The height of a cone with semi-vertical angle | vedclass

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Question

(C) Let $h$ be the height and $r$ be the radius of the cone. The semi-vertical angle $\alpha = \pi / 3$. We know that $	an(\alpha) = r / h$,so $r = h

Vedclass · Accepted Answer

$\sqrt{3}$

Vedclass · Answer

(C) Let $h$ be the height and $r$ be the radius of the cone. The semi-vertical angle $\alpha = \pi / 3$. We know that $	an(\alpha) = r / h$,so $r = h 	an(\pi / 3) = h \sqrt{3}$. The volume of the cone is $V = \frac{1}{3} \pi r^2 h$. Substituting $r = h \sqrt{3}$,we get $V = \frac{1}{3} \pi (h \sqrt{3})^2 h = \pi h^3$. Since the volume $V$ is fixed,differentiating with respect to time $t$ gives $\frac{dV}{dt} = 3 \pi h^2 \frac{dh}{dt} = 0$. However,the problem implies $V$ is constant at a specific instant. From $V = \frac{1}{3} \pi r^2 h$,differentiating with respect to $t$: $\frac{dV}{dt} = \frac{1}{3} \pi (2rh \frac{dr}{dt} + r^2 \frac{dh}{dt}) = 0$. Thus,$2rh \frac{dr}{dt} = -r^2 \frac{dh}{dt}$,which simplifies to $\frac{dr}{dt} = -\frac{r}{2h} \frac{dh}{dt}$. Given $\frac{dh}{dt} = 2$ and $r = h \sqrt{3}$,we have $\frac{dr}{dt} = -\frac{h \sqrt{3}}{2h} (2) = -\sqrt{3}$. The rate at which the radius is decreasing is $\sqrt{3} 	ext{ units/min}$.

The height of a cone with semi-vertical angle $\pi / 3$ is increasing at the rate of $2 \text{ units/min}$. The rate at which the radius of the cone is to be decreased so as to have a fixed volume always is

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