Parametric Equations

 

An equation in two variables (like $\,y = x^2\,$) describes a set of points in the plane—the set of points $\,(x,y)\,$ that make the equation true. However, it's a static collection of points, with no concept of the curve being ‘swept out’ (traced/drawn) in any particular way.

Parametric equations describe a curve that is swept out in a particular way!

To illustrate:
The four graphs below have precisely the same set of points, but are swept out in different ways.
Measure time $\,t\,$ in (say) seconds.
Then, each arrow below (connecting green points) represents a part of the curve traced out in one second.

(1)
$x(t) = t\,$
$\,y(t) = t^2$
(2)
$x(t) = 2t\,$
$\,y(t) = (2t)^2 = 4t^2$
(3)
$x(t) = \frac 12t\,$
$\,y(t) = (\frac 12t)^2 = \frac{t^2}4$
(4)
$x(t) = -t\,$
$\,y(t) = (-t)^2 = t^2$
swept out from left to right swept from left to right,
faster than in (1)
swept from left to right,
slower than in (1)
swept from right to left,
same speed as in (1)
(Note: For better-sized images, the horizontal and vertical scales on each individual graph are different.
However, all four graphs use precisely the same scales.)

What is a parameter?

A parameter is a variable used in a special way: to identify particular members of a family of mathematical objects.

For example, consider the familiar equation $\,y = mx + b\,$.
Four variables appear in this equation ($\,x\,$, $\,y\,$, $\,m\,$ and $\,b\,$), but only two of them ($\,m\,$ and $\,b\,$) are parameters. Why?

The single equation ‘$\,y = mx + b\,$’ defines an entire family of equations.
Choices for $\,m\,$ and $\,b\,$ determine which member of this ‘family’ you get!
For example:

choosing these parameters ... ... gives this equation
$\,m = 2\,$, $\,b = -1$ $\,y = 2x - 1\,$
$\,m = -3\,$, $\,b = 0$ $\,y = -3x\,$
$\,m = 0\,$, $\,b = 7\,$ $\,y = 7\,$
The list goes on and on!

As you'll learn below, parametric equations define a family of points instead of a family of equations, but the foundational idea is the same.

What are parametric equations?

Describe a curve in the following way:

  1. Choose a parameter.
    The variable $\,t\,$ is a common parameter choice, since it is often viewed as time.
  2. Give formulas, as functions of $\,t\,$, for both the $x$-values and $y$-values of points on the curve.
    That is (using function notation) write: $$ \begin{alignat}{2} x(t) &= \text{some function of $\,t\,$}&\qquad&\text{(read aloud as: ‘$\,x\,$ of $\,t\,$ is some function of $\,t\,$’)}\cr y(t) &= \text{some function of $\,t\,$}&\qquad&\text{(read aloud as: ‘$\,y\,$ of $\,t\,$ is some function of $\,t\,$’)}\cr \end{alignat} $$
Each of these equations is called a parametric equation.
Together, these parametric equations describe a curve in a plane that is swept out in a particular way.

What is ‘parametric’ about parametric equations?

When a curve is described by parametric equations, you naturally get a family of points: $$ \text{all points of the form } \bigl(\,x(t)\,,\,y(t)\,\bigr) \text{ for the allowable values of $\,t\,$} $$ The parameter $\,t\,$ is used to identify each particular point in this family!

Note:

Some Interesting Parametric Equations

Here are some parametric equations that are more interesting than the simple examples at the top of this page.
Arrows and gradually-changing color have been added, so you can see how the curve is traced out.
Curves (except for Einstein) begin at the green point, and end at the red point.
For a given image, each arrow represents a part of the curve traced out in a fixed amount of time—so longer arrows means that piece is traced out faster!
The first three examples have additional parameters (other than $\,t\,$), so each describes an entire family of curves.


