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Math Help - Rate of how fast a shadow grows as you walk

  1. #1
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    Rate of how fast a shadow grows as you walk



    There is a lamp post 15 feet tall casting a shadow 'B' ft. long of a person that is 6 feet tall standing 'A' ft. from the lamp post. If the person moves away from the lamp post at 5 feet per second, how fast does the shadow lengthen?
    So here I thought it might be something to formulate a triangle, or actually two:



    So the problem is asking me to find the derivative of the shadow's length with respect to time: \frac{dB}{dt}

    I think the triangles are similar:

    \frac{15}{A+B} = \frac{6}{B} \\<br />
\\<br />
\cdots <br />
\\<br />
B = \frac{2}{3}A

    So we have B and I will attempt to take the derivative of this:

    \frac{dB}{dt} = [\frac{2}{3}A]'

    Constant/Chain rules:

    \frac{dB}{dt} = \frac{2}{3} \cdot [A'] \\<br />
= \frac{2}{3} \cdot \frac{dA}{dt}

    Now how do I find the derivative of 'A'? It must have something to do with the movement velocity because it was given information and I haven't used it yet. Unless I am being tricked and it is extraneous/unnecessary information.
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  2. #2
    Super Member ebaines's Avatar
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    Re: Rate of how fast a shadow grows as you walk

    You're almost there! The term dA/dt is the rate of change of length A per unit time, which is precisely the velocity of point A. You were told the velocity of A is 5 ft/s, so there you go:

    dB/dt = 2/3 dA/dt = 2/3 (5 ft/s) = 10/3 ft/s.
    Last edited by ebaines; July 10th 2012 at 08:51 AM.
    Thanks from tom@ballooncalculus and daigo
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    Re: Rate of how fast a shadow grows as you walk

    Quote Originally Posted by daigo View Post
    Now how do I find the derivative of 'A'? It must have something to do with the movement velocity because it was given information and I haven't used it yet.
    Yes... what is the meaning of A' ...?

    The derivative of A wrt t...

    the rate of change of distance wrt time...

    what we normally call...
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