Graphs of Polar Functions (College Board AP® Precalculus): Study Guide

On exam questions, keep an eye out for graphs showing curves with 'inner loops' like in the second graph above.

• This is a clear sign that the function being graphed, $r=f\left(\theta \right)$, outputs negative values of $r$ for some values of $\theta$

• In a multiple choice question this may allow you immediately to rule out certain of the answer options

When sketching or identifying a polar graph, always make a small table of values first.

• Pick several angles in the domain (e.g. $0, \frac{\pi }{4}, \frac{\pi }{2}, \frac{3\pi }{4}, \pi ,\dots$)

• and compute $r$ for each

Be especially careful with values of $\theta$ where $f\left(\theta \right)$ is negative

• The resulting point appears on the opposite side of the origin from the terminal ray, not on the terminal ray itself

• Forgetting this is one of the most common errors in identifying polar graphs

A polar function is defined by $f\left(\theta \right)=2\sin \theta +1$. Which of the following is the graph of the polar function $r=f\left(\theta \right)$ in the polar coordinate system for $0\le \theta \le 2\pi$?

(A)

(B)

(C)

(D)

Answer:

Start examining the value of $r=f\left(\theta \right)$ for different values of $\theta$

When $\theta =0$

$f\left(0\right)=2\sin 0+1=1$

• This rules out option (B), which doesn't go through $\left(r,\theta \right)=\left(1,0\right)$

• Note as well that the inner loop on the option (A) graph does not make it immediately clear whether the point at $r=1$ on the polar axis actually corresponds to $\theta =0$

  • It could also come from $\theta =\pi$ with a negative $r$ value

When $\theta =\frac{\pi }{2}$

$f\left(\frac{\pi }{2}\right)=2\sin \frac{\pi }{2}+1=3$

• This rules out options (A) and (D), neither of which goes through $\left(r,\theta \right)=\left(3,\frac{\pi }{2}\right)$

• Option (B) goes through that point, but that option was already ruled out in the previous step

That only leaves option (C), which must be the correct answer

• If in doubt you could confirm this by computing $r=f\left(\theta \right)$ for other values of $\theta$

• If you did that you would find that $r<0$ for $\frac{7\pi }{6}<\theta <\frac{11\pi }{6}$, which is what creates the inner loop on the graph of (C)

The correct answer is (C)

The figure shows the graph of the polar function $r=f\left(\theta \right)$, where $f\left(\theta \right)=5\cos \left(2\theta \right)$, in the polar coordinate system for $0\le \theta \le 2\pi$. There are five points labeled $K, L, M, N$, and $O$, where $O$ is the origin. If the domain of $f$ is restricted to $\pi \le \theta \le \frac{3\pi }{2}$, the portion of the given graph that remains consists of two pieces. One of those pieces is the portion of the graph in Quadrant III from $K$ to $O$. Which of the following describes the other remaining piece?

(A) The portion of the graph in Quadrant I from $O$ to $L$

(B) The portion of the graph in Quadrant I from $O$ to $M$

(C) The portion of the graph in Quadrant III from $O$ to $N$

(D) The portion of the graph in Quadrant IV from $O$ to $M$

Answer:

Note at the start that the terminal rays for $\pi \le \theta \le \frac{3\pi }{2}$ are all in Quadrant III

• So points corresponding to those $\theta$ values can only lie in Quadrant III (if values of $r=f\left(\theta \right)$ are positive) or Quadrant I (if values of $r=f\left(\theta \right)$ are negative)

• This instantly rules out answer option (D)

Trace the values of $\theta$ across the restricted domain $\pi \le \theta \le \frac{3\pi }{2}$

At $\theta =\pi$

$r=5\cos \left(2\pi \right)=5$

• The terminal ray for $\theta =\pi$ points left from the origin (in the opposite direction to the polar axis)

• and because $r$ is positive, the point lies on that ray at a distance of 5 from the origin

  • This is point $K$

• Note that $\cos \left(2\pi \right)=\cos \left(0\right)=1$

At $\theta =\frac{5\pi }{4}$

$r=5\cos \left(2·\frac{5\pi }{4}\right)=5\cos \left(\frac{5\pi }{2}\right)=0$

• This is point $O$ at the origin

• Note that $\cos \left(\frac{\pi }{2}\right)=0$, and cosine has a period of $2\pi$

  • so $\cos \left(\frac{\pi }{2}+2\pi \right)=\cos \left(\frac{5\pi }{2}\right)$ is also equal to $0$

So going between $\theta =\pi$ and $\theta =\frac{5\pi }{4}$ traces out "the portion of the graph in Quadrant III from $K$ to $O$" that is mentioned in the question:

For the portion of the graph between $\theta =\frac{5\pi }{4}$ and $\theta =\frac{3\pi }{2}$ there are only two possibilities:

• If $r=f\left(\theta \right)$ is positive then this portion of the graph will be in Quadrant III, tracing out the portion of the graph between $O$ and $N$

• If $r=f\left(\theta \right)$ is negative then this portion of the graph will be in Quadrant I (in the 'slice' of Quadrant I opposite the slice in Quadrant III between $\theta =\frac{5\pi }{4}$ and $\theta =\frac{3\pi }{2}$), tracing out the portion of the graph between $O$ and $L$

At $\theta =\frac{3\pi }{2}$

$r=5\cos \left(2·\frac{3\pi }{2}\right)=5\cos \left(3\pi \right)=-5$

• The terminal ray for $\theta =\frac{3\pi }{2}$ points down from the origin

• However $r$ is negative, so the point lies on the ray opposite to that ray (i.e., up from the origin) at a distance of 5 from the origin

  • This is point $L$

• Note that $\cos \left(\pi \right)=-1$, and cosine has a period of $2\pi$

  • so $\cos \left(\pi +2\pi \right)=\cos \left(3\pi \right)$ is also equal to $-1$

This means that the second of the two possibilities mentioned above is the correct one

(A) The portion of the graph in Quadrant I from $O$ to $L$