# Tag Archives: CAS

## Stats Exploration Yields Deeper Understanding

or “A lesson I wouldn’t have learned without technology”

Last November, some of my AP Statistics students were solving a problem involving a normal distribution with an unknown mean.  Leveraging the TI Nspire CAS calculators we use for all computations, they crafted a logical command that should have worked.  Their unexpected result initially left us scratching heads.  After some conversations with the great folks at TI, we realized that what at first seemed perfectly reasonable for a single answer, in fact had two solutions.  And it took until the end of this week for another student to finally identify and resolve the mysterious results.  This ‘blog post recounts our journey from a questionable normal probability result to a rich approach to confidence intervals.

THE INITIAL PROBLEM

I had assigned an AP Statistics free response question about a manufacturing process that could be manipulated to control the mean distance its golf balls would travel.  We were told that the process created balls with a normally distributed distance of 288 yards and a standard deviation of 2.8 yards.  The first part asked students to find the probability of balls traveling more than an allowable 291.2 yards.  This was straightforward.  Find the area under a normal curve with a mean of 288 and a standard deviation of 2.8 from 291.2 to infinity.  The Nspire (CAS and non-CAS) syntax for this is:

[Post publishing note: See Dennis’ comment below for a small correction for the non-CAS Nspires.  I forgot that those machines don’t accept “infinity” as a bound.]

As 12.7% of the golf balls traveling too far is obviously an unacceptably high percentage, the next part asked for the mean distance needed so only 99% of the balls traveled allowable distances.  That’s when things got interesting.

A “LOGICAL” RESPONSE RESULTS IN A MYSTERY

Their initial thought was that even though they didn’t know the mean, they now knew the output of their normCdf command.  Since the balls couldn’t travel a negative distance and zero was many standard deviations from the unknown mean, the following equation with representing the unknown mean should define the scenario nicely.

Because this was an equation with a single unknown, we could now use our CAS calculators to solve for the missing parameter.

Something was wrong.  How could the mean distance possibly be just 6.5 yards?  The Nspires are great, reliable machines.  What happened?

I had encountered something like this before with unexpected answers when a solve command was applied to a Normal cdf with dual finite bounds .  While it didn’t seem logical to me why this should make a difference, I asked them to try an infinite lower bound and also to try computing the area on the other side of 291.2.  Both of these provided the expected solution.

The caution symbol on the last line should have been a warning, but I honestly didn’t see it at the time.  I was happy to see the expected solution, but quite frustrated that infinite bounds seemed to be required.  Beyond three standard deviations from the mean of any normal distribution, almost no area exists, so how could extending the lower bound from 0 to negative infinity make any difference in the solution when 0 was already $\frac{291.2}{2.8}=104$ standard deviations away from 291.2?  I couldn’t make sense of it.

My initial assumption was that something was wrong with the programming in the Nspire, so I emailed some colleagues I knew within CAS development at TI.

GRAPHS REVEAL A HIDDEN SOLUTION

They reminded me that statistical computations in the Nspire CAS were resolved through numeric algorithms–an understandable approach given the algebraic definition of the normal and other probability distribution functions.  The downside to this is that numeric solvers may not pick up on (or are incapable of finding) difficult to locate or multiple solutions.  Their suggestion was to employ a graph whenever we got stuck.  This, too, made sense because graphing a function forced the machine to evaluate multiple values of the unknown variable over a predefined domain.

It was also a good reminder for my students that a solution to any algebraic equation can be thought of as the first substitution solution step for a system of equations.  Going back to the initially troublesome input, I rewrote normCdf(0,291.2,x,2.8)=0.99 as the system

y=normCdf(0,291.2,x,2.8)
y=0.99

and “the point” of intersection of that system would be the solution we sought.  Notice my emphasis indicating my still lingering assumptions about the problem.  Graphing both equations shone a clear light on what was my persistent misunderstanding.

I was stunned to see two intersection solutions on the screen.  Asking the Nspire for the points of intersection revealed BOTH ANSWERS my students and I had found earlier.

