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April 1, 2015

April 1, 2015

Posted by **Greg** on Friday, March 2, 2012 at 3:07pm.

- Math -
**Steve**, Friday, March 2, 2012 at 4:42pmLuckily, 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

- Math -
**Emory**, Thursday, July 3, 2014 at 4:58pmThe 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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