Find the semi-vertical angle of a right circu | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let $l$ be the slant height and $\alpha$ be the semi-vertical angle of the right circular cone. Let $h$ be the height and $r$ be the radius of the

Vedclass · Accepted Answer

$	an^{-1}(\sqrt{2})$

Vedclass · Answer

(B) Let $l$ be the slant height and $\alpha$ be the semi-vertical angle of the right circular cone. Let $h$ be the height and $r$ be the radius of the base.Then $h = l \cos \alpha$ and $r = l \sin \alpha$.The volume $V$ of the cone is given by $V = \frac{1}{3} \pi r^2 h$.Substituting the values of $r$ and $h$,we get $V = \frac{1}{3} \pi (l \sin \alpha)^2 (l \cos \alpha) = \frac{1}{3} \pi l^3 \sin^2 \alpha \cos \alpha$.To find the maximum volume,differentiate $V$ with respect to $\alpha$:$\frac{dV}{d\alpha} = \frac{1}{3} \pi l^3 [\sin^2 \alpha (-\sin \alpha) + \cos \alpha (2 \sin \alpha \cos \alpha)]$$\frac{dV}{d\alpha} = \frac{1}{3} \pi l^3 [-\sin^3 \alpha + 2 \sin \alpha \cos^2 \alpha]$.Setting $\frac{dV}{d\alpha} = 0$ for critical points:$-\sin^3 \alpha + 2 \sin \alpha (1 - \sin^2 \alpha) = 0$$\sin \alpha (2 - 3 \sin^2 \alpha) = 0$.Since $\alpha 
eq 0$,we have $2 - 3 \sin^2 \alpha = 0$,which implies $\sin^2 \alpha = \frac{2}{3}$.Using $\cos^2 \alpha = 1 - \sin^2 \alpha = 1 - \frac{2}{3} = \frac{1}{3}$,we find $	an^2 \alpha = \frac{\sin^2 \alpha}{\cos^2 \alpha} = \frac{2/3}{1/3} = 2$.Thus,$	an \alpha = \sqrt{2}$,or $\alpha = 	an^{-1}(\sqrt{2})$.Checking the second derivative $\frac{d^2V}{d\alpha^2}$ confirms that the volume is maximum at this value.

Find the semi-vertical angle of a right circular cone of a given slant height,if the volume of the cone is maximum.

Similar Questions

The least area of a circle circumscribing any right triangle of area $S$ is

The maximum area of the rectangle that can be inscribed in a circle of radius $r$ is

$A$ solid hemisphere is mounted on a solid cylinder,both having equal radii. If the whole solid is to have a fixed surface area and the maximum possible volume,then the ratio of the height of the cylinder to the common radius is

If the function $f(x) = (\frac{1}{x})^{2x}$ for $x > 0$ attains the maximum value at $x = \frac{1}{e}$,then:

The condition for the function $f(x) = x^3 + px^2 + qx + r$ $(x \in R)$ to have no extreme value is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module