Related Rates? A water tank is shaped as an inverted cone with a heigh 10 ft and the radius of the circle at the top 10 ft. Volume of water drains from bottom at rate proportional to height of water level.

Question

anonymous · Answer

So $$\frac{ dV }{ dt }=c \times h(t)$$ where V(t) is water in the tank, t is time, and c is constant, and h(t) is height of water at time t. When the tank is full, the drain is opened and measurement shows that water level drops at instantaneous rate of 1 ft/hr. What is the instantaneous rate of change for the volume of the water when the drain is first opened?

anonymous · Answer

$$V=\frac{ \pi r^2h }{ 3 }$$

anonymous · Answer

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At any given time, the $r$ leg of the triangle will be the same length as the $h$ leg, so
$$V=\frac{\pi r^2h}{3}=\frac{\pi r^3}{3}=\frac{\pi h^3}{3}$$
Let's use the volume expression as a function of the water level $h$. At time $t=0$, you know that the rate of change of the water level is $\dfrac{dh}{dt}=-1$ (negative because the tank is being drained).

Differentiating the volume expression with respect to time gives
$$\frac{dV}{dt}=\pi h^2\frac{dh}{dt}=-\pi h^2$$
Also, at time $t=0$, since the tank is full, you know that $h=10$. Simple computation from here. Oh, and keep track of your units.