9. Parametric Equations

Parametric Equations

Understanding Parametric Equations

Parametric equations are a way to represent the coordinates of points on a curve using a set of equations. These equations express the coordinates in terms of one or more parameters, typically denoted by 't'. By varying the parameter 't', we can trace out the curve.

A set of parametric equations can be written as:

$\begin{array}{rcl} x & = & f (t) \\ y & = & g (t) \end{array}$

where f(t) and g(t) are functions of the parameter 't'.

Example 1:

A classic example of a parametric equation is the representation of a circle:

$\begin{array}{rcl} x & = & a \cos t \\ y & = & a \sin t \end{array}$

Here, 'a' is the radius of the circle, and 't' varies from 0 to 2π. As 't' changes, the values of x and y trace out a circle with radius 'a'.

Eliminating the Parameter

To find the Cartesian equation of a curve from its parametric equations, you need to eliminate the parameter 't'. This can be done by expressing 't' in terms of 'x' or 'y' and substituting it back into one of the parametric equations.

Example 2:

Consider the following parametric equations:

$\begin{array}{rcl} x & = & 2 t + 1 \\ y & = & t^{2} - 1 \end{array}$

To eliminate 't', solve the first equation for 't':

$t = \frac{x - 1}{2}$

Substitute this expression for 't' into the second equation:

$y = {(\frac{x - 1}{2})}^{2} - 1$

Simplify the equation to obtain the Cartesian equation:

$\begin{array}{rcl} y & = & \frac{x^{2} - 2 x + 1}{4} - 1 \\ = & \frac{1}{4} x^{2} - \frac{1}{2} x - \frac{3}{4} \end{array}$

More information: Eliminating the parameter of parametric equations

Differentiation and Integration of Parametric Equations

To differentiate or integrate a parametric equation, you need to use the chain rule. This involves differentiating both the 'x' and 'y' equations with respect to 't' and then using these derivatives to find $\frac{d y}{d x}$ or the integral of y with respect to x.

Example 3:

Differentiate the parametric equations from Example 2:

$\begin{array}{rcl} \frac{d x}{dt} & = & 2 \\ \frac{dy}{dt} & = & 2 t \end{array}$

To find $\frac{d y}{d x}$ , divide $\frac{d y}{d t}$ by $\frac{d x}{d t}$ :

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

Example 4:
Integrating the parametric equations from Example 2:

$\begin{array}{rcl} \int y d x & = & \int y \frac{d x}{d t} d t \\ = & \int (t^{2} - 1) \times 2 d t \\ = & \frac{2}{3} t^{3} - 2 t + c \end{array}$

Parametric Equation Exam Tips

Be familiar with common parametric representations (for example, a circle mentioned above)
Trigonometric identities can often help to eliminate parameters
Pay attention to any specific domain of 't' which determines the range of values it can take

Parametric Equations Practice Questions

To help you master parametric equations, try these parametric equation topic questions.

Edexcel A Level Maths: Pure

Revision Notes