# Monthly Archives: June 2012

## 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!

## Odyssey Robotics

One of the coolest programs in which I’ve participated over the years is the five weeks I spend every June & July teaching for the Odyssey Transformers robotics program for rising sophomores from the Atlanta Public Schools.  Yesterday we took a field trip to Atlanta’s St. Joseph Hospital Visconti Center for Robotic Surgery.  This is our 3rd or 4th trip there, and it continues to be one of the best experiences for the group each year.

Each student in the program got to sit at the controls

and drive the \$2 million surgical robot that has made dramatic advances in heart, urological, prostate, and gynecologic surgeries.

This last image shows the robotic “hands” as one of the students tried to manipulate a dime and a suturing needle.

If you ever have any surgical needs, you really should investigate the promises of robotic-assisted surgery.  We learned that there are medical facilities throughout the country offering this, and St. Joseph’s Hospital in Atlanta is a leading training facility.  AMAZING field trip.

We also had a spectacular visit last week to Factory Automation, a full-service systems integration company, and got to see an automotive manufacturing robot Factory Automation had customized for a portion of an axle construction.  Their comments about what the students needed to do to work as part of a team on complicated projects, reinforcement of the team and robotics concepts my Odyssey partner and I have been daily repeating to the students, how they needed to prepare themselves in high school, and the MANY collegiate paths one could take to pursue robotics as a career were invaluable.  Overall, this was probably the best all-around connected field trip I’ve ever attended.  The only bummer for the students was that they didn’t actually get to drive anything.  On the other hand, the owners and engineers went WAY out of their way to reach out to the students and show them all aspects of engineering an integrated robotics approach to manufacturing.

I am deeply grateful for what I’ve learned from both Factory Automation and St. Joseph’s Hospital.  I’m even more thankful for the extremely valuable time both took out of their busy days to talk with my students.

## Math Humor, II

Here’s two more funny math t-shirts.  The first image a high school classmate  shared with me (Thanks, Tammy!), and the second I found at the MIT bookstore.  Enjoy!

## What would you teach?

I’d love some insights for the statistics content of a non-AP senior math course.

BACKGROUND: I’ve been asked to step into a combination terminal course for seniors covering an introduction to statistics in the fall semester and an introduction to calculus in the spring.  The course has no prerequisite and most of its students have seen precious little high school statistics or probability beyond model regressions to bivariate data in 10th grade.  The class is populated almost entirely by students who are very smart, but who’ve never experienced an honors math course.  A large portion have been frustrated and unmotivated by their previous math courses; for some, this is the last math course they ever take.

The very broad purpose is to introduce students to both branches to learn the fundamentals of what each does.  The department’s pitch for the course when it was created last year acknowledged that we could not know whether our students would be required to take calculus or statistics once they got to college, so this was the “best” way we could prepare students for college mathematics.  It meets 4 days/week for 55 minutes/session.

REFINED QUESTION:   Imagine you were to teach statistics for exactly one semester with no external limitations beyond what has already been described.  What would you make the key focal points for your class?  Why?

## In Plain Sight, but Unseen

Thanks to a comment from Doug Kuhlmann on my last post, I’ve got a few new cool connections on the transformational effects of the parameters in $y=a\cdot x^2+b\cdot x+c$ on its graph.  This is exactly why I share.  Thanks, Doug!!

THE NEW PATTERN:  Use this GeoGebraTube Web document to model the problem.  Set the value of b to any non-zero value and vary a.  The parabola’s vertex moves along a line, as shown below.

As with the changes in the b parameter, define that line and prove your claims.

[The GeoGebra link above does not produce the vertex geometry trace footprints shown in the image.  If you want to create these, download GeoGebra and create this simple document for yourself.  It is FREE.  If anyone wants explicit instructions for how to do this, email me and I’ll post instructions on the ‘blog.]

Another option to see the line is to use Geogebra’s locus tool.  It requires two inputs:  which object is the locus following, and which variable driving the variation.  After selecting the locus tool, click on the vertex and then the slider for a.  You get the next image.

SOLUTION ALERT!  Don’t read further if you want to solve the problem for yourself.

I knew the line contained the vertex and noticed that it seemed to pass through y-intercept.  Predicting the y-intercept was c, all I needed was the slope.  With my prediction of the two generic points, I could compute that, too.  I enjoy symbol manipulation for the mental exercise.  The symbols (to me) weren’t all that complicated, so I took a brief moment of fun solving that by hand.  But this is another of those situations where the symbol manipulation isn’t the point, so using my CAS is 100% legitimate.  It is also a great leveler of ability for those intimidated for any reason by algebraic manipulations.

The next image is a great use of CAS commands to find the line’s slope.  In particular, notice the use of a function definition to minimize the algebraic clutter through function notation.

Lovely and surprisingly simple.  That means the line the parabola’s vertex follows when a varies for non-zero b is $y=\frac{b}{2}\cdot x+c$.

Students often overlook the domain warning.  It doesn’t matter for the creation of the line, but ultimately lies at the heart of the unequal spacing of the vertex footprints in the first image and explains the unique behavior of the parabola’s movement.

If a student didn’t use the vertex and y-intercept to derive the linear equation, a CAS solve command could legitimately be used to show that those two generic points were always on the line.

