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Calculus

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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 her shadow lengthening when she is 30 ft from the base of the pole?

  • Calculus -

    I hope you made a diagram.
    let her distance from the pole be x ft
    let the length of her shadow be y ft

    by similar triangles,
    6/y = 17/(x+y)
    simplifying,
    11y = 6x
    11 dy/dt = 6 dx/dt
    but dx/dt = 6
    so dy/dt = 6/11 ft/sec

    notice the 30 ft is irrelevant.

    Also be careful to notice the wording.
    You asked, "how fast is her shadow lengthening" and that answer is 6/11

    Had you asked, " how fast is her shadow moving", we would have had to add her own velocity to it, namely 6 + 6/11 ft/sec

  • Calculus -

    There is 1 error in the above answer.

    11dy/dt = 6 dx/dt is true
    dx/dt = 6 is also true
    but dy/dt is not 6/11 ft/sec
    the 6 must be multiplied by the 6 from dx/dt giving dy/dt = 36/11

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