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Math Help - Related Rate models question

  1. #1
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    Related Rate models question

    I'm getting the biggest headache trying to solve this question.. and its probably the easiest question ever but I just can't get the right answer and its driving me nuts. And i have a test on monday on this stuff.. so any help would be appreciated beyond belief.

    A man 2 m tall walks away from a lamppost whose light is 5 m above the ground. If he walks at a speed of 1.5 m/s, at what rate is his shadow increasing when he is 10 m from the lamppost.

    The answer is supposed to be 1 m/s, but I keep on getting 2.5 m/s, and I have no clue what I'm doing wrong.

    Thanks for the help!!!
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  2. #2
    Eater of Worlds
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    Let x=shadow length and y=distance bewtween man and lamppost.

    Similar triangles:

    \frac{x}{2}=\frac{x+y}{5}

    x=\frac{2y}{3}

    Differentiate:

    \frac{dx}{dt}=\frac{2}{3}\frac{dy}{dt}

    We know that dy/dt=3/2

    We have:

    \frac{dx}{dt}=\frac{2}{3}\cdot\frac{3}{2}=1
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  3. #3
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    Hello, jmailloux!

    A man 2 m tall walks away from a lamppost whose light is 5 m above the ground.
    If he walks at a speed of 1.5 m/s, at what rate is his shadow increasing
    when he is 10 m from the lamppost.

    The answer is supposed to be 1 m/s, but I keep on getting 2.5 m/s,
    and I have no clue what I'm doing wrong. . I know!

    You made a very common error.
    I know ... I still do it from time to time.
    Code:
          *
          |   *
          |       *
          |           *
         5|           |   *
          |          2|       *
          |           |           *
          * - - - - - + - - - - - - - *
          :     x     :       s       :

    I bet you let s equal the whole bottom of the diagram.
    That would give us the rate at which the tip of his shadow is moving.
    . . But they want the rate at which the length of his shadow is changing.

    From the similar right triangles, we have: . \frac{s}{2}\:=\:\frac{x+s}{5}

    Then: . 5s\:=\:2x + 2s\quad\Rightarrow\quad 3s \,= \,2x\quad\Rightarrow\quad s \,= \,\frac{2}{3}x

    Differentiate with respect to time: . \frac{ds}{dt}\:=\:\frac{2}{3}\cdot\frac{dx}{dt}


    The man's speed is 1.5 m/sec: . \frac{dx}{dt} = 1.5

    Therefore: . \frac{ds}{dt}\:=\:\frac{2}{3}(1.5) \:=\:1\text{ m/s}

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  4. #4
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    thank you soo much

    it makes a lot more sense after you explained it, and i did make s equal to the whole bottom of the diagram.

    anyways.. thanks for the help!
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