calculus
posted by rachel .
the cross section of a 5ft long trough is an isosceles trapezoid with a 2 foot lower base, a 3foot 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?

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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