Parametric equations

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We discuss the basics of parametric curves.

The idea of parametric equations

Think back to when you first learned how to graph a function. I’m pretty sure you used a so-called “T-chart,” and if $y = x^2$ , I bet it looked something like this: $\begin {array}{c|c} x & y = x^2\\\hline 0 & 0 \\ 1 & 1\\ -1 & 1\\ 2 & 4\\ -2 & 4 \end {array}$ With a parametric plot, both $x$ and $y$ are now functions of a third parameter, we’ll call it $t$ , often thought of as time: $\begin {array}{c|c|c} t & x = 2-3t & y = -1+4t\\\hline 0 & 2 & -1 \\ 1 & -1 & 3 \\ 2 & -4 & 7 \\ 3 & -7 & 11\\ 4 & -10& 15 \end {array}$ If $x=t$ , then there isn’t much difference between a parametric plot and a regular plot. On the other hand, with parametric functions, we can generate plots that fail the vertical line test! Check out this graph of

$\begin{align*} x(t) &= \cos(t)\\ y(t) &= \sin(t) \end{align*}$ as $t$ runs from $0$ to $2\pi$ :

Do the parametric equations $\begin{align*} x(t) &= \cos(t)\\ y(t) &= \sin(t) \end{align*}$ as $t$ runs from $0$ to $2\pi$ define a function of $t$ ?

No, because the graph does not pass the vertical line test. Yes, it is a function of $t$ , because for each input $t$ , there is exactly one output value, an ordered pair.

For the graph of a circle, $y$ is not a function of $x$ . However, it can be a function of $t$ that maps $\begin{align*} \R &\to \R\times \R\\ t &\mapsto (x,y) \end{align*}$ where the domain is $[0,2\pi )$ and the elements of the range consist of ordered pairs.

Famous parametric equations

We’ll discuss some basic parametric equations.

Circles

The standard form for a circle centered at a point $(a,b)$ with radius $r$ is given by $(x-a)^2 + (y-b)^2 = r^2.$ One problem with the standard form for a circle is that it is somewhat difficult to find points on the circle. A parametric equation representing a circle solves this problem.

Give a parametric equation representing the circle $(x-1)^2 + (y-2)^2 = 3^2$ and explain why your answer is correct.

This is the circle of radius $3$ centered at the point $(1,2)$ . Here we set $\begin{align*} x(t) &= \answer[given]{1} + \answer[given]{3}\cos(t)\\ y(t) &= \answer[given]{2} + \answer[given]{3}\sin(t) \end{align*}$ as $t$ runs from $0$ to $2\pi$ . To see that our answer is correct, we can “plug” it back into the implicit equation for the circle. Write with me: $\begin{align*} (x-1)^2 + (y-2)^2 &= (x(t)-1)^2 + (y(t)-2)^2\\ &= (1 + 3\cos(t)-1)^2 + (2 + 3\sin(t)-2)^2\\ &= (3\cos(t))^2 + (3\sin(t))^2\\ &= 3^2\cos^2(t) + 3^2\sin^2(t)\\ &= 3^2(\cos^2(t) + \sin^2(t))\\ &= 3^2 \end{align*}$ by the Pythagorean identity. Since our functions $(x(t),y(t))$ satisfy the form of the circle, our solution is correct.

One way to think about parametric formulas for circles is to imagine $\begin{align*} x(t) &= \cos(t)\\ y(t) &= \sin(t) \end{align*}$ as “drawing” the unit circle as $t$ changes. Make a table showing how the circle is being plotted as $t$ runs from $0$ to $2\pi$ : $\begin {array}{c|c|c} t & x(t) = \cos (t) & y(t) = \sin (t)\\ \hline 0 & \answer {1} & \answer {0}\\ \pi /4 & \answer {\sqrt {2}/2} & \answer {\sqrt {2}/2}\\ \pi /2 & \answer {0} & \answer {1}\\ 3\pi /4 & \answer {-\sqrt {2}/2} & \answer {\sqrt {2}/2}\\ \pi & \answer {-1} & \answer {0}\\ 5\pi /4 & \answer {-\sqrt {2}/2} & \answer {-\sqrt {2}/2}\\ 3\pi /2 & \answer {0} & \answer {-1}\\ 7\pi /4 & \answer {\sqrt {2}/2} & \answer {-\sqrt {2}/2}\\ 2\pi & \answer {1} & \answer {0} \end {array}$

