MHB Dexter's question at Yahoo Answers regarding related rates

[MarkFL]

Here is the question:

View attachment 886

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

 

[MarkFL]

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:

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

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

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

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

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

By Pythagoras, we know:

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

and thus we may state:

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

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

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

And so to answer the questions, we find:

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

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