The semi-vertical angle of a cone is 45°. If the height of the cone is 20.025, then its approximate lateral surface area, is

Solution:

Let $r$ be the radius, $h$ be the height and $I$ be the slant

height of a cone of semi-vertical angle 45°. Then,

$r=h$ and $I=\sqrt{2h}$

Let $h=20\text{ and }h+\mathrm{\Delta }h=20.025\Rightarrow \mathrm{\Delta }h=0.025$

Let $S$ be the lateral surface. Then,

$\begin{array}{l}S=\sqrt{2}\pi h^2 [\because S=\pi rl\Rightarrow S=\pi h\times \sqrt{2}h]\\ \Rightarrow \frac{dS}{dh}=2\sqrt{2}\pi h\\ \Rightarrow {\left(\frac{dS}{dh}\right)}_{h=20}=40\sqrt{2}\pi \\ \therefore \mathrm{\Delta }S=\frac{dS}{dh}\mathrm{\Delta }h\Rightarrow \mathrm{\Delta }S=40\sqrt{2}\pi \times 0.025=\sqrt{2}\pi \end{array}$

Also, at $h=20,$ we have $S=400\sqrt{2\mathrm{\pi }}$

$\therefore S+\mathrm{\Delta }S=400\sqrt{2}\pi +\sqrt{2}\pi =401\sqrt{2}\pi$