JEE Main 2022 (Online) 27th July Evening ShiftQuestion 1
A water tank has the shape of a right circular cone with axis vertical and vertex downwards. Its semi-vertical angle is $\tan^{-1}\frac34$. 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 ________.
Differentiation and its Application
Answer$5$
Solution
Concept. Express both the volume and the wet curved area in terms of the depth $h$, then link the rates. Formulas used. (1) $\tan\alpha=\frac34$: $r=\frac34h$, slant $l=\frac54h$(2) $V=\frac13\pi r^2h$, curved area $S=\pi rl$Step 1 - The cone.Step 2 - $h'$. By (1), (2), $V=\frac{3\pi h^3}{16}$, so $$\frac{dV}{dt}=\frac{9\pi h^2}{16}\cdot\frac{dh}{dt}=6.$$ At $h=4$: $9\pi\frac{dh}{dt}=6$, so $\frac{dh}{dt}=\frac{2}{3\pi}$. Step 3 - $S'$. $S=\pi\cdot\frac34h\cdot\frac54h=\frac{15\pi}{16}h^2$, so $$\frac{dS}{dt}=\frac{15\pi}{8}h\,\frac{dh}{dt}=\frac{15\pi}{8}\cdot4\cdot\frac{2}{3\pi}=5.$$ Answer: 5