problem help

[butterflyy]

A right circular cone tank is filled with water at a rate of 5ft^3 per minute, but water is also flowing out at a rate of 1ft^3 per minute. The tank's 60 ft deep and 40 ft across the top.

How deep is the water when the tank is 1/2 full? How quickly is the depth of the water changing when the tan is 1/2 full?

V=pr^2h/3 p(20)^2(60)/3 = 8000p(1/2) = 4000p

Find h when v=4000
h=3(4000)/p(20)^2 = 12000/400p = 30pft

I'm unsure how to start the 2nd question. I know that dv/dt=4 and r=h/3

 

[tkhunny]

\(\displaystyle V(t) = \frac{1}{3} \pi r(t)^{2} \cdot h(t)\)

You got that.

\(\displaystyle dV/dt = \frac{1}{3} \pi (r^{2}*dh/dt + h*2*r*dr/dt)\)

Did you find this?

It's a good thought question. is the change int he volume constant? Right at the beginning, with an empty tank, is the drain working at full capacity?