(5)
play with Lissajous curves
$x(t) = A\sin(at + \delta)$
$y(t) = B\sin(bt)$
Here:
$\,a = 5\,$, $\,b = A = B = 4\,$, $\,\delta = 0\,$
$\,t\in [0,\frac{5\pi}4]$
Note: it slows down going around curves

(6)
Witch of Agnesi
$x(t) = at$
$\displaystyle y(t) = a/(1+ t^2)$
Here:
$\,a = 1$
$\,t\in [-10,10]$

(7)
Astroid
$x(t) = a\,\cos^3 t$
$y(t) = a\,\sin^3 t$
Here:
$\,a = 5\,$
$\,t\in [0,2\pi]$
(start and end points are the same)

(8)
an Albert Einstein curve

After the linked page fully loads,
scroll down to look at the equations
for $\,x(t)\,$ and $\,y(t)\,$.
Pretty impressive!

Notes about Parametric Equations

Constructing a Parametric Curve with Desired Properties

Suppose we want to travel between two fixed points, $\,P_1(x_1,y_1)\,$ and $\,P_2(x_2,y_2)\,$, in a fixed time.
More specifically, we want to be at $\,P_1\,$ at time $\,t_1\,$, and at $\,P_2\,$ at time $\,t_2\,$ (with $\,t_1\ne t_2\,$).

Of course, there are many different ways to travel between two points!
The simplest solution is a straight line path between the two.

Using function notation, we want:

$$ \begin{align} x(t_1) = x_1\cr x(t_2) = x_2\cr y(t_1) = y_1\cr y(t_2) = y_2 \end{align} $$

Using point-slope form in the $\,tx\,$-plane, we get: $$ \begin{gather} x(t) - x_1 = \frac{x_2-x_1}{t_2-t_1}(t - t_1)\cr\cr x(t) = x_1 + \frac{x_2-x_1}{t_2-t_1}(t - t_1)\tag{9a} \end{gather} $$ Similarly, $$ y(t) = y_1 + \frac{y_2-y_1}{t_2-t_1}(t - t_1)\tag{9b} $$ Equations (9a) and (9b) are the desired parametric equations.
Together, they give a line through points $\,P_1(x_1,y_1)\,$ and $\,P_2(x_2,y_2)\,$ satisfying:

EXAMPLE:

The line at right is at
  • $\,P_1(-2,5)\,$ at time $\,t = 1\,$
  • $\,P_2(3,-4)\,$ at time $\,t = 11\,$
It was generated by substituting the following values into equations (9a) and (9b): $$ x_1 = -2\,,\ \ \ y_1 = 5\,,\ \ \ t_1 = 1\,,\ \ \ x_2 = 3\,,\ \ \ y_2 = -4\,,\ \ \ t_2 = 11 $$ giving $$ \begin{gather} x(t) = -2 + \frac{1}{2}(t - 1)\cr\cr y(t) = 5 + \frac{-9}{10}(t - 1) \end{gather} $$ Measuring $\,t\,$ in (say) seconds, each arrow represents the part of the curve traced in two seconds.

Eliminating the Parameter:
Going from a Parametric Curve to $\,y = f(x)\,$ (when possible)

Suppose a parametric curve passes a vertical line test: like examples 1, 2, 3, 4, 6, 9ab (but not 5, 7, 8).
Then, you may be able to get a description of the curve in the form ‘$\,y = f(x)\,$’.
In the ‘$\,y = f(x)\,$’ form, you lose information about how the curve is traced out, but this single-equation form may be easier to work with for some applications.

When we want to emphasize that an $x$-value depends on (say) $\,t\,$, we can write ‘$\,x = x(t)\,$’.
Read this aloud as: ‘$\,x\,$ equals $\,x\,$ of $\,t\,$’, meaning that the $x$-value is a function of $\,t\,$.

In going from a pair of parametric equations ‘$\,x = x(t)\,$ and $\,y = y(t)\,$’ to a single equation ‘$\,y = f(x)\,$’,
the parameter $\,t\,$ has been eliminated—it is gone! Therefore, this process is typically referred to as ‘eliminating the parameter’.

The process of eliminating the parameter is often used to help identify the graph of parametric equations.
It may be easier to recognize a curve in ‘$\,y = f(x)\,$’ form than in parametric equation form.

