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April 20, 2014

April 20, 2014

Posted by **Justine** on Monday, March 12, 2012 at 10:16pm.

- math -
**Reiny**, Monday, March 12, 2012 at 10:54pmLet the rate of the input source be x cm^3 per minute

So the effective dV/dt = (x- 700) cm^3/min

let the height of the water level be h cm

let the radius of the water level be r cm

h/r = 10/1.75

10r = 1.75h

r = .175 h

V = (1/3)π r^2 h

= (1/3)π(.030625h^2)(h)

= .010208333πh^3

dV/dt = .030625πh^2 dh/dt

x - 700 = .030625π(4.5)^2 (26)

x = 750.655 cm^3/min

check my arithmetic.

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