If the semi-vertical angle of a cone is 45^{c | vedclass

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(A) Let $r$ be the radius,$h$ be the height,and $l$ be the slant height of a cone with semi-vertical angle $\alpha = 45^{\circ}$.Then $r = h 	an(45^{

Vedclass · Accepted Answer

$401 \pi \sqrt{2}$

Vedclass · Answer

(A) Let $r$ be the radius,$h$ be the height,and $l$ be the slant height of a cone with semi-vertical angle $\alpha = 45^{\circ}$.Then $r = h 	an(45^{\circ}) = h$ and $l = \sqrt{r^2 + h^2} = \sqrt{h^2 + h^2} = h\sqrt{2}$.The curved surface area $S$ is given by $S = \pi r l = \pi (h) (h\sqrt{2}) = \sqrt{2} \pi h^2$.Let $h = 20$ and $h + \Delta h = 20.025$,so $\Delta h = 0.025$.The derivative of $S$ with respect to $h$ is $\frac{dS}{dh} = 2\sqrt{2} \pi h$.At $h = 20$,$\frac{dS}{dh} = 2\sqrt{2} \pi (20) = 40\sqrt{2} \pi$.The approximate change in surface area is $\Delta S \approx \frac{dS}{dh} \Delta h = (40\sqrt{2} \pi) (0.025) = \sqrt{2} \pi$.The initial surface area at $h = 20$ is $S = \sqrt{2} \pi (20)^2 = 400\sqrt{2} \pi$.Therefore,the approximate surface area is $S + \Delta S = 400\sqrt{2} \pi + \sqrt{2} \pi = 401\sqrt{2} \pi$.

If the semi-vertical angle of a cone is $45^{\circ}$ and its height is $20.025 \text{ cm}$,then the approximate value of its curved surface area (in $\text{sq. cm}$) is

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