Example: Eliminating the Parameter in Examples 1–4

Procedure:

Note:

(1) $$ \begin{alignat}{2} x(t) = t\,,\ \ y(t) = t^2&\qquad&&\text{original parametric equations}\cr\cr x = t\,,\ \ y = t^2 &&&\text{drop function notation}\cr\cr t = x&&&\text{solve for $\,t\,$ in simplest equation}\cr\cr y = t^2 = x^2&&&\text{substitute into remaining equation}\cr\cr y = x^2&&&\text{$\,y = f(x)\,$ form} \end{alignat} $$ (2) $$ \begin{alignat}{2} x(t) = 2t\,,\ \ y(t) = 4t^2&\qquad&&\text{original parametric equations}\cr\cr x = 2t\,,\ \ y = 4t^2 &&&\text{drop function notation}\cr\cr t = \frac x2&&&\text{solve for $\,t\,$ in simplest equation}\cr\cr y = 4t^2 = 4\bigl(\frac x2\bigl)^2 = \frac{4x^2}{4} = x^2&&&\text{substitute into remaining equation}\cr\cr y = x^2&&&\text{$\,y = f(x)\,$ form} \end{alignat} $$
(3) $$ \begin{alignat}{2} x(t) = \frac 12t\,,\ \ y(t) = \frac{t^2}4&\qquad&&\text{original parametric equations}\cr\cr x = \frac 12t\,,\ \ y = \frac{t^2}4 &&&\text{drop function notation}\cr\cr t = 2x&&&\text{solve for $\,t\,$ in simplest equation}\cr\cr y = \frac{t^2}4 = \frac{(2x)^2}4 = \frac{4x^2}4 = x^2&&&\text{substitute into remaining equation}\cr\cr y = x^2&&&\text{$\,y = f(x)\,$ form} \end{alignat} $$ (4) $$ \begin{alignat}{2} x(t) = -t\,,\ \ y(t) = t^2&\qquad&&\text{original parametric equations}\cr\cr x = -t\,,\ \ y = t^2 &&&\text{drop function notation}\cr\cr t = -x&&&\text{solve for $\,t\,$ in simplest equation}\cr\cr y = t^2 = (-x)^2 = x^2&&&\text{substitute into remaining equation}\cr\cr y = x^2&&&\text{$\,y = f(x)\,$ form} \end{alignat} $$

Example: Eliminating the Parameter in 9ab

To eliminate the parameter, you don't have to solve for (just) $\,t\,$.
As illustrated in this example, it's sometimes easier to solve for a different expression (involving $\,t\,$) that appears in both $\,x(t)\,$ and $\,y(t)\,$.
Here, we solve for $\,t - t_1\,$:

$$ \begin{alignat}{2} x(t) = x_1 + \frac{x_2-x_1}{t_2-t_1}(t - t_1)\,,\ \ \ \ y(t) = y_1 + \frac{y_2-y_1}{t_2-t_1}(t - t_1) &\qquad&&\text{original parametric equations}\cr \cr x = x_1 + \frac{x_2-x_1}{t_2-t_1}(t - t_1)\,,\ \ \ \ y = y_1 + \frac{y_2-y_1}{t_2-t_1}(t - t_1) &&&\text{drop function notation}\cr \cr t-t_1 = \frac{(x-x_1)(t_2-t_1)}{x_2-x_1}&&&\text{solve for $\,t-t_1\,$ in the equation for $\,x\,$}\cr \cr y = y_1 + \frac{y_2-y_1}{t_2-t_1}\cdot\frac{(x-x_1)(t_2-t_1)}{x_2-x_1} &&&\text{substitute into remaining equation}\cr\cr y = y_1 + \frac{y_2-y_1}{x_2-x_1}(x - x_1)&&&\text{cancel; re-arrange factors; this is $\,y = f(x)\,$ form}\cr \cr y - y_1 = \frac{y_2-y_1}{x_2-x_1}(x - x_1)&&&\text{point-slope form for line through $\,(x_1,y_1)\,$ and $\,(x_2,y_2)\,$} \end{alignat} $$

Parametric Equations at WolframAlpha

Hop up to WolframAlpha and type in (say):

parametric plot (0.5 + cos(t) , 0.5*tan(t) + sin(t)), t=0 to 2pi

(You can cut-and-paste.)

Voila! How easy is that?
Master the ideas from this section
by practicing the exercise at the bottom of this page.

Congratulations on finishing Precalculus!!
Next up—Calculus!


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