Applications of Derivatives
Related Rates
MMTS_Full_Test_13
Grade 12
Question:
A water tank has the shape of an inverted circular cone with axis vertical. Semi-vertical angle is $\tan^{-1}(0.5)$. Water is poured at 5 cubic metres per hour. Rate at which water level is rising when depth is 4m is (Take $\pi=22/7$)
$1$ m/h
$\dfrac{30}{17}$ m/h
$\dfrac{70}{88}$ m/h
$\dfrac{35}{88}$ m/h
Step-by-Step Solution
Key Concept: $r=h/2$; $V=\pi r^2h/3=\pi h^3/12$; $dV/dt=\pi h^2/4\cdot dh/dt$
$r=h/2$. $V=\frac{\pi h^3}{12}$. $\frac{dV}{dt}=\frac{\pi h^2}{4}\frac{dh}{dt}$. At $h=4$: $5=\frac{22/7\cdot 16}{4}\frac{dh}{dt}\Rightarrow\frac{dh}{dt}=\frac{5\cdot 7}{22\cdot 4}=\frac{35}{88}$.
Correct Answer: 4