# Tag Archives: precalculus

## Transformations III

My last post interpreted the determinants of a few transformation matrices.  For any matrix of a transformation, $[T]$, $det[T]$ is the area scaling factor from the pre-image to the image (addressing the second half of CCSSM Standard NV-M 12 on page 61 here), and the sign of $det[T]$ indicates whether the pre-image and image have the same or opposite orientation.

These are not intuitively obvious, in my opinion, so it’s time for some proof accessible to middle and high school students.

Setting Up:  Take the unit square defined clockwise by vertices (0,0), (0,1), (1,1), and (1,0) under a generic transformation $[T]= \left[ \begin{array}{cc} A & C \\ B & D \end{array}\right]$ where A, B, C, and D are real constants.  Because the unit square has area 1, the area of the image is also the area scaling factor from the pre-image to the image.

As before, the image of the unit square under T is determined by

$\left[ \begin{array}{cc} A & C \\ B & D \end{array}\right] \cdot$ $\left[ \begin{array}{cccc} 0 & 0 & 1 & 1 \\ 0 & 1 & 1 & 0 \end{array}\right] =$ $\left[ \begin{array}{cccc} 0 & C & A+C & A \\ 0 & D & B+D & B \end{array}\right]$.

So the origin is its own image, (0,1) becomes (C,D), (1,1) becomes (A+C,B+D), and (1,0) becomes (A,B).  As $[T]$ is a generic transformation matrix, nothing can be specifically known about the sign or magnitude of its components, but the image below shows one possible case of the image that maintains the original orientation.

When I was working on this problem the first time, I did not expect the image of the unit square to become a parallelogram under every possible $[T]$ (remember that all of its components are assumed constant), but that can be verified by comparing coordinates.  To confirm the area scale change claim, I need to know the generic parallelogram’s area.  I’ll do this two ways.  The first is more elegant, but it invokes vectors–likely a precalculus topic.  The second should be accessible to middle school students.

Area (Method 1):  A parallelogram can be defined using two vectors.  In the image above, the “left side” from the origin to (C,D) is $$–the 3rd dimensional component is needed to compute a cross product.  Likewise, the “bottom side” can be represented by vector $$. The area of a parallelogram is the magnitude of the cross product of the two vectors defining the parallelogram (an explanation of this fact is here).  Because $ \times = <0,0,AD-BC>$ ,

$| \text{Area of Parallelogram} | = |AD-BC|$.

Cross products are not commutative, but reversing the order gives $ \times = -<0,0,AD-BC>$ , which has the same magnitude.  Either way, Claim #1 is true.

Area (Method 2):  Draw a rectangle around the parallelogram with two sides on the coordinate axes, one vertex at the origin, and another at (A+C,B+D).  As shown below, the area interior to the rectangle, but exterior to the parallelogram can be decomposed into right triangular and rectangular regions.

$\Delta I \cong \Delta IV$ with total area $A\cdot B$, and $\Delta III \cong \Delta VI$ with total area $C\cdot D$.  Finally, rectangles II and V are congruent with total area $2B\cdot C$ .  Together, these lead to an indirect computation of the parallelogram’s area.

$|Area|=\left| (A+C)(B+D)-AB-CD-2BC \right| =|AD-BC|$

The absolute values are required because the magnitudes of the constants are unknown.  This is exactly the same result obtained above.  While I used a convenient case for the positioning of the image points in the graphics above, that positioning is irrelevant.  No matter what the sign or relative magnitudes of the constants in $[T]$, the parallelogram area can always be computed indirectly by subtracting the areas of four triangles and two rectangles from a larger rectangle, giving the same result.

Whichever area approach works for you, Claim #1 is true.

Establishing Orientation:  The side from the origin to (A,B) in the parallelogram is a segment on the line $y=\frac{B}{A} x$.  The position of (C,D) relative to this line can be used to determine the orientation of the image parallelogram.

• Assuming the two cases for $A>0$ shown above, the image orientation remains clockwise iff vertex (C,D) is above $y=\frac{B}{A} x$.  Algebraically, this happens if $D>\frac{B}{A}\cdot C \Longrightarrow AD-BC>0$ .

