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A street light is at the top of a 17 ft tall pole. A woman 6 ft tall walks away from the pole with a speed of 6 ft/sec along a straight path. How fast is the tip of her shadow moving when she is 40 ft from the base of the pole?

  • calculus -

    Draw a right-angled triangle ABC, B the right angle, and the vertical AB = 17 feet

    Somewhere between B and C draw a vertical DE = 6 feet, the height of the woman

    let BD = x and DC = y, the length of her shadow

    given: dx/dt = 6 ft/s
    find: dy/dt

    (actually d(x+y)/dt, I'll come back to that later)

    by similar triangles 6/y = 17/(x+y)
    11y = 6x
    then 11dy/dt = 6dx/dt
    dy/dt = 6(6)/11 = 36/11 ft/s

    so no matter where she is, the length of her shadow is increasing at 36/11 ft/s

    but the woman herself is moving at 6 ft/sec

    so her shadow is moving at 6+36/11 or
    102/11 ft/s

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