Differentiating Parametric Equations is Simple

Recall: The chain rule:

\dfrac{dy}{dx}=\dfrac{dy}{dt}\dfrac{dt}{dx}

By flipping the last fraction:

\dfrac{dy}{dx}=\dfrac{dy}{dt}\div\dfrac{dx}{dt}

So all we need to do is differentiate x and y with respect to t.

Example: A parametric equation is y=t^{2}, x=3t+6. Find \dfrac{dy}{dx} in terms of t.

\dfrac{dy}{dt}=2t

\dfrac{dx}{dt}=3

\begin{aligned}\dfrac{dy}{dx}&=2t\div3\\[1.2em]&=\dfrac{2t}{3}\end{aligned}

Tangents and Normals of Parametric Equations

You could be asked to find the gradient of the tangent or the normal of a parametric equation. The gradient of the tangent is just \dfrac{dy}{dx}, while the gradient of the normal is -1 divided by \dfrac{dy}{dx}. You will need to evaluate these gradients for specific values of t, that will either be given to you in the question or that you will be required to work out from a given x or y value.

Example: Find the gradient of the tangent to x=t^{2}, y=t^{2}+6t-7, at t=12.

\dfrac{dy}{dx}=\dfrac{dy}{dt}\div\dfrac{dx}{dt}

\dfrac{dy}{dt}=2t+6

\dfrac{dx}{dt}=2t

\begin{aligned}\dfrac{dy}{dx}&=(2t+6)\div(2t)\\[1.2em]&=\dfrac{2t+6}{2t}\\[1.2em]&=\dfrac{t+3}{t}\end{aligned}

Substitute in t=12 to get tangent.

\begin{aligned}\dfrac{dy}{dx}&=\dfrac{12+3}{12}\\[1.2em]&=\dfrac{15}{12}\\[1.2em]&=\dfrac{5}{4}\end{aligned}