Good Question 9

Posted on February 14, 2016 by Lin McMullin

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 “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 on 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 does the distance between the cars change when the cars are w feet apart?

Notice:

That the cars could be travelling 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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Implicit Relations & Related Rates

Posted on March 13, 2013 by Lin McMullin

AP Type Questions 7

Implicitly defined relations and implicit differentiation

What students should know how to do

Know how to find the first derivative of an implicit relation using the product rule, quotient rule, the chain rule, etc.
Know how to find the second derivative, including substituting for the first derivative.
Know how to evaluate the first and second derivative by substituting both coordinates of the point. (Note: If all that is needed is the numerical value of the derivative then the substitution is often easier if done before solving for dy/dx or d²y/dx².)
Analyze the derivative to determine where the relation has horizontal and/or vertical tangents.
Write and work with lines tangent to the relation.
Find extreme values. It may also be necessary to show that the point where the derivative is zero is actually on the graph.

Related Rates

What students should know how to do

Set up and solve related rate problems.
Be familiar with the standard type of related rate situations, but also be able to adapt to different contexts.
Know how to differentiate with respect to time, using any of the differentiation techniques.
Interpret the answer in the context of the problem.
Unit analysis.

Shorter questions on both these concepts appear in the multiple-choice sections. As always, look over as many questions of this kind from past exams as you can find.

For some previous posts on related rate see October 8, and 10, 2012 and for implicit relation s see November 14, 2012

Painting a Point

Posted on February 4, 2013 by Lin McMullin

Accumulation 7: An application (of paint)

Suppose you started with a point, the origin to be specific, and painted it. You put on layers and layers of paint until your point grows to a sphere with radius r. Let’s stop and admire your work part way through the job; at this point the radius is ${{x}_{i}}$ and $0\le {{x}_{i}}\le r$ .

How much paint will you need for the next layer?

Easy: you need an amount equal to the surface area of the sphere, $4\pi {{x}_{i}}^{2}$ , times the thickness of the paint. As everyone knows paint is thin, specifically $\Delta x$ thin. So we add an amount of paint to the sphere equal to $4\pi {{x}^{2}}\left( \Delta x \right)$ .

The volume of the final sphere must be the same as the total amount of paint. The total amount of paint must be the (Riemann) sum of all the layers or $\displaystyle \sum\limits_{i=1}^{n}{4\pi {{x}_{i}}^{2}\left( \Delta x \right)}$ . As usual $\Delta x$ is very thin, tending to zero as a matter of fact, so the amount of paint must be

$\displaystyle \underset{\Delta x\to 0}{\mathop{\lim }}\,\sum\limits_{i=1}^{n}{4\pi {{x}_{i}}^{2}\left( \Delta x \right)=\int_{0}^{r}{4\pi {{x}^{2}}}dx=\left. \tfrac{4}{3}\pi {{x}^{3}} \right|_{0}^{r}=\tfrac{4}{3}\pi {{r}^{3}}}$ .

A standard related rate problem is to show that the rate of change of the volume of a sphere is proportional to its surface area – the constant of proportionality is dr/dt. So it should not be a surprise that the integral of this rate of change of the surface area is the volume. The integral of a rate of change is the amount of change.

Interestingly this approach works other places as long as you properly define “radius:”

A circle centered at the origin with radius x and perimeter of $2\pi x$ , gains area at a rate equal to its perimeter times the “thickness of the edge” $\Delta x$ : $\displaystyle A=\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{k=1}^{n}{2\pi x\Delta x}=\int_{0}^{r}{2\pi x\,dx=\pi {{r}^{2}}}$

A square centered at the origin with “radius” x with sides whose length are s =2 x, gains area at a rate equal to its perimeter (8x) times the “thickness of the edge” $\Delta x$ : $\displaystyle A=\underset{n\to \infty }{\mathop{\text{lim}}}\,\sum\limits_{k=1}^{n}{8x\Delta x=\int_{0}^{x}{8x\,dx}}=4{{x}^{2}}={{(2x)}^{2}}={{s}^{2}}$

A cube centered at the origin with “radius” x and edges of length 2x, gains volume at a rate equal to its surface area, $6(4{{x}^{2}})$ , times the “thickness of the face” $\Delta x$ – think paint again: $V=\underset{n\to \infty }{\mathop{\text{lim}}}\,\sum\limits_{k=1}^{n}{\text{6(4}{{\text{x}}^{2}})\Delta x}=\int_{0}^{x}{24{{x}^{2}}dx}=8{{x}^{3}}={{(2x)}^{3}}={{s}^{3}}$

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Implicitly defined relations and implicit differentiation

Related Rates

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