Name: Concavity and Point of Inflection of a Function
Uploaded: 2019-07-19
Duration: 23 min 41 s
Description: This video talks about concavity and the point of inflection of a function with sketching a graph. Let's learn one of the most important applications of deri...

Question

Show that the semi vertical angle of a cone of given slant height is

\tan^{- 1} \sqrt{2}

when its volume is maximum.

Accepted Answer

Let $θ$ be the semi-vertical angle of the cone.

It is clear that $θ \in [0, \frac{π}{2}]$ .

Let $r, h$ and $l$ be the radius, height and slant height of the cone respectively.

The slant height of the cone is given as constant.

Now, $r = l \sin θ$ and $h = l \cos θ$

The volume $(V)$ of the cone is given by, $V = \frac{1}{3} π r^{2} h$

$= \frac{1}{3} π (l^{2} \sin^{2} θ) (l \cos θ)$

$= \frac{1}{3} π l^{3} \sin^{2} θ \cos θ$

$∴ \frac{d V}{d θ} = \frac{l^{3} π}{3} [\sin^{2} θ (- \sin θ) + \cos θ (2 \sin θ \cos θ)]$

$= \frac{l^{3} π}{3} [- \sin^{3} + 2 \sin θ \cos^{2} θ]$

$\frac{d^{2} V}{d θ^{2}} = \frac{l^{3} π}{3} [- 3 \sin^{2} θ \cos θ + 2 \cos^{3} θ - 4 \sin^{2} θ \cos θ]$

$= \frac{l^{3} π}{3} [2 \cos^{3} θ - 7 \sin^{2} θ \cos θ]$

Now, $\frac{d V}{d θ} = 0$

$\Rightarrow \sin^{3} θ = 2 \sin θ \cos^{2} θ$

$\Rightarrow \tan^{2} θ = 2$

$\Rightarrow \tan θ = \sqrt{2}$

$\Rightarrow θ = \tan^{- 1} \sqrt{2}$

Now, when $θ = \tan^{- 1} \sqrt{2}$ , then $\tan^{2} θ = 2$ or $\sin^{2} θ = 2 \cos^{2} θ$

Then, we have $\frac{d^{2} V}{d θ^{2}} = \frac{l^{3} π}{3} [2 \cos^{3} θ - 14 \cos^{3} θ]$

$= - 4 π l^{3} \cos^{3} θ < 0$ for $θ \in [0, \frac{π}{2}]$

By second derivative test, the volume $(V)$ is the maximum when $θ = \tan^{- 1} \sqrt{2}$ .

Hence, for a given slant height, the semi-vertical angle of the cone of the maximum volume is $\tan^{- 1} \sqrt{2}$ .

Hence, shown.

Show that the semi vertical angle of a cone of given slant height is $\tan^{- 1} \sqrt{2}$ when its volume is maximum.

Important Questions on Application of Derivatives

Learn and Prepare for any exam you want !

Show that the semi vertical angle of a cone of given slant height is tan-12 when its volume is maximum.

Important Questions on Application of Derivatives

Learn and Prepare for any exam you want !

Show that the semi vertical angle of a cone of given slant height is $\tan^{- 1} \sqrt{2}$ when its volume is maximum.