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Math Help - Related Rates

  1. #1
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    Related Rates

    A man 6 ft tall is walking toward a building at the rate of 5ft/sec. If there is a light on the ground 50ft from the building, how fast is the man's shadow on the building growing shorter when he is 30 ft from the building?
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  2. #2
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    ^_^Engineer_Adam^_^!
    i
    A man 6 ft tall is walking toward a building at the rate of 5ft/sec.
    If there is a light on the ground 50ft from the building, how fast is the man's shadow
    on the building growing shorter when he is 30 ft from the building?

    First, make a sketch . . .
    Code:
          |
        D *
          |   *
          |       *   B
          |           *
        s |           |   *
          |           |       *
          |           |6          *
          |           |               *
          |     x     |       50-x        * 
        E * - - - - - + - - - - - - - - - - - * A
                      C
          : - - - - - - - 50  - - - - - - - - :

    The light is at A; the man is: BC = 6.

    The wall is at the far left; the length of his shadow is: s = DE.

    Let x = EC, then CA = 50 - x.
    . . We are given: dx/dt = -5 ft/sec


    Right triangle DEA is similar to right triangle BCA.
    . . . . . . . . . . . . .s . . . . . .6
    . . so we have: . --- . = . -------
    . . . . . . . . . . . . 50 . . . .50 - x

    . . which simplifies to: .s .= .300 / (50 - x)


    Differentiate with respect to time: .ds/dt .= .300(dx/dt) / (50 - x)


    When x = 30 and dx/dt = -5: .ds/dt .= .300(-5) / (50- 30) .= .-15/4


    Therefore, the length of the shadow is decreasing at 3 feet/sec.

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