A man 5 feet tall walks at the rate 4 feet per second directly away from a street light wich is 20 feet above the street. B) At what rate is the length of his shadow changing? A) Is the length increasing or decreasing?

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amistre64 · Answer

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amistre64 · Answer

hmm

amistre64 · Answer

we might be able to use similar triangles to compare with

amistre64 · Answer

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amistre64 · Answer

$$\frac AB=\frac ab$$
B = walking + b, and lets change b to s for shadow
$$\frac A{w+s}=\frac as$$
$$As=a({w+s})$$
$$A's+As'=a'({w+s})+a({w+s})'$$
since A and a are constants, their derivatives go to zero
$$As'=a(w+s)'$$
$$As'=aw'+as'$$
$$As'-as'=aw'$$
$$s'(A-a)=aw'$$
$$s'=\frac{aw'}{(A-a)}$$
so with any luck
$$s'=\frac{5(4)}{(20-5)}$$

anonymous · Answer

Thank you so much! That helped a lot.

amistre64 · Answer

youre welcome, and with any luck it was right :)  is there a way to check?

anonymous · Answer

Yes it was right. I have the answer but I didn't know how to solve it. Thanks :)