# Tag Archives: calculus

## Implicit Derivative Question

Yesterday, this question posted within the AP Calculus Community about implicit derivatives.

Below, I argue why the derivatives MUST be the same, show how four different variations can all be shown to give the same derivative, and provide a final conclusion.

INITIAL INTUITION

The Desmos graph of the given relation, is $y^2 = \frac{x-1}{x+1}$, is shown below.  Logically, it seems that even when the terms of the relation are algebraically rearranged, the graph should be invariant.   The other two forms mentioned in the Community post are on lines 2 and 3.  Lines 4, 5, and 6 show three other variations.  Here is the link to my Desmos graph allowing you to change between the forms to visually confirm the graphical invariance intuition.

If calculus “works”, it also shouldn’t matter how one calculates a derivative.   While the forms of the derivative certainly could LOOK different,  because any point on the invariant graph has the same tangent line no matter what the form of its equation, and the derivative of a relation at a point is the slope of that invariant tangent line, then the derivative also MUST be invariant.

CALCULATING “DIFFERENT” DERIVATIVES

To show the derivatives are fundamentally all the same (as suspected by the initial post), I calculate the derivatives of the equations on lines 1 and 3 given in the initial post as well as my variations on lines 4 and 6.

LINE 1:

Using the Chain Rule on the left and the Quotient Rule on the right gives

LINE 3:

This version is more complicated, requiring the Product Rule in addition to the earlier Chain and Quotient Rules.  In the penultimate line, I used the original equation to substitute for $y^2$ to transform the derivative into the same form as line 1.

LINE 4:

This time, differentiation requires only the Chain and Product Rules.

After the usual substitution for $y^2$, I multiplied both sides by $(x+1)$ to clear the denominator and solved for $y'$, returning the same result.

LINE 6:

This time, the relation is solved for x, resulting in a much more complicated Quotient+Chain Rule calculation, but substituting for $y^2$ and changing the form leads once again to the same answer.

Hopefully this is convincing evidence that all derivative forms can be shown to be equivalent.   If you’re still learning implicit differentiation, I encourage you to show the derivatives from the lines 2 and 5 variations are also equivalent.

CONCLUSION

So which approach is “best”?  In my opinion, it all depends on your personal comfort with algebraic manipulations.  Some prefer to just take a derivative from the given form of $y^2 = \frac{x-1}{x+1}$.  I avoid the more complicated quotient rule whenever I can, so the variation from line 4 would have been my approach.

The cool part is that it doesn’t matter what approach you use, so long as your algebraic manipulations are sound.  You don’t have to accept the form in which a problem is given; CHANGE IT to a form that works for you!

## Many Roads Give Same Derivative

A recent post in the AP Calculus Community expressed some confusion about different ways to compute $\displaystyle \frac{dy}{dx}$ at (0,4) for the function $x=2ln(y-3)$.  I share below the two approaches suggested in the original post, proffer two more, and a slightly more in-depth activity I’ve used in my calculus classes for years.  I conclude with an alternative to derivatives of inverses.

### Two Approaches Initially Proposed

1 – Accept the function as posed and differentiate implicitly.

$\displaystyle \frac{d}{dx} \left( x = 2 ln(y-3) \right)$

$\displaystyle 1 = 2*\frac{1}{y-3} * \frac{dy}{dx}$

$\displaystyle \frac{dy}{dx} = \frac{y-3}{2}$

Which gives $\displaystyle \frac{dy}{dx} = \frac{1}{2}$ at (x,y)=(0,4).

2 – Solve for y and differentiate explicitly.

$\displaystyle x = 2ln(y-3) \longrightarrow y = 3 + e^{x/2}$

$\displaystyle \frac{dy}{dx} = e^{x/2} * \frac{1}{2}$

Evaluating this at (x,y)=(0,4) gives $\displaystyle \frac{dy}{dx} = \frac{1}{2}$ .

### Two Alternative Approaches

3 – Substitute early.

The question never asked for an algebraic expression of $\frac{dy}{dx}$, only the numerical value of this slope.  Because students tend to make more silly mistakes manipulating algebraic expressions than numeric ones, the additional algebra steps are unnecessary, and potentially error-prone.  Admittedly, the manipulations are pretty straightforward here, in more algebraically complicated cases, early substitutions could significantly simplify work. Using approach #1 and substituting directly into the second line gives

$\displaystyle 1 = 2 * \frac{1}{y-3} * \frac{dy}{dx}$.

At (x,y)=(0,4), this is

$\displaystyle 1 = 2 * \frac{1}{4-3}*\frac{dy}{dx}$

$\displaystyle \frac{dy}{dx} = \frac{1}{2}$

The numeric manipulations on the right side are obviously easier than the earlier algebra.

4 – Solve for $\frac{dx}{dy}$ and reciprocate.

There’s nothing sacred about solving for $\frac{dy}{dx}$ directly.  Why not compute the derivative of the inverse and reciprocate at the end? Differentiating first with respect to y eventually leads to the same solution.

$\displaystyle \frac{d}{dy} \left( x = 2 ln(y-3) \right)$

$\displaystyle \frac{dx}{dy} = 2 * \frac{1}{y-3}$

At (x,y)=(0,4), this is

$\displaystyle \frac{dx}{dy} = \frac{2}{4-3} = 2$, so

$\displaystyle \frac{dy}{dx} = \frac{1}{2}$.

### Equivalence = A fundamental mathematical concept

I sometimes wonder if teachers should place much more emphasis on equivalence.  We spend so much time manipulating expressions in mathematics classes at all levels, changing mathematical objects (shapes, expressions, equations, etc.) into a different, but equivalent objects.  Many times, these manipulations are completed under the guise of “simplification.”  (Here is a brilliant Dan Teague post cautioning against taking this idea too far.)

But it is critical for students to recognize that proper application of manipulations creates equivalent expressions, even if when the resulting expressions don’t look the same.   The reason we manipulate mathematical objects is to discover features about the object in one form that may not be immediately obvious in another.

For the function $x = 2 ln(y-3)$, the slope at (0,4) must be the same, no matter how that slope is calculated.  If you get a different looking answer while using correct manipulations, the final answers must be equivalent.