If both solutions were correct, then there really were two different normal pdfs that could solve the finite bounded problem.  Graphing these two pdfs finally explained what was happening.

By equating the normCdf result to 0.99 with FINITE bounds, I never specified on which end the additional 0.01 existed–left or right.  This graph showed the 0.01 could have been at either end, one with a mean near the expected 284 yards and the other with a mean near the unexpected 6.5 yards.  The graph below shows both normal curves with the 6.5 solution having an the additional 0.01 on the left and the 284 solution with the 0.01 on the right.

The CAS wasn’t wrong in the beginning.  I was.  And as has happened several times before, the machine didn’t rely on the same sometimes errant assumptions I did.  My students had made a very reasonable assumption that the area under the normal pdf for the golf balls should start only 0 (no negative distances) and inadvertently stumbled into a much richer problem.

A TEMPORARY FIX

The reason the infinity-bounded solutions didn’t give the unexpected second solution is that it is impossible to have the unspecified extra 0.01 area to the left of an infinite lower or upper bound.

To avoid unexpected multiple solutions, I resolved to tell my students to use infinite bounds whenever solving for an unknown parameter.  It was a little dissatisfying to not be able to use my students’ “intuitive” lower bound of 0 for this problem, but at least they wouldn’t have to deal with unexpected, counterintuitive results.

Surprisingly, the permanent solution arrived weeks later when another student shared his fix for a similar problem when computing confidence interval bounds.

A PERMANENT FIX FROM AN UNEXPECTED SOURCE

I really don’t like the way almost all statistics textbooks provide complicated formulas for computing confidence intervals using standardized z- and t-distribution critical scores.  Ultimately a 95% confidence interval is nothing more than the bounds of the middle 95% of a probability distribution whose mean and standard deviation are defined by a sample from the overall population.  Where the problem above solved for an unknown mean, on a CAS, computing a confidence interval follows essentially the same reasoning to determine missing endpoints.

My theme in every math class I teach is to memorize as little as you can, and use what you know as widely as possible.  Applying this to AP Statistics, I never reveal the existence of confidence interval commands on calculators until we’re 1-2 weeks past their initial introduction.  This allows me to develop a solid understanding of confidence intervals using a variation on calculator commands they already know.

For example, assume you need a 95% confidence interval of the percentage of votes Bernie Sanders is likely to receive in Monday’s Iowa Caucus.  The CNN-ORC poll released January 21 showed Sanders leading Clinton 51% to 43% among 280 likely Democratic caucus-goers.  (Read the article for a glimpse at the much more complicated reality behind this statistic.)  In this sample, the proportion supporting Sanders is approximately normally distributed with a sample p=0.51 and sample standard deviation of p of $\sqrt((.51)(.49)/280)=0.0299$.  The 95% confidence interval is the defined by the bounds containing the middle 95% of the data of this normal distribution.

Using the earlier lesson, one student suggested finding the bounds on his CAS by focusing on the tails.

giving a confidence interval of (0.45, 0.57) for Sanders for Monday’s caucus, according to the method of the CNN-ORC poll from mid-January.  Using a CAS keeps my students focused on what a confidence interval actually means without burying them in the underlying computations.

That’s nice, but what if you needed a confidence interval for a sample mean?  Unfortunately, the t-distribution on the Nspire is completely standardized, so confidence intervals need to be built from critical t-values.  Like on a normal distribution, a 95% confidence interval is defined by the bounds containing the middle 95% of the data.  One student reasonably suggested the following for a 95% confidence interval with 23 degrees of freedom.  I really liked the explicit syntax definition of the confidence interval.

Alas, the CAS returned the input.  It couldn’t find the answer in that form.  Cognizant of the lessons learned above, I suggested reframing the query with an infinite bound.

That gave the proper endpoint, but I was again dissatisfied with the need to alter the input, even though I knew why.

That’s when another of my students spoke up to say that he got the solution to work with the initial commands by including a domain restriction.

Of course!  When more than one solution is possible, restrict the bounds to the solution range you want.  Then you can use the commands that make sense.

