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Calculus ~ Related Rates

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A man 2 meters tall walks at the rate of 2 meters per second toward a streetlight that's 5 meters above the ground. At what rate is the tip of his shadow moving?We've already set this up part of the way. We know that dx/dt = - 2 meters per second, and we're looking for dy/dt.

  • Calculus ~ Related Rates - ,

    From what you said, I will assume you defined variables as such....
    x is the distance between the man and the streetligh
    y is the length of his shadow.
    then dx/dt = -2 , find dy/dt

    by similar triangles:
    2/y = 5/(x+y)
    5y = 2x + 2y
    3y = 2x
    3dy/dt = 2dxdt
    dy/dt = 2(-2)/3 = -4/3

    so his shadow is decreasing at a rate of 4/3 m/s

    (notice , since I said "decreasing", I do not use the negative sign, since the word decreasing implies it)

  • Calculus ~ Related Rates - ,

    -10/3

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