Question

Water is flowing out at the rate of $6m^3/min$ from a reservoir shaped like a hemispherical bowl of radius$R=13$ m The volume of water in the hemispherical bowl is given by $v=\frac{\pi }{3}.y^2(3R-y)$ when the water is $y$ meter deep Find at what rate is the water level changing when the water is $8$ m deep.

• A. $-\frac{1}{12\pi }m/min$
• B. $-\frac{1}{18\pi }m/min$
• C. $-\frac{1}{24\pi }m/min$
• D. $-\frac{1}{30\pi }m/min$

Answer 1

Answer: (C)

$-\frac{1}{24\pi }m/min$

$v=\frac{\pi }{3}{y}^2(3R-y)$
Using $\frac{dv}{dt}=\left(\frac{dv}{dy}\right)\left(\frac{dy}{dt}\right)$
$\Rightarrow \frac{dv}{dt}=6\frac{m^2}{min}$
$\Rightarrow \frac{dv}{dy}=\frac{\pi }{3}(6Ry-{3y}^2)$
We need to find $\frac{dy}{dt}$ if $y=8m$
So, $6=\frac{\pi }{3}[6\left(13\right)\left(8\right)-3{\left(8\right)}^2]\left(\frac{dy}{dt}\right)$
$\Rightarrow 6=144\pi \frac{dy}{dt}$

$\Rightarrow \frac{dy}{dt}=\frac{1}{24\pi }m/min$