Answer to Question #299418 in Calculus for Ari

Question

Waterisbeingpouredattherateof2⇡m3/min.intoaninvertedconicaltankthat is 12-meter deep with a radius of 6 meters at the top. If the water level is rising at the rate of 61 m/min and there is a leak at the bottom of the tank, how fast is the water leaking when the water is 6-meter deep?

Answer 1

For the conical tank with radius R=6 and height h=12

$$\dfrac{R}{h}=\dfrac{1}{2}\Rightarrow R= \dfrac{1}{2}\cdot h$$

Whatever the water level, the water in the tank is a cone

similar in shape to the tank itself, with  $R= \dfrac{1}{2}h$

The volume of water in the tank when the water is h meter deep is

$$V=\dfrac{1}{3}\pi R^2h=\dfrac{1}{3}\pi (\frac{1}{2}h)^2h=\dfrac{\pi}{12}h^3 \ \ m^3$$

The volume in the tank increases at a rate

$$\dfrac{dV}{dt}=(\dfrac{\pi}{12})\cdot 3h^2(\frac{dh}{dt})=(\dfrac{\pi}{4})\cdot h^2(\frac{dh}{dt})$$

When the water depth is h=6 m and the rate of water volume increases is

$$\dfrac{dV}{dt}=(\dfrac{\pi}{4})\cdot 6^2\cdot (\dfrac{1}{6})=\dfrac{3\pi}{12} \ \ m^3/min$$

With water entering the tank at 2$\pi$ $\ \ m^3/min$ and water leaking out at unknown rate of $x\ \ m^3/min$

The water is increasing at a rate of $(\dfrac{3\pi}{12})\ \ m^3/min$

That means that

$$\dfrac{3\pi}{12}=2\pi-x\\\Rightarrow x=2\pi-\dfrac{3\pi}{12}\\\Rightarrow x=\dfrac{7\pi}{4}\ m^3/min$$