CAS header

 

Find this content useful?
Think it should stay ad-free?

Parametric Curves

Parametric Curves

A curve is called a parametric curve when the coordinates of the points belonging to the curve are specified in terms of another variable (usually $t,\;\lambda,\;\theta$ etc).

For example, the quadratic curve $C$ with equation $y=x^2$ could be described parametrically as $\left(t,t^2\right),\; t\in \mathbb{R}$. Equally however, it could be specified as something a little more strange such as $\left(\tan t, \sec^2t - 1\right),\; -\frac{\pi}{2}\lt t \lt \frac{\pi}{2}$

Parametric curves can be very useful for describing curves that don't have equations of the form $y = f(x)$.

circle with t valuesA circle, centre $(2,3)$ and radius $4$ can be expressed parametrically as

$$x(t) = 2 +4 \sin(t),\; y(t) = 3 + 4\cos(t),\; 0\lt t \lt 2\pi$$

The image shows the parametric curve with points highlighted for $t = 0,\frac{\pi}{4}, \frac{\pi}{2}, \dots, \frac{7\pi}{4}$. 

Note that the direction of increasing t is anticlockwise around the circle, from the highest point!

We can find the Cartesian equation of the curve by determining what $\sin(t)$ and $\cos(t)$ are in terms of $x$ and $y$ and then applying the Pythagorean identities:

$$\sin(t) = \frac{x-2}{4},\;\cos(t) = \frac{y-3}{4}$$

By application of the Pythagorean identities, we have

$$\left(\frac{x-2}{4}\right)^2 + \left(\frac{y-3}{4}\right)^2= 1$$ 

This can be rewritten as $(x-2)^2 + (y-3)^2 = 4^2$ which is the classic form for the equation of a circle.

Conversion between parametric and cartesian equations can however be quite tricky (and sometimes effectively impossible), so a number of classical techniques are demonstrated in the examples that follow.

Parametric Curves 1

The curve $C$ can be defined parametrically by $$x(t) = \frac{1}{1-t},\; y(t) = \frac{2t}{1-t}, 0 \lt t \lt 2$$ Determine the Cartesian equation of $C$
solution - press button to display
To solve this, we first note the common denominator of the $x(t)$ and $y(t)$ functions.  $$x(t) = \frac{1}{1-t},\; y(t) = \frac{2t}{1-t} $$ linear combinations of $x(t)$ and $y(t)$ will therefore potentially simplify the problem.

$$x(t) - \frac{1}{2}y(t) = \frac{1}{1-t} - \frac{t}{1-t} = \frac{1-t}{1-t} = 1$$ Hence the Cartesian equation is $y = 2x-2$

However, the question is not quite complete because the domain specified for $t$ limits what part of the line is defined. Since $0\lt t \lt 2$ and $x(t) = \frac{1}{1-t}$, then $x \gt 1$ or $x \lt -1$

Parametric Curves 2

The ellipse $C$ is defined parametrically by the equations $$ x(\theta) =1 + 4\sin(\theta),\;y(\theta) = 2 - 2\cos(\theta) $$ Sketch the curve C and determine its Cartesian equation

solution - press button to display

The following diagram highlights the main properties of the curve

ellipse

To determine the Cartesian equation of the curve, we note that the functions $\sin(\theta)$ and $\cos(\theta)$ feature in the equation and so we make these functions the subject of their respective equations

$$ \begin{align} x &= 1 + 4\sin\theta\\ \frac{x-1}{4} &= \sin\theta \end{align} $$ $$ \begin{align} y &= 2 -2\cos\theta\\ \frac{y-2}{-2} &= \cos\theta \end{align} $$

We now apply the Pythagorean identity: $$\sin^2(\theta) + \cos^2(\theta) \equiv 1$$

Substitution of our two equations above gives: $$ \left(\frac{x-1}{4}\right)^2 + \left(\frac{y-2}{-2}\right)^2 = 1 $$