Tag Archives: polar

A Generic Approach to Arclength in Calculus

Earlier this week, a teacher posted in the College Board’s AP Calculus Community a request for an explanation of computing the arclength of a curve without relying on formulas.

The following video is my proposed answer to that question.  In it, I derive the fundamental arclength relationship before computing the length of y=x^2 from x=0 to x=3 four different ways:

  • As a function of x,
  • As a function of y,
  • Parametrically, and
  • As a polar function.

In summary, the length of any differentiable curve can be thought of as


where a and b are the bounds of the curve, the square root is just the local linearity application of the Pythagorean Theorem, and the integral sums the infinitesimal roots over the length of the curve.

To determine the length of any differentiable curve, factor out the form of the differential that matches the independent variable of the curve’s definition.

Clever math

Here are what I think are three clever uses of math by students.

This was a gift certificate I received from a student last month from a graduating senior.


In my last week of classes at my former school in May, 2013, my entire Honors Precalculus class showed up wearing these shirts designed by one of my students, M.  The back listed all of the students and a lovely “We will miss you.”  As much as I liked the use of a polar function, I loved that M opted not for the simplest possible version of the equation (r=2-2sin(\theta )), but for a rotation–a perfect use of my transformations theme for the course.


Now for a throwback.  When I was a graduate student and TA at Syracuse from 1989-1990, one of my fellow grad students designed this shirt for all of the math and math ed students.  I don’t remember who designed it, but I’ve always loved this shirt.


A Student’s Powerful Polar Exploration

I posted last summer on a surprising discovery of a polar function that appeared to be a horizontal translation of another polar function.  Translations happen all the time, but not really in polar coordinates.  The polar coordinate system just isn’t constructed in a way that makes translations appear in any clear way.

That’s why I was so surprised when I first saw a graph of \displaystyle r=cos \left( \frac{\theta}{3} \right).



It looks just like a 0.5 left translation of r=\frac{1}{2} +cos( \theta ) .


But that’s not supposed to happen so cleanly in polar coordinates.  AND, the equation forms don’t suggest at all that a translation is happening.  So is it real or is it a graphical illusion?

I proved in my earlier post that the effect was real.  In my approach, I dealt with the different periods of the two equations and converted into parametric equations to establish the proof.  Because I was working in parametrics, I had to solve two different identities to establish the individual equalities of the parametric version of the Cartesian x- and y-coordinates.

As a challenge to my precalculus students this year, I pitched the problem to see what they could discover. What follows is a solution from about a month ago by one of my juniors, S.  I paraphrase her solution, but the basic gist is that S managed her proof while avoiding the differing periods and parametric equations I had employed, and she did so by leveraging the power of CAS.  The result was that S’s solution was briefer and far more elegant than mine, in my opinion.

S’s Proof:

Multiply both sides of r = \frac{1}{2} + cos(\theta ) by r and translate to Cartesian.

r^2 = \frac{1}{2} r+r\cdot cos(\theta )
x^2 + y^2 = \frac{1}{2} \sqrt{x^2+y^2} +x
\left( 2\left( x^2 + y^2 -x \right) \right) ^2= \sqrt{x^2+y^2} ^2

At this point, S employed some CAS power.


[Full disclosure: That final CAS step is actually mine, but it dovetails so nicely with S’s brilliant approach. I am always delightfully surprised when my students return using a tool (technological or mental) I have been promoting but hadn’t seen to apply in a particular situation.]

S had used her CAS to accomplish the translation in a more convenient coordinate system before moving the equation back into polar.

Clearly, r \ne 0, so

4r^3 - 3r = cos(\theta ) .

In an attachment (included below), S proved an identity she had never seen, \displaystyle cos(\theta) = 4cos^3 \left( \frac{\theta }{3} \right) - 3cos \left( \frac{\theta }{3} \right) , which she now applied to her CAS result.

