Derivative

[Vempy]

Here is the question:

The amount of cleaning fluid in a partially filled, spherical tank, is given by V= piRh^2-(pih^3)/3, where R is the radius and h is the depth. If cleaning fluid is being pumped into the tank, of radius 12m, at a rate of 8 m^3/min., how fast is the fluid rising when the tank is 3/4 full?

Okay, so I went through and graphed it and stuff to find that h at 3/4 full is approx. 16.2m. And then I found the derivative of 3/4 multipled by the equation in the question and subbed in h and R, but I got a negative rate of change for dh/dt, which can't be right.

V= (3/4)[piRh^2-(pih^3)/3]
V= (9piRh^2-3pih^3)/12
V= [(3pih^2)(3R-h)]/12

The derivative I ended up with for that is:

dv/dt= [18piRh.dh/dt-6pih^2.dh/dt+9Rpih^2-3pih^2.dh/dt]/12

I'm quite sure I did something wrong.

 

[skeeter]

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

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

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

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

try it from here.