# Water is leaking out of an inverted conical tank at a rate of 10,000 cm3/min at the same time water is being pumped into the tank at a constant rate If the tank has a height of 6m and the diameter at the top is 4 m and if the water level is rising at a rate of 20 cm/min when the height of the water is 2m, how do you find the rate at which the water is being pumped into the tank?

Let $V$ be the volume of water in the tank, in $c {m}^{3}$; let $h$ be the depth/height of the water, in cm; and let $r$ be the radius of the surface of the water (on top), in cm. Since the tank is an inverted cone, so is the mass of water. Since the tank has a height of 6 m and a radius at the top of 2 m, similar triangles implies that $\setminus \frac{h}{r} = \setminus \frac{6}{2} = 3$ so that $h = 3 r$.
The volume of the inverted cone of water is then $V = \setminus \frac{1}{3} \setminus \pi {r}^{2} h = \setminus \pi {r}^{3}$.
Now differentiate both sides with respect to time $t$ (in minutes) to get $\setminus \frac{\mathrm{dV}}{\mathrm{dt}} = 3 \setminus \pi {r}^{2} \setminus \cdot \setminus \frac{\mathrm{dr}}{\mathrm{dt}}$ (the Chain Rule is used in this step).
If ${V}_{i}$ is the volume of water that has been pumped in, then $\setminus \frac{\mathrm{dV}}{\mathrm{dt}} = \setminus \frac{{\mathrm{dV}}_{i}}{\mathrm{dt}} - 10000 = 3 \setminus \pi \setminus \cdot {\left(\setminus \frac{200}{3}\right)}^{2} \setminus \cdot 20$ (when the height/depth of water is 2 meters, the radius of the water is $\setminus \frac{200}{3}$ cm).
Therefore $\setminus \frac{{\mathrm{dV}}_{i}}{\mathrm{dt}} = \setminus \frac{800000 \setminus \pi}{3} + 10000 \setminus \approx 847758 \setminus \setminus \frac{\setminus m b \otimes {\left\{c m\right\}}^{3}}{\min}$.