The vertical angle of a right circular cone i | vedclass

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(C) Let $r$ be the radius and $h$ be the height of the water level in the cone at any time $t$.The vertical angle is $60^{\circ}$, so the semi-vertica

Vedclass · Accepted Answer

$\frac{1}{9 \pi}$

Vedclass · Answer

(C) Let $r$ be the radius and $h$ be the height of the water level in the cone at any time $t$.The vertical angle is $60^{\circ}$, so the semi-vertical angle is $30^{\circ}$.From the geometry of the cone, $\frac{r}{h} = 	an 30^{\circ} = \frac{1}{\sqrt{3}}$, which implies $h = \sqrt{3}r$.The volume $V$ of the water in the cone is given by $V = \frac{1}{3} \pi r^2 h$.Substituting $h = \sqrt{3}r$, we get $V = \frac{1}{3} \pi r^2 (\sqrt{3}r) = \frac{\sqrt{3}}{3} \pi r^3 = \frac{1}{\sqrt{3}} \pi r^3$.Differentiating with respect to $t$, we get $\frac{dV}{dt} = \frac{1}{\sqrt{3}} \pi (3r^2) \frac{dr}{dt} = \sqrt{3} \pi r^2 \frac{dr}{dt}$.Given $\frac{dV}{dt} = \frac{1}{\sqrt{3}}$, we have $\sqrt{3} \pi r^2 \frac{dr}{dt} = \frac{1}{\sqrt{3}}$, so $\frac{dr}{dt} = \frac{1}{3 \pi r^2}$.Since $h = \sqrt{3}r$, when $h = 3 	ext{ m}$, $r = \frac{3}{\sqrt{3}} = \sqrt{3} 	ext{ m}$.Substituting $r = \sqrt{3}$ into the expression for $\frac{dr}{dt}$, we get $\frac{dr}{dt} = \frac{1}{3 \pi (\sqrt{3})^2} = \frac{1}{3 \pi (3)} = \frac{1}{9 \pi} 	ext{ m/min}$.

The vertical angle of a right circular cone is $60^{\circ}$. If water is being poured into the cone at the rate of $\frac{1}{\sqrt{3}} \text{ m}^3/\text{min}$, then the rate $(\text{m/min})$ at which the radius of the water level is increasing when the height of the water level is $3 \text{ m}$ is

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