FIXING THE INITIAL APPROACH

That small fix finally gave me the solution to the earlier syntax issue with the golf ball problem.  There were two solutions to the initial problem, so if I bounded the output, they could use their intuitive approach and get the answer they needed.

If a mean of 288 yards and a standard deviation of 2.8 yards resulted in 12.7% of the area above 291.2, then it wouldn’t take much of a left shift in the mean to leave just 1% of the area above 291.2. Surely that unknown mean would be no lower than 3 standard deviations below the current 288, somewhere above 280 yards.  Adding that single restriction to my students’ original syntax solved their problem.

Perfection!

CONCLUSION

By encouraging a deep understanding of both the underlying statistical content AND of their CAS tool, students are increasingly able to find creative solutions using flexible methods and expressions intuitive to them.  And shouldn’t intellectual strength, creativity, and flexibility be the goals of every learning experience?

## Best Algebra 2 Lab Ever

This post shares what I think is one of the best, inclusive, data-oriented labs for a second year algebra class.  This single experiment produces linear, quadratic, and exponential (and logarithmic) data from a lab my Algebra 2 students completed this past summer.  In that class, I assigned frequent labs where students gathered real data, determined models to fit that data, and analyzed goodness of the models’ fit to the data.   I believe in the importance of doing so much more than just writing an equation and moving on.

For kicks, I’ll derive an approximation for the coefficient of gravity at the end.

THE LAB:

On the way to school one morning last summer, I grabbed one of my daughters’ “almost fully inflated” kickballs and attached a TI CBR2 to my laptop and gathered (distance, time) data from bouncing the ball under the Motion Sensor.  NOTE:  TI’s CBR2 can connect directly to their Nspire and TI84 families of graphing calculators.  I typically use computer-based Nspire CAS software, so I connected the CBR via my laptop’s USB port.  It’s crazy easy to use.

One student held the CBR2 about 1.5-2 meters above the ground while another held the ball steady about 20 cm below the CBR2 sensor.  When the second student released the ball, a third clicked a button on my laptop to gather the data:  time every 0.05 seconds and height from the ground.  The graphed data is shown below.  In case you don’t have access to a CBR or other data gathering devices, I’ve uploaded my students’ data in this Excel file.

Remember, this is data was collected under far-from-ideal conditions.  I picked up a kickball my kids left outside on my way to class.  The sensor was handheld and likely wobbled some, and the ball was dropped on the well-worn carpet of our classroom floor.  It is also likely the ball did not remain perfectly under the sensor the entire time.  Even so, my students created a very pretty graph on their first try.

For further context, we did this lab in the middle of our quadratics unit that was preceded by a unit on linear functions and another on exponential and logarithmic functions.  So what can we learn from the bouncing ball data?

LINEAR 1:

While it is very unlikely that any of the recorded data points were precisely at maximums, they are close enough to create a nice linear pattern.

As the height of a ball above the ground helps determine the height of its next bounce (height before –> energy on impact –> height after), the eight ordered pairs (max height #n, max height #(n+1) ) from my students’ data are shown below

This looks very linear.  Fitting a linear regression and analyzing the residuals gives the following.

The data seems to be close to the line, and the residuals are relatively small, about evenly distributed above and below the line, and there is no apparent pattern to their distribution.  This confirms that the regression equation, $y=0.673x+0.000233$, is a good fit for the = height before bounce and = height after bounce data.

NOTE:  You could reasonably easily gather this data sans any technology.  Have teams of students release a ball from different measured heights while others carefully identify the rebound heights.

The coefficients also have meaning.  The 0.673 suggests that after each bounce, the ball rebounded to 67.3%, or 2/3, of its previous height–not bad for a ball plucked from a driveway that morning.  Also, the y-intercept, 0.000233, is essentially zero, suggesting that a ball released 0 meters from the ground would rebound to basically 0 meters above the ground.  That this isn’t exactly zero is a small measure of error in the experiment.

