Question

At what rate is the radius of the water surface changing when the water is 8 m deep

• A. $-\frac{5}{288\pi }m/min$
• B. $-\frac{1}{288\pi }m/min$
• C. $-\frac{5}{2\pi }m/min$
• D. $-\frac{5}{\pi }m/min$

Answer 1

Answer: (A)

$-\frac{5}{288\pi }m/min$

$v=\frac{\pi }{3}{y}^2(3R-y)$
Using $\frac{dv}{dt}=\left(\frac{dv}{dy}\right)\left(\frac{dy}{dt}\right)$
$\frac{dv}{dt}=6\frac{m^2}{min}$
$\frac{dv}{dy}=\frac{\pi }{3}(6Ry-{3y}^2)$
we need to find $\frac{dy}{dt}$ if $y=8m$
So $6\frac{m^2}{min}=\frac{\pi }{3}[6\left(13\right)\left(8\right)-3{\left(8\right)}^2]\left(\frac{dy}{dt}\right)$
$\Rightarrow 6=144\pi \frac{dy}{dt}\Rightarrow \frac{dy}{dt}=\frac{1}{24\pi }\frac{m}{min}$
Radius of water's surface
$r=\sqrt{R^2-{(R-y)}^2}$ if $R=13$
$r=\sqrt{169-{(13-y)}^2}\Rightarrow r=\sqrt{26y-y^2}$
Now we have $\frac{dr}{dt}=\left(\frac{dr}{dy}\right)\left(\frac{dy}{dt}\right)$
$\Rightarrow \frac{dr}{dy}=\frac{13-y}{\sqrt{26y-y^2}}=\frac{13-8}{\sqrt{26\left(8\right)-{\left(8\right)}^2}}=\frac{5}{12}$
$\Rightarrow \frac{dr}{dt}=\left(\frac{5}{12}\right)\left(0.0133\frac{m}{min}\right)=\frac{-5}{288}\frac{m}{min}$