Yesterday, I was thinking about some changes I could introduce to a unit on polar functions. Realizing that almost all of the polar functions traditionally explored in precalculus courses have graphs that are complete over the interval , I wondered if there were any interesting curves that took more than units to graph.

My first attempt was which produced something like a merged double limaçon with loops over its period.

Trying for more of the same, I graphed guessing (without really thinking about it) that I’d get more loops. I didn’t get what I expected at all.

Wow! That looks exactly like the image of a standard limaçon with a loop under a translation left of 0.5 units.

Further exploration confirms that completes its graph in units while requires units.

As you know, in mathematics, it is never enough to claim things look the same; proof is required. The acute challenge in this case is that two polar curves (based on angle rotations) appear to be separated by a horizontal translation (a rectangular displacement). I’m not aware of any clean, general way to apply a rectangular transformation to a polar graph or a rotational transformation to a Cartesian graph. But what I can do is rewrite the polar equations into a parametric form and translate from there.

For , becomes . Sliding this a unit to the right makes the parametric equations .

This should align with the standard limaçon, , whose parametric equations for are .

The only problem that remains for comparing and is that their domains are different, but a parameter shift can handle that.

If , then becomes and becomes .

Now that the translation has been applied and both functions operate over the same domain, the two functions must be identical iff and . It’s time to prove those trig identities!

Before blindly manipulating the equations, I take some time to develop some strategy. I notice that the equations contain only one type of angle–double angles of the form –while the equations contain angles of two different types, and . It is generally easier to work with a single type of angle, so my strategy is going to be to turn everything into trig functions of double angles of the form .

Proving that the *x* expressions are equivalent. Now for the *y*s

Therefore the graph of is exactly the graph of slid unit left. Nice.

If there are any students reading this, know that it took a few iterations to come up with the versions of the identities proved above. Remember that published mathematics is almost always cleaner and more concise than the effort it took to create it. One of the early steps I took used the substitution to clean up the appearance of the algebra. In the final proof, I decided that the 2 extra lines of proof to substitute in and then back out were not needed. I also meandered down a couple unnecessarily long paths that I was able to trim in the proof I presented above.

Despite these changes, my proof still feels cumbersome and inelegant to me. From one perspective–Who cares? I proved what I set out to prove. On the other hand, I’d love to know if someone has a more elegant way to establish this connection. There is ** always **room to learn more. Commentary welcome.

In the end, it’s nice to know these two polar curves are identical. It pays to keep one’s eyes eternally open for unexpected connections!

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