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Dexter's question at Yahoo! Answers regarding related rates

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MarkFL

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Feb 24, 2012
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Here is the question:

dexter.jpg

I have posted a link to this topic so the OP can see my work.
 
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MarkFL

Administrator
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Feb 24, 2012
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Hello Dexter,

I would let $E$ be the elevation of the observer, $h$ be the height of the elevator above the ground, and $w$ be the horizontal distance of the observer from the elevator shaft.

From the diagram, we see we mat then state:

\(\displaystyle \tan(\theta)=\frac{h-E}{w}\)

Differentiating with respect to time $t$ (recognizing that $\theta$ and $h$ are the only variables that change with time), we find:

\(\displaystyle \sec^2(\theta)\frac{d\theta}{dt}=\frac{1}{w}\frac{dh}{dt}\)

Multiplying through by \(\displaystyle \cos^2(\theta)\), we get:

\(\displaystyle \frac{d\theta}{dt}=\frac{\cos^2(\theta)}{w}\frac{dh}{dt}\)

By Pythagoras, we know:

\(\displaystyle \cos^2(\theta)=\frac{w^2}{(h-E)^2+w^2}\)

and thus we may state:

\(\displaystyle \frac{d\theta}{dt}=\frac{w}{(h-E)^2+w^2}\frac{dh}{dt}\)

Plugging in the given data, we have (in radians per second):

\(\displaystyle \frac{d\theta}{dt}=\frac{54}{(h-21)^2+18^2}\)

And so to answer the questions, we find:

\(\displaystyle \left.\frac{d\theta}{dt} \right|_{h=15}=\frac{54}{(15-21)^2+18^2}=\frac{3}{20}\)

\(\displaystyle \left.\frac{d\theta}{dt} \right|_{h=39}=\frac{54}{(39-21)^2+18^2}=\frac{1}{12}\)