### Another Example

A similar question appeared on the AP Calculus email list-server almost a decade ago right at the moment I was introducing implicit differentiation.  A teacher had tried to find $\displaystyle \frac{dy}{dx}$ for

$\displaystyle x^2 = \frac{x+y}{x-y}$

using implicit differentiation on the quotient, manipulating to a product before using implicit differentiation, and finally solving for y in terms of x to use an explicit derivative.

1 – Implicit on a quotient

Take the derivative as given:\$

$\displaystyle \frac{d}{dx} \left( x^2 = \frac{x+y}{x-y} \right)$

$\displaystyle 2x = \frac{(x-y) \left( 1 + \frac{dy}{dx} \right) - (x+y) \left( 1 - \frac{dy}{dx} \right) }{(x-y)^2}$

$\displaystyle 2x * (x-y)^2 = (x-y) + (x-y)*\frac{dy}{dx} - (x+y) + (x+y)*\frac{dy}{dx}$

$\displaystyle 2x * (x-y)^2 = -2y + 2x * \frac{dy}{dx}$

$\displaystyle \frac{dy}{dx} = \frac{-2x * (x-y)^2 + 2y}{2x}$

2 – Implicit on a product

Multiplying the original equation by its denominator gives

$x^2 * (x - y) = x + y$ .

Differentiating with respect to x gives

$\displaystyle 2x * (x - y) + x^2 * \left( 1 - \frac{dy}{dx} \right) = 1 + \frac{dy}{dx}$

$\displaystyle 2x * (x-y) + x^2 - 1 = x^2 * \frac{dy}{dx} + \frac{dy}{dx}$

$\displaystyle \frac{dy}{dx} = \frac{2x * (x-y) + x^2 - 1}{x^2 + 1}$

3 – Explicit

Solving the equation at the start of method 2 for y gives

$\displaystyle y = \frac{x^3 - x}{x^2 + 1}$.

Differentiating with respect to x gives

$\displaystyle \frac{dy}{dx} = \frac {\left( x^2+1 \right) \left( 3x^2 - 1\right) - \left( x^3 - x \right) (2x+0)}{\left( x^2 + 1 \right) ^2}$

Equivalence

Those 3 forms of the derivative look VERY DIFFERENT.  Assuming no errors in the algebra, they MUST be equivalent because they are nothing more than the same derivative of different forms of the same function, and a function’s rate of change doesn’t vary just because you alter the look of its algebraic representation.

Substituting the y-as-a-function-of-x equation from method 3 into the first two derivative forms converts all three into functions of x.  Lots of by-hand algebra or a quick check on a CAS establishes the suspected equivalence.  Here’s my TI-Nspire CAS check.

Here’s the form of this investigation I gave my students.

### Final Example

I’m not a big fan of memorizing anything without a VERY GOOD reason.  My teachers telling me to do so never held much weight for me.  I memorized as little as possible and used that information as long as I could until a scenario arose to convince me to memorize more.  One thing I managed to avoid almost completely were the annoying derivative formulas for inverse trig functions.

For example, find the derivative of $y = arcsin(x)$ at $x = \frac{1}{2}$.

Since arc-trig functions annoy me, I always rewrite them.  Taking sine of both sides and then differentiating with respect to x gives.

$sin(y) = x$

$\displaystyle cos(y) * \frac{dy}{dx} = 1$

I could rewrite this equation to give $\frac{dy}{dx} = \frac{1}{cos(y)}$, a perfectly reasonable form of the derivative, albeit as a less-common  expression in terms of y.  But I don’t even do that unnecessary algebra.  From the original function, $x=\frac{1}{2} \longrightarrow y=\frac{\pi}{6}$, and I substitute that immediately after the differentiation step to give a much cleaner numeric route to my answer.

$\displaystyle cos \left( \frac{\pi}{6} \right) * \frac{dy}{dx} = 1$

$\displaystyle \frac{\sqrt{3}}{2} * \frac{dy}{dx} = 1$

$\displaystyle \frac{dy}{dx} = \frac{2}{\sqrt{3}}$

And this is the same result as plugging $x = \frac{1}{2}$ into the memorized version form of the derivative of arcsine.  If you like memorizing, go ahead, but my mind remains more nimble and less cluttered.

One final equivalent approach would have been differentiating $sin(y) = x$ with respect to y and reciprocating at the end.

### CONCLUSION

There are MANY ways to compute derivatives.  For any problem or scenario, use the one that makes sense or is computationally easiest for YOU.  If your resulting algebra is correct, you know you have a correct answer, even if it looks different.  Be strong!

## CAS and Normal Probability Distributions

My presentation this past Saturday at the 2015 T^3 International Conference in Dallas, TX was on the underappreciated applicability of CAS to statistics.  This post shares some of what I shared there from my first year teaching AP Statistics.

MOVING PAST OUTDATED PEDAGOGY

It’s been decades since we’ve required students to use tables of values to compute by hand trigonometric and radical values.  It seems odd to me that we continue to do exactly that today for so many statistics classes, including the AP.  While the College Board permits statistics-capable calculators, it still provides probability tables with every exam.  That messaging is clear:  it is still “acceptable” to teach statistics using outdated probability tables.

In this, my first year teaching AP Statistics, I decided it was time for my students and I to completely break from this lingering past.  My statistics classes this year have been 100% software-enabled.  Not one of my students has been required to use or even see any tables of probability values.

My classes also have been fortunate to have complete CAS availability on their laptops.  My school’s math department deliberately adopted the TI-Nspire platform in part because that software looks and operates exactly the same on tablet, computer, and handheld platforms.  We primarily use the computer-based version for learning because of the speed and visualization of the large “real estate” there.  We are shifting to school-owned handhelds in our last month before the AP Exam to gain practice on the platform required on the AP.

The remainder of this post shares ways my students and I have learned to apply the TI-Nspire CAS to some statistical questions around normal distributions.

FINDING NORMAL AREAS AND PROBABILITIES

Assume a manufacturer makes golf balls whose distances traveled under identical testing conditions are approximately normally distributed with a mean 295 yards with a standard deviation of 3 yards.  What is the probability that one such randomly selected ball travels more than 300 yards?

