}$ , $t \ne 0$ . Show that $\dfrac{dy}{dx} = \dfrac{t^2 + 1}{t^2 - 1}$ .
Level 3 · 3 marksTry it

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