\displaystyle 4r^3 - 3r = 4cos^3 \left( \frac{\theta }{3} \right) - 3cos \left( \frac{\theta }{3} \right)

So, \displaystyle r = cos \left( \frac{\theta }{3} \right)

Therefore, \displaystyle r = cos \left( \frac{\theta }{3} \right) is the image of \displaystyle r = \frac{1}{2} + cos(\theta ) after translating \displaystyle \frac{1}{2} unit left.  QED

Simple. Beautiful.

Obviously, this could have been accomplished using lots of by-hand manipulations.  But, in my opinion, that would have been a horrible, potentially error-prone waste of time for a problem that wasn’t concerned at all about whether one knew some Algebra I arithmetic skills.  Great job, S!

S’s proof of her identity, \displaystyle cos(\theta) = 4cos^3 \left( \frac{\theta }{3} \right) - 3cos \left( \frac{\theta }{3} \right) :


Controlling graphs and a free online calculator

When graphing functions with multiple local features, I often find myself wanting to explain a portion of the graph’s behavior independent of the rest of the graph.  When I started teaching a couple decades ago, the processor on my TI-81 was slow enough that I could actually watch the pixels light up sequentially.  I could see HOW the graph was formed.  Today, processors obviously are much faster.  I love the problem-solving power that has given my students and me, but I’ve sometimes missed being able to see function graphs as they develop.

Below, I describe the origins of the graph control idea, how the control works, and then provide examples of polynomials with multiple roots, rational functions with multiple intercepts and/or vertical asymptotes, polar functions, parametric collision modeling, and graphing derivatives of given curves.

BACKGROUND:  A colleague and I were planning a rational function unit after school last week wanting to be able to create graphs in pieces so that we could discuss the effect of each local feature.  In the past, we “rigged” calculator images by graphing the functions parametrically and controlling the input values of t.  Clunky and static, but it gave us useful still shots.  Nice enough, but we really wanted something dynamic.  Because we had the use of sliders on our TI-nSpire software, on Geogebra, and on the Desmos calculator, the solution we sought was closer than we suspected.

REALIZATION & WHY IT WORKS: Last week, we discovered that we could use g(x)=\sqrt \frac{\left | x \right |}{x} to create what we wanted.  The argument of the root is 1 for x<0, making g(x)=1.  For x>0, the root’s argument is -1, making g(x)=i, a non-real number.  Our insight was that multiplying any function y=f(x) by an appropriate version of g wouldn’t change the output of f if the input to g is positive, but would make the product ungraphable due to complex values if the input to g is negative.

If I make a slider for parameter a, then g_2(x)=\sqrt \frac{\left | a-x \right |}{a-x} will have output 1 for all x<a.  That means for any function y=f(x) with real outputs only, y=f(x)\cdot g_2(x) will have real outputs (and a real graph) for x<a only.  Aha!  Using a slider and g_2 would allow me to control the appearance of my graph from left to right.

NOTE:  While it’s still developing, I’ve become a big fan of the free online Desmos calculator after a recent presentation at the Global Math Department (join our 45-60 minute online meetings every Tuesday at 9PM ET!).  I use Desmos for all of the following graphs in this post, but obviously any graphing software with slider capabilities would do.

EXAMPLE 1:  Graph y=(x+2)^3x^2(x-1), a 6th degree polynomial whose end behavior is up for \pm \infty, “wiggles” through the x-axis at -2, then bounces off the origin, and finally passes through the x-axis at 1.

Click here to access the Desmos graph that created the image above.  You can then manipulate the slider to watch the graph wiggle through, then bounce off, and finally pass through the x-axis.

EXAMPLE 2:  Graph y=\frac{(x+1)^2}{(x+2)(x-1)^2}, a 6th degree polynomial whose end behavior is up for \pm \infty, “wiggles” through the x-axis at -2, then bounces off the origin, and finally passes through the x-axis at 1.

Click here to access the Desmos graph above and control the creation of the rational function’s graph using a slider.

EXAMPLE 3:  I believe students understand polar graphing better when they see curves like the  limacon r=2+3cos(\theta ) moving between its maximum and minimum circles.  Controlling the slider also allows users to see the values of \theta at which the limacon crosses the pole. Here is the Desmos graph for the graph below.

