Posted by **Swk13** on Sunday, December 3, 2006 at 5:18pm.

Im doing some practice questions and im stumped on this one.

The amount of cleaning fluid in a partially filled, spherical tank is given by V = (pi)R(h^2)-(pi(h^3))/3

Where R is the radius of the rank, and h is the depth of the center of the tank. If cleaning fluid is being pumped into the tank, of radius 12m, at a rare of 8 m^3/min, how fast is the fluid rising when the tank is 3/4 full?

Thanks

Take the derivative with respect to time of V = (pi)R(h^2)-(pi(h^3))/3

dV/dt= PI*R*2h dh/dt - PI 3h dh/dt /3

you are given dv/Dt as 8 m^3/min, so solve for dh/dt

- Math - Calculus -
**Anonymous**, Friday, January 1, 2016 at 5:11pm
0.5m/min

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