Traditional statistics courses teach students to transform the 300 yards into a z-score to look up in a probability table.  That approach obviously works, but with appropriate technology, I believe there will be far less need to use or even compute z-scores in much the same way that always converting logarithms to base-10 or base-to use logarithmic tables is anachronistic when using many modern scientific calculators.

TI calculators and other technologies allow computations of non-standard normal curves.  Notice the Nspire CAS calculation below the curve uses both bounds of the area of interest along with the mean and standard deviation of the distribution to accomplish the computation in a single step.

So the probability of a randomly selected ball from the population described above going more than 300 yards is 4.779%.

GOING BACKWARDS

Now assume the manufacturing process can control the mean distance traveled.  What mean should it use so that no more than 1% of the golf balls travel more than 300 yards?

Depending on the available normal probability tables, the traditional approach to this problem is again to work with z-scores.  A modified CAS version of this is shown below.

Therefore, the manufacturer should produce a ball that travels a mean 293.021 yards under the given conditions.

The approach is legitimate, and I shared it with my students.  Several of them ultimately chose a more efficient single line command:

But remember that the invNorm() and normCdf() commands on the Nspire are themselves functions, and so their internal parameters are available to solve commands.  A pure CAS, “forward solution” still incorporating only the normCdf() command to which my students were first introduced makes use of this to determine the missing center.

DIFFERENTIATING INSTRUCTION

While calculus techniques definitely are NOT part of the AP Statistics curriculum, I do have several students jointly enrolled in various calculus classes.  Some of these astutely noted the similarity between the area-based arguments above and the area under a curve techniques they were learning in their calculus classes.  Never being one to pass on a teaching moment, I pulled a few of these to the side to show them that the previous solutions also could have been derived via integration.

I can’t recall any instances of my students actually employing integrals to solve statistics problems this year, but just having the connection verified completely solidified the mathematics they were learning in my class.

CONFIDENCE INTERVALS

The mean lead level of 35 crows in a random sample from a region was 4.90 ppm and the standard deviation was 1.12 ppm.  Construct a 95 percent confidence interval for the mean lead level of crows in the region.

Many students–mine included–have difficulty comprehending confidence intervals and resort to “black box” confidence interval tools available in most (all?) statistics-capable calculators, including the TI-Nspire.

As n is greater than 30, I can compute the requested z-interval by filling in just four entries in a pop-up window and pressing Enter.

Convenient, for sure, but this approach doesn’t help the confused students understand that the confidence interval is nothing more than the bounds of the middle 95% of the normal pdf described in the problem, a fact crystallized by the application of the tools the students have been using for weeks by that point in the course.

Notice in the solve+normCdf() combination commands that the unknown this time was a bound and not the mean as was the case in the previous example.

EXTENDING THE RULE OF FOUR

I’ve used the “Rule of Four” in every math class I’ve taught for over two decades, explaining that every mathematical concept can be explained or expressed four different ways:  Numerically, Algebraically, Graphically (including graphs and geometric figures), and Verbally.  While not the contextual point of his quote, I often cite MIT’s Marvin Minsky here:

“You don’t understand anything until you learn it more than one way.”

Learning to translate between the four representations grants deeper understanding of concepts and often gives access to solutions in one form that may be difficult or impossible in other forms.

After my decades-long work with CAS, I now believe there is actually a 5th representation of mathematical ideas:  Tools.  Knowing how to translate a question into a form that your tool (in the case of CAS, the tool is computers) can manage or compute creates a different representation of the problem and requires deeper insights to manage the translation.

I knew some of my students this year had deeply embraced this “5th Way” when one showed me his alternative approach to the confidence interval question:

I found this solution particularly lovely for several reasons.

• The student knew about lists and statistical commands and on a whim tried combining them in a novel way to produce the desired solution.
• He found the confidence interval directly using a normal distribution command rather than the arguably more convenient black box confidence interval tool.  He also showed explicitly his understanding of the distribution of sample means by adjusting the given standard deviation for the sample size.
• Finally, while using a CAS sometimes involves getting answers in forms you didn’t expect, in this case, I think the CAS command and list output actually provide a cleaner, interval-looking result than the black box confidence interval command much more intuitively connected to the actual meaning of a confidence interval.
• While I haven’t tried it out, it seems to me that this approach also should work on non-CAS statistical calculators that can handle lists.

(a very minor disappointment, quickly overcome)

Returning to my multiple approaches, I tried using my student’s newfound approach using a normCdf() command.

Alas, my Nspire returned the very command I had entered, indicating that it didn’t understand the question I had posed.  While a bit disappointed that this approach didn’t work, I was actually excited to have discovered a boundary in the current programming of the Nspire.  Perhaps someday this approach will also work, but my students and I have many other directions we can exploit to find what we need.

Leaving the probability tables behind in their appropriate historical dust while fully embracing the power of modern classroom technology to enhance my students’ statistical learning and understanding, I’m convinced I made the right decision to start this school year.  They know more, understand the foundations of statistics better, and as a group feel much more confident and flexible.  Whether their scores on next month’s AP exam will reflect their growth, I can’t say, but they’ve definitely learned more statistics this year than any previous statistics class I’ve ever taught.

COMPLETE FILES FROM MY 2015 T3 PRESENTATION

If you are interested, you can download here the PowerPoint file for my entire Nspired Statistics and CAS presentation from last week’s 2015 T3 International Conference in Dallas, TX.  While not the point of this post, the presentation started with a non-calculus derivation/explanation of linear regressions.  Using some great feedback from Jeff McCalla, here is an Nspire CAS document creating the linear regression computation updated from what I presented in Dallas.  I hope you found this post and these files helpful, or at least thought-provoking.

## Calculus Humor

Completely frivolous post.

OK, my Halloween “costume” at school this year was pretty lame, but I actually did put a minute amount of thought into it.

In case you can’t read the sign, it says $\int 3(ice)^2d(ice)$.  If you remember some calculus and treat ice as your variable, that works out to $ice^3$–An Ice Cube!  Ha!

