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Topic: [ap-calculus] Units of RADIANS per Second--Related Rates Problem
Replies: 1   Last Post: Oct 20, 2012 4:20 PM

 Richard Sisley Posts: 4,189 Registered: 12/6/04
Re: [ap-calculus] Units of RADIANS per Second--Related Rates Problem
Posted: Oct 20, 2012 4:20 PM

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On Oct 19, 2012, at 6:00 PM, Leanne Wied-Brusky wrote:

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> I have a question about the units in a related rates problem. I insist on using units throughout a related rates problem to reinforce the proper set-up and understanding of an application problem.
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Units are important in any question about how calculus can be used as a tool to model physical events, even when those "events" are as contrived as the common slipping ladder questions.

> When dealing with a rate of change of angle problem, how do units of radians/second suddenly appear?

Here is one way to make sense of the use of units in the question about which you are asking. First, for a few preliminaries. We can explicitly introduce two functions with the same domain of time measures. The first can be called function "d" and the values of "d" are the changing distances of the end of the ladder on the ground from the base of a wall along which the other end of the down which the ladder is sliding. The second is a function we can call "A." The domain of "A" is the same unspecified set of time measures as the domain of "d." But how to describe the values of "A"? Imagine a family of circles of radius 1 foot whose centers are the horizontally moving end of the ladder. Then imagine the changing arcs of those circles that have one end at the intersection of the ladder and one of those circles and the other end at the intersection of that circle with the ground. The value of function "A" at time some time "t" (A(t)) is the length of that arc in radii. That means that the values of "A" are arc lengths in feet.

Now consider what how the original question can be phrased. There is some unique time "m" when the distance of the horizontally moving foot of the ladder is five feet from the base of the wall. The question is "What is A'(m)?". Since the values of A are arc lengths in feet, the unit designation for for A'(m) will be feet per second.

All we need to answer the question is some equation that related d(t) to A(t) for each value of t in the shared domain of functions "d" and "A."

We can modify the equation you gave to include explicit mentions of the values of "A" and "d" and write: cos(A(t))=d(t)/13. Then use implicit differentiation to get a relationship between A(t), A'(t), and d'(t) valid for any time. Now use the fact that d(m) = 5 to get the value of -sin(A(m)) and then the value of A'(m).

I think that if this kind of explicit function is used for related rates problems both the question and the road to a solution are more clear. It does provide an analysis that is consistent with the excellent idea of being conscious of the units for the values of the functions and their derivatives.

Sincerely,

Richard Sisley
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