What are parametric equations?

Parametric equations express the coordinates x and y of a curve as functions of a third independent variable known as a parameter (usually t). They are often used to describe complex curves and motion in calculus and physics.

Parametric Equations – Definitions, Examples & Solutions

Parametric equations are presented with examples and their solutions. More questions with solutions are included.

In what follows, \( C \) represents the constant of integration where applicable.

Examples

Click on each example to view its detailed step-by-step solution.

Example 1

Some curves are best described using parametric equations \( x \) and \( y \) in terms of a parameter [1] [2].
The following is an example of parametric equations \( x(t) \) and \( y(t) \) in terms of the parameter \( t \):

\[ \left\{ \begin{aligned} x(t) &= t \\ y(t) &= 0.5 t^2 \qquad \text{for} \; t \in [0, 3] \end{aligned} \right. \]

Show Solution and Analysis for Example 1

The plot of the curve is made by finding the \( x \) and \( y \) coordinates for different values of the parameter \( t \):

Table of Values of Parametric Equations — Table 1. Table of Values for \( x(t) = t \) and \( y(t) = 0.5 t^2 \)

Each value of \( t \) determines a point \( (x, y) \) plotted in a coordinate system. As \( t \) varies, the point \( (x(t), y(t)) \) traces out a curve called a parametric curve. The red arrows show the direction of increase of the parameter \( t \).

Example of Parametric Equations — Fig. 1. Plot of the Parametric Curve \( x(t) = t \) and \( y(t) = 0.5 t^2 \)

Note that the curve obtained is part of a parabola that can be obtained by eliminating \( t \):
Solve the first equation \( x(t) = t \) for \( t \): \( t = x \).
Substitute \( t \) into the second equation \( y(t) = 0.5 t^2 \):

\[ y = 0.5 x^2 \]

When taking values of \( t \) in the range \( [0, 3] \), only part of this parabola is shown.

Example 2

This example shows the advantage of using simple parametric equations to describe complex curves. The parametric equations below describe Lissajous curves used in electrical engineering.
Let:

\[ \left\{ \begin{aligned} x(t) &= \sin t + 1 \\ y(t) &= \sin 2t + 2 \qquad \text{for} \; t \in [0, 2\pi] \end{aligned} \right. \]

Show Solution and Analysis for Example 2

Table of Values of Lissajous Curves — Table 2. Table of Values for \( x(t) = \sin t + 1 \) and \( y(t) = \sin 2t + 2 \)

Plot of Lissajous Curves — Fig. 2. Plot of Lissajous Parametric Curve

Let us find an equation of the curve in rectangular coordinates by eliminating the parameter \( t \):

The first equation gives:

\[ \sin t = x - 1 \]

Using the trigonometric identity \( \sin(2t) = 2\sin t \cos t \), rewrite the second equation:

\[ y = 2\sin t \cos t + 2 \]

Since \( \cos t = \pm\sqrt{1 - \sin^2 t} \):

\[ y = \pm 2\sin t \sqrt{1 - \sin^2 t} + 2 \]

Substitute \( \sin t = x - 1 \):

\[ y = \pm 2(x - 1)\sqrt{1 - (x - 1)^2} + 2 \implies y - 2 = \pm 2(x - 1)\sqrt{1 - (x - 1)^2} \]

Squaring both sides and simplifying yields the single rectangular equation:

\[ (y - 2)^2 = 4(x - 1)^2(2x - x^2) \]

Note: This demonstrates the simplicity of parametric equations in representing complex curves that are cumbersome to describe in rectangular coordinates.

Questions

Find an equation in rectangular coordinates to describe the curve for each set of parametric equations below and plot the resulting curves.

\[ \left\{ \begin{aligned} x(t) &= t^3 - 1 \\ y(t) &= \dfrac{1}{t - 1} \qquad \text{for} \; t \in [-1, 1.5] \end{aligned} \right. \]

\[ \left\{ \begin{aligned} x(t) &= 2\sin t - 3 \\ y(t) &= 2\cos t + 4 \qquad \text{for} \; t \in [0, 2\pi] \end{aligned} \right. \]

\[ \left\{ \begin{aligned} x(t) &= 2\cos(2t) \\ y(t) &= \sin t - 1 \qquad \text{for} \; t \in [0, 2\pi] \end{aligned} \right. \]

Solutions to the Above Exercises

Solution to Question 1

Show Solution to Question 1

Solve \( x = t^3 - 1 \) for \( t \):

\[ t = \sqrt[3]{x + 1} \]

Substitute \( t \) into \( y \):

\[ \boxed{y = \dfrac{1}{\sqrt[3]{x + 1} - 1}} \]

Solution to Question 2

Show Solution to Question 2

Rewrite the equations as:

\[ \sin t = \dfrac{x + 3}{2}, \quad \cos t = \dfrac{y - 4}{2} \]

Substitute into the identity \( \sin^2 t + \cos^2 t = 1 \):

\[ \left(\dfrac{x + 3}{2}\right)^2 + \left(\dfrac{y - 4}{2}\right)^2 = 1 \]

Simplify to obtain the standard circle equation:

\[ \boxed{(x + 3)^2 + (y - 4)^2 = 2^2} \]

Note: This is a circle centered at \( (-3, 4) \) with radius \( 2 \).

Solution to Question 3

Show Solution to Question 3

Use the trigonometric identity \( \cos(2t) = 1 - 2\sin^2 t \):

\[ x = 2(1 - 2\sin^2 t) \]

From \( y = \sin t - 1 \), we get \( \sin t = y + 1 \). Substitute this into the equation for \( x \):

\[ \boxed{x = 2 - 4(y + 1)^2} \]

Note: This represents a parabola with a horizontal axis and vertex at \( (2, -1) \).

Plot of Curves Given by Parametric Equations — Fig. 3. Plot of the Curves for All Parametric Equations Given Above

More References

University Calculus - Early Transcendentals - Joel Hass, Maurice D. Weir, George B. Thomas, Jr., Christopher Heil - ISBN-13: 978-0134995540
Calculus - Early Transcendentals - James Stewart - ISBN-13: 978-0-495-01166-8
Derivative of Parametric Equations and their Applications
Calculus Problems and Tutorials
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