But it gets better.  As there weren’t any bounds, adding the random constant of integration makes it $ice^3+C$ or “Ice Cube + C”, or maybe “Ice Cube + Sea”–I was really dressed up as an Iceberg.  Ha! Ha!  Having no idea how to dress like an iceberg, I wore a light blue shirt for the part of the iceberg above the water and dark blue pants for the part below the water.  I tried to be clever even if the underlying joke was just “punny”.

Then a colleague posted another integral joke I’d seen sometime before.  It has some lovely extensions, so I’ll share that, too.

What is $\int \frac{d(cabin)}{cabin}$?

At first glance, it’s a “log cabin.”  Funny.

But notationally, the result is actually $ln(cabin)$, so the environmentalists out there will appreciate that the answer is really a “natural log cabin.”  Even funnier.

The most correct solution is $ln(cabin)+C$.  If you call the end “+ Sea”, then the most clever answer is that $\int \frac{d(cabin)}{cabin}$ is a “Houseboat”.  Ha!

Hope you all had some fun.

## Traveling Dots, Parabolas, and Elegant Math

Toward the end of last week, I read a description a variation on a paper-folding strategy to create parabolas.  Paraphrased, it said:

1. On a piece of wax paper, use a pen to draw a line near one edge.  (I used a Sharpie on regular copy paper and got enough ink bleed that I’m convinced any standard copy or notebook paper will do.  I don’t think the expense of wax paper is required!)
2. All along the line, place additional dots 0.5 to 1 inch apart.
3. Finally, draw a point F between 0.5 and 2 inches from the line roughly along the midline of the paper toward the center of the paper.
4. Fold the paper over so one of the dots on line is on tope of point F.  Crease the paper along the fold and open the paper back up.
5. Repeat step 4 for every dot you drew in step 2.
6. All of the creases from steps 4 & 5 outline a curve.  Trace that curve to see a parabola.

I’d seen and done this before, I had too passively trusted that the procedure must have been true just because the resulting curve “looked like a parabola.”  I read the proof some time ago, but I consumed it too quickly and didn’t remember it when I was read the above procedure.  I shamefully admitted to myself that I was doing exactly what we insist our students NEVER do–blindly accepting a “truth” based on its appearance.  So I spent part of that afternoon thinking about how to understand completely what was going on here.

What follows is the chronological redevelopment of my chain of reasoning for this activity, hopefully showing others that the prettiest explanations rarely occur without effort, time, and refinement.  At the end of this post, I offer what I think is an even smoother version of the activity, freed from some of what I consider overly structured instructions above.

CONIC DEFINITION AND WHAT WASN’T OBVIOUS TO ME

A parabola is the locus of points equidistant from a given  point (focus) and line (directrix).

What makes the parabola interesting, in my opinion, is the interplay between the distance from a line (always perpendicular to some point C on the directrix) and the focus point (theoretically could point in any direction like a radius from a circle center).

What initially bothered me about the paper folding approach last week was that it focused entirely on perpendicular bisectors of the Focus-to-C segment (using the image above).  It was not immediately obvious to me at all that perpendicular bisectors of the Focus-to-C segment were 100% logically equivalent to the parabola’s definition.

1. I knew without a doubt that all parabolas are similar (there is a one-to-one mapping between every single point on any parabola and every single point on any other parabola), so I didn’t need to prove lots of cases.  Instead, I focused on the simplest version of a parabola (from my perspective), knowing that whatever I proved from that example was true for all parabolas.
2. I am quite comfortable with my algebra, geometry, and technology skills.  Being able to wield a wide range of powerful exploration tools means I’m rarely intimidated by problems–even those I don’t initially understand.  I have the patience to persevere through lots of data and explorations until I find patterns and eventually solutions.

I love to understand ideas from multiple perspectives, so I rarely quit with my initial solution.  Perseverance helps me re-phrase ideas and exploring them from alternative perspectives until I find prettier ways of understanding.

In my opinion, it is precisely this willingness to play, persevere, and explore that formalized education is broadly failing to instill in students and teachers.  “What if?” is the most brilliant question, and the one we sadly forget to ask often enough.

ALGEBRAIC PROOF

While I’m comfortable handling math in almost any representation, my mind most often jumps to algebraic perspectives first.  My first inclination was a coordinate proof.

PROOF 1:  As all parabolas are similar, it was enough to use a single, upward facing parabola with its vertex at the origin.  I placed the focus at $(0,f)$, making the directrix the line $y=-f$.  If any point on the parabola was $(x_0,y_0)$, then a point C on the directrix was at $(x_0,-f)$.

From the parabola’s definition, the distance from the focus to P was identical to the length of CP:

$\sqrt{(x_0-0)^2-(y_0-f)^2}=y_0+f$

Squaring and combining common terms gives

$x_0 ^2+y_0 ^2-2y_0f+f^2=y_0 ^2+2y_0f+f^2$
$x_0 ^2=4fy$

But the construction above made lines (creases) on the perpendicular bisector of the focus-to-C segment.  This segment has midpoint $\displaystyle \left( \frac{x_0}{2},0 \right)$ and slope $\displaystyle -\frac{2f}{x_0}$, so an equation for its perpendicular bisector is $\displaystyle y=\frac{x_0}{2f} \left( x-\frac{x_0}{2} \right)$.

Finding the point of intersection of the perpendicular bisector with the parabola involves solving a system of equations.

$\displaystyle y=\frac{x_0}{2f} \left( x-\frac{x_0}{2} \right)=\frac{x^2}{4f}$
$\displaystyle \frac{1}{4f} \left( x^2-2x_0x+x_0 ^2 \right) =0$
$\displaystyle \frac{1}{4f} \left( x-x_0 \right) ^2 =0$

So the only point where the line and parabola meet is at $\displaystyle x=x_0$–the very same point named by the parabola’s definition.  QED

Proof 2:  All of this could have been brilliantly handled on a CAS to save time and avoid the manipulations.

Notice that the y-coordinate of the final solution line is the same $y_0$ from above.

MORE ELEGANT GEOMETRIC PROOFS

I had a proof, but the algebra seemed more than necessary.  Surely there was a cleaner approach.