Is the circle “drawn” in a clockwise or counterclockwise fashion?

clockwise counterclockwise

One day while trying to graph a unit circle, you accidentally write down $\begin{align*} x(t) &= \sin(t)\\ y(t) &= \cos(t) \end{align*}$ What happens now? Do you still get a circle? How is this different from what we did in the previous question?

you still plot a unit circle in a counterclockwise fashion, with the same starting and ending points you plot a unit circle but in a clockwise fashion, with the same starting and ending points you still plot a unit circle in a counterclockwise fashion, but the starting and ending points are different you plot a unit circle but in a clockwise fashion, but the starting and ending points are different this no longer plots a circle

In mathematics, when parameterizing closed curves (like circles), the convention is to draw them in a “counterclockwise” direction. This is called the positive orientation.

If you parameterize your closed curves in a clockwise direction, you may find your “answers” are off by a factor of $-1$ .

What should $x(t)$ and $y(t)$ be to parameterize the circle $(x-a)^2 + (y-b)^2 = r^2.$ in a counterclockwise fashion, with $t=0$ corresponding to $(1,0)$ ? $\begin{align*} x(t) &= \answer{a} + \answer{r}\cdot \cos(t)\\ y(t) &= \answer{b} + \answer{r}\cdot \sin(t) \end{align*}$

Lines

Suppose you want a parametric equation for a line that goes through the point $(2,1)$ with a certain direction:

where the $\mathbf {direction}$ is imagined as starting at $(0,0)$ and going to the point $(u,v)$ :

So to understand the direction of the arrow above, we need to move it back to the origin.

In the graph above, what is the direction of the arrow? The direction is given by $(u,v) = \left (\answer {4},\answer {3}\right )$ .

At this point we can give a very useful representation for a line:

$\begin{align*} \l(t) &= \mathbf{point} + t\cdot \mathbf{direction}\\ &= (a,b) + t \cdot(u,v) \end{align*}$ where the $\mathbf {direction}$ is imagined as starting at $(0,0)$ and going to the point $(u,v)$ . Another way of writing this is $\l (t) = \left \{\begin {aligned} x(t) &= a + t\cdot u\\ y(t) &= b + t\cdot v \end {aligned}\right .$

Suppose you have the parametric formula for a line $\l (t) = (2,1) + t \cdot (4,3)$ What is $\l (0)$ ? $\l (0) = \left (\answer {2},\answer {1}\right )$

We’re at the starting point!

What is $\l (1)$ ? $\l (1) = \left (\answer {6},\answer {4}\right )$

We’ve moved exactly the “direction” from the point!

Other equations and other plots

One thing that can be confusing about parametric plots is that there can be multiple representations of the same plot:

Which of the following parametric equations draw the circle $(x-1)^2 + (y-2)^2 = 3^2$ ?

$x(t) = 1 + 3\cos (t)$ and $y(t) = 2 + 3\sin (t)$ for $0 \le t \le 2\pi$ $x(t) = 1 + 3\cos (t)$ and $y(t) = 2 + 3\sin (t)$ for $2\pi \le t \le 4\pi$ $x(t) = 1 + 3\sin (t)$ and $y(t) = 2 + 3\cos (t)$ for $0 \le t \le 2\pi$ $x(t) = 2 + 3\sin (t)$ and $y(t) = 1 + 3\cos (t)$ for $0 \le t \le 2\pi$ $x(t) = 1 + 3\cos (e^t)$ and $y(t) = 2 + 3\sin (e^t)$ for $0 \le t \le \ln (2\pi )$ $x(t) = 1 + 3\cos (e^t)$ and $y(t) = 2 + 3\sin (e^t)$ for $\ln (2\pi ) \le t \le \ln (4\pi )$

Which of the following parametric equations draw the line $y=x$ ?

$x(t) = t$ and $y(t) = t$ for $-\infty < t < \infty$ $x(t) = t-5$ and $y(t) = t-5$ for $-\infty < t < \infty$ $x(t) = \sin (t)$ and $y(t) = \sin (t)$ for $-\infty < t < \infty$ $x(t) = \tan (t)$ and $y(t) = \tan (t)$ for $-\pi /2 < t < \pi /2$ $x(t) = t^2$ and $y(t) = t^2$ for $-\infty < t < \infty$ $x(t) = t^3$ and $y(t) = t^3$ for $-\infty < t < \infty$

Parametric plots allow us to make some pretty crazy plots.