EXPONENTIAL:

Using the same idea, consider data of the form (x,y) = (bounce number, bounce height). the graph of the nine points from my students’ data is:

This could be power or exponential data–something you should confirm for yourself–but an exponential regression and its residuals show

While something of a pattern seems to exist, the other residual criteria are met, making the exponential regression a reasonably good model: $y = 0.972 \cdot (0.676)^x$.  That means bounce number 0, the initial release height from which the downward movement on the far left of the initial scatterplot can be seen, is 0.972 meters, and the constant multiplier is about 0.676.  This second number represents the percentage of height maintained from each previous bounce, and is therefore the percentage rebound.  Also note that this is essentially the same value as the slope from the previous linear example, confirming that the ball we used basically maintained slightly more than 2/3 of its height from one bounce to the next.

And you can get logarithms from these data if you use the equation to determine, for example, which bounces exceed 0.2 meters.

So, bounces 1-4 satisfy the requirement for exceeding 0.20 meters, as confirmed by the data.

A second way to invoke logarithms is to reverse the data.  Graphing x=height and y=bounce number will also produce the desired effect.

Each individual bounce looks like an inverted parabola.  If you remember a little physics, the moment after the ball leaves the ground after each bounce, it is essentially in free-fall, a situation defined by quadratic movement if you ignore air resistance–something we can safely assume given the very short duration of each bounce.

I had eight complete bounces I could use, but chose the first to have as many data points as possible to model.  As it was impossible to know whether the lowest point on each end of any data set came from the ball moving up or down, I omitted the first and last point in each set.  Using (x,y) = (time, height of first bounce) data, my students got:

What a pretty parabola.  Fitting a quadratic regression (or manually fitting one, if that’s more appropriate for your classes), I get:

Again, there’s maybe a slight pattern, but all but two points are will withing  0.1 of 1% of the model and are 1/2 above and 1/2 below.  The model, $y=-4.84x^2+4.60x-4.24$, could be interpreted in terms of the physics formula for an object in free fall, but I’ll postpone that for a moment.

LINEAR 2:

If your second year algebra class has explored common differences, your students could explore second common differences to confirm the quadratic nature of the data.  Other than the first two differences (far right column below), the second common difference of all data points is roughly 0.024.  This raises suspicions that my student’s hand holding the CBR2 may have wiggled during the data collection.

Since the second common differences are roughly constant, the original data must have been quadratic, and the first common differences linear. As a small variation for each consecutive pair of (time, height) points, I had my students graph (x,y) = (x midpoint, slope between two points):

If you get the common difference discussion, the linearity of this graph is not surprising.  Despite those conversations, most of my students seem completely surprised by this pattern emerging from the quadratic data.  I guess they didn’t really “get” what common differences–or the closely related slope–meant until this point.

Other than the first three points, the model seems very strong.  The coefficients tell an even more interesting story.

GRAVITY:

The equation from the last linear regression is $y=4.55-9.61x$.  Since the data came from slope, the y-intercept, 4.55, is measured in m/sec.  That makes it the velocity of the ball at the moment (t=0) the ball left the ground.  Nice.

The slope of this line is -9.61.  As this is a slope, its units are the y-units over the x-units, or (m/sec)/(sec).  That is, meters per squared second.  And those are the units for gravity!  That means my students measured, hidden within their data, an approximation for coefficient of gravity by bouncing an outdoor ball on a well-worn carpet with a mildly wobbly hand holding a CBR2.  The gravitational constant at sea-level on Earth is about -9.807 m/sec^2.  That means, my students measurement error was about $\frac{9.807-9.610}{9.807}=2.801%$.  And 2.8% is not a bad measurement for a very unscientific setting!

CONCLUSION:

Whenever I teach second year algebra classes, I find it extremely valuable to have students gather real data whenever possible and with every new function, determine models to fit their data, and analyze the goodness of the model’s fit to the data.  In addition to these activities just being good mathematics explorations, I believe they do an excellent job exposing students to a few topics often underrepresented in many secondary math classes:  numerical representations and methods, experimentation, and introduction to statistics.  Hopefully some of the ideas shared here will inspire you to help your students experience more.

