Water is running into an inverted cone at the rate of `pi` cubic metres per minute. The height of the cone is 10 metres, and the radius of its base is 5m. How fast the water level is rising when the water stands 7.5 m below the base.

Explanation

`V=frac{1}{3} pi r^{2} h`


`Rightarrow frac{d V}{d t}=frac{1}{3} pi r^{2} frac{d h}{d t}+frac{2}{3} pi r h frac{d r}{d t}`


Now,


`frac{h}{r}=frac{10}{5}` or `r=frac{h}{2}` and `frac{d h}{d t}=2 frac{d r}{d t}`


`Rightarrow frac{d V}{d t}=frac{1}{3} pi(frac{h}{2})^{2} frac{d h}{d t}+frac{2}{3} pi(frac{h}{2}) h frac{1}{2} frac{d h}{d t}`


`Rightarrow frac{d V}{d t}=frac{pi}{3}[frac{h^{2}}{4} frac{d h}{d t}+frac{h^{2}}{2} frac{d h}{d t}]`


`Rightarrow frac{d V}{d t}=frac{pi}{3} times frac{3 h^{2} d h}{4 d t}`


`Rightarrow frac{d V}{d t}=frac{pi h^{2}}{4} frac{d h}{d t}`


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