EXAMPLE 4:  Object A leaves (2,3) and travels south at 0.29 units/second.  Object B leaves (-2,1) traveling east at 0.45 units/second.  The intersection of their paths is (2,1), but which object arrives there first?  Here is the live version.

OK, I know this is an overly simplistic example, but you’ll get the idea of how the controlling slider works on a parametrically-defined function.  The $latex \sqrt{\frac{\left | a-x \right |}{a-x}}$ term only needs to be on one of parametric equations.  Another benefit of the slider approach is the ease with which users can identify the value of t (or time) when each particle reaches the point of intersection or their axes intercepts.  Obviously those values could be algebraically determined in this problem, but that isn’t always true, and this graphical-numeric approach always gives an alternative to algebraic techniques when investigating parametric functions.

ASIDE 1–Notice the ease of the Desmos notation for parametric graphs.  Enter [r,s] where r is the x-component of the parametric function and s is the y-component.  To graph a point, leave r and s as constants.  Easy.

EXAMPLE 5:  When teaching calculus, I always ask my students to sketch graphs of the derivatives of functions given in graphical forms.  I always create these graphs one part at a time.  As an example, this graph shows y=x^3+2x^2 and allows you to get its derivative gradually using a slider.

ASIDE 2–It is also very easy to enter derivatives of functions in the Desmos calculator.  Type “d/dx” before the function name or definition, and the derivative is accomplished.  Desmos is not a CAS, so I’m sure the software is computing derivatives numerically.  No matter.  Derivatives are easy to define and use here.

I’m hoping you find this technology tip as useful as I do.

Trig Identities with a Purpose

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 0\le\theta\le 2\pi, I wondered if there were any interesting curves that took more than 2\pi units to graph.

My first attempt was r=cos\left(\frac{\theta}{2}\right) which produced something like a merged double limaçon with loops over its 4\pi period.

Trying for more of the same, I graphed r=cos\left(\frac{\theta}{3}\right) 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 r=cos\left(\frac{\theta}{3}\right) completes its graph in 3\pi units while r=\frac{1}{2}+cos\left(\theta\right) requires 2\pi 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 0\le\theta\le 3\pi , r=cos\left(\frac{\theta}{3}\right) becomes \begin{array}{lcl} x_1 &= &cos\left(\frac{\theta}{3}\right)\cdot cos\left (\theta\right) \\ y_1 &= &cos\left(\frac{\theta}{3}\right)\cdot sin\left (\theta\right) \end{array} .  Sliding this \frac{1}{2} a unit to the right makes the parametric equations \begin{array}{lcl} x_2 &= &\frac{1}{2}+cos\left(\frac{\theta}{3}\right)\cdot cos\left (\theta\right) \\ y_2 &= &cos\left(\frac{\theta}{3}\right)\cdot sin\left (\theta\right) \end{array} .

This should align with the standard limaçon, r=\frac{1}{2}+cos\left(\theta\right) , whose parametric equations for 0\le\theta\le 2\pi  are \begin{array}{lcl} x_3 &= &\left(\frac{1}{2}+cos\left(\theta\right)\right)\cdot cos\left (\theta\right) \\ y_3 &= &\left(\frac{1}{2}+cos\left(\theta\right)\right)\cdot sin\left (\theta\right) \end{array} .

The only problem that remains for comparing (x_2,y_2) and (x_3,y_3) is that their domains are different, but a parameter shift can handle that.

If 0\le\beta\le 3\pi , then (x_2,y_2) becomes \begin{array}{lcl} x_4 &= &\frac{1}{2}+cos\left(\frac{\beta}{3}\right)\cdot cos\left (\beta\right) \\ y_4 &= &cos\left(\frac{\beta}{3}\right)\cdot sin\left (\beta\right) \end{array} and (x_3,y_3) becomes \begin{array}{lcl} x_5 &= &\left(\frac{1}{2}+cos\left(\frac{2\beta}{3}\right)\right)\cdot cos\left (\frac{2\beta}{3}\right) \\ y_5 &= &\left(\frac{1}{2}+cos\left(\frac{2\beta}{3}\right)\right)\cdot sin\left (\frac{2\beta}{3}\right) \end{array} .