In the image above, F is the focus, and I is a point on the parabola.  If D is the midpoint of $\overline{FC}$, can I conclude $\overline{ID} \perp \overline{FC}$, proving that the perpendicular bisector of $\overline{FC}$ always intersects the parabola?

PROOF 3:  The definition of the parabola gives $\overline{FI} \cong \overline{IC}$, and the midpoint gives $\overline{FD} \cong \overline{DC}$.  Because $\overline{ID}$ is self-congruent, $\Delta IDF \cong \Delta IDC$ by SSS, and corresponding parts make the supplementary $\angle IDF \cong \angle IDC$, so both must be right angles.  QED

PROOF 4:  Nice enough, but it still felt a little complicated.  I put the problem away to have dinner with my daughters and when I came back, I was able to see the construction not as two congruent triangles, but as the single isosceles $\Delta FIC$ with base $\overline{FC}$.  In isosceles triangles, altitudes and medians coincide, automatically making $\overline{ID}$ the perpendicular bisector of $\overline{FC}$.  QED

Admittedly, Proof 4 ultimately relies on the results of Proof 3, but the higher-level isosceles connection felt much more elegant.  I was satisfied.

TWO DYNAMIC GEOMETRY SOFTWARE VARIATIONS

Thinking how I could prompt students along this path, I first considered a trace on the perpendicular lines from the initial procedure above (actually tangent lines to the parabola) using to trace the parabolas.  A video is below, and the Geogebra file is here.

http://vimeo.com/89759785

It is a lovely approach, and I particularly love the way the parabola appears as a digital form of “string art.”  Still, I think it requires some additional thinking for users to believe the approach really does adhere to the parabola’s definition.

I created a second version allowing users to set the location of the focus on the positive y-axis and using  a slider to determine the distances and constructs the parabola through the definition of the parabola.  [In the GeoGebra worksheet (here), you can turn on the hidden circle and lines to see how I constructed it.]  A video shows the symmetric points traced out as you drag the distance slider.

A SIMPLIFIED PAPER PROCEDURE

Throughout this process, I realized that the location and spacing of the initial points on the directrix was irrelevant.  Creating the software versions of the problem helped me realize that if I could fold a point on the directrix to the focus, why not reverse the process and fold F to the directrix?  In fact, I could fold the paper so that F touched anywhere on the directrix and it would work.  So, here is the simplest version I could develop for the paper version.

1. Use a straightedge and a Sharpie or thin marker to draw a line near the edge of a piece of paper.
2. Place a point F roughly above the middle of the line toward the center of the paper.
3. Fold the paper over so point F is on the line from step 1 and crease the paper along the fold.
4. Open the paper back up and repeat step 3 several more times with F touching other parts of the step 1 line.
5. All of the creases from steps 3 & 4 outline a curve.  Trace that curve to see a parabola.

This procedure works because you can fold the focus onto the directrix anywhere you like and the resulting crease will be tangent to the parabola defined by the directrix and focus.  By allowing the focus to “Travel along the Directrix”, you create the parabola’s locus.  Quite elegant, I thought.

As I was playing with the different ways to create the parabola and thinking about the interplay between the two distances in the parabola’s definition, I wondered about the potential positions of the distance segments.

1. What is the shortest length of segment CP and where could it be located at that length?  What is the longest length of segment CP and where could it be located at that length?
2. Obviously, point C can be anywhere along the directrix.  While the focus-to-P segment is theoretically free to rotate in any direction, the parabola definition makes that seem not practically possible.  So, through what size angle is the focus-to-P segment practically able to rotate?
3. Assuming a horizontal directrix, what is the maximum slope the focus-to-P segment can achieve?
4. Can you develop a single solution to questions 2 and 3 that doesn’t require any computations or constructions?

CONCLUSIONS

I fully realize that none of this is new mathematics, but I enjoyed the walk through pure mathematics and the enjoyment of developing ever simpler and more elegant solutions to the problem.  In the end, I now have a deeper and richer understanding of parabolas, and that was certainly worth the journey.

## The Value of Counter-Intuition

Numberphile caused quite a stir when it posted a video explaining why

$\displaystyle 1+2+3+4+...=- \frac{1}{12}$

Doug Kuhlman recently posted a great follow-up Numberphile video explaining a broader perspective behind this sum.

It’s a great reminder that there are often different ways of thinking about problems, and sometimes we have to abandon tradition to discover deeper, more elegant connections.

For those deeply bothered by this summation result, the second video contains a lovely analogy to the “reality” of $\sqrt{-1}$.  From one perspective, it is absolutely not acceptable to do something like square roots of negative numbers.  But by finding a way to conceptualize what such a thing would mean, we gain a far richer understanding of the very real numbers that forbade $\sqrt{-1}$ in the first place as well as opening the doors to stunning mathematics far beyond the limitations of real numbers.

On the face of it, $\displaystyle 1+2+3+...=-\frac{1}{12}$ is obviously wrong within the context of real numbers only.  But the strange thing in physics and the Zeta function and other places is that $\displaystyle -\frac{1}{12}$ just happens to work … every time.  Let’s not dismiss this out of hand.  It gives our students the wrong idea about mathematics, discovery, and learning.

There’s very clearly SOMETHING going on here.  It’s time to explore and learn something deeper.  And until then, we can revel in the awe of manipulations that logically shouldn’t work, but somehow they do.

May all of our students feel the awe of mathematical and scientific discovery.  And until the connections and understanding are firmly established, I hope we all can embrace the spirit, boldness, and fearless of Euler.

## Base-x Numbers and Infinite Series

In my previous post, I explored what happened when you converted a polynomial from its variable form into a base-x numerical form.  That is, what are the computational implications when polynomial $3x^3-11x^2+2$ is represented by the base-x number $3(-11)02_x$, where the parentheses are used to hold the base-x digit, -11, for the second power of x?

So far, I’ve explored only the Natural number equivalents of base-x numbers.  In this post, I explore what happens when you allow division to extend base-x numbers into their Rational number counterparts.

