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diffirential calculus

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a woman 6ft tall walks away from a light 10ft above the ground. if her shadow lengthens at the rate of 2ft/sec. how fast is she walking?

  • diffirential calculus - ,

    did you make a diagram?
    let her distance from the lightpole be x ft
    let the length of her shadow be y ft
    by similar triangles ...
    6/y = 10/(x+y)
    6x + 6y = 10y
    6x = 4y
    6 dx/dt = 4 dy/dt

    we are given: dx/dt + dy/dt = 2 ft/sec
    dx/dt = 2 - dy/dt
    so
    6(2 - dy/dt) = 4 dy/dt
    12 - 6dy/dt = 4dy/dt
    12 = 10dy/dt
    dy/dt = 12/10 = 1.2 ft/sec

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