## Recentering Normal Curves, revisited

I wrote here about using a CAS to determine a the new mean of a recentered normal curve from an AP Statistics exam question from the last decade.  My initial post shared my ideas on using CAS technology to determine the new center.  After hearing some of my students’ attempts to solve the problem, I believe they took a simpler, more intuitive approach than I had proposed.

REVISITING:

In the first part of the problem, solvers found the mean and standard deviation of the wait time of one train: $\mu = 30$ and $\sigma = \sqrt{500}$, respectively.  Then, students computed the probability of waiting to be 0.910144.

The final part of the question asked how long that train would have to be delayed to make that wait time 0.01.  Here’s where my solution diverged from my students’ approach.  Being comfortable with transformations, I thought of the solution as the original time less some unknown delay which was easily solved on our CAS.

STUDENT VARIATION:

Instead of thinking of the delay–the explicit goal of the AP question–my students  sought the new starting time.  Now that I’ve thought more about it, knowing the new time when the train will leave does seem like a more natural question and avoids the more awkward expression I used for the center.

The setup is the same, but now the new unknown variable, the center of the translated normal curve, is newtime.  Using their CAS solve command, they found

It was a little different to think about negative time, but they found the difference between the new time difference (-52.0187 minutes) and the original (30 minutes) to be 82.0187 minutes, the same solution I discovered using transformations.

CONCLUSION:

This is nothing revolutionary, but my students’ thought processes were cleaner than mine.  And fresh thinking is always worth celebrating.

## Confidence Intervals via graphs and CAS

Confidence intervals (CIs) are a challenging topic for many students, a task made more challenging, in my opinion, because many (most?) statistics texts approach CIs via z-scores.  While repeatedly calculating CI endpoints from standard deviations explains the underlying mathematical structure, it relies on an (admittedly simple) algebraic technique that predates classroom technology currently available for students on the AP Statistics exam.

Many (most?) statistics packages now include automatic CI commands.  Unfortunately for students just learning what a CI means, automatic commands can become computational “black boxes.”  Both CAS and graphing techniques offer a strong middle ground–enough foundation to reinforce what CIs mean with enough automation to avoid unnecessary symbol manipulation time.

In most cases, this is accomplished by understanding a normal cumulative distribution function (cdf) as a function, not just as an electronic substitute for normal probability tables of values.  In this post, I share two alternatives each for three approaches to determining CIs using a TI-Nspire CAS.

SAMPLE PROBLEM:

In 2010, the mean ACT mathematics score for all tests was 21.0 with standard deviation 5.3.  Determine a 90% confidence interval for the math ACT score of an individual chosen at random from all 2010 ACT test takers.

METHOD 1a — THE STANDARD APPROACH:

A 90% CI excludes the extreme 5% on each end of the normal distribution.  Using an inverse normal command gives the z-scores at the corresponding 5% and 95% locations on the normal cdf.

Of course, utilizing symmetry would have required only one command.  To find the actual boundary points of the CI, standardize the endpoints, x, and equate that to the two versions of the z-scores.

$\displaystyle \frac{x-21.0}{5.3} = \pm 1.64485$

Solving these rational equations for x gives $x=12.28$ and $x=29.72$, or $CI = \left[ 12.28,29.72 \right]$ .

Most statistics software lets users avoid this computation with optional parameters for the mean and standard deviation of non-standard normal curves.  One of my students last year used this in the next variation.

METHOD 1b — INTRODUCING LISTS:

After using lists as shortcuts on our TI-Nspires last year for evaluating functions at several points simultaneously, one of my students creatively applied them to the inverse normal command, entering the separate 0.05 and 0.95 cdf probabilities as a single list.  I particularly like how the output for this approach outputs looks exactly like a CI.

METHOD 2a — CAS:

The endpoints of a CI are just endpoints of an interval on a normal cdf, so why not avoid the algebra and additional inverse normal command and determine the endpoints via CAS commands?  My students know the solve command from previous math classes, so after learning the normal cdf command, there are very few situations for them to even use the inverse.

This approach keeps my students connected to the normal cdf and solving for the bounds quickly gives the previous CI bounds.

