# Application of derivatives: related rates problem

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.

• I drew a diagram and converted 50L/min to 50000cu. cm/min and thats all i have. I dont know which equation relating variables to use or how to find the the rate at which water is pouring... – Laura Apr 15 '14 at 15:10
• Think of the water as a cone itself. Its change in volume is constant. Start with the equation for the volume of a cone, and differentiate to get $\frac{dV}{dt}$. Then solve for that with what's given. – user137794 Apr 15 '14 at 15:17
• I have tried that but i do not understand how to use the rate of 4cm/min when the greatest depth is 3m – Laura Apr 15 '14 at 15:23
• What did you get for your equation of $dV\over dt$? – user137794 Apr 15 '14 at 15:45
• 50000 = (5pi/19)(h^2)(dh/dt) – Laura Apr 15 '14 at 15:47

## 1 Answer

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. $$\frac{dV}{dt}\text{going in}-50\frac{\text{L}}{\text{min}}=\frac{25}{36}\pi h^2\frac{dh}{dt}$$