# Differentiating Parametric Equations

A LevelAQAEdexcelOCRAQA 2022Edexcel 2022

## Differentiating Parametric Equations

Recall: Parametric equations are equations that are written as $x=f(t)$, $y=g(t)$, rather than $y=f(x)$.

On the face of it, differentiating them might seem difficult. However, it is made easier by again treating $\dfrac{dy}{dx}$ as a regular fraction.

A Level

## 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}

A Level

## 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}

A Level

## Example Questions

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

$\dfrac{dy}{dt}=2t+3$

$\dfrac{dx}{dt}=12t^{2}+30t+18$

\begin{aligned}\dfrac{dy}{dx}&=\dfrac{2t+3}{12t^{2}+30t+18}\\[1.2em]&=\dfrac{2t+3}{6(2t^{2}+5t+3)}\\[1.2em]&=\dfrac{2t+3}{6(2t+3)(t+1)}\\[1.2em]&=\dfrac{1}{6(t+1)}\end{aligned}
$\dfrac{dy}{dx}=\dfrac{dy}{dt}\div\dfrac{dx}{dt}$

$\dfrac{dy}{dt}=\cos(\theta)$

$\dfrac{dx}{dt}=-\sin(\theta)$

$\dfrac{dy}{dx}=-\dfrac{\cos(\theta)}{\sin(\theta)}$

At $\theta=\dfrac{\pi}{6}$:

\begin{aligned}\dfrac{dy}{dx}&=-\dfrac{\cos\left(\dfrac{\pi}{6}\right)}{\sin\left(\dfrac{\pi}{6}\right)}\\[1.2em]&=-\dfrac{\dfrac{\sqrt{3}}{2}}{\dfrac{1}{2}}\\[1.2em]&=-\sqrt{3}\end{aligned}
$\dfrac{dy}{dx}=\dfrac{dy}{dt}\div\dfrac{dx}{dt}$

$\dfrac{dy}{dt}=6t$

$\dfrac{dx}{dt}=e^{t}$

$\dfrac{dy}{dx}=\dfrac{6t}{e^{t}}$

Hence: $\text{normal}=\dfrac{-e^{t}}{6t}$

Find value of $t$

$x=5$

$e^{t}=5$

$t=\ln(5)$

Substitute in to find normal:

\begin{aligned}\text{normal}&=\dfrac{-e^{\ln(5)}}{6\ln(5)}\\[1.2em]&=\dfrac{-5}{6\ln(5)}\end{aligned}

A Level

A Level

A Level

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