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

## Answer this Question

## Related Questions

- math - A tank was 3/5 filled with water. After removing 120 litres of water from...
- science - Each month the Speedy Dry Cleaning Company buys one barrel (0.160 m3...
- Environmental science - Each month the Speedy Dry Cleaning Company buys one ...
- Calculus - A cylindrical tank, with height of 15 m and diameter 4m, is being ...
- Fluid Mechanics - A cylindrical tank 2m diameter and 4m long with its axis ...
- Calculus 2 - A circular plate of radius r feet is submerged vertically in a tank...
- math - a spherical fishtank has radius of 25 cm. The tank is filled with water ...
- Calculus - Fuel is flowing into a storage tank which can be filled to a depth of...
- Calculus - Fuel is flowing into a storage tank which can be filled to a depth of...
- maths - 3.An 1145 L tank contains 295 L of water and is being filled with brine ...