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Math Help - Word Problem-Related Rates

  1. #1
    Junior Member
    Joined
    Sep 2007
    Posts
    37

    Cool Word Problem-Related Rates

    You are standing under a 20 foot lamppost. Your 5 foot friend is 10 feet away from you and sprints away at a speed of 20 feet per second.
    (a) how fast is the tip of her shadow moving
    (b) at that same instant, how fast is the length of her shadow changing.

    Drawing if it helps

    b
    .-
    . ----
    . -----
    . -------
    . ---------
    . -----------
    . -------------
    .....................-
    a --------------c

    ab is the light post

    thanks in advance
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  2. #2
    Senior Member
    Joined
    Dec 2007
    From
    Anchorage, AK
    Posts
    276

    Similar triangles

    Let x be the distance from the lamppost to your friend, and y the distance from the lamppost to the tip of her shadow. Then y-x is the length of her shadow, and by similar triangles, \frac{y}{20}=\frac{y-x}{5}

    B
    |**-
    |****-
    |******-
    |********|
    |********|*-
    |********|***-
    |********|*****-
    A----------C--------P
    \----x-----/\-(y-x)-/


    Solving for y:
    \frac{y}{20}=\frac{y-x}{5}

    5y=20(y-x)
    20x=15y
    y=\tfrac{4}{3}x

    For (a), we want \frac{dy}{dt}. We use the chain rule \frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}, along with y in terms of x above and the fact that \frac{dx}{dt}=20 from the problem.

    For (b), we remember that the shadow length is y-x, and so we want \frac{d}{dt}(y-x)=\frac{dy}{dt}-\frac{dx}{dt}, which should follow easily from the answer to (a).


    --Kevin C.
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