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### Parametric Functions

#### Definition

A parametric equation is one where the $x$ and $y$ coordinates of the curve are both written as functions of another variable called a parameter; this is usually given the letter $t$ or $\theta$.

For example, $x = t^2,\; y=2t, \qquad x = \sin^2\theta, \; y=\cos^2\theta$

$\theta$ tends to be used when working with trigonometric funtions.

#### Plotting Graphs

##### Definition

The easiest way to plot a curve from parametric equations is to make a table and plot the points. Alternatively, it can be transformed into a single cartesian equation first and then plotted.

###### Worked Example

Plot the curve given by $x=t$ and $y=t^2$.

###### Solution

Make a table and for every value of $t$, work out the corresponding $x$ and $y$ values.

t

-3

-2

-1

0

1

2

3

x

-3

-2

-1

0

1

2

3

y

9

4

1

0

1

4

9

Now we have a table of $x$ and $y$ coordinates which can be easily plotted on a graph.

#### Cartesian Equation

##### Definition

A cartesian equation is only in terms of $x$ and $y$, so we must eliminate the parameter. To do this either, rearrange one of the parametric functions to get $t=\dotso$ and then substitute this into the other equation, or use a trigonometric identity to eliminate the $\theta$.

##### Worked Examples
###### Worked Example

Express the parametric equations $x=t^2$ and $y=2t$ in Cartesian form.

###### Solution

Rearrange the first equation to make $t$ the subject

$t = \sqrt{x}.$

Sub this into the second equation to eliminate the parameter $t$

$y = 2\sqrt{x}.$

So the Cartesian form of these parametric equations is

$y = 2\sqrt{x}.$

###### Worked Example

Express the parametric equations $x = 3\sin\theta$ and $y=4\cos\theta$ in Cartesian form.

###### Solution

Using the identity $\cos^2\theta+\sin^2\theta=1$.

Find $\sin^2\theta$.

\begin{align} x &= 3\sin\theta\\\\ \frac{x}{3} &= \sin\theta\\\\ \sin^2\theta &= \frac{x^2}{9} \end{align}

And $\cos^2\theta$.

\begin{align} y &= 4\cos\theta\\\\ \frac{y}{4} &= \cos\theta\\\\ \cos^2\theta &= \frac{y^2}{16} \end{align}

Subbing these into the identity will eliminate $\theta$ and give the Cartesian form.

$\frac{y^2}{16} + \frac{x^2}{9} = 1$

#### Finding the Gradient

##### Definition

To find the gradient, we use the Chain Rule. We differentiate both our equations and use the rule: $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}y}{\mathrm{d}t} \frac{\mathrm{d}t}{\mathrm{d}x}$ Alternatively, the parametric equations can be transformed into cartesian equations first and then differentiated as normal.

##### Worked Example

Find the gradient of the curve given by the parametric equations $x=t^2$ and $y=2t$.

###### Solution

Using $\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\mathrm{d}y}{\mathrm{d}t} \dfrac{\mathrm{d}t}{\mathrm{d}x}$.

First find $\dfrac{\mathrm{d}y}{\mathrm{d}t}$.

$y = 2t \; \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}t} = 2$

Then find $\dfrac{\mathrm{d}x}{\mathrm{d}t}$.

$x = t^2 \; \Rightarrow \dfrac{\mathrm{d}x}{\mathrm{d}t} = 2t$

Finally, substitute these into the chain rule formula given above. Don't forget to flip $\dfrac{\mathrm{d}x}{\mathrm{d}t}$ as we need $\dfrac{\mathrm{d}t}{\mathrm{d}x}$ in the formula.

$\frac{\mathrm{d}y}{\mathrm{d}x} = 2 \times \frac{1}{2t}$

$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2}{2t}$

$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{t}$

#### Workbook

This is a workbook on graphs of functions and parametric form produced by HELM.

This is a workbook on parametric curves produced by HELM.

#### More Support

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