Can you give a parametric formula for this cool spiral that starts at the origin and runs to $(8\pi ,0)$ ?

$\begin{align*} x(t) = \answer{t}\cdot \cos(t)\\ y(t) = \answer{t}\cdot \sin(t) \end{align*}$

One important class of parametric curves are Lissajous figures. These are curves of the form

$\begin{align*} x(t) &= A \sin(at + \delta)\\ y(t) &= B\sin(bt) \end{align*}$ Here is a plot of a Lissajous curve where $A=B=1$ , $\delta = \pi /2$ , $a=5$ and $b=4$

These figures come up a lot in electrical engineering. Do yourself a favor and play around with Lissajous figures for differing values of $\delta$ , $a$ and $b$ .

Converting to parametric equations

If you are given $y = f(x)$ it is really easy to convert this to a parametric function, just write

$\begin{align*} x(t) &= t\\ y(t) &= f(t). \end{align*}$

Can you use the technique described immediately above to express $y = e^x$ as a parametric function? $\begin{align*} x(t) &= \answer{t}\\ y(t) &= \answer{e^t} \end{align*}$

Converting from parametric equations

On the other hand, if you are given a parametric function, to express $y$ as function of $x$ can be much more difficult. Here are the basic strategies to try:

Solve for $t$ .
Solve for a function of $t$ .
Use a trigonometric identity.

In each case the process that we are using is called elimination of a parameter.

We’ll give several examples of how one actually eliminates a parameter.

Solving for the variable

In the first example, we’ll solve for $t$ .

Let $\begin{align*} x(t) &= -4 t^2\\ y(t) &= 3t-1. \end{align*}$ Eliminate a parameter to express this curve purely in terms of $x$ and $y$ .

Here we will solve for $t$ . Since it is easier, we will solve for $t$ in this equation: $\begin{align*} y &= 3t-1\\ y+1 &= \answer[given]{3t}\\ \frac{y+1}{3} & = \answer[given]{t}. \end{align*}$ Now plug this into $x(t) = -4 t^2$ , and write $x = \answer [given]{-4 \left (\frac {y+1}{3}\right )^2}.$

Solving for a common function

In our next example we’ll solve for a function of $t$ that is common to both $x(t)$ and $y(t)$ .

Let $\begin{align*} x(t) &= -5e^t\\ y(t) &= e^{3t}. \end{align*}$ Eliminate a parameter to express this curve purely in terms of $x$ and $y$ .

Here we will solve for a function of $t$ . Write $\begin{align*} x &= -5e^t\\ \frac{x}{-5} &= e^t \end{align*}$ Now we can rewrite $y(t)$ as $\begin{align*} y(t) &= e^{3t}\\ y(t) &= \left(e^{t}\right)^{\answer[given]{3}}. \end{align*}$ Now we see that $y = \answer [given]{\left (\frac {x}{-5}\right )^3}.$

Solving for related functions

In our final example, we will use a trigonometric identity.

Let $\begin{align*} x(t) &= 3\cos(t)\\ y(t) &= 1+4\sin(t). \end{align*}$ Eliminate a parameter to express this curve purely in terms of $x$ and $y$ .

The basic idea is to try an use the Pythagorean identity: $\cos ^2(t) + \sin ^2(t) = 1$ So first isolate cosine and sine, and square the equations. With $x(t)$ we have $\begin{align*} x &= 3 \cos(t) \\ \answer[given]{\frac{x}{3}} &= \cos(t)\\ \answer[given]{\left(\frac{x}{3}\right)^2} &= \cos^2(t) \end{align*}$ and with $y(t)$ we have $\begin{align*} y &= 1 + 4\sin(t)\\ y-1 &= 4\sin(t)\\ \frac{y-1}{4} &= \sin(t)\\ \left(\frac{y-1}{4}\right)^2 &= \sin^2(t). \end{align*}$ Plugging this back into the Pythagorean identity, we see: $\begin{align*} \cos^2(t) + \sin^2(t) &= 1\\ \left(\frac{x}{3}\right)^2 + \left(\frac{y-1}{4}\right)^2 &= 1. \end{align*}$