Level 5–Infinite Series:

Numbers can have decimals, so what’s the equivalence for base-x numbers?  For starters, I considered trying to get a “decimal” form of $\displaystyle \frac{1}{x+2}$.  It was “obvious” to me that $12_x$ won’t divide into $1_x$.  There are too few “places”, so some form of decimals are required.  Employing division as described in my previous post somewhat like you would to determine the rational number decimals of $\frac{1}{12}$ gives

Remember, the places are powers of x, so the decimal portion of $\displaystyle \frac{1}{x+2}$ is $0.1(-2)4(-8)..._x$, and it is equivalent to

$\displaystyle 1x^{-1}-2x^{-2}+4x^{-3}-8x^{-4}+...=\frac{1}{x}-\frac{2}{x^2}+\frac{4}{x^3}-\frac{8}{x^4}+...$.

This can be seen as a geometric series with first term $\displaystyle \frac{1}{x}$ and ratio $\displaystyle r=\frac{-2}{x}$.  It’s infinite sum is therefore $\displaystyle \frac{\frac{1}{x}}{1-\frac{-2}{x}}$ which is equivalent to $\displaystyle \frac{1}{x+2}$, confirming the division computation.  Of course, as a geometric series, this is true only so long as $\displaystyle |r|=\left | \frac{-2}{x} \right |<1$, or $2<|x|$.

I thought this was pretty cool, and it led to lots of other cool series.  For example, if $x=8$,you get $\frac{1}{10}=\frac{1}{8}-\frac{2}{64}+\frac{4}{512}-...$.

Likewise, $x=3$ gives $\frac{1}{5}=\frac{1}{3}-\frac{2}{9}+\frac{4}{27}-\frac{8}{81}+...$.

I found it quite interesting to have a “polynomial” defined with a rational expression.

Boundary Convergence:

As shown above, $\displaystyle \frac{1}{x+2}=\frac{1}{x}-\frac{2}{x^2}+\frac{4}{x^3}-\frac{8}{x^4}+...$ only for $|x|>2$.

At $x=2$, the series is obviously divergent, $\displaystyle \frac{1}{4} \ne \frac{1}{2}-\frac{2}{4}+\frac{4}{8}-\frac{8}{16}+...$.

For $x=-2$, I got $\displaystyle \frac{1}{0} = \frac{1}{-2}-\frac{2}{4}+\frac{4}{-8}-\frac{8}{16}+...=-\frac{1}{2}-\frac{1}{2}-\frac{1}{2}-\frac{1}{2}-...$ which is properly equivalent to $-\infty$ as $x \rightarrow -2$ as defined by the convergence domain and the graphical behavior of $\displaystyle y=\frac{1}{x+2}$ just to the left of $x=-2$.  Nice.

I did find it curious, though, that $\displaystyle \frac{1}{x}-\frac{2}{x^2}+\frac{4}{x^3}-\frac{8}{x^4}+...$ is a solid approximation for $\displaystyle \frac{1}{x+2}$ to the left of its vertical asymptote, but not for its rotationally symmetric right side.  I also thought it philosophically strange (even though I understand mathematically why it must be) that this series could approximate function behavior near a vertical asymptote, but not near the graph’s stable and flat portion near $x=0$.  What a curious, asymmetrical approximator.

Maclaurin Series:

Some quick calculus gives the Maclaurin series for $\displaystyle \frac{1}{x+2}$ :  $\displaystyle \frac{1}{2}-\frac{x}{4}+\frac{x^2}{8}-\frac{x^3}{16}+...$, a geometric series with first term $\frac{1}{2}$ and ratio $\frac{-x}{2}$.  Interestingly, the ratio emerging from the Maclaurin series is the reciprocal of the ratio from the “rational polynomial” resulting from the base-x division above.

As a geometric series, the interval of convergence is  $\displaystyle |r|=\left | \frac{-x}{2} \right |<1$, or $|x|<2$.  Excluding endpoint results, the Maclaurin interval is the complete Real number complement to the base-x series.  For the endpoints, $x=-2$ produces the right-side vertical asymptote divergence to $+ \infty$ that $x=-2$ did for the left side of the vertical asymptote in the base-x series.  Again, $x=2$ is divergent.

It’s lovely how these two series so completely complement each other to create clean approximations of $\displaystyle \frac{1}{x+2}$ for all $x \ne 2$.

Other base-x “rational numbers”

Because any polynomial divided by another is absolutely equivalent to a base-x rational number and thereby a base-x decimal number, it will always be possible to create a “rational polynomial” using powers of $\displaystyle \frac{1}{x}$ for non-zero denominators.  But, the decimal patterns of rational base-x numbers don’t apply in the same way as for Natural number bases.  Where $\displaystyle \frac{1}{12}$ is guaranteed to have a repeating decimal pattern, the decimal form of $\displaystyle \frac{1}{x+2}=\frac{1_x}{12_x}=0.1(-2)4(-8)..._x$ clearly will not repeat.  I’ve not explored the full potential of this, but it seems like another interesting field.

CONCLUSIONS and QUESTIONS

Once number bases are understood, I’d argue that using base-x multiplication might be, and base-x division definitely is, a cleaner way to compute products and quotients, respectively, for polynomials.

The base-x division algorithm clearly is accessible to Algebra II students, and even opens the doors to studying series approximations to functions long before calculus.

Is there a convenient way to use base-x numbers to represent horizontal translations as cleanly as polynomials?  How difficult would it be to work with a base-$(x-h)$ number for a polynomial translated h units horizontally?

As a calculus extension, what would happen if you tried employing division of non-polynomials by replacing them with their Taylor series equivalents?  I’ve played a little with proving some trig identities using base-x polynomials from the Maclaurin series for sine and cosine.

What would happen if you tried to compute repeated fractions in base-x?

It’s an open question from my perspective when decimal patterns might terminate or repeat when evaluating base-x rational numbers.

I’d love to see someone out there give some of these questions a run!

## Number Bases and Polynomials

About a month ago, I was working with our 5th grade math teacher to develop some extension activities for some students in an unleveled class.  The class was exploring place value, and I suggested that some might be ready to explore what happens when you allow the number base to be something other than 10.  A few students had some fun learning to use their basic four algorithms in other number bases, but I made an even deeper connection.