METHOD 2b (Alas, not yet) — CAS and LISTS:

Currently, the numerical techniques the TI-Nspire family uses to solve equations with statistics commands don’t work well with lists in all situations.  Curiously, the Nspire currently can’t handle the solve+lists equivalent of the inverse normal+lists approach in METHOD 1b.

But, I’ve also learned that problems not easily solved in an Nspire CAS calculator window typically crack pretty easily when translated to their graphical equivalents.

METHOD 3a — GRAPHING:

This approach should work for any CAS or non-CAS graphing calculator or software with statistics commands.

Remember the “f” in cdf.  A cumulative distribution function is a function, and graphing calculators/software treats them as such.  Replacing the normCdf upper bounds with an x for standard graphing syntax lets one graph the normal cdf (below).

Also remember that any algebraic equation can be solved graphically by independently graphing each side of the equation and treating the resulting pair of equations as a system of equations.  In this case, graphing $y=0.05$ and $y=0.95$ and finding the points of intersection gives the values of the CI.

METHOD 3b — GRAPHING and LISTS:

SIDENOTE:  While lists didn’t work with the CAS in the case of METHOD 2b, the next screen shows the syntax to graph both ends of the CI using lists with a single endpoint equation.

The lists obviously weren’t necessary here, but the ability to use lists is a very convenient feature on the TI-Nspire that I’ve leveraged countless times to represent families of functions.  In my opinion, using them in METHOD 3b again leverages that same idea, that the endpoints you seek are different aspects of the same family–the CI.

CONCLUSION:

There are many ways for students in their first statistics courses to use what they already know to determine the endpoints of a confidence interval.  And keeping students attention focused on new ways to use old information solidifies both old and new content.  Eliminating unnecessary computations that aren’t the point of most of introductory statistics anyway is an added bonus.

Happy learning everyone…

## Recentering a Normal Curve with CAS

Sometimes, knowing how to ask a question in a different way using appropriate tools can dramatically simplify a solution.  For context, I’ll use an AP Statistics question from the last decade about a fictitious railway.

THE QUESTION:

After two set-up questions, students were asked to compute how long to delay one train’s departure to create a very small chance of delay while waiting for a second train to arrive.  I’ll share an abbreviated version of the suggested solution before giving what I think is a much more elegant approach using the full power of CAS technology.

BACKGROUND:

Initially, students were told that X was the normally distributed time Train B took to travel to city C, and Y was the normally distributed time Train D took to travel to C.  The first question asked for the distribution of Y-X if the mean and standard deviation of X are respectively 170 and 20, and the mean and standard deviation of Y are 200 and 10, respectively.  Knowing how to transform normally distributed variables quickly gives that Y-X is normally distributed with mean 30 and standard deviation $\sqrt{500}$.

Due to damage to a part of the railroad, if Train B arrived at C before Train D, B would have to wait for D to clear the tracks before proceeding.  In part 2, you had to find the probability that B would wait for D.  Translating from English to math, if B arrives before D, then $X \le Y$.  So the probability of Train B waiting on Train D is equivalent to $P(0 \le Y-X)$.  Using the distribution information in part 1 and a statistics command on my Nspire, this probability is

FIX THE DELAY:

Under the given conditions, there’s about a 91.0% chance that Train B will have to wait at C for Train D to clear the tracks.  Clearly, that’s not a good railway management situation, setting up the final question.  Paraphrasing,

How long should Train B be delayed so that its probability of being delayed is only 0.01?

FORMAL PROPOSED SOLUTION:

A delay in Train B says the mean arrival time of Train D, Y, will remain unchanged at 200, while the mean of arrival time of Train B, X, is increased by some unknown amount.  Call that new mean of X, $\hat{X}=170+delay$.  That makes the new mean of the difference in arrival times

$Y - \hat{X} = 200-(170+delay) = 30-delay$

As this is just a translation, the distribution of $Y - \hat{X}$ is congruent to the distribution of $Y-X$, but recentered.  The standard deviation of both curves is $\sqrt{500}$.  You want to find the value of delay so that $P \left(0 \le Y - \hat{X} \right) = 0.01$.  That’s equivalent to knowing the location on the standard normal distribution where the area to the right is 0.01, or equivalently, the area to the left is 0.99.  One way that can be determined is with an inverse normal command.

