Water is flowing at the rate of 50 m^3/min into a holding tank shaped like an cone, sitting vertex down. The tank’s base diameter is 40 m and a height of 10 m. What would the expression for the rate of change of the water level with respect to time, in terms of h (the water’s height in the tank) be?

Question

anonymous · Answer

Any ideas on this? I don't rally know where to start?

anonymous · Answer

Too easy...............
$$Volume.......V=\frac{1}{3} {pi }r^2h$$

anonymous · Answer

$$\frac{r}{h}=constant...........    $$
As angle of cone is constant
You can find this constant using data given
lets say it is k
r=kh
Volume becomes
$$V=\frac{1}{3}pi.h^3k^2$$
$$\frac{dV}{dt}=50=pi*h^2*k^2*\frac{dh}{dt}$$

anonymous · Answer

How would I then represent h as a function of t then? (Also I guess we'd be assuming that at t=0 the water level is zero yes?)

anonymous · Answer

And then how would I a when the water is 8 feet deep, I am supposed to find the rate of change of the radius. Would that be dr/dt?