When writing something like 512 in expanded form ($5\cdot 10^2+1\cdot 10^1+2\cdot 10^0$), I realized that if the 10 was an x, I’d have a polynomial.  I’d recognized this before, but this time I wondered what would happen if I applied basic math algorithms to polynomials if I wrote them in a condensed numerical form, not their standard expanded form.  That is, could I do basic algebra on $5x^2+x+2$ if I thought of it as $512_x$–a base-x “number”?  (To avoid other confusion later, I read this as “five one two base-x“.)

Following are some examples I played with to convince myself how my new notation would work.  I’m not convinced that this will ever lead to anything, but following my “what ifs” all the way to infinite series was a blast.  Read on!

If I wanted to add $(3x+5)$$(2x^2+4x+1)$, I could think of it as $35_x+241_x$ and add the numbers “normally” to get $276_x$ or $2x^2+7x+6$.  Notice that each power of x identifies a “place value” for its characteristic coefficient.

If I wanted to add $3x-7$ to itself, I had to adapt my notation a touch.  The “units digit” is a negative number, but since the number base, x, is unknown (or variable), I ended up saying $3x-7=3(-7)_x$.  The parentheses are used to contain multiple characters into a single place value.  Then, $(3x-7)+(3x-7)$ becomes $3(-7)_x+3(-7)_x=6(-14)_x$ or $6x-14$.  Notice the expanding parentheses containing the base-x units digit.

The last example also showed me that simple multiplication would work.  Adding $3x-7$ to itself is equivalent to multiplying $2\cdot (3x-7)$.  In base-x, that is $2\cdot 3(-7)_x$.  That’s easy!  Arguably, this might be even easier that doubling a number when the number base is known.  Without interactions between the coefficients of different place values, just double each digit to get $6(-14)_x=6x-14$, as before.

What about $(x^2+7)+(8x-9)$?  That’s equivalent to $107_x+8(-9)_x$.  While simple, I’ll solve this one by stacking.

and this is $x^2+8x-2$.  As with base-10 numbers, the use of 0 is needed to hold place values exactly as I needed a 0 to hold the $x^1$ place for $x^2+7$. Again, this could easily be accomplished without the number base conversion, but how much more can we push these boundaries?

Level 3–Multiplication & Powers:

Compute $(8x-3)^2$.  Stacking again and using a modification of the multiply-and-carry algorithm I learned in grade school, I got

and this is equivalent to $64x^2-48x+9$.

All other forms of polynomial multiplication work just fine, too.

From one perspective, all of this shifting to a variable number base could be seen as completely unnecessary.  We already have acceptably working algorithms for addition, subtraction, and multiplication.  But then, I really like how this approach completes the connection between numerical and polynomial arithmetic.  The rules of math don’t change just because you introduce variables.  For some, I’m convinced this might make a big difference in understanding.

I also like how easily this extends polynomial by polynomial multiplication far beyond the bland monomial and binomial products that proliferate in virtually all modern textbooks.  Also banished here is any need at all for banal FOIL techniques.

Level 4–Division:

What about $x^2+x-6$ divided by $x+3$? In base-x, that’s $11(-6)_x \div 13_x$. Remembering that there is no place value carrying possible, I had to be a little careful when setting up my computation. Focusing only on the lead digits, 1 “goes into” 1 one time.  Multiplying the partial quotient by the divisor, writing the result below and subtracting gives

Then, 1 “goes into” -2 negative two times.  Multiplying and subtracting gives a remainder of 0.

thereby confirming that $x+3$ is a factor of $x^2+x-6$, and the other factor is the quotient, $x-2$.

Perhaps this could be used as an alternative to other polynomial division algorithms.  It is somewhat similar to the synthetic division technique, without its  significant limitations:  It is not limited to linear divisors with lead coefficients of one.

For $(4x^3-5x^2+7) \div (2x^2-1)$, think $4(-5)07_x \div 20(-1)_x$.  Stacking and dividing gives

So $\displaystyle \frac{4x^3-5x^2+7}{2x^2-1}=2x-2.5+\frac{2x+4.5}{2x^2-1}$.

CONCLUSION

From all I’ve been able to tell, converting polynomials to their base-x number equivalents enables you to perform all of the same arithmetic computations.  For division in particular, it seems this method might even be a bit easier.

In my next post, I push the exploration of these base-x numbers into infinite series.

## Old school integral

This isn’t going to be one of my typical posts, but I just cracked a challenging indefinite integral and wanted to share.

I made a mistake solving a calculus problem a few weeks ago and ended up at an integral that looked pretty simple.  I tried several approaches and found many dead ends before finally getting a breakthrough.  Rather than just giving a proof, I thought I’d share my thought process in hopes that some students just learning integration techniques might see some different ways to attack a problem and learn to persevere through difficult times.

In my opinion, most students taking a calculus class would never encounter this problem.  The work that follows is clear evidence why everyone doing math should have access to CAS (or tables of integrals when CAS aren’t available).

Here’s the problem:

Integrate $\int \left( x^2 \cdot \sqrt{1+x^2} \right) dx$.

For convenience, I’m going to ignore in this post the random constant that appears with indefinite integrals.

While there’s no single algebraic technique that will work for all integrals, sometimes there are clues to suggest productive approaches.  In this case, the square root of a binomial involving a constant and a squared variable term suggests a trig substitution.

From trig identities, I knew $tan^2 \theta + 1 = sec^2 \theta$, so my first attempt was to let $x=tan \theta$, which gives $dx=sec^2 \theta d\theta$.  Substituting these leads to

$(tan \theta)'=sec^2 \theta$, claiming two secants for the differential in a reversed chain rule, but left a single secant in the expression, so I couldn’t make the trig identities work because odd numbers of trigs don’t convert easily using Pythagorean identities.  Then I tried using $(sec \theta)'=sec \theta \cdot tan \theta$, leaving a single tangent after accounting for the potential differential–the same problem as before.  A straightforward trig identity wasn’t going to do the trick.

