# Category Archives: Technology

## Algebra Explorations Before Algebra

Here’s a short post to share two great tools for students to learn algebra–long before any formal algebra course and without seeming like you’re even learning algebra.

ACTIVITY ONE:  The first is a phenomenal recent set of posts on Imbalance Problems (here and here) from Paul Salomon.  If you’re on Twitter, interested in math or math education, and haven’t already, you should definitely follow him  (@lostinrecursion).

Paul is (or was) hosting an Imbalance Problem competition (mentioned at the end of his first post on  imbalance problems), but the following image from his post gives the general idea. Can you figure these out?  Better yet, can you write some of your own?  Can you encourage your students (or kids) to create some?  The process of thinking about the values of the unknown measures of the circles, squares, and triangles necessary to create these puzzles lies at the very heart of the concept of unknown variables that is so critical to algebraic reasoning.  Best of all, this feels like a game, and puzzle solvers don’t even realize they’re learning algebra.

ACTIVITY TWO:  Several months ago, a Westminster alumnus (Thanks, Phillip!) suggested an iPad app that my 3rd grade daughter instantly fell in love with–Dragon Box.  The app is available for $5.99 on both iOS and Android platforms. Jonathan H. Liu (@jonathanhliu) wrote such a great overview of the play on DragonBox for Wired. Rather than trying to imitate his post, I recommend you read his review: DragonBox: Algebra Beats Angry Birds. The play of the game is great for teaching algebraic skills, again without ever seeming like it’s teaching as much as it is. As a parent, I was THRILLED to see an educational game pull my daughter in so effectively and completely. A high point came when I was driving carpool last week and my daughter recommended to a friend who was wondering what app would be fun to play for the ride, “Try DragonBox. It’s fun!” As a math teacher, I certainly can appreciate the game’s end-of-level challenges to get the box alone (solve for a variable) in the right number of moves (efficiency) using the right number of cards (also efficiency). Still, there were a few times when I noted that a particular scenario could have been solved using a different sequence of equally-efficient moves that were not appropriately acknowledged as such by the software. My teacher side wasn’t particularly pleased with the only-one-way-earns-top-recognition approach of the app, especially when other alternatives are equally valid. Too many times, I fear students are faced with similar scenarios in their math classes. Efficiency and elegance are certainly valuable skills in mathematics, but I think we too often try to impose rigor on young learners long before they have achieved basic understanding. My grousing aside, my daughter and her friends weren’t bothered at all by the rare times where I had identified alternatives–I don’t think they even noticed. Back to my point about too much emphasis on “the ‘right’ way” too early, I decided not to mention it to them. As Mark Twain noted, I decided not “to let schooling get in the way of [their] education.” CONCLUSION: I hope you get a kick out of Paul’s imbalance problems (no matter what your age) and DragonBox if you have some younger kids around. As always, make learning fun and not obvious–your charges will learn in spite of themselves! Advertisements ## New Nspire Apps PLUS Weekend Savings TI finally converted its Nspire calculators to the iPad platform and through this weekend only in celebration of 25 years of Teachers Teaching with Technology, they’re offering both of their Nspire apps at$25 off their usual $29.99, or$4.99 each.  This is a GREAT deal, especially considering everything the Nspire can do!  Clicking on either of the images below will take you to a description page for that app.

In my opinion, if you’re going to get one of these, I’d grab the CAS version.  It does EVERYTHING the non-CAS version does plus great CAS tools.  Why pay the same money for the non-CAS and get less?  You aren’t required to use the CAS tools, but I’d rather have a tool and not need it than the other way around.  If you read my ‘blog, though, you know I strongly advocate for CAS use for anyone exploring mathematics.

Now, on to my brief review of the new apps.

MY REVIEW:  From my experimentations the last few days, this app appears to do EVERYTHING the corresponding handheld calculators can do.  I wouldn’t be surprised if there are a few things the computer version can do that the app can’t, but I haven’t been able to find it yet.  In a few places, I actually like the iPad app better than either the handheld or computer versions.  Here are a few.

• When you start the app, your home page shows all of the documents available that have been created on the app.  It’s easy enough to navigate there on the handheld or computer, but it’s a nice touch (to me) to see all of my files easily arranged when I start up.

