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calculus

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A woman 5.5 ft tall walks at a rate of 6 ft/sec toward a streetlight that is 22 ft above the ground. At what rate is the length of her shadow changing when she is 15 ft from the base of the light?

  • calculus -

    If x is her horizontal distance from the street light, the shadow length s is given by
    s = (5.5/22)*(s + x)
    so
    (16.5/22)s = (5.5/22) x
    s = x/3

    Therefore ds/dt = (1/3) dx/dt
    = (2.0 ft/s)*(dx/dt)

  • calculus -

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