Then I recognized that the derivative of the root’s interior is $2x$.  It was not the exterior $x^2$, but perhaps integration by parts would work.  I tried $u=x \longrightarrow u'=dx$ and $v'=x\sqrt{1+x^2} dx \longrightarrow v=\frac{1}{2} \left( 1+x^2 \right)^{3/2} \cdot \frac{2}{3}$.  Rewriting the original integral gave

The remaining integral still suggested a trig substitution, so I again tried $x =tan \theta$ to get

but the odd number of secants led me to the same dead end from trigonometric identities that stopped my original attempt.  I tried a few other variations on these themes, but nothing seemed to work.  That’s when I wondered if the integral even had a closed form solution.  Lots of simple looking integrals don’t work out nicely; perhaps this was one of them.  Plugging the integral into my Nspire CAS gave the following.

OK, now I was frustrated.  The solution wasn’t particularly pretty, but a closed form definitely existed.  The logarithm was curious, but I was heartened by the middle term I had seen with a different coefficient in my integration by parts approach.  I had other things to do, so I employed another good problem solving strategy:  I quit working on it for a while.  Sometimes you need to allow your sub-conscious to chew on an idea for a spell.  I made a note about the integral on my To Do list and walked away.

As often happens to me on more challenging problems, I woke this morning with a new idea.  I was still convinced that trig substitutions should work in some way, but my years of teaching AP Calculus and its curricular restrictions had blinded me to other possibilities.  Why not try a hyperbolic trig substitution? In many ways, hyperbolic trig is easier to manipulate than circular trig.  I knew

$\frac{d}{dt}cosh(t)=sinh(t)$ and $\frac{d}{dt}sinh(t)=cosh(t)$,

and the hyperbolic identity

$cosh^2t - sinh^2t=1 \longrightarrow cosh^2t=1+sinh^2t$.

(In case you haven’t worked with hyperbolic trig functions before, you can prove these for yourself using the definitions of hyperbolic sine and cosine:  $cosh(x)=\frac{1}{2}\left( e^x + e^{-x} \right)$ and $sinh(x)=\frac{1}{2}\left( e^x - e^{-x} \right)$.)

So, $x=sinh(A) \longrightarrow dx=cosh(A) dA$, and substitution gives

Jackpot!  I was down to an even number of (hyperbolic) trig functions, so Pythagorean identities should help me revise my latest expression into some workable form.

To accomplish this, I employed a few more hyperbolic trig identities:

1. $sinh(2A)=2sinh(A)cosh(A)$
2. $cosh(2A)=cosh^2(A)+sinh^2(A)$
3. $cosh^2(A) = \frac{1}{2}(cosh(2A)+1)$
4. $sinh^2(A) = \frac{1}{2}(cosh(2A)-1)$

(All of these can be proven using the definitions of sinh and cosh above.  I encourage you to do so if you haven’t worked much with hyperbolic trig before.  I’ve always liked the close parallels between the forms of circular and hyperbolic trig relationships and identities.)

If you want to evaluate $\int x^2 \sqrt{x^2+1} dx$ yourself, do so before reading any further.

Using equations 3 & 4, expanding, and then equation 3 again turns the integral into something that can be integrated directly.

The integral was finally solved!  I then used equations 1 & 2 to rewrite the expression back into hyperbolic functions of A only.

The integral was solved using the substitution $x=sinhA \longrightarrow A=sinh^{-1}x$ and (using $cosh^2A-sinh^2A=1$), $coshA=\sqrt{x^2+1}$.  Substituting back gave:

but that didn’t match what my CAS had given.  I could have walked away, but I had to know if I had made an error someplace or just had found a different expression for the same quantity.  I knew the inverse sinh could be replaced with a logarithm via a quadratic expression in $e^x$.

Well, that explained the presence of the logarithm in the CAS solution, but I was still worried by the cubic in my second term and the fact that my first two terms were a sum whereas the CAS’s solution’s comparable terms were a difference.  But as a former student once said, “If you take care of the math, the math will take care of you.”  These expressions had to be the same, so I needed to complete one more identity–algebraic this time.  Factoring, rewriting, and re-expanding did the trick.

What a fun problem (for me) this turned out to be.  It’s absolutely not worth the effort to do this every time when a CAS or integral table can drop the solution so much more quickly, but it’s also deeply satisfying to me to know why the form of the solution is what it is.  It’s also nice to know that I found not one, but three different forms of the solution.

Morals:  Never give up.  Trust your instincts. Never give up. Try lots of variations on your instincts. And never give up!

## Volumes of Revolution and Differentials

Here’s classic volume of revolution problem:

Determine the volume of the solid created when the region defined by $y=x^3$, the line $x=2$, and the x-axis is rotated about the x-axis.

Typical calculus classes present TWO ways to compute such volumes of revolution–orthogonal and parallel to the axis of rotation.  In this post, I argue that with substitutions involving differentials, you can actually get your choice of FOUR integrals to solve.  Hopefully one of them will be easy to solve.  Here’s how.

Slicing the volume ORTHOGONAL to the x-axis creates discs with radius $y$ and thickness $dx$ giving an original (orthogonal) volume integral $\int \pi y^2 dx$.

Method 1) I think the approach most familiar to most students is to substitute using $y=x^3$ to get

Method 2) Here’s my suggestion that I don’t see too many people use.  Consider substituting for the differential using $x=y^{1/3} \longrightarrow dx=\frac{1}{3} y^{-2/3} dy$ to get

Using the method of cylindrical shells, slicing the volume PARALLEL to the axis of rotation creates a series of annuli, each with radius $y$, thickness $dy$, and height $(2-x)$, giving an original (parallel) volume integral $\int 2\pi y \cdot (2-x) dy$.

Method 3) Substitute $x=y^{1/3}$ into this to get

Method 4) Or you could use the other differential, $y=x^3 \longrightarrow dy=3x^2 dx$, to get

CONCLUSION:

No matter what approach you take, it’s important for students to understand that there is no single approach that will will work well for all volumes.  Increasing the number of representations available for manipulation generally should increase your chances of finding one that’s easy (or at least easier) to solve.

In the end, I just think it’s pretty that all four of these integrals are exactly equivalent even though they really don’t look much alike.