• A BRILLIANT addition is the ability to export your working files to share with others.  Using the standard export button common to all iPad apps with export features, you get the ability to share your current doc via email or iTunes.
• The calculator history items can now be accessed using a simple tap instead of just arrow key or mouse navigation.

• Personally, I find it much easier to access the menus and settings with conveniently located app buttons.  I prefer having my tools available on a tap rather than buried in menus.  A nice touch, from my perspective.

• Moving objects is easy.  I was easily able to graph $y=x$ and the generic $y=a\cdot x^2+b\cdot x+c$ with sliders for each parameter.  It’s easy to drag the slider values, and after a brief tap-and-hold, a pop-up gives you an option to animate, change settings, move, or delete your slider.

• Also notice on the left side of the three previous screens that you have thumbnails of your currently open windows.  With a quick tap, you can quickly change between windows.
• One of the best features of the Nspire has always been its ability to integrate multiple representations of mathematical ideas.  That continues here.  As I said, the app appears to be a fully-functional variation of the pre-existing handheld and computer versions.
• The 3D-graphing option from a graphing page seems much easier to use on the iPad app.  Being able to use my finger to rotate a graph the way I want just seems much more intuitive than using my mouse.  As with the computer software, you can define your 3D surfaces and curves in Cartesian function form or parametrically.

• A lovely touch on the iPad version is the ability to use finger pinch and spread maneuvers to zoom in and out on 2D and 3D graphs.  Dragging your finger over a 2D graph easily repositions it.  Combined, these options make it incredibly easy to obtain good graphing windows.

For now, I see two drawbacks, but I can easily deal with both given the other advantages.

1. This concern has been resolved.  See my response here. At the bottom of the 3rd screenshot above, you can see that variable x is available in the math entry keyboard, but variables y and t are not.  You can easily grab a y through the alpha keyboard.  It won’t matter for most, I guess, but entering parametric equations on a graph page and solving systems of equations on a calculator page both require flipping between multiple screens to get the variable names and math symbols.  I get issues with space management, but making parametric equation entry and CAS use more difficult is a minor frustration.
2. I may not have looked hard enough, but I couldn’t find an easy way to adjust the computation scales for 3D graphs.  I can change the graph scales, but I was not able to get my graph of $z=sin \left( x^2 + y^2 \right)$ to look any smoother.

As I said, these are pretty minor flaws.

CONCLUSION:  It looks like strong, legitimate math middle and high school math-specific apps are finally entering the iPad market, and I know of others in development.  TI’s Nspire apps are spectacular (and are even better if you can score one for the current deeply discounted price).

## Polar Derivatives on TI-Nspire CAS

The following question about how to compute derivatives of polar functions was posted on the College Board’s AP Calculus Community bulletin board today.

From what I can tell, there are no direct ways to get derivative values for polar functions.  There are two ways I imagined to get the polar derivative value, one graphically and the other CAS-powered.  The CAS approach is much  more accurate, especially in locations where the value of the derivative changes quickly, but I don’t think it’s necessarily more intuitive unless you’re comfortable using CAS commands.  For an example, I’ll use $r=2+3sin(\theta )$ and assume you want the derivative at $\theta = \frac{\pi }{6}$.

METHOD 1:  Graphical

Remember that a derivative at a point is the slope of the tangent line to the curve at that point.  So, finding an equation of a tangent line to the polar curve at the point of interest should find the desired result.

Create a graphing window and enter your polar equation (menu –> 3:Graph Entry –> 4:Polar).  Then drop a tangent line on the polar curve (menu –> 8:Geometry –> 1:Points&Lines –> 7:Tangent).  You would then click on the polar curve once to select the curve and a second time to place the tangent line.  Then press ESC to exit the Tangent Line command.

To get the current coordinates of the point and the equation of the tangent line, use the Coordinates & Equation tool (menu –> 1:Actions –> 8:Coordinates and Equations).  Click on the point and the line to get the current location’s information.  After each click, you’ll need to click again to tell the nSpire where you want the information displayed.

To get the tangent line at $\theta =\frac{\pi }{6}$, you could drag the point, but the graph settings seem to produce only Cartesian coordinates.  Converting $\theta =\frac{\pi }{6}$ on $r=2+3sin(\theta )$ to Cartesian gives

$\left( x,y \right) = \left( r \cdot cos(\theta ), r \cdot sin(\theta ) \right)=\left( \frac{7\sqrt{3}}{4},\frac{7}{4} \right)$ .

