Polar function plots may leave your head spinning like you got off the Tilt-a-Whirl ride at the amusement park. But using them, it's possible to model that Tilt-a-Whirl ride, and they make some of the sweetest looking plots.
We need to place some bounds on the number of times the plot goes around the origin. These things are complicated, we will still need a calculator to help us plot. Think of it like were placing a polar bear on a leash. You can try to tame it, but it may take you for a walk.
Sometimes wel need to know which bounds on θ give a particular piece of a polar graph. The easiest way to find these bounds is to graph the function on the calculator and play with the bounds until we find the right piece of the graph.
In the example, we need to be careful to find the petal shown in the graph. Other choices for bounds might give us a single petal, but the wrong petal. If we take
π / 4 ≤ θ ≤ 3π/4
we find one petal, but not the one asked for:

Depending on the problem, it might be important to find the specific petal mentioned.
It can also be important to find bounds on θ that trace out the graph exactly once.
Sample Problem
The graph of r = cos θ$ for 0 ≤ θ ≤ 2π looks like this:

However, the graph of r = cos θ for 0 ≤ θ ≤ π looks the same.

If asked what values of θ are needed to describe the whole graph of r = cos θ, we would take the narrower bounds:
0 ≤ θ ≤ π.
When asked what bounds on θ give a particular portion of the graph, there are multiple correct answers. We know that r = cos(2θ) for - π/4 ≤ θ ≤ π/4 looks like this:

If instead we take 7π/4 ≤ θ ≤ 9π/4, we find the same piece of the graph:
How do we know if we've found correct bounds for θ? Put them in the calculator and draw the graph. As long as we find the correct portion of the graph, and the calculator traces it only once, then we've found correct bounds for θ.
Practice:
What bounds on θ produce the single petal of the graph r = cos(2θ) shown below? PICTURE: graph for $-\frac{pi}{4}≤ θ ≤ \frac{pi}{4}$ | |
We'll start out with something that seems reasonable, like 0 ≤ θ ≤ π, and see what we find. PICTURE: graph for 0 ≤ θ ≤ π That's definitely not right. We didn't find all of the petal we wanted, and we got a bunch of other junk. See if we can decrease the upper bound on θ until we find the top half of the petal we want. How about 0 ≤ θ ≤ \frac{π}{2} ? PICTURE: graph for 0 ≤ θ ≤ \frac{π}{2} Still not right. We have the upper half of the petal, but we also have half a petal that we don't want: PICTURE polar funcs 29 This seems like we have twice as much graph as we want, we'll try cutting the upper bound of θ in half again. If 0 ≤ θ ≤ \frac{π}{4} we find PICTURE : graph it That's the upper half of the petal we want. To find the lower half of the petal, let the lower bound for θ reflect its upper bound. When -\frac{π}{4} ≤ θ ≤ \frac{π}{4} we find the piece of the graph we want: PICTURE: graph it | |
Find bounds on θ that trace out the specified piece of function only once.
- PICTURE: graph r = sin(3θ) for \frac{π}{3}≤ θ ≤ \frac{2π}{3}. Label graph r = sin(3θ) but don't say what the bounds are.
Answer
- Since the function uses sin(3θ), it's a reasonable guess that values of θ in the $\frac{π}{3}$ family will be useful. See what we find for0 ≤ θ ≤ \frac{π}{3}:
PICTURE: graph sin(3θ) for 0 ≤ θ ≤ \frac{π}{3}
That's a full petal, but not the one we want. Try \frac{π}{3}≤ θ≤\frac{2π}{3}:
PICTURE: graph sin(3θ) for \frac{π}{3}≤ θ≤\frac{2π}{3}
That's the part we want. The bounds for θ are
\frac{π}{3}≤ θ≤\frac{2π}{3}.
Find bounds on θ that trace out the specified piece of function only once.
- PICTURE: graph r = sin(\frac{θ}{4}) for 0 ≤ θ ≤ 2π. Label function, not bounds.
Answer
- Start with 0 ≤ θ ≤ π, since that seems to work a lot:PICTURE: graph r = sin(\frac{θ}{4}) for 0≤ θ≤ π
That's not enough graph. Try expanding the bounds for θ. How about $0≤θ≤2π$?
PICTURE: graph r = sin(\frac{θ}{4}) for 0 ≤ θ≤ 2π
That's the right piece of the graph. The bounds we want are
0≤θ≤2π.
Find bounds on θ that trace out the specified piece of function only once.
- PICTURE: graph r = cos(θ + \frac{π}{6}) for 0 ≤ θ ≤ \frac{π}{3}. Label func.
Answer
- Try 0 ≤ θ≤ π:PICTURE: graph it
No, that's too much graph. Since there's a $\frac{π}{6}$ in the argument to cos, try knocking the upper bound down to $\frac{π}{6}.$
If 0≤ θ ≤ \frac{π}{6} we find this graph:
PICTURE: graph it
That's not enough graph. Try doubling the upper bound of θ, we have 0 ≤ θ ≤ \frac{π}{3}:
PICTURE: graph it
That's what we want. The final answer is
0 ≤ θ ≤ \frac{π}{3}.