Name: Rate of Change of Function: Example
Uploaded: 2019-07-19
Duration: 3 min 57 s
Description: In this video, we will use the rate of change to find the speed of the lower end of the ladder when it is sliding on the ground. Watch this video to know more.

Question

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

Accepted Answer

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

And let $θ$ be the semi-vertical angle of the cone

$\Rightarrow l^{2} = h^{2} + r^{2}$

And volume, $V = \frac{1}{3} π r^{2} h$

$\Rightarrow V = \frac{1}{3} π h (l^{2} - h^{2})$

$\Rightarrow V = \frac{π}{3} (l^{2} h - h^{3})$

$\Rightarrow \frac{d V}{d h} = \frac{π}{3} (l^{2} - 3 h^{2})$

$\Rightarrow \frac{d^{2} V}{d h^{2}} = \frac{π}{3} (- 6 h)$

For critical point, $\frac{d V}{d h} = 0$

$\Rightarrow \frac{π}{3} (l^{2} - 3 h^{2}) = 0$

$\Rightarrow l^{2} = 3 h^{2}$

$\Rightarrow h = \frac{l}{\sqrt{3}}$

$\frac{d^{2} V}{d h^{2}}$ at $h = \frac{l}{\sqrt{3}} = \frac{π}{3} (- 6 \frac{l}{\sqrt{3}}) < 0$

So, volume of cone is maximum, when $h = \frac{l}{\sqrt{3}}$

$\Rightarrow l^{2} = 3 h^{2}$

$\Rightarrow h^{2} + r^{2} = 3 h^{2}$

$\Rightarrow r = \sqrt{2} h$

$\Rightarrow \frac{r}{h} = \sqrt{2}$

$\Rightarrow \tan θ = \sqrt{2}$

$\Rightarrow θ = \tan^{- 1} \sqrt{2}$

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

Important Questions on Application of Derivatives

Learn and Prepare for any exam you want !

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

Important Questions on Application of Derivatives

Learn and Prepare for any exam you want !

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