# Related Rate Problems II

If you look in most textbooks for related rate questions you will find pretty much the same related rate problems: ladders sliding down walls, people walking away from lamppost, water running into or out of cone shaped tanks, etc.

Here are two somewhat different related problems you may like.

1. A girl starts riding down an escalator at the same time a boy starts riding up a parallel escalator. The escalators are 5 meter apart, 30 meters long, and they both move at the rate of 1 m/sec.
1. How fast is the distance between the kids changing 10 seconds later? (Answer: -1.789 m/sec.)
2. Variation: Suppose that the girl is moving at 1 m/sec and the boy at 3 m/sec. The boy reaches the end of the escalator first and stands there. Now how fast is the distance between them changing 8 seconds after they start?  (1.486 m/sec)
3. Variation: With the situation given in part b., write and graph a piece-wise defined function that gives the distance between the kids.  Find where this graph is continuous but not differentiable. Why does this happen? (Answer: the boy reaches the top in 10 seconds and stands still. Up to this point the distance between them is represented by a hyperbola; here it now becomes a different parabola with an abrupt change in the graph. At this point, the function is not differentiable. This second parabola appears almost linear from the 10 second point on.
1. A 60 foot long rope is attached to a pulley 36 feet above the ground. A lantern his attached to one end of the rope and a man holds the other end on the top of his head 6 feet off the ground. He walks away at the rate of 5 feet/sec.
1. Find the rate at which the distance between the pulley and the top of the man’s head is changing when he is 40 feet from the point directly under the pulley? (Answer: 4 ft./sec.)
2. Find the rate of change of the length of the man’s shadow when he is 40 feet from a point directly under the pulley? (Answer:  -0.90 ft./sec.)
3. (Extension – extreme value problem, rather difficult) When the man starts walking the lantern is at the height of his head and his shadow is infinitely long. When is the tip of the man’s shadow closest to the point directly under the pulley and how far away is it? (Answer: At t = 6.654 sec the tip of the shadow is 39.658 feet from the point under the pulley.)

These two questions are from Audrey Weeks’ Calculus in Motion. This is a really good package of dynamic illustrations of calculus concepts and AP Exams free-response question that runs on Geometer’s Sketchpad. For more information click here. The related rate sections include both standard and non-standard problems.

## 2 thoughts on “Related Rate Problems II”

1. Marika McFal says:

Hi Lin! I have been enjoying your blog and resources since discovering them about 6 months ago. I have also appreciated your feedback so much on the AP Calc forum. When working through these Related Rates II problems above (to do with my students), I ran into a few difficulties.

In the escalator problem (#1 part 3), I wrote the piecewise function and found the value where it was not differentiable rather easily, I thought, but the statement that it “then becomes linear” did not make sense to me. I got another hyperbola: sqrt (25+t^2) for 10<t<30

Also, for the lantern problem, part 2, I got 6.5 ft/sec. I'd be happy to share my work in some way if you'd like. I can not get the answer that you posted. Could you please forward me the work if possible??? or at least the set up so I can compare?

Thank you! and thank you for all of your wonderful work! I wish you were closer to Cincinnati so I could do a workshop with you!!!!!

Marika McFall
Anderson High School
Cincinnati, Ohio

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• Marika
Whoops, looks like a did make a few mistakes.
In the escalator problem the second part is a hyperbola. I was looking at a graph and it appeared very linear.
I also have a mistake in the lantern problem. I do not agree with your answer (either). I got -0.90 feet/sec. My computation is this:

Let x= distance to man from point under pulley. $\frac{dx}{dt}=5$ (Given)
Let the length of the rope from the pulley to the man be

$z=\sqrt{{{30}^{2}}+{{x}^{2}}}$.

When x = 40, z = 50 so at this point

$\displaystyle \frac{dz}{dt}=\frac{2x\left( dx/dt \right)}{2\sqrt{900+{{x}^{2}}}}=\frac{5x}{\sqrt{900+{{x}^{2}}}}=\frac{5\left( 40 \right)}{\sqrt{900+{{40}^{2}}}}=4$

Let s = length of shadow. Z – 30 is the distance the lantern has moved upward from its initial position. By similar triangles

$\displaystyle \frac{s}{6}=\frac{x}{z-30}$, $\displaystyle s=\frac{6x}{z-30}$

$\displaystyle \frac{ds}{dt}=\frac{\left( z-30 \right)\left( 6 \right)\left( dx/dt \right)-6x\left( dz/dt \right)}{{{\left( z-30 \right)}^{2}}}=\frac{20\left( 6 \right)\left( 5 \right)-6\left( 40 \right)\left( 4 \right)}{{{20}^{2}}}=-0.90$

Thanks for catching these. The disappointing thing is that in two years no one caught them sooner.

I did a workshop in Cincinnati many years ago. The College Board hires us by region and I do not live in the Cincinnati region. I would be happy to come if someone is setting up a conference sometime or to do a workshop in your district. Check the “schedule” tab occasionally maybe I’ll be close some time.

Thanks again for writing.

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