So the x-coordinate is $\frac{7\sqrt{3}}{4} \approx 3.031$.  Drag the point to find the approximate slope, $\frac{dy}{dx} \approx 8.37$.  Because the slope of the tangent line changes rapidly at this location on this polar curve, this value of 8.37 will be shown in the next method to be a bit off.

Unfortunately, I tried to double-click the x-coordinate to set it to exactly $\frac{7\sqrt{3}}{4}$, but that property is also disabled in polar mode.

METHOD 2:  CAS

Using the Chain Rule, $\displaystyle \frac{dy}{dx} = \frac{dy}{d\theta }\cdot \frac{d\theta }{dx} = \frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}$.  I can use this and the nSpire’s ability to define user-created functions to create a $\displaystyle \frac{dy}{dx}$ polar differentiator for any polar function $r=a(\theta )$.  On a Calculator page, I use the Define function (menu –> 1:Actions –> 1:Define) to make the polar differentiator.  All you need to do is enter the expression for a as shown in line 2 below.

This can be evaluated exactly or approximately at $\theta=\frac{\pi }{6}$ to show $\displaystyle \frac{dy}{dx} = 5\sqrt{3}=\approx 8.660$.

Conclusion:

As with all technologies, getting the answers you want often boils down to learning what questions to ask and how to phrase them.

## Invariable Calculus Project II

As Rocky hinted in his comment to my last post, $\displaystyle f(x)=\frac{k}{x}$  also has the constant area property.  Following a lead from Cohen, et al’s Student Research Projects in Calculus, I discovered long ago that with nothing more than knowledge of the product rule, the quotient and/or chain rule, and a healthy dose of some patient algebra manipulations, students could actually determine all functions which have the property that right triangles formed by the x– and y-axes and tangent lines to those functions have constant area.  This morning, I discovered a nice CAS approach that makes the problem accessible to far more students.

This makes another great project for calculus students who’ve just learned algebraic rules for differentiation:

Determine an equation for any twice-differentiable function–$y=g(x)$–with the property that all tangent lines to g in Quadrant I, along with the x- and y-axes, form triangles of constant area.

(Very minor) HINT:  At some point in solving this problem, you’ll need to make and use some assumptions about the values of $a, g(a), g'(a)$, and $g''(a)$.

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

Assumptions:  Let $(a,g(a))$ be any arbitrary point on $y=g(x)$ in Quadrant I.  This makes $a>0$ and $g(a)>0$.  I also know $g'(a)<0$ because otherwise both of the x– and y-intercepts of the tangent line would not be positive, making the triangle’s area negative.  Finally, if $g''(a)=0$, then g would be a linear function, and there would be only one triangle.  To keep the problem interesting, I’m going to assume $g''(a)\ne 0$.

Setting up:  We no longer have a specific function, so everything must be in generalities.  A generalized equation for a tangent line to any function $y=g(x)$ at $x=a$ is

$y-g(a)=g'(a)\cdot (x-a)$.

From here, the generalized x-intercept is $\displaystyle a-\frac{g(a)}{g'(a)}$, and the y-intercept is $g(a)-a\cdot g'(a)$.  [Side note, the x-intercept is also the same form used in Newton’s Method for root approximations, a connection I’ll make later in the term when I’m teaching AP Calculus.]  Combining the generalized intercepts, I can write a generic area formula.

$\displaystyle Area = \frac{1}{2} \cdot \left( a - \frac{g(a)}{g'(a)} \right) \cdot \left( g(a) - a \cdot g'(a) \right)$

Differentiating and Cleaning Up:  Whenever I’ve used this problem in the past, my students and I have always used paper & pencil techniques.  While I’m quite comfortable with my algebraic manipulation skills, the significant majority of my students struggle with this part of the assignment.  As I was writing this post today, I finally had a technology insight that I should have years ago, given my long interest in CAS.  So, I’ll show the next portion in two different approaches, first traditional, and then via CAS.

APPROACH 1 – TRADITIONAL PAPER & PENCIL:  Trying to keep some brevity in this expanding post, I skip a few algebra steps below while providing some guiding explanations.  Finding a common denominator in the Area equation and recognizing a common factor leads to

$\displaystyle Area(a) = - \frac{1}{2} \cdot \frac{\left( a \cdot g'(a) - g(a) \right) ^2}{g'(a)}$ .