MOTION ALONG THE LINE:  One of the interesting parts of this problem is how the parabola moves along $y=\frac{b}{2}\cdot x+c$.  After some play with the GeoGebra document, you can see that as $|a|\rightarrow 0$ the parabolas’ vertices move infinitely far away from the y-axis, and as $|a|\rightarrow\infty$ the vertices approach the y-axis. This also can be seen numerically from the generic x-coordinate of the vertex, $-\frac{b}{2a}$.  For a fixed, non-zero value of b, the fraction representing the x-coordinate of the vertex increases in magnitude as $|a|\rightarrow 0$ and decreases in magnitude toward 0 as $|a|\rightarrow\infty$.

The vertex trace points in the first image above are separated by $\Delta a=.01$.  The reason for the differences in distances between the points noted above is because $-\frac{b}{2a}$ does not change linearly when a changes linearly.  As $|a|\rightarrow\infty$, $-\frac{b}{2a}\rightarrow 0$ slower and slower, explaining the increasing density of the vertex trace points near the y-axis.

When $a=0$, the x-coordinate of the vertex is undefined.  At that moment, the generic quadratic, $y=a\cdot x^2+b\cdot x+c$, becomes the degenerate $y=b\cdot x+c$, a line.  Graphing that line (the red dashed line below) against a trace of all possible parabolas as a varies, the degenerate parabola resulting when $a=0$ is precisely the tangent line to all of these parabolas at their y-intercept, $(0,c)$–a pretty extension on a connection suggested by Dave Radcliffe on a cross-posting of my initial post.  Nice.

A FINAL NOTE:  My memory suggests that I’ve seen this pattern before in some of the numerous times I’ve presented the b-variation of this problem in conferences and assigned it in classes.  Despite all the times I must have seen it, the pattern never rose to my active conscience.  Serendipitously, I’m currently reading Tina Seelig’s inGenius:  A Crash Course on Creativity. I offer two quotes from her Are You Paying Attention? chapter:

• We think we understand the world and look for the patterns that we already recognize. (p. 71)
• We focus predominantly on things that are at our eye level rather than looking around more broadly.  In addition, we pay attention to objects that we expect to find and ignore those things that don’t fit. (p. 71)

The moral:  Even after all of my attempts and success at finding unique patterns, I missed this one until Doug pointed it out to me.  I suspect my focus on what I knew about b‘s effect blinded me to the a effect.  This is a great reminder to me to always hold myself ready to see beauty and pattern in unexpected places.

Summer is giving me some time to tie up loose ends that inevitably get dropped during the busy-ness of the school year. Here’s one of those.

I hinted in a post from several months ago about a cool underlying pattern in the quadratic function family.  Most algebra students know how the coefficients a and c control the graph of $y=a\cdot x^2+b\cdot x+c$, but what does b do?

I wrote a Geogebra worksheet to allow exploration of a, b, and c.  On this page, there are sliders on the right side that allow users to vary the values of these three coefficients while the graph of $y=a\cdot x^2+b\cdot x+c$ changes live to reflect those values.

Over the past decade, I’ve become increasingly enamored with this approach to exploring the behavior of function families in advance of more formal analyses.  “Seeing” the effects of parameter can inform and guide the work you do later.  Students quickly recognize that c vertically changes the parabola’s position; closer inspection notes that c is the y-intercept.

Most also note that a changes the “width” of the parabola.  This is true enough, but (in my opinion) a clearer description is that a changes the quadratic’s height.  For any value of x, the y-values of $y=2x^2$ ($a=2$) are twice the y-values of $y=x^2$.  If you attempt to quantify the width, then $a=2$ means the corresponding points are $\displaystyle \frac{1}{\sqrt{2}}$ “wider”.  That just isn’t intuitive to anyone I know, and describing lead coefficients as vertical scale changes is an idea that applies to all functions. I eventually refocus these descriptions to vertical scale changes, but that’s not the point right now.

So what happens when you change b?  If you don’t already know the answer, I encourage you to explore the Geogebra worksheet before reading further.  Try to be precise.

SOLUTION ALERT!  Don’t read further if you want to solve this first.

Like c, varying b changes the position of the parabola, but not its shape.  The difference is that b moves the parabola both horizontally and vertically.   Closer observation suggests that the motion might be along a parabolic path.  Using a new Geogebra worksheet, I placed a trace on the vertex to record the “footprints” of the vertex as b changed.

That’s pretty compelling evidence.  The challenge students face at this point is defining an equation for the suspected parabola. Following the vertex of a parabola is a good proxy for following the entire parabola.

FAST FORWARD:  Through lots of trial-and-error, students eventually propose $y=-a\cdot x^2+c$.  That’s nice, but writing an equation isn’t a proof.  One of the most elegant proofs I’ve seen solves the system of equations defined by the original generic quadratic family and the proposed path of the vertex. A CAS is obviously an appropriate tool in this situation.

There are two solutions:  $(\frac{-b}{2\cdot a},\frac{4\cdot a\cdot c-b^2}{4\cdot a})$ and $(0,c)$ implying two graphical intersections, a fact verified by the vertex trace image above.  The proof lies in the first ordered pair–the generic form of the coordinates of the vertex of $y=a\cdot x^2+b\cdot x+c$–clearly establishing that the generic vertex always travels on the proposed path.  Nice.

What amazed me most about this problem is that I had been teaching quadratic equations for years and remembered from my time as a student what a and c did to the graph.  How is it that I had never explored b ?  How could such a pretty result have been overlooked?  No longer.  This is a project every time I teach an algebra class.