Water runs into an inverted conical tent at the rate of $20$ cubic feet per minute and leaks out at the rate of $5$ cubic feet per minute. The height of the water is three times the radius of the water's surface. The radius of the water surface is increasing when the radius is $5$ feet, is

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$\frac{1}{10 π} ft . / \min$

$\frac{1}{15 π} ft . / \min$

$\frac{1}{5 π} ft . / \min$

answer is A.

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Detailed Solution

$\frac{d V}{d t} = 15; h = 3 r V = \frac{1}{3} π x^{2} y; \frac{dx}{dt} = ? when x = 5 \frac{x}{y} = \frac{r}{h} = \frac{1}{3} V = \frac{2}{3} π r^{2} 3 x = {πx}^{3} \frac{dV}{dt} = 3 {πx}^{2} \frac{dx}{dt} \Rightarrow 15 = 3 π \cdot 25 \frac{dx}{dt} \Rightarrow \frac{dx}{dt} = \frac{1}{5 π}$