Applying the quotient rule with respect to a gives

$\frac{d(Area(a))}{da} = -\frac{1}{2} \cdot \frac{g'(a)\cdot 2(a\cdot g'(a)-g(a))(1\cdot g'(a)+a\cdot g''(a) - g'(a))- (a\cdot g'(a)-g(a))^2\cdot g''(a)}{(g'(a))^2}$ .

Remember that I seek functions whose tangent lines create constant area triangles, so $\displaystyle \frac{d(Area(a))}{da} = 0$.  Using this on the left and canceling some terms on the right gives

$\displaystyle 0 = -\frac{1}{2} \cdot \frac{2a\cdot g'(a)\cdot g''(a)(a\cdot g'(a)-g(a))-(a\cdot g'(a)-g(a))^2 \cdot g''(a)}{(g'(a))^2}$ .

Pulling out common factors and cleaning up a little more turns this into a completely factored form.

$\displaystyle 0 = - \frac{1}{2} \cdot \frac{(a\cdot g'(a)-g(a))\cdot g''(a)\cdot (a\cdot g'(a)+g(a))}{(g'(a))^2}$

APPROACH 2 – CAS:  Because the algebra was going to be complicated enough to make viewing on a handheld calculator very difficult, I used my TI-nSpire CAS Computer software to tackle the problem.  In the image below, I defined the area function in line 1 and differentiated with respect to a in line 2.  Equating the final equation to 0 gives the last algebraic line above from APPROACH 1.

Notice that I did not need to define a differentiation technique or to manipulate the factoring.  The results on both lines automatically accomplish the factoring I worked so carefully to establish earlier in APPROACH 1.  This is a beautiful example of what I see as a central benefit of CAS:  Keeping users focused on the mathematics of the problem situation.

Some students might actually be curious about how the challenge of differentiating line 1 could end up as relatively “clean” as the result in line 2.  GOOD!  CAS also inspires creative thinking.

Other than the differentiation step, everything else in APPROACH 1 was simple algebra.  Complicated, perhaps, but simple.  In fact, I don’t think it’s mathematics at all; it’s algebraic arithmetic.  I’m not disparaging the work or the approach, but I see mathematics as pattern recognition and big thinking.  I think CAS is completely justified in this problem.

Applying the Zero Product Property:  Our initial assumptions clear the denominator because $g'(a)<0$.  Because $g''(a)\ne 0$, I can eliminate that term, too.  With a and $g(a)$ both positive and $g'(a)<0$, the $(a\cdot g'(a)-g(a))$ term must be negative and therefore can be eliminated.  That drops the initially complicated differential equation to

$\displaystyle 0 = a\cdot g'(a)+g(a)$.

Finally–the Solution:  Depending on how much your students know, this last equation can be solved three different ways:  A) recognizing differentiation rules, B) solving a separable differentiable equation, or C) using a CAS solver.  I typically assign this problem so early in a calculus course that they have no idea what a differential equation is, making  the first approach the only available technique.  But this is also a great problem to introduce after learning about separable DEs.

APPROACH A:  If you look carefully, you can recognize the right side as the result of the product rule applied to $a\cdot g(a)$.  (In my experience, most students need some time, encouragement, and occasionally some hints to “see” this.)  Because the product rule result equals zero, the original expression must have equalled a constant.  That means $a\cdot g(a) = C$ for any constant, C.  Solving gives $\displaystyle g(a)=\frac{C}{a}$.  That means Rocky’s suggested family of functions at the top of this post, $\displaystyle f(x)=\frac{k}{x}$ not only produces triangles of constant area, it’s the only family of functions that does!  Very cool!

APPROACH B:  Rewriting the result of the Zero Product Property simplification using xs and ys gives $\displaystyle 0=x\cdot \frac{dy}{dx} +y$.  The variables can be rearranged to give $\displaystyle -\frac{dx}{x}=\frac{dy}{y}$.  Integration gives $-ln(x)+ln(C)=ln(y)$ for any random constant, $ln(C)$.  Logarithm properties lead to $y=\displaystyle \frac{C}{x}$, as before.

APPROACH C:  While I like the pattern recognition insights from the previous two approaches, the solution can also be found using a CAS.