Now that the translation has been applied and both functions operate over the same domain, the two functions must be identical iff x_4 = x_5 and y_4 = y_5 .  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 (x_5, y_5) equations contain only one type of angle–double angles of the form 2\cdot\frac{\beta}{3} –while the (x_4, y_4) equations contain angles of two different types, \beta and \frac{\beta}{3} .  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 2\cdot\frac{\beta}{3} .

\displaystyle \begin{array}{lcl} x_4 &= &\frac{1}{2}+cos\left(\frac{\beta}{3}\right)\cdot cos\left (\beta\right) \\  &= &\frac{1}{2}+cos\left(\frac{\beta}{3}\right)\cdot cos\left (\frac{\beta}{3}+\frac{2\beta}{3} \right) \\  &= &\frac{1}{2}+cos\left(\frac{\beta}{3}\right)\cdot\left( cos\left(\frac{\beta}{3}\right) cos\left(\frac{2\beta}{3}\right)-sin\left(\frac{\beta}{3}\right) sin\left(\frac{2\beta}{3}\right)\right) \\  &= &\frac{1}{2}+\left[cos^2\left(\frac{\beta}{3}\right)\right] cos\left(\frac{2\beta}{3}\right)-\frac{1}{2}\cdot 2cos\left(\frac{\beta}{3}\right) sin\left(\frac{\beta}{3}\right) sin\left(\frac{2\beta}{3}\right) \\  &= &\frac{1}{2}+\left[\frac{1+cos\left(2\frac{\beta}{3}\right)}{2}\right] cos\left(\frac{2\beta}{3}\right)-\frac{1}{2}\cdot sin^2\left(\frac{2\beta}{3}\right) \\  &= &\frac{1}{2}+\frac{1}{2}cos\left(\frac{2\beta}{3}\right)+\frac{1}{2} cos^2\left(\frac{2\beta}{3}\right)-\frac{1}{2} \left( 1-cos^2\left(\frac{2\beta}{3}\right)\right) \\  &= & \frac{1}{2}cos\left(\frac{2\beta}{3}\right) + cos^2\left(\frac{2\beta}{3}\right) \\  &= & \left(\frac{1}{2}+cos\left(\frac{2\beta}{3}\right)\right)\cdot cos\left(\frac{2\beta}{3}\right) = x_5  \end{array}

Proving that the x expressions are equivalent.  Now for the ys

\displaystyle \begin{array}{lcl} y_4 &= & cos\left(\frac{\beta}{3}\right)\cdot sin\left(\beta\right) \\  &= & cos\left(\frac{\beta}{3}\right)\cdot sin\left(\frac{\beta}{3}+\frac{2\beta}{3} \right) \\  &= & cos\left(\frac{\beta}{3}\right)\cdot\left( sin\left(\frac{\beta}{3}\right) cos\left(\frac{2\beta}{3}\right)+cos\left(\frac{\beta}{3}\right) sin\left(\frac{2\beta}{3}\right)\right) \\  &= & \frac{1}{2}\cdot 2cos\left(\frac{\beta}{3}\right) sin\left(\frac{\beta}{3}\right) cos\left(\frac{2\beta}{3}\right)+\left[cos^2 \left(\frac{\beta}{3}\right)\right] sin\left(\frac{2\beta}{3}\right) \\  &= & \frac{1}{2}sin\left(2\frac{\beta}{3}\right) cos\left(\frac{2\beta}{3}\right)+\left[\frac{1+cos \left(2\frac{\beta}{3}\right)}{2}\right] sin\left(\frac{2\beta}{3}\right) \\  &= & \left(\frac{1}{2}+cos\left(\frac{2\beta}{3}\right)\right)\cdot sin\left (\frac{2\beta}{3}\right) = y_5  \end{array}

Therefore the graph of r=cos\left(\frac{\theta}{3}\right) is exactly the graph of r=\frac{1}{2}+cos\left(\theta\right) slid \frac{1}{2} 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 \gamma =\frac{\beta}{3} 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!

Polar Graphing Surprise

Nurfatimah Merchant 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.