Typically, we draw a graph by finding a direct relationship between x and y. Sometimes, though, it is more natural to think of x and y as both functions of some third variable t (think time in physical applications), and then see how x and y vary as t varies.

In other words, instead of having a y be a function of x (\(y=f(x)\)), we will write x and y as functions of a **parameter** t:
\[\begin{align}
x &= f(t)\\
y &= g(t)
\end{align}\]

We can still draw a graph in the xy plane; now for each value of t, we calculate values for x and y and plot the point \((x,y)=(f(t),g(t))\).

Graph the parametric curve for the following set of equations. \[x=2t,\ \ y=\dfrac{1}{2}t^2-4\ \ \textrm{ for } 0 \leq t \leq 8\]

To draw this graph, let \(t=0\), \(t=1\), etc. until \(t=8\) (since t is limited to values from 0 to 8). For each value of t, calculate x and y and plot the corresponding point. The final graph looks like the following.

Note the red arrows; these indicate the direction of the graph as t increases.

We can also *eliminate the parameter* to get a direct relationship between x and y. To do so, solve for t in one of the equations (if you can) and substitute this expression for t into the other equation.

Eliminate the parameter t in the following set of equations. \[x=2t,\ \ y=\dfrac{1}{2}t^2-4\ \ \textrm{ for } 0 \leq t \leq 8\]

Since \(x=2t\), \(t=\dfrac{1}{2}x\), and we can substitute this into the equation for y:

\[\begin{align} &y = \dfrac{1}{2}t^2-4 = \dfrac{1}{2}\left(\dfrac{1}{2}x\right)^2-4\\ &\ans{y = \dfrac{1}{8}x^2-4} \end{align}\]The following set of parametric equations describe a circle: \[\begin{align} x &= 4 \cos 2\pi t \ \ \ \ \ \ \ \ 0 \leq t \leq 1\\ y &= 4 \sin 2\pi t \end{align}\] Note: \[\begin{align} x^2+y^2 &= (4\cos 2\pi t)^2 + (4\sin 2\pi t)^2\\ &= 16(\cos^2 2\pi t + \sin^2 2\pi t)\\ &= 16 \end{align}\] and \(x^2+y^2=16\) is the equation of a circle centered at the origin with radius 4.

Notice that as t increases, the graph moves counterclockwise.

Also note (and this is important) that the same equation \(x^2+y^2=16\) could have been written as the set of parametric equations
\[\begin{align}
x &= 4 \cos t \ \ \ \ \ \ \ \ 0 \leq t \leq 2\pi\\
y &= 4 \sin t
\end{align}\]
The key here is that *parametrization is not unique*. It is possible to parametrize a curve in different ways; one way may be easier than others, though.

An example of a line in the xy plane is written parametrically as \[\begin{align} x &= 2+3t \ \ \ \ \ \ \ \ -\infty \leq t \leq \infty\\ y &= -1-2t \end{align}\]

Of course, to graph a line, we only need to plot two points and connect them, so the following graph shouldn't be hard to obtain:

Notice what's happening here: when \(t=0\), we get a starting point (the point \((2,-1)\) in this case), and then every time t increases by 1, x increases by 3 and y decreases by 2 (the coefficients of t in the parametric equations). Therefore, the slope of this line (rise over run) will be -2/3.

In general, the equations \[\begin{align} x &= x_0+at \ \ \ \ \ \ \ \ -\infty < t < \infty\\ y &= y_0+bt \end{align}\] where \(a \neq 0\), describe a line in the xy plane passing through the point \((x_0,y_0)\) with slope \(\dfrac{b}{a}\).

Eliminate the parameter t in the following set of equations. \[\begin{align} x &= -2+3t \ \ \ \ \ \ \ \ -\infty < t < \infty\\ y &= 4-6t \end{align}\]

Since \(x=-2+3t\), \(t=\dfrac{x+2}{3}\), and we can substitute this into the equation for y:

\[\begin{align} &y = 4-6t = 4-6\left(\dfrac{x+2}{3}\right) = 4-2x-4\\ &\ans{y = -2x} \end{align}\]We could also do this using what we know from above about the parametric equations of a line (if we recognize it). Here, \((x_0,y_0)=(-2,4)\) and the slope is \(b/a = -6/3 = -2\). Thus, we can use the point-slope formula (or the slope intercept form) to find the equation of this line.

\[\begin{align} y - y_0 &= m(x-x_0)\\ y-4 &= -2(x+2)\\ y -4 &= -2x-4\\ &\longrightarrow \ans{y = -2x} \end{align}\]Find two pairs of parametric equations for the line with slope \(\dfrac{2}{5}\) passing through the point \((-1,4)\).

