Water is leaking out of an inverted conical tank at a rate of 10,000 cm3/min...

Let V be the volume of water in the tank, in ${\displaystyle c{m}^3}$; 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 ${\displaystyle \frac{h}{r}=\frac{6}{2}=3}$ so that ${\displaystyle h=3r}$
The volume of the inverted cone of water is then ${\displaystyle V=\frac{1}{3}\pi r^{2}h=\pi r^3}$
Now differentiate both sides with respect to time ${\displaystyle t}$ (in minutes) to get ${\displaystyle \frac{dV}{dt}=3\pi r^2\cdot \frac{dr}{dt}}$ (the Chain Rule is used in this step).
If ${\displaystyle V_i}$ is the volume of water that has been pumped in, then ${\displaystyle \frac{dV}{dt}=\frac{d{V}_i}{dt}-10000=3\pi \cdot {\left(\frac{200}{3}\right)}^2\cdot 20}$ (when the height/depth of water is 2 meters, the radius of the water is ${\displaystyle \frac{200}{3}}$ cm).
Hence ${\displaystyle \frac{d{V}_i}{dt}=\frac{800000\pi }{3}+10000\approx 847758 \frac{{\text{cm}}^3}{min}}$