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January 30, 2015

January 30, 2015

Posted by **rachel** on Sunday, October 16, 2011 at 7:56pm.

- calculus -
**Steve**, Monday, October 17, 2011 at 12:35pmAnother 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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