Here again, we're seeing that parametric equations are not unique. There are infinitely many pairs of parametric equations that we could use to describe this line; any multiple of the slope will give us new values for a and b (although b/a will remain constant), and we can use any point on the line as the initial point \((x_0,y_0)\). In this problem, we'll two different multiples of the slope, since that's a bit easier.

- \(\dfrac{b}{a} = \dfrac{2}{5} \longrightarrow a=5, b=2\) \[\ans{x=-1+5t,\ \ \ \ y=4+2t,\ \ \ \ \ \ \ \ -\infty < t < \infty}\]
- \(\dfrac{b}{a} = \dfrac{4}{10} \longrightarrow a=10, b=4\) \[\ans{x=-1+10t,\ \ \ \ y=4+4t,\ \ \ \ \ \ \ \ -\infty < t < \infty}\]

Find parametric equations for the line segment starting at \((0,7)\) and ending at \((3,-2)\).

The slope is \[m=\dfrac{y_2-y_1}{x_2-x_1} = \dfrac{-2-7}{3} = -3,\] so one choice is \(a=1\) and \(b=-3\). Therefore, \[\ans{x=t,\ \ \ \ y=7-3t,\ \ \ \ \ \ \ \ 0 \leq t \leq 3}\]

Notice that here t does not extend to infinity, since we're only dealing with a line *segment*.

We'll wrap up our discussion of parametric equations with a quick mention of how to differentiate them.

Recall the Chain Rule: \[\dfrac{dy}{dt} = \dfrac{dy}{dx} \cdot \dfrac{dx}{dt}\] Therefore, \[\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt},\ \ \textrm{if } \dfrac{dx}{dt} \neq 0.\]

Find \(\dfrac{dy}{dx}\) if \[\begin{align} x &= 4\cos t \ \ \ \ \ \ \ \ 0 \leq t \leq 2\pi\\ y &= 16\sin t \end{align}\]

The derivative is \[\ans{\dfrac{dy}{dx}=\dfrac{16 \cos t}{-4\sin t} = -4\cot t}\]

Recall how polar coordinates are defined.

\[\begin{align} x &= r \cos \theta\\ y &= r \sin \theta\\ \\ r &= \sqrt{x^2+y^2}\\ \theta &= \tan^{-1} \dfrac{y}{x} \end{align}\]

We can use this to graph polar equations.

Graph \(r=3\).

This describes a curve with a constant radius (i.e. a circle centered at the origin with radius 3).

Graph \(r=4 \cos \theta\).

Use the definition of polar coordinates to find x and y:

\[\begin{align} x &= r \cos \theta = 4 \cos^2 \theta\\ y &= r \sin \theta = 4 \cos \theta \sin \theta \end{align}\]Build a table of values to draw the graph:

\[\begin{array}{c c c} \theta & x & y\\ \hline 0 & 4 & 0\\ \dfrac{\pi}{4} & 2 & 2\\ \dfrac{\pi}{2} & 0 & 0\\ \dfrac{3\pi}{4} & 2 & -2\\ \pi & 4 & 0 \end{array}\]This graph ends up being a circle centered at \((2,0)\) with radius 2.

In general,

- \(r=a\) is a circle of radius \(|a|\) centered at the origin.
- \(r=2a \sin \theta\) is a circle of radius \(|a|\) centered at \((0,a)\).
- \(r=2a \cos \theta\) is a circle of radius \(|a|\) centered at \((a,0)\).

For example, to graph \(r=1+\sin \theta\), follow these steps:

>> theta = 0:0.01:2*pi; >> r = 1+sin(theta); >> polarplot(theta,r)

- \(r=3\sin 2\theta\):
>> theta = 0:0.01:2*pi; >> r = 3*sin(2*theta); >> polarplot(theta,r)

- \(r=\cos \dfrac{2\theta}{5}\), \(0 \leq \theta \leq 5\pi\):
>> theta = 0:0.01:5*pi; >> r = cos(2*theta/5); >> polarplot(theta,r)

- \(r=\sin 2\theta\ \cos 2\theta\):
>> theta = 0:0.01:2*pi; >> r = sin(2*theta).*cos(2*theta); %Note the use of .* (component-wise multiplication) >> polarplot(theta,r)