On the TI-nSpire, c1 represents any random constant, so the DE solver again gives the same results.

Conclusion:  No matter what approaches you take, this problem shows that the only functions that have the property of their tangent lines producing constant area triangles.

Shiny.

## Exploring Sequences and Lines

Here’s another favorite problem that could be used for any middle or high school students who’ve been exposed to both arithmetic sequences and linear equations.

There is a family of lines, $Ax+By+C=0$, for which $\left\{ A, B, C \right\}$ can be any arithmetic sequence.  What do all members of this family have in common?

As with most great math problems, the problem is simply stated and can be approached from many different perspectives.  In the end, this one isn’t that difficult to crack, but the solution (at least to me) is not intuitively obvious from the problem statement. If you’ve not seen this before, please try it before reading further.

WARNING:  Problem Solution Follows

MOST COMMON STUDENT APPROACH:  Given the problem’s phrasing, most begin by writing out a few possible equations using different arithmetic sequences, typically with positive common differences.  After a few tries, most get a little frustrated as no obvious pattern emerges from the equations.

It is this point that is typically the most difficult for me as a teacher.  I want to help anyone who struggles, but “help” often means allowing others the room to struggle and to find ways of solving problems on their own.  Only when a student or group gets really frustrated do I sometimes ask, “Are there any other ways you can look at this problem or at your work?”

Eventually, most decide to graph their equations to see if anything pops out.  The following graph was submitted this past semester by one of my precalculus students using the free online Desmos calculator.

Two lines intersecting in a point is common.  Three or more in a single point almost always indicates something interesting.  Seven lines through a single point is screaming for attention!  From this graph, all lines in this family apparently contain the point (1,-2).  That seems a bit random until you investigate further, but pattern recognition is just half of the battle for a mathematician.  Now that something interesting has been discovered, a mathematician wants to know if this is a fluke or something inherent to all members of the family.

NOTE:  All graphs my students have produced over the years have always contained specific equations.  I don’t know that it’s any more enlightening, but I did create on Desmos a slider version of the graphs of this family with one slider for the initial term (A) and a second for its common difference (d).

UNIQUE SOLUTION METHODS FROM STUDENTS:

All successful solutions eventually rewrite the sequence $\left\{ A,B,C \right\}$ to $\left\{ A,A+d,A+2d \right\}$ where d is the common difference for a generic arithmetic sequence with initial term A.

Method I: After discovering the common point of intersection, most plug it into the left side of the equation and simplify to get

$Ax+By+C=A\cdot 1+\left( A+d\right)\cdot -2+\left( A+2d \right)=0$.

Because the left side reduces to zero for all generic arithmetic sequences, $\left\{ A,A+d,A+2d \right\}$, (1,-2) must be common to all members of this family.

A few students aren’t comfortable dealing with 0=0, so these tend to plug in $x=1$ and solve for y to get $y=-2$, proving that the y-coordinate for $x=1$ for all members of this family is always -2.

Method II:  A few students every year work algebraically from properties of arithmetic sequences.  For any arithmetic sequence, $\left\{ A,B,C \right\}$, $\frac{A+C}{2}=B$.  This rewrites to $1\cdot A-2\cdot B+C=0$, so whenever $\left( x,y \right)=\left(1,-2 \right)$, then $Ax+By+C=0$ is a fundamental property of all arithmetic sequences.

Personally, I think this method gets closest to explaining why the point (1,-2) is the common characteristic of this family.

Method III:  This year, I had a student take an approach I’d never seen before.  She defined one arithmetic sequence as $\left\{ a,a+d,a+2d \right\}$ and another as $\left\{ m,m+n,m+2n \right\}$ for any real values of a,d,m, and n.  This leads to a system of equations: $a\cdot x+(a+d)\cdot y+(a+2d)=0$ and $m\cdot x+(m+n)\cdot y+(m+2n)=0$ .  If you have some younger students or if all the variables make you nervous, the solution is available via Wolfram Alpha.

Still, this system is pretty easy to manipulate.  If you multiply the first equation by m and the second by a, the x-terms will eliminate with subtraction, giving

$m\cdot((a+d)\cdot y+(a+2d))-a\cdot((m+n)\cdot y+(m+2n))=0$.

Solving for y shows that all of the coefficients simplify surprisingly easily.

