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Math Help - Another related rate model

  1. #1
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    Another related rate model

    I posted a question almost the same as this yesterday.. and I understand how to do them. I followed the way the people who answered my post solved the problem, but it's not working for this question.

    A 1.7 m tall man walks away from a 9-m high streetlight at 0.8 m/s. How fast is the end of his shadow moving when he is 20 m from the lamppost.

    My textbook says the answer is 0.986 m/s, but I keep on getting 0.186 m/s.

    PLEASE HELP, i have a test tommorrow, and I know a question like this is going to be on it
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  2. #2
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    That's because you found the rate at which the shadow length is changing.

    The tip of the shadow is x+y feet from the light. Therefore, the rate it is

    changing is dx/dt+dy/dt. You have 0.8 + 0.186 = 0.986
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  3. #3
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    Hello, jmailloux!

    You did it again . . . you made s the wrong part.


    A 1.7-m tall man walks away from a 9-m high streetlight at 0.8 m/s.
    How fast is the end of his shadow moving when he is 20 m from the lamppost?
    Code:
          *
          |     *
          |           *
        9 |           |     *
          |        1.7|           *
          |     x     |       s-x       *
      ----*-----------+-----------------------*----
          : - - - - - - - - s - - - - - - - - :

    This time, we make s be the entire distance,
    . . from the lampost to the end of his shadow.

    From the similar right triangles, we have: . \frac{s - x}{1.7} \:=\:\frac{s}{9}

    Then: . 9s - 9x\:=\:1.7s\quad\Rightarrow\quad s\:=\:\frac{9}{7.3}x

    Differentiate with respect to time: . \frac{ds}{dt}\:=\:\frac{9}{7.3}\left(\frac{dx}{dt}  \right)

    Since \frac{dx}{dt} = 0.8\text{ m/s}:\;\frac{dx}{dt} \:=\:\frac{9}{7.3}(0.8) \:\approx\:0.986\text{ m/s}

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