Posted by Swk13 on .
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,
0.5m/min