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Math Help - Related Rate

  1. #1
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    Related Rate

    A mountain climber on one edge of a deep canyon 800 ft wide sees a large rock fall from the opposite edge at time=0. As he watches the rock plummet downward, his eyes first move slowly, then faster, and than more slowly again. Let a be the angle of depression of his line of sight below the horizontal. At what angle a would seem to be moving the most rapidly. That is, would da/dt be maximal? Remember the rocks acceleration would be -32 ft/sec^2
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  2. #2
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    Quote Originally Posted by BarlowBarlow1 View Post
    A mountain climber on one edge of a deep canyon 800 ft wide sees a large rock fall from the opposite edge at time=0. As he watches the rock plummet downward, his eyes first move slowly, then faster, and than more slowly again. Let a be the angle of depression of his line of sight below the horizontal. At what angle a would seem to be moving the most rapidly. That is, would da/dt be maximal? Remember the rocks acceleration would be -32 ft/sec^2
    Take the climber's height be takes as the height reference, that is we take
    a coordinate system with the climber at h=0.

    Let the initial height of the rock he h1, then as it falls its height is:

    s=-16 t^2 + h1.

    We also have the angle of depression a satisfies the equation:

    tan(a) = s/800 = [16 t^2 + h1]/800.

    Now you may be expected to assume that h1=0, but I don't see that in
    the question as asked.

    Now the rock appears to be moving most rapidly when da/dt is a maximum,
    which will be a point at which d^2a/dt^2 = 0.

    RonL
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