The proposed solution used z-scores to suggest finding the value of delay by solving

$\displaystyle \frac{0-(30-delay)}{\sqrt{500}} = 2.32635$

A little algebra gives $delay=82.0187$, so the railway should delay Train B just a hair over 82 minutes.

CAS-ENHANCED ALTERNATIVE SOLUTION:

From part 2, the initial conditions suggest Train B has a 91.0% chance of delay, and part 3 asks for the amount of recentering required to change that probability to 0.01.  Rephrasing this as a CAS command (using TI-Nspire syntax), that’s equivalent to solving

Notice that this is precisely the command used in part 2, re-expressed as an equation with a variable adjustment to the mean.  And since I’m using a CAS, I recognize the left side of this equation as a function of delay, making it something that can easily be “solved”.

Notice that I got exactly the same solution without the algebraic manipulation of a rational expression.

My big point here is not that use of a CAS simplifies the algebra (that wasn’t that hard in the first place), but rather that DEEP KNOWLEDGE of the mathematical situation allows one to rephrase a question in a way that enables incredibly efficient use of the technology.  CAS aren’t replacements for basic algebra skills, they are enhancements for mathematical thinking.

I DON”T HAVE CAS IN MY CLASSROOM.  NOW WHAT????

The CAS solve command is certainly nice, but many teachers and students don’t yet have CAS access, even though it is 100% legal for the PSAT, SAT, and all AP math exams.   But that’s OK.  If you recognize the normCdf command as a function, you can essentially use a graph menu to accomplish the same end.

Too often, I think teachers and students think of normCdf and invNorm commands as nothing more than glorified “lookup commands”–essentially nothing more than electronic versions of the probability tables they replaced.  But, when one of the parameters is missing, replacing it with X makes it graphable.  In fact, whenever you have an equation that is difficult (or impossible to solve), graph both sides and find the intersection, just like a solution to a system of equations.  Using this strategy, graphing $y=normCdf(0,\infty,30-X,\sqrt{500})$ and $y=0.01$ and finding the intersection gives the required solution.

CONCLUSION

Whether you can access a CAS or not, think more deeply about what questions ask and find creative alternatives to symbolic manipulations.

## A Generic Approach to Arclength in Calculus

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

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

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

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

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

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

## Chemistry, CAS, and Balancing Equations

Here’ s a cool application of linear equations I first encountered about 20 years ago working with chemistry colleague Penney Sconzo at my former school in Atlanta, GA.  Many students struggle early in their first chemistry classes with balancing equations.  Thinking about these as generalized systems of linear equations gives a universal approach to balancing chemical equations, including ionic equations.

This idea makes a brilliant connection if you teach algebra 2 students concurrently enrolled in chemistry, or vice versa.

FROM CHEMISTRY TO ALGEBRA

Consider burning ethanol.  The chemical combination of ethanol and oxygen, creating carbon dioxide and water:

$C_2H_6O+3O_2 \longrightarrow 2CO_2+3H_2O$     (1)

But what if you didn’t know that 1 molecule of ethanol combined with 3 molecules of oxygen gas to create 2 molecules of carbon dioxide and 3 molecules of water?  This specific set coefficients (or multiples of the set) exist for this reaction because of the Law of Conservation of Matter.  While elements may rearrange in a chemical reaction, they do not become something else.  So how do you determine the unknown coefficients of a generic chemical reaction?

Using the ethanol example, assume you started with

$wC_2H_6O+xO_2 \longrightarrow yCO_2+zH_2O$     (2)

for some unknown values of w, x, y, and z.  Conservation of Matter guarantees that the amount of carbon, hydrogen, and oxygen are the same before and after the reaction.  Tallying the amount of each element on each side of the equation gives three linear equations:

Carbon:  $2w=y$
Hydrogen:  $6w=2z$
Oxygen:  $w+2x=2y+z$

where the coefficients come from the subscripts within the compound notations.  As one example, the carbon subscript in ethanol ( $C_2H_6O$ ) is 2, indicating two carbon atoms in each ethanol molecule.  There must have been 2w carbon atoms in the w ethanol molecules.

