the cross section of a 5-ft long trough is an isosceles trapezoid with a 2 foot lower base, a 3-foot upper base, and an altitude of 2 feet. water is running into the trough at a rate of 1 cubic foot per minute. how fast is the water level rising when the water is 1 foot deep?

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  1. Another trapezoid problem?? Oh well, here goes:

    Volume is area of base * length

    When the water is h feet deep The trapezoid has lower base = 2, upper base = 2 + h/2 (draw a diagram to see why this is so)

    So, the volume is 5 * (2 + 2+h/2)/2 * h
    = 5/4 * (h^2 + 8h)

    dV = 5/4 * (2h + 8) dh
    1 = 5/4 * (2+8) * dh
    dh = 1/12.5 ft/min

    This makes sense. The base of the trough has area 5x2 = 10 ft^2. If it had straight sides, the height would rise 1/10 ft/min.

    The top of the trough has area 15 ft^2. If it had straight sides from the top, the height would rise 1/15 ft/min

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