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A swimming pool is 40 feet long, 20 feet wide, 8 feet deep at the deep end, and 3 feet deep at the shallow end; the bottom is rectangular. If the pool is filled by pumping water into it at the rate of 40 cubic feet per minute, how fast is the water level rising when it is 3 feet deep at the deep end?

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4 answers

  1. Luckily, all we are working with is a triangular section, until the water is 5' deep; then we have to work with a trapezoid.

    Anyway, looking at the side of the pool, when the water is y' deep, we have a triangular prism with a cross-section that is y by y/8 (since the pool bottom is a line with slope 5/40 = 1/8

    So, the volume is
    v = 1/2 * y*y/8 * 20 = 5y^2/4
    dv/dt = 5y/2 dy/dt
    40 = 5*3/2 dy/dt
    dy/dt = 16/3 ft/min

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  2. The error in Steve's answer is that v = 1/2 * y * 8y * 20. The height of the line of the pool bottom is determined by h = 1/8 *l, so the length of the cross section is 8h, not h/8. The correct answer is 1/12 ft^3/min.

    1/12 ft^3/min is the selected answer in Varberg Early Transcendentals for this problem. Please check answers carefully.

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  3. The error in Emory's answer is that the units should be ft/min since it is asking how fast the depth of the water is increasing. But otherwise the 1/12 is correct!

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  4. IDK any math ahahahahah but I like cheese!

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