A woman 5.5 ft walks at a rate of 6 ft/sec towards a street light that is 22 ft above the ground. At what rate is the length of her shadow c

Question

A woman 5.5 ft walks at a rate of 6 ft/sec towards a street light 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?

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Adela 4 years 2021-08-26T19:47:50+00:00 2 Answers 9 views 0

Answers ( )

  1. Ben
    0
    2021-08-26T19:49:41+00:00
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    Answer:

    Explanation:

    By applying similar triangles rule,

    22/5.5 = (x + s)/s

    s = 5.5/22 × (x + s)

    ds/dt = 5.5/22 × (dx/dt + ds/dt)

    16.5/22 × ds/dt = 5.5/22 × dx/dt

    16.5/22 × ds/dt = 5.5/22 × -6

    ds/st = -2 ft/s.

  2. thanhha
    0
    2021-08-26T19:49:48+00:00
    Reply

    Answer:

    The length of her shadow is changing at the rate  -2 m/s

    Explanation:

    Let the height oh the street light, h = 22 ft

    Let the height of the woman, w = 5.5 ft

    Horizontal distance to the street light = l

    length of shadow = x

    h/w = (l + x)/x

    22/5.5 =  (l + x)/x

    4x = l + x

    3x = l

    x = 1/3 l

    taking the derivative with respect to t of both sides

    dx/dt = 1/3 dl/dt

    dl/dt = -6 ft/sec ( since the woman is walking towards the street light, the value of l is decreasing with time)

    dx/dt = 1/3 * (-6)

    dx/dt = -2 m/s

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Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )