Differential Calculus
Rate of change of surface area; conical vessel
MMTS_Full_Test_15
Grade 12
Question:
Water is filled at rate $\pi$ cm$^3$/s in right circular conical vessel (vertex up) of height 5 cm and diameter 8 cm. When water height is 3 cm, rate of increase of wet conical surface area is (cm$^2$/s)
(A) $\dfrac{2\pi}{3}$
(B) $\dfrac{\sqrt{41}}{12}\pi$
(C) $\dfrac{\sqrt{41}}{18}\pi$
(D) none of these
Step-by-Step Solution
Key Concept: $r/h=4/5$ (similarity). $V=\pi r^2h/3=16\pi h^3/75$. $dV/dt=16\pi h^2/25\cdot dh/dt=\pi\Rightarrow dh/dt=25/(16\cdot9)$. Lateral SA $=\pi rl=\pi r\sqrt{r^2+h^2}=\pi(4h/5)\sqrt{16h^2/25+h^2}=\frac{4\pi h^2}{5}\cdot\frac{\sqrt{41}}{5}$. $dS/dt=(8\sqrt{41}\pi h)/(25)\cdot dh/dt=\sqrt{41}\pi/12$.
$\dfrac{\sqrt{41}}{12}\pi$.
Correct Answer: (B) $\dfrac{\sqrt{41}}{12}\pi$