$((ma+md)-(am+an))\cdot y=-(ma+2md)+(am+2an)$
$(md-an)\cdot y = -2\cdot (md-an)\Longrightarrow y=-2$

From here, determining $x=1$ is easy, proving the relationship.

SOLUTIONS and APPROACHES NOT YET OFFERED BY STUDENTS:

Approach A:  High school students don’t often think about ways to simplify problem situations, especially at the beginning of problems.  One way I did that for this problem in later class discussions was to recognize that it one of the terms in the arithmetic sequence was 0, you didn’t need to deal with nearly as many terms. For example, if your sequence was ${1,0,-1}$, the linear equation would be $x-1=0$.  Similarly, the sequence $\left\{ 01,2 \right\}$ leads to $y+2=0$.  Obviously, the only thing these two lines have in common is the point (1,-2).  A proof of the property must still be established, but this is one of the fastest ways I’ve seen to identify the central property.

Approach B:  A purely algebraic approach to this problem could redefine the arithmetic sequence as $\left\{ a,a+d,a+2d\right\}$as before, giving:

$a\cdot x+(a+d)\cdot y+(a+2d)=0$

Collecting like terms gives

$(x+y+1)\cdot a+(y+2)\cdot d=0$.

The values of a and d must remain as parameters to include all possible arithmetic sequences.  Because the equation always equals 0, the coefficients of a and d are both 0, making $y=-2$ (for the coefficient of d) and therefore $x=1$.

EXTENSION:

We once had a test question at the end of the unit containing this exercise.  Basically, it reminded students that they had discovered that all lines $Ax+By+C=0$, for which $\left\{ A, B, C \right\}$ can be any arithmetic sequence contained the point (1,-2).  It then asked for an equation of a family of linear functions using the same arithmetic $\left\{ A, B, C \right\}$ that all contained the point (1,2).

The two most common responses we’ve seen involve a reflection or a vertical translation.  (1,-2) can become (1,2) by reflecting over the x-axis, so making the y-values negative would do the trick:  $Ax-By+C=0$.  Similarly, (1,-2) can become (1,2) by translating up 4 units, giving $Ax+B(y-4)+C=0$.

## Exponentials and Transformations

Here’s an old and (maybe) a new way to think about equations of exponential functions.  I suspect you’ve seen the first approach.  If you understand what exponentials functions are, my second approach using transformations is much faster and involves no algebra!

Members of the exponential function family can be written in the form $y=a\cdot b^x$ for real values of a and postive real values of b.  Because there are only two parameters, only two points are required to write an equation of any exponential.

EXAMPLE 1: Find an exponential function through the points (2,5) and (4,20).

METHOD 1:  Plug the points into the generic exponential equation to get a 2×2 system of equations.  It isn’t necessary, but to simplify the next algebra step, I always write the equation with the larger exponent on top.

$\left\{\begin{matrix} 20=a\cdot b^4 \\ 5=a\cdot b^2 \end{matrix}\right.$

If the algebra isn’t the point of the lesson, this system could be solved with a CAS.  Users would need to remember that $b>0$ to limit the CAS solutions to just one possibility.

If you want to see algebra, you could use substitution, but I recommend division.  Students’ prior experience with systems typically involved only linear functions for which they added or subtracted the equations to eliminate variables.  For exponentials, the unknown parameters are multiplied, so division is a better operational choice.  Using the system above, I get $\displaystyle \frac{20}{5}=\frac{a\cdot b^4}{a\cdot b^2}$.  The fractions must be equivalent because their numerators are equal and their denominators are equal.

Simplifying gives $4=b^2\rightarrow b=+2$ (because $b>0$ for exponential functions) and $a=\frac{5}{4}$.

This approach is nice because the a term will always cancel from the first division step, leaving a straightforward constant exponent to undo, a pretty easy step.

METHOD 2:  Think about what an exponential function is and does.  Then use transformations.

Remember that linear functions ($y=m\cdot x+b$) “start” with a y-value of b (when $x=0$) and add m to y every time you add 1 to x.  The only difference between linear and exponential functions is that exponentials multiply while linears add.  Therefore, exponential functions ($y=a\cdot b^x$) “start” with a y-value of a when $x=0$ and multiply by b every time 1 is added to x.

