# Good Question 9

This is a good question that leads to other good questions, both mathematical and philosophical. A few days ago this question was posted on a private Facebook page for AP Calculus Readers. The problem and illustration were photographed from an un-cited textbook.

Player 1 runs to first base [from home plate] at a speed of 20 ft/s while player 2 runs from second base to third base a speed of 15 ft/s. Let s be the distance between the two players. How fast is s changing when player 1 is 30 feet from home plate and Player 2 is 60 feet from second base. [A figure was given showing that the distance between the bases is 90 feet.]

Some commenters indicated some possible inconsistencies in the question, such as assuming the Player 2 is on second base when Player 1 leaves home plate. In this case the numbers don’t make sense. So someone suggested this must be a hit-and-run situation. To which someone else replied that with a lead of that much it’s really a stolen base situation. So the first thing to be learned here is that even writing a simple problem like this you need to take some of the real aspects into consideration. But this doesn’t change the mathematical aspects of the problem.

One of the things I noticed before I attempted to work out the solution was that Player 2 is the same distance from third base as Player 1 is from home plate. I verbalized this as the “players are directly across the field from each other.” I filed this away since it didn’t seem to matter much. Wrong!

Then I worked the problem two ways. These are shown in the appendix at the end of this post. I discovered (twice) that s’ = 0; at the moment suggested in the question the distance is not changing.

Then it hit me. Doh! – I didn’t have to do all that. So I posted this solution (which I now notice someone beat me to):

At the time described, the players are directly across the field from each other (90 feet apart). This is the closest they come. The distance between them has been decreasing and now starts to increase. So at this instant s is not changing (s‘ = 0).

The Philosophical Question

Then the original poster asked for someone “to post [actual] work done in calculus” and “to see some related rates.” So I posted some “calculus” and got to thinking – the philosophical question – isn’t my first answer calculus?

I think it is. It makes use of an important calculus concept, namely that as things change, at the minimum place, the derivative is zero. Furthermore, the justification (that the distance changes from decreasing to increasing at the minimum implies the derivative is zero) is included.* Why do you need variables?

Also, this solution is approached as an extreme value (max/min) problem rather than a related rate problem. This shows a nice connection between the two types of problems.

The Related (but not related rate) Good Question

So here is another calculus question with none of the numbers we’ve grown to expect:

Two cars travel on parallel roads. The roads are w feet apart. At what rate is the distance between the cars changing when the cars are w feet apart?

Notice:

• That the cars could be traveling in the same or opposite directions.
• Their speeds are not given.
• You don’t know when or where they started; only that at some time they are opposite each other (w feet apart).
• In fact, they could start opposite each other and travel in the same direction at the same speed remaining always w feet apart.
• One car could be standing still and the other just passes it.

But you can still answer the question.

(*Continuity and differentiability are given (or at least implied) in the original statement of the problem.)

Appendix

My first attempt was to set up a coordinate system with the origin at third base as shown below.

Then, taking the time indicated in the problem as t = 0, the position of Player 1 is (90, 30 + 20t) and the position of player 2 is (0, 30 – 15t). Then the distance between them is

$s=\sqrt{{{90}^{2}}+{{\left( 30-15t-\left( 30+20t \right) \right)}^{2}}}=\sqrt{{{90}^{2}}+{{\left( -35t \right)}^{2}}}$

and then

$\displaystyle {s}'\left( t \right)=\frac{2\left( -35t \right)\left( -35 \right)}{2s}\text{ and }{s}'\left( 0 \right)=0$

This is correct, but for some reason I was suspicious probably because zeros can hide things. So I re-stated this time taking t = 0 to be one second before the situation described in the problem. Now player 1’s position is (90, 10+20t) and player 2’s position is (0, 45-15t).

$s\left( t \right)=\sqrt{{{90}^{2}}+{{\left( 45-15t-\left( 10+20t \right) \right)}^{2}}}=\sqrt{{{90}^{2}}+{{\left( 30-35t \right)}^{2}}}$

$\displaystyle {s}'\left( t \right)=\frac{2\left( 35-35t \right)\left( -35 \right)}{2s}\text{ and }{s}'\left( 1 \right)=0$

.

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## 5 thoughts on “Good Question 9”

Mr. McMullin – Thanks for posts like these. I’ve benefitted greatly from the material you put out. I was actually trying to get on your main website today, and it appears to be down. Is this just a temporary issue? Or did you move your content somewhere else? Thanks for any help you can offer on that!

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• Thank you. If you are referring to my old website at linmcmullin.net: I have closed it. Most of the material that was there is now on this blog site. Click the drop down menu at the top labeled “Website.”

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