Question

The height of a cone increases by \[k\% \] its semi vertical angle remaining the same. What is the approximate percent increase in the volume assuming that \[k\] is small?

Answer 1

Hint:
Here we need to find the percent increase in the volume of cones. For that, we will first assume the semi vertical angle of the cone to be any variable. Then we will find the radius of the cone in terms of height and the semi vertical angle. Then we will use the conditions given in the question one by one. After simplifying all the equations, we will get the required answer.

Complete step by step solution:
Let the radius of the cone be \[r\] , semi vertical angle be \[\alpha \] and height be \[h\] .
Now, we will draw the figure of a cone.

In, we have
\[ \Rightarrow \tan \alpha = \dfrac{{OB}}{{OA}}\]
Now, we will substitute the values here.
\[ \Rightarrow \tan \alpha = \dfrac{R}{H}\]
On cross multiplying the terms, we get
\[ \Rightarrow R = H\tan \alpha \] …………. \[\left( 1 \right)\]
Now, we will find the volume of cube.
\[ \Rightarrow V = \dfrac{1}{3} \cdot \pi \cdot {R^2} \cdot H\]
We know the value \[R\] from equation 1. So we will substitute that value here.
\[ \Rightarrow V = \dfrac{1}{3} \cdot \pi \cdot {\left( {H\tan \alpha } \right)^2} \cdot H\]
Applying exponent on the bases, we get
\[ \Rightarrow V = \dfrac{1}{3} \cdot \pi \cdot {\tan ^2}\alpha \cdot {H^3}\] ………. \[\left( 2 \right)\]
Now, we will differentiate both sides with respect to \[H\] and \[{\tan ^2}\alpha \] is constant here.
\[ \Rightarrow \dfrac{{dV}}{{dH}} = \dfrac{d}{{dH}}\left( {\dfrac{1}{3} \cdot \pi \cdot {{\tan }^2}\alpha \cdot {H^3}} \right)\]
On differentiating the terms, we get
\[ \Rightarrow \dfrac{{dV}}{{dH}} = \dfrac{1}{3} \cdot \pi \cdot {\tan ^2}\alpha \cdot 3{H^2}\]
On cross multiplication, we get
\[ \Rightarrow dV = \left( {\dfrac{1}{3} \cdot \pi \cdot {{\tan }^2}\alpha \cdot 3{H^2}} \right)dH\] ……….. \[\left( 3 \right)\]
It is given that:
\[dH = k\% \] of \[H\]
Therefore, we can write it as
\[dH = \dfrac{k}{{100}} \times H\]
Now, we will substitute this value in equation 3.
\[ \Rightarrow dV = \left( {\dfrac{1}{3} \cdot \pi \cdot {{\tan }^2}\alpha \cdot 3{H^2}} \right)\dfrac{k}{{100}} \times H\]
On further simplification, we get
\[ \Rightarrow dV = 3 \times \dfrac{k}{{100}} \times \left( {\dfrac{1}{3} \cdot \pi \cdot {{\tan }^2}\alpha \cdot {H^3}} \right)\]
We know from equation 2 that \[V = \dfrac{1}{3} \cdot \pi \cdot {\tan ^2}\alpha \cdot {H^3}\]
So we will substitute this value here.
\[ \Rightarrow dV = 3 \times \dfrac{k}{{100}} \times V\]
Therefore, we can write it as
\[dV = 3k\% \] of \[V\]

Thus, volume increased by \[3k\% \] of \[V\].

Note:
Cone is a three dimensional geometric figure such that its base is a circle. Volume of any object is defined as the amount of