Calc practice question help.

[Dave J]

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

 

[stapel]

You are given that R = 12 and dV/dt = 8. Naturally, dR/dt = 0.

I believe you can use the fact that the volume of the entire sphere is (4/3)(pi)R³ to find the height of the fluid at the time in question. (If the take in 3/4 full, then the volume, at that time t, is (3/4)V = (formula they gave you), which can be solved for "h".)

Then differentiate the volume formula with respect to time t, plug in the known values, and solve for dh/dt.

If you get stuck, please reply showing all of your steps. Thank you.

Eliz.

 

[galactus]

One thing good, they gave you the formula.

\(\displaystyle \L\\V={\pi}Rh^{2}-\frac{{\pi}h^{3}}{3}\)

We want dh/dt when the tank is 3/4 full. We can use the given formula to find the height when it's 3/4 full, therefore, we have h.

\(\displaystyle \L\\\underbrace{1728{\pi}}_{\text{3/4 volume}}={\pi}(12)h^{2}-\frac{{\pi}h^{3}}{3}\)

Solve for h.

Differentiate V:

\(\displaystyle \L\\\frac{dV}{dt}=2{\pi}Rh\frac{dh}{dt}-{\pi}h^{2}\frac{dh}{dt}\)

You have dV/dt=8, h=from above, R=12. Solve for dh/dt.