Question

Prove that the semi-vertical angle of the right circular cone of given volume and least curved surface area is ${{\operatorname{Cot}}^{-1}}\sqrt{2}$ .

Answer 1

We are given a right circular cone, draw the cone to get the idea about angles. Since, we are given a condition that curved surface area is least so we will first evaluate this area in terms of volume and radius only and then use the definition of minima, that area can be least where the differentiation of it is 0. Using this, we will get a condition and hence get the desired results.

Complete step by step answer:
We are given a right circular cone with given volume,
We know that the volume of a cone is
$\begin{aligned} & V=\dfrac{1}{3}\pi {{r}^{2}}h \\\ & \Rightarrow h=\dfrac{3V}{\pi {{r}^{2}}} \\\ \end{aligned}$
Now, since we have to find a condition on a least curved surface area,
So, first consider the Curved surface are of a cone
We know that the curves surface area of a cone is
$A=\pi r\sqrt{{{r}^{2}}+{{h}^{2}}}$
Now, putting the value of $h$ , that we have evaluated earlier, using the volume of a cone, we get,
$\begin{aligned} & A=\pi r\sqrt{{{r}^{2}}+{{\left( \dfrac{3V}{\pi {{r}^{2}}} \right)}^{2}}} \\\ & =\pi r\sqrt{{{r}^{2}}+\left( \dfrac{9{{V}^{2}}}{{{\pi }^{2}}{{r}^{4}}} \right)} \\\ & =\sqrt{{{\pi }^{2}}{{r}^{4}}+\dfrac{9{{V}^{2}}}{{{r}^{2}}}} \end{aligned}$
Now, since we are interested in least surface area, so we will use the definition of minima here.
Now, differentiating $A$ with respect to $''r''$ and putting that equal to 0, we get,

& \dfrac{dA}{dr}=\dfrac{1}{2\sqrt{{{\pi }^{2}}{{r}^{4}}+\dfrac{9{{V}^{2}}}{{{r}^{2}}}}}\left( 4{{\pi }^{2}}{{r}^{3}}-\dfrac{18{{V}^{2}}}{{{r}^{3}}} \right)=0 \\\ & \Rightarrow 4{{\pi }^{2}}{{r}^{3}}-\dfrac{18{{V}^{2}}}{{{r}^{3}}}=0 \\\ & \Rightarrow 4{{\pi }^{2}} {{r}^{6}}=18{{V}^{2}} \\\ \end{aligned}$$ Now, putting the value of $V$ ,we get, $\begin{aligned} & 4{{\pi }^{2}}{{r}^{6}}=18{{\left( \dfrac{1}{3}\pi {{r}^{2}}h \right)}^{2}} \\\ & \Rightarrow 2{{r}^{2}}={{h}^{2}} \\\ & \Rightarrow {({\dfrac{h}{r}})^{2}}=2 \\\ & \Rightarrow \dfrac{h}{r}=\sqrt{2} \\\ \end{aligned}$  We, can see the right circular cone and we have to find the angle ABC, Using the definition of trigonometric functions, $$\begin{aligned} & \operatorname{Cot}\theta =\dfrac{h}{r}=\sqrt{2} \\\ & \Rightarrow \theta ={{\operatorname{Cot}}^{-1}}\sqrt{2} \\\ \end{aligned}$$ **Hence, the vertical angle of this right circular cone with least curved surface area is $${{\operatorname{Cot}}^{-1}}\sqrt{2}$$** **Note:** Note that the perpendicular and base of a triangle depends on the angle here . the angle we are considering here, the side opposite to that angle is considered as perpendicular and the other one as base. The possible mistake is that a student could have considered the opposite and hence could have answered $$\theta ={{\operatorname{Cot}}^{-1}}\dfrac{1}{\sqrt{2}}$$ , which is wrong .