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(B) Let $h$ be the height, $r$ be the radius, and $l$ be the slant height of the water at any time $t$. Given the semi-vertical angle $\alpha = 30^{\c

Vedclass · Accepted Answer

$2 \pi$

Vedclass · Answer

(B) Let $h$ be the height, $r$ be the radius, and $l$ be the slant height of the water at any time $t$. Given the semi-vertical angle $\alpha = 30^{\circ}$.From the geometry of the cone, $r = h 	an 30^{\circ} = \frac{h}{\sqrt{3}}$.The volume $V$ of the water is $V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi \left(\frac{h}{\sqrt{3}}ight)^2 h = \frac{\pi h^3}{9}$.Given $V = \frac{8 \pi}{\sqrt{3}}$, we have $\frac{\pi h^3}{9} = \frac{8 \pi}{\sqrt{3}} \Rightarrow h^3 = \frac{72}{\sqrt{3}} = 24 \sqrt{3} = (2 \sqrt{3})^3$, so $h = 2 \sqrt{3} \ ft$.The slant height $l = \sqrt{h^2 + r^2} = \sqrt{h^2 + \frac{h^2}{3}} = \sqrt{\frac{4h^2}{3}} = \frac{2h}{\sqrt{3}}$.Differentiating with respect to $t$, $\frac{dl}{dt} = \frac{2}{\sqrt{3}} \frac{dh}{dt}$.Given $\frac{dl}{dt} = -\frac{1}{\sqrt{3}}$ (decreasing), we have $-\frac{1}{\sqrt{3}} = \frac{2}{\sqrt{3}} \frac{dh}{dt} \Rightarrow \frac{dh}{dt} = -\frac{1}{2} \ ft/min$.The rate of change of volume is $\frac{dV}{dt} = \frac{d}{dt} \left(\frac{\pi h^3}{9}ight) = \frac{\pi}{3} h^2 \frac{dh}{dt}$.Substituting $h = 2 \sqrt{3}$ and $\frac{dh}{dt} = -\frac{1}{2}$, we get $\frac{dV}{dt} = \frac{\pi}{3} (2 \sqrt{3})^2 \left(-\frac{1}{2}ight) = \frac{\pi}{3} (12) \left(-\frac{1}{2}ight) = -2 \pi \ cu. \ ft/min$.Thus, the volume is decreasing at a rate of $2 \pi \ cu. \ ft/min$.

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