Application of derivatives: related rates problem

Question

Water is leaking out of an inverted conical tank at a rate of 50L/min the tank is 10m in diameter at the top. It is 6m deep at the deepest point, which is the vertex of the cone and lies on the vertical centre axis of the tank. If the water level os raising at the rate of 4cm/min when the greatest depth is 3m, find the rate at which water is being pouring into the tank.

Answer 1

For the volume of a cone,

$$V=\frac13\pi r^2h$$

You can see for this cone $r=(5/6)h$

$$V=\frac13\pi \left(\frac56h\right)^2h=\frac{25}{108}\pi h^3$$

Then

$$\frac{dV}{dt}=\frac{25}{36}\pi h^2\frac{dh}{dt}$$

Change in volume is equal to the water pouring in minus water pouring out. $$\frac{dV}{dt}\text{going in}-50\frac{\text{L}}{\text{min}}=\frac{25}{36}\pi h^2\frac{dh}{dt}$$