• When $A<0$, the image orientation remains clockwise iff vertex (C,D) is below $y=\frac{B}{A} x$.  Algebraically, this happens if $D<\frac{B}{A}\cdot C \Longrightarrow AD-BC>0$ .
• When $A=0$ and $B<0$, the image is clockwise when $C>0$, again making $AD-BC>0$ .  The same is true for $A=0$, $B>0$, and $C<0$.

In all cases, the pre-image and image have the identical orientation when $AD-BC=det[T]>0$ and are oppositely oriented when $det[T]<0$.

Q.E.D.

## 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.

## Transforming inverse trig graphs

It all started when I tried to get an interesting variation on graphs of inverse trigonometric functions.  Tiring of constant scale changes and translations of inverse trig graphs, I tried $i(x)=x*tan^{-1}x$ , thinking that this product of odd functions leading to an even function would be a nice, but minor, extension for my students.

I reasoned that because the magnitude of arctangent approached $\frac{\pi}{2}$ as $x\rightarrow\infty$, the graph of $i(x)=x*tan^{-1}x$ must approach $y=\frac{\pi}{2}|x|$ .  As shown and to my surprise, $y=i(x)$ seemed to parallel the anticipated absolute value function instead of approaching it.  Hmmmm…..

If this is actually true, then the gap between $i(x)=x*tan^{-1}x$ and $y=\frac{\pi}{2}|x|$ must be constant.  I suspected that this was probably beyond the abilities of my precalculus students, but with my CAS in hand, I (and they) could compute that limit anyway.

Now that was just too pretty to leave alone.  Because the values of x are positive for the limit, this becomes $\displaystyle y=\frac{\pi}{2}|x|-x*tan^{-1}x=\frac{\pi}{2}x-x*tan^{-1}x=x*(\frac{\pi}{2}-tan^{-1}x)$ .

So, four things my students should see here (with guidance, if necessary) are

1. $i(x)=x*tan^{-1}x$ actually approaches $y=\frac{\pi}{2}|x|-1$,
2. the limit can be expressed as a product,
3. each of the terms in the product describes what is happening to the individual terms of the factors of $i(x)$ as x approaches infinity, and
4. (disturbingly) this limit seems to approach $\infty*0$.  A less-obvious recognition  is that as $x\rightarrow\infty$, $\frac{\pi}{2}-tan^{-1}x$  must behave exactly like $\frac{1}{x}$ because its product with x becomes 1,

But what do I do with this for my precalculus students?

NOTE:  As a calculus teacher, I immediately recognized the $\infty*0$ product as a precursor to L’Hopital’s rule.

$\displaystyle\lim_{x\to\infty} [x*(\frac{\pi}{2}-tan^{-1}x)]\rightarrow\infty*0$
$=\displaystyle\lim_{x\to\infty}\frac{\frac{\pi}{2}-tan^{-1}x}{\frac{1}{x}}\rightarrow\frac{0}{0}$
and this form permits L’Hopital’s rule
$=\displaystyle\lim_{x\to\infty}\frac{\frac{-1}{\displaystyle 1+x^2}}{\frac{-1}{\displaystyle x^2}}$
$\displaystyle=\lim_{x\to\infty}\displaystyle\frac{x^2}{1+x^2}=1$

OK, that proves what the graph suggests and the CAS computes.  Rather than leaving students frustrated with a point in a problem that they couldn’t get past (determining the gap between the suspected and actual limits), the CAS kept the problem within reach.  Satisfying enough for some, I suspect, but I’d love suggestions on how to make this particular limit more attainable for students without invoking calculus.  Ideas, anyone?

## March 2011 Conference Presentations

T3 Regional Conference – Suwanee, GA – Saturday, March 19, 2011.

PreCalculus:  Transformed & Nspired

This workshop offers an innovative understanding of pre-calculus concepts through nonstandard transformations, allowing functions and concepts to be unified by a handful of underlying mathematical structures. It provides approaches that dramatically simplify many initially complicated-looking problems. CAS-enhanced ideas are presented.  (Co-presented with Nurfatimah Merchant)

Conics within Conics

This session presents the family of conic sections by connecting their algebraic and graphical representations, showing how each section can evolve from the others. The conclusion is a surprisingly elegant conic property and a 9th grader’s proof submitted for publication.