Differentiating Parametric Equations

Differentiating Parametric Equations is Simple

Recall: The chain rule:

$dfrac{dy}{dx}=dfrac{dy}{dt}dfrac{dt}{dx}$

By flipping the last fraction:

$dfrac{dy}{dx}=dfrac{dy}{dt}divdfrac{dx}{dt}$

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 $t$.

$dfrac{dy}{dt}=2t$

$dfrac{dx}{dt}=3$

$begin{aligned}dfrac{dy}{dx}&=2tdiv3[1.2em]&=dfrac{2t}{3}end{aligned}$

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 $t=12$.

$dfrac{dy}{dx}=dfrac{dy}{dt}divdfrac{dx}{dt}$

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

$dfrac{dx}{dt}=2t$

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

Substitute in $t=12$ to get tangent.

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