A tank containing water is in the form of a cone with vertex C. The axis is vertical and the semi-vertical angle is 60°.
At time t= 0, the tank is full and the depth of water is H. At this instant, a tap at C is opened and water begins to flow out.
The volume of water in the tank decreases at a rate proportional to $\sqrt h$, where h is the depth of water at time t.
The tank becomes empty when t=$60$. Find h at time t in terms of H and t.
Hint.
$$ \frac{dV}{dt} = k\sqrt{h} $$ $$ r = h\tan\theta $$
$$ dV = \pi r^2 dh = \pi(h\tan\theta)^2dh \therefore \frac{dV}{dt}= \pi(h\tan\theta)^2\frac{dh}{dt} = k\sqrt{h} $$