Water is dripping out from a conical funnel of semi-vertical angle `pi/4` at the uniform rate of `2c m^2//s e c` in its surface area through a tiny hole at the vertex in the bottom. When the slant height of the water is 4cm, find the rate of decrease of the slant height of the water.

Explanation

`sin 45^{circ}=frac{D E}{A E} Rightarrow frac{1}{sqrt{2}}=frac{r}{l} `


`cos 45^{circ}=frac{A D}{A E} Rightarrow frac{1}{sqrt{2}}=frac{h}{l}`


`r=frac{1}{sqrt{2}} text { and } {h}=frac{1}{sqrt{2}} ldots text { (ii) }`



`therefore` the equation (i) can be rewritten as :



`V=frac{1}{3} pi times(frac{I}{sqrt{2}})^{2} times frac{I}{sqrt{2}}=frac{pi}{3 times 2 times sqrt{2}} `


`V=frac{pi}{6 sqrt{2}} I^{3} ldots text {.(iii) } `


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