Transforming Polar Functions

This is cross-posted on CAS Musings.

Nurfatimah and I were playing around with polar graphs, trying to find something that would stretch students beyond simple circles and types of limacons while still being within the conceptual reach of those who had just been introduced to polar coordinates roughly two weeks earlier.

We remembered that Cartesian graphs of trigonometric functions are much more “interesting” with different center lines.  That is, the graph of y=cos(x)+3 is nothing more than a standard cosine graph oscillating around y=3.

Likewise, the graph of y=cos(x)+0.5x is a standard cosine graph oscillating around y=0.5x.

We teach polar graphing the same way.  To graph r=3+cos(2\theta ), we encourage our students to “read” the function as a cosine curve of period \pi oscillating around the polar function r=3.  Because of its period, this curve will complete a cycle in 0\le\theta\le\pi.  The graph begins this interval at \theta =0 (the positive x-axis) with a cosine graph 1 unit “above” r=3, moving to 1 unit “below” the “center line” at \theta =\frac{\pi}{2}, and returning to 1 unit above the center line at \theta =\pi.  This process repeats for \pi\le\theta\le 2\pi.

Our students graph polar curves far more confidently since we began using this approach (and a couple extensions on it) than those we taught earlier in our careers.  It has become a matter of understanding what functions do and how they interact with each other and almost nothing to do with memorizing particular curve types.

So, now that our students are confidently able to graph polar curves like r=3+cos(2\theta ), we wondered how we could challenge them a bit more.  Remembering variable center lines like the Cartesian y=cos(x)+0.5x, we wondered what a polar curve with a variable center line would look like.  Not knowing where to start, I proposed r=2+cos(\theta )+sin(\theta), thinking I could graph a period 2\pi sine curve around the limacon r=2+cos(\theta ).

There’s a lot going on here, but in its most simplified version, we thought we would get a curve on the center line at \theta =0, 1 unit above at \theta =\frac{\pi}{2}, on at \theta =\pi, 1 unit below at \theta =\frac{3\pi}{2}, and returning to its starting point at \theta =2\pi.  We had a very rough “by hand” sketch, and were quite surprised by the image we got when we turned to our grapher for confirmation.  The oscillation behavior we predicted was certainly there, but there was more!  What do you see in the graph of r=2+cos(\theta )+sin(\theta) below?

This looked to us like some version of a cardioid.  Given the symmetry of the axis intercepts, we suspected it was rotated \frac{\pi}{4} from the x-axis.  An initially x-axis symmetric polar curve rotated \frac{\pi}{4} would contain the term cos(\theta-\frac{\pi}{4}) which expands using a trig identity.

\begin{array}{ccc} cos(\theta-\frac{\pi}{4})&=&cos(\theta )cos(\frac{\pi}{4})+cos(\theta )cos(\frac{\pi}{4}) \\ &=&\frac{1}{\sqrt{2}}(cos(\theta )+sin(\theta )) \end{array}

Eureka!  This identity let us rewrite the original polar equation.

\begin{array}{ccc} r=2+cos(\theta )+sin(\theta )&=&2+\sqrt{2}\cdot\frac{1}{\sqrt{2}} (cos(\theta )+sin(\theta )) \\ &=&2+\sqrt{2}\cdot cos(\theta -\frac{\pi}{4}) \end{array}

And this last form says our original polar function is equivalent to r=2+\sqrt{2}\cdot cos(\theta -\frac{\pi}{4}), or a \frac{\pi}{4} rotated cosine curve of amplitude \sqrt{2} and period 2\pi oscillating around center line r=2.

This last image shows a cosine curve starting at \theta=\frac{\pi}{4} beginning \sqrt{2} above the center circle r=2, crossing the center circle \frac{\pi}{2} later at \theta=\frac{3\pi}{4}, dropping to \sqrt{2} below the center circle at \theta=\frac{5\pi}{4}, back to the center circle at \theta=\frac{7\pi}{4} before finally returning to the starting point at \theta=\frac{9\pi}{4}.  Because the radius is always positive, this also convinced us that this curve is actually a rotated limacon without a loop and not the cardioid that drove our initial investigation.

So, we thought we were departing into some new territory and found ourselves looking back at earlier work from a different angle.  What a nice surprise!

One more added observation:  We got a little lucky in guessing the angle of rotation, but even if it wasn’t known, it is always possible to compute an angle of rotation (or translation in Cartesian) for a sum of two sinusoids with identical periods.  This particular topic is covered in some texts, including Precalculus Transformed.

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3 comments on “Transforming Polar Functions

  1. pat ballew says:

    “Likewise, the graph of y=cos(x)+0.5x is a standard cosine graph oscillating around y=0.5x” I don’t think that is exactly what you mean, student’s might assume that this is the same as a cosine wave rotated by arctan (.5) and I don’t believe that’s true.

  2. chrisharrow says:

    How one oscillates depends on the coordinate system in which one graphs. In Cartesian coordinates, oscillation is always vertical as the graph moves left and right. In Polar coordinates, oscillation is towards and away from the pole (origin) as the graph rotates around the pole.

    Pat, you are absolutely correct in your interpretation of non-rotation of y=cos(x)+0.5x, but I think the oscillation perspective is valid.

    Think of y=0.5x as the center line for y=cos(x)+0.5x. Then the final graph is exactly 1 unit above the center line at x=0, is on the center line at x=\frac{\pi}{2}, is one below the center line at x=\pi, returns to the center line at x=\frac{3\pi}{2}, and returns to 1 unit above at x=2\pi. This behavior is confirmed by the graph, showing that the oscillation in this case is entirely VERTICAL, as one would expect from a Cartesian function. From this perspective, y=cos(x)+0.5x indeed is a standard cosine graph oscillating around y=0.5x.

  3. […] anyone is interested in learning more about how transformations like those described in our Transforming Polar Functions post can dramatically simplify existing secondary mathematics while opening doors to other, richer ideas […]

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