AP Precalculus 3.14: Polar Function Graphs

A polar function $r=f(\theta )$ is a function like any other: one input, one output. The input is an angle, measured from the polar axis. The output is a radius. Each input and its output make the point $(r,\theta )$, and the graph is all of those points.

Take $r=3+2\cos \theta$ and make a table, the way you would for any function.

To plot a pair, turn to the angle first, then go out along that ray by $r$. Join enough points in order of $\theta$ and the curve appears. This one is wide on the right, where $r$ is near 5, and tucked close to the origin on the left, where $r$ is near 1.

Watch the order inside the pair. The input is $\theta$, but polar coordinates write the radius first: $(r,\theta )$. Swapping them, or putting the angle's value where the radius goes, is the slip to watch for.

You already know how to graph $y=1+2\cos x$ on ordinary axes. That graph is the fastest way to the polar graph of $r=1+2\cos \theta$, and it is the left panel of the diagram: $\theta$ across, $r$ up.

Read it with two rules. Moving right along the $\theta$-axis means the ray turns counterclockwise from the polar axis; going from 0 to $\frac{\pi }{2}$ is a quarter turn. The height of the graph is how far out along that ray the point sits. When the rectangular graph climbs, the point slides out along the turning ray. When it falls toward the axis, the point slides in.

For $r=1+2\cos \theta$, from $\theta =0$ to $\frac{\pi }{2}$ the height falls from 3 to 1 while the ray turns from the polar axis to straight up. On the polar graph, the point moves from P, 3 units out on the polar axis, around to Q, 1 unit up.

Where the rectangular graph touches the $\theta$-axis, $r=0$: zero units out, so the polar graph passes through the origin. For this function that happens at $\theta =\frac{2\pi }{3}$ and $\theta =\frac{4\pi }{3}$.

Between $\frac{2\pi }{3}$ and $\frac{4\pi }{3}$ the rectangular graph of $r=1+2\cos \theta$ dips below the axis, as low as $-1$ at $\theta =\pi$. The output is a signed radius. You still turn to $\theta$, but a negative $r$ sends you backward through the origin, onto the ray that points the opposite way, $|r|$ units out.

At $\theta =\pi$ the ray points left and $r=-1$, so you step 1 unit to the right. That is R in the diagram, the rectangular point $(1,0)$: on the right of the origin, although the angle points left.

This is where the small inner loop comes from. As $\theta$ runs from $\frac{2\pi }{3}$ to $\frac{4\pi }{3}$, the ray sweeps across the left side of the plane, but every point lands on the opposite ray, on the right side. The dashed piece in each panel is that stretch.

So the sign of $r$ tells you which way along the line to go, and $|r|$ tells you how far.

The full graph comes from letting $\theta$ run over a whole interval, often $0\le \theta \le 2\pi$. To draw only one piece, restrict $\theta$. The endpoints of the restriction are the angles at which the piece starts and ends, so find them from the points: for each end, ask which $\theta$, with which $r$, produces that point.

With $r=1+2\cos \theta$, say you want only the inner loop. It starts and ends at the origin, where $r=0$, so restrict to $\frac{2\pi }{3}\le \theta \le \frac{4\pi }{3}$. Want the piece from P to Q instead? P comes from $\theta =0$ with $r=3$, and Q from $\theta =\frac{\pi }{2}$ with $r=1$, so use $0\le \theta \le \frac{\pi }{2}$.

An endpoint is an angle, never a radius. Writing $0\le \theta \le 3$ because P is 3 units out mixes the two up. And when $r$ is negative, the direction you see the point in is not its $\theta$. R sits on the polar axis, but it comes from $\theta =\pi$, not $\theta =0$.

One more thing to watch: a longer interval is not always a bigger picture. Some curves are finished before $\theta$ reaches $2\pi$, and the rest of the turn draws over them again.