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Rate of Change question

  1. Nov 22, 2009 #1
    A man 1.8 m tall walks at speed of 10m/s towards a street lamp which is 7m above the ground. How fast is the length of the man's shadow decreasing?

    My attempt:

    Let the man's shadow and the lamp's length be l, and the distance be d and time be t. Given, dd/dt = 10.

    I am supposed to find dl/dt.

    dd/dt= dl/dt X dd/dl

    So how do I find dd/dl?:confused:
  2. jcsd
  3. Nov 22, 2009 #2
    You can imagine the man walking on the x-axis towards the y-axis. Call his position x and the length of his shadow s.
    The key to these types of problems is using similar triangles. So we can get a relation between s and x.

    (1.8/s) = (7/x+s)

    Get s on one side and differentiate with respect to t. And remember dx/dt = -10 since x is decreasing.

    I hope i didn't make any mistake. Check the book to make sure if you have the answer.
  4. Nov 22, 2009 #3


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    Draw a picture and think "similar triangles".
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