What makes the given points a bit annoying is that neither is a y-intercept.  No problem.  If you don’t like the way a problem is phrased, CHANGE IT!    (Just remember to change the answer back to the original conditions!)

If you slide the given points left 2 units, you get (0,5) and (2,20).  It would also be nice if the points were 1 x-unit apart, so halving the x-values gives (0,5) and (1,20).  Because the y-intercept is now 5, and the next point multiplies that by 4, an initial equation for the exponential is $y = 5\cdot 4^x$ . To change this back to the original points, undo the transformations at the start of this paragraph:  stretch horizontally by 2 and then move right 2.  This gives $y = 5\cdot 4^\frac{x-2}{2}$.

This is algebraically equivalent to the $y=\frac{5}{4}\cdot 2^x$ found early.  Obviously, my students prove this.

One student asked why we couldn’t make the (4,20) point the y-intercept.  Of course we can!  To move more quickly through the set up, starting at (4,20) and moving to (2,5) means my initial value is 20 and I multiply by $\frac{1}{4}$ if the x-values move left 2 from an initial x-value of 4.  This gives $y = 20\cdot\left( \frac{1}{4} \right) ^\frac{x+4}{-2}$.  Of course, this 3rd equation is algebraically equivalent to the first two.

Here’s one more example to illustrate the speed of the transformations approach, even when the points aren’t convenient.

EXAMPLE 2: Find an exponential function through (-3,7) and (12,13).

Starting at (-3,7) and moving to (12,13) means my initial value is 7, and I multiply by $\frac{13}{7}$ if the x-values move right 15 from an initial x-value of -3.  This gives $y = 7\cdot\left( \frac{13}{7} \right) ^\frac{x-3}{15}$.

Equivalently, starting at (12,13) and moving to (-3,7) means my initial value is 13, and I multiply by $\frac{7}{13}$ if the x-values move left 15 from an initial x-value of 12.  This gives $y = 13\cdot\left( \frac{7}{13} \right) ^\frac{x+3}{-15}$.

If you get transformations, exponential equations require almost no algebraic work, no matter how “ugly” the coordinates.  I hope this helps give a different perspective on exponential function equations and helps enhance the importance of the critical math concept of equivalence.

## Calling for More CAS in Statistics

When you allow your students to solve problems in ways that make the most sense to them, interesting and unexpected results sometimes happen.  On a test in my senior, non-AP statistics course earlier this semester, we posed this.

A child is 40 inches tall, which places her in the top 10% of all children of similar age. The heights for children of this age form an approximately normal distribution with a mean of 38 inches. Based on this information, what is the standard deviation of the heights of all children of this age?

From the question, you likely deduced that we had been exploring normal distributions and the connection between areas under such curves and their related probabilities and percentiles.  Hoping to get students to think just a little bit, we decided to reverse a cookbook question (given or derive a z-score and compute probability) and asked instead for standard deviation.  My fellow teacher and I saw the question as a simple Algebra I-level manipulation, but our students found it a very challenging revision.  Only about 5% of the students actually solved it the way we thought.  The majority employed a valid (but not always justified) trial-and-error approach.  And then one of my students invoked what I thought to be a brilliant use of a CAS command that I should have imagined myself.  Unfortunately, it did not work out, even though it should.  I’m hoping future iterations of CAS software of all types will address this shortcoming.

What We Thought Would Happen

The problem information can be visually represented as shown below.

Given x-values, means, and standard deviations, our students had practice with many problems which gave the resulting area.  They had also been given areas under normal curves and worked backwards to z-scores which could be re-scaled and re-centered to corresponding points on any normal curve.  We hoped they would be able to apply what they knew about normal curves to make this a different, but relatively straightforward question.

Given the TI-nSpire software and calculators each of our students has, we’ve completely abandoned statistics tables.  Knowing that the given score was at the 90th percentile, the inverse normal TI-nSpire command quickly shows that this point corresponds to an approximate z-score of 1.28155.  Substituting this and the other givens into the x-value to z-score scaling relationship, $z=\frac{x-\overline{x}}{s}$ , leads to an equation with a single unknown which easily can be solved by hand or by CAS to find the unknown standard deviation, $s \approx 1.56061$ .  Just a scant handful of students actually employed this.

