A water tank has the shape of a right circular cone with axis vertical and vertex downwards. Its semi-vertical angle is $\tan^{- 1} \frac{3}{4}$ . Water is poured in it at a constant rate of 6 cubic meter per hour. The rate (in square meter per hour), at which the wet curved surface area of the tank is increasing, when the depth of water in the tank is 4 meters, is ……

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Detailed Solution

Let h be the height and V be the volume of the water in the tank at some instant t.
$\frac{d V}{d t} = 6 m^{3} / h r$
But $V = \frac{1}{3} π r^{2} h$ and $r = \frac{3 h}{4} = h \tan θ$
$V = \frac{1}{3} π \frac{9 h^{2}}{16} h = \frac{3 π h^{3}}{16} \frac{d V}{d t} = \frac{9 π}{16} h^{2} \frac{d h}{d t} = 6 \Rightarrow \frac{d h}{d t} |_{h = 4} = \frac{2}{3 π} m / h r$
Let curved surface area at $t = π r l = π \frac{3 h}{4} \sqrt{h^{2} + {(\frac{3 h}{4})}^{2}}$
$S = \frac{15 π h^{2}}{16}$
$\frac{d s}{d t} = \frac{15 π}{8} h \frac{d h}{d t}$
At $h = 4 m, \frac{d s}{d t} = \frac{15 π}{8} .4. \frac{2}{3 π} = 5 m^{2} / h r$