This system of 3 equations in 4 variables won’t have a unique solution, but let’s see what my Nspire CAS says.  (NOTE:  On the TI-Nspire, you can solve for any one of the four variables.  Because the presence of more variables than equations makes the solution non-unique, some results may appear cleaner than others.  For me, w was more complicated than z, so I chose to use the z solution.)

All three equations have y in the numerator and denominators of 2.  The presence of the y indicates the expected non-unique solution.  But it also gives me the freedom to select any convenient value of y I want to use.  I’ll pick $y=2$ to simplify the fractions.  Plugging in gives me values for the other coefficients.

Substituting these into (2) above gives the original equation (1).

VARIABILITY EXISTS

Traditionally, chemists write these equations with the lowest possible natural number coefficients, but thinking of them as systems of linear equations makes another reality obvious.  If 1 molecule of ethanol combines with 3 molecules of hydrogen gas to make 2 molecules of carbon dioxide and 3 molecules of water, surely 10 molecule of ethanol combines with 30 molecules of hydrogen gas to make 20 molecules of carbon dioxide and 30 molecules of water (the result of substituting $y=20$ instead of the $y=2$ used above).

You could even let $y=1$ to get $z=\frac{3}{2}$, $w=\frac{1}{2}$, and $x=\frac{3}{2}$.  Shifting units, this could mean a half-mole of ethanol and 1.5 moles of hydrogen make a mole of carbon dioxide and 1.5 moles of water.  The point is, the ratios are constant.  A good lesson.

ANOTHER QUICK EXAMPLE:

Now let’s try a harder one to balance:  Reacting carbon monoxide and hydrogen gas to create octane and water.

$wCO + xH_2 \longrightarrow y C_8 H_{18} + z H_2 O$

Setting up equations for each element gives

Carbon:  $w=8y$
Oxygen:  $w=z$
Hydrogen:  $2x=18y+2z$

I could simplify the hydrogen equation, but that’s not required.  Solving this system of equations gives

Nice.  No fractions this time.  Using $y=1$ gives $w=8$, $x=17$, and $z=8$, or

$8CO + 17H_2 \longrightarrow C_8 H_{18} + 8H_2 O$

Simple.

EXTENSIONS TO IONIC EQUATIONS:

Now let’s balance an ionic equation with unknown coefficients a, b, c, d, e, and f:

$a Ba^{2+} + b OH^- + c H^- + d PO_4^{3-} \longrightarrow eH_2O + fBa_3(PO_4)_2$

In addition to writing equations for barium, oxygen, hydrogen, and phosphorus, Conservation of Charge allows me to write one more equation to reflect the balancing of charge in the reaction.

Barium:  $a = 3f$
Oxygen:  $b +4d = e+8f$
Hydrogen:  $b+c=2e$
Phosphorus:  $d=2f$
CHARGE (+/-):  $2a-b-c-3d=0$

Solving the system gives

Now that’s a curious result.  I’ll deal with the zeros in a moment.  Letting $d=2$ gives $f=1$ and $a=3$, indicating that 3 molecules of ionic barium combine with 2 molecules of ionic phosphate to create a single uncharged molecule of barium phosphate precipitate.

The zeros here indicate the presence of “spectator ions”.  Basically, the hydroxide and hydrogen ions on the left are in equal measure to the liquid water molecule on the right.  Since they are in equal measure, one solution is

$3Ba^{2+}+6OH^- +6H^-+2PO_4^{3-} \longrightarrow 6H_2O + Ba_3(PO_4)_2$

CONCLUSION:

You still need to understand chemistry and algebra to interpret the results, but combining algebra (and especially a CAS) makes it much easier to balance chemical equations and ionic chemical equations, particularly those with non-trivial solutions not easily found by inspection.

The minor connection between science (chemistry) and math (algebra) is nice.

As many others have noted, CAS enables you to keep your mind on the problem while avoiding getting lost in the algebra.