What the Majority Did

Recognizing the problem as a twist on their previous work, most invoked a trial-and-error approach.  From their work, I could see that most essentially established bounds around the potential standard deviation and employed an interval bisection approach (not that any actually formally named their technique).

If you know the bounds, mean, and standard deviation of a portion of a normal distribution, you can find the percentage area using the nSpire’s normal Cdf command.  Knowing that the percentage area was 0.1, most tried a standard deviation of 1, and saw that not enough area (0.02275) was captured.  Then they tried 2, and goth too much area (0.158655).  A reproduction of one student’s refinements leading to a standard deviation of $s \approx 1.56$ follows.

THE COOL PART:  Students who attempted this approach got to deal directly with the upper 10% of the area; they weren’t required to adapt this to the “left-side area” input requirement of the inverse normal command.  While this adjustment is certainly minor, being able to focus on the problem parameters–as defined–helped some of my students.

THE PROBLEM:  As a colleague at my school told me decades ago when I started teaching, “Remember that a solution to every math problem must meet two requirements.  Show that the solution(s) you find is (are) correct, and show that there are no other solutions.”

Given a fixed mean and x-value, it seems intuitive to me that there is a one-to-one correspondence between the standard deviation of the normal curve and the area to the right of that point.  This isn’t a class for which I’d expect rigorous proof of such an assertion, but I still hoped that some might address the generic possibility of multiple answers and attempt some explanation for why that can’t happen here.  None showed anything like that, and I’m pretty certain that not a single one of my students in this class considered this possibility.  They had found an answer that worked and walked away satisfied.  I tried talking with them afterwards, but I’m not sure how many fully understood the subtle logic and why it was mathematically important.

The Creative Solution

Exactly one student remembered that he had a CAS and that it could solve equations.  Tapping the normal Cdf command used by the majority of his peers, he set up and tried to solve an equation as shown below.

Sadly, this should have worked for my student, but it didn’t.  (He ultimately fell back on the trial-and-error approach.) The equation he wrote is the heart of the trial-and-error process, and there is a single solution.  I suspect the programmers at TI simply hadn’t thought about applying the CAS commands from one area of their software to the statistical functions in another part of their software.  Although I should have, I hadn’t thought about that either.

Following my student’s lead, I tried a couple other CAS approaches (solving systems, etc.) to no avail.  Then I shifted to a graphical approach.  Defining a function using the normal Cdf command, I was able to get a graph.

Graphing $y=0.1$ showed a single point of intersection which could then be computed numerically in the graph window to give the standard deviation from earlier.

What this says to me is that the CAS certainly has the ability to solve my student’s equation–it did so numerically in the graph screen–but for some reason this functionality is not currently available on the TI-nSpire CAS’s calculator screens.

Extensions

My statistics students just completed a unit on confidence intervals and inferential reasoning.  Rather than teaching them the embedded TI syntax for confidence intervals and hypothesis testing, I deliberately stayed with just the normal Cdf and inverse normal commands–nothing more.  A core belief of my teaching is

Memorize as little as possible and use it as broadly as possible.

By staying with these just two commands, I continued to reinforce what normal distributions are and do, concepts that some still find challenging.  What my student taught me was that perhaps I could limit these commands to just one, the normal Cdf.

For example, if you had a normal distribution with mean 38 and standard deviation 2, what x-value marks the 60th percentile?

Now that’s even more curious.  The solve command doesn’t work (even in an approximation mode), but now the numerical solve gives the solution confirmed by the inverse normal command.

What if you wanted the bounds on the same normal distribution that defined the middle 80% of the area?  As shown below, I failed to solve then when I asked the CAS to compute directly for any of the equivalent distance of the bounds from the mean (line 1), the z-scores (line 2), or the location of the upper x-value (line 3).

But reversing the problem to define the 10% area of the right tail does give the desired result (line 2) (note that nsolve does work even though solve still does not) with the solution confirmed by the final two lines.

Conclusion

Admittedly, there’s not lots of algebra involved in most statistics classes–LOADS of arithmetic and computation, but not much algebra.  I’m convinced, though, that more attention to some algebraic thinking could benefit students.  The different statistics packages out there do lots of amazing things, especially the TI-nSpire, but it would be very nice if these packages could align themselves better with CAS features to permit students to ask their questions more “naturally”.  After all, such support and scaffolding are key features that make CAS and all technology so attractive for those of us using them in the classroom.