Name: Orthogonal Curves
Uploaded: 2019-07-19
Duration: 11 min 48 s
Description: Have you heard the term 'orthogonal curves'? Can you find angles between two curves? Do watch this video and find all such answers!! Have fun!!

Question

If the curved surface of right circular cylinder inscribed in a sphere of radius

R

is maximum, show that the height of the cylinder is

\sqrt{2} R

.

Accepted Answer

let $r$ be the radius and $h$ be the height of the cylinder.

From $Δ OAB, {OA}^{2} + {AB}^{2} = {OB}^{2}$

$\Rightarrow r^{2} + \frac{h^{2}}{4} = R^{2}; r^{2} = R^{2} - \frac{h^{2}}{4}$

Curved surface area $= 2 π h = 2 π h \sqrt{R^{2} - \frac{h^{2}}{4}}$

$= π h \sqrt{4 R^{2} - h^{2}}$

Let $f (h) = π h \sqrt{4 R^{2} - h^{2}}$

$\sqrt{2} R$

For maxima or minima $f^{'} (h) = 0$

$\Rightarrow \frac{2 π (2 R^{2} - h^{2})}{\sqrt{4 R^{2} - h^{2}}} = 0$

$\Rightarrow h^{2} = 2 R^{2} \Rightarrow h = \sqrt{2} R$

$\Rightarrow \sqrt{4 R^{2} - h^{2}} (- 2 h) + (2 R^{2} - h^{2})$

and $f^{''} (h) = 2 π \frac{\frac{d}{d h} \sqrt{4 R^{2} - h^{2}}}{4 R^{2} - h^{2}} = - \frac{4 π h + 0}{\sqrt{4 R^{2} -^{2}}} < 0 [when h = \sqrt{2} R]$

$f (h)$ is greatest when $h = \sqrt{2} R$

Therefore, Height of the cylinder $= \sqrt{2} R$

If the curved surface of right circular cylinder inscribed in a sphere of radius $R$ is maximum, show that the height of the cylinder is $\sqrt{2} R$ .

Important Questions on Applications of Derivatives

Learn and Prepare for any exam you want !

If the curved surface of right circular cylinder inscribed in a sphere of radius R is maximum, show that the height of the cylinder is 2R.

Important Questions on Applications of Derivatives

Learn and Prepare for any exam you want !

If the curved surface of right circular cylinder inscribed in a sphere of radius $R$ is maximum, show that the height of the cylinder is $\sqrt{2} R$ .