# Differentiating Parametric Equations

## Differentiating Parametric Equations Revision

**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**.

**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}

## Differentiating Parametric Equations Example Questions

**Question 1: **A parametric curve is y=t^{2}+3t+4, x=4t^{3}+15t^{2}+18t. Find \dfrac{dy}{dx}.

**[2 marks]**

\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}

**Question 2: **A parametric equation is y=\sin(\theta), x=\cos(\theta). Find the gradient at \theta=\dfrac{\pi}{6}.

**[3 marks]**

\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}

**Question 3: **What is the gradient of the normal to y=3t^{2}, x=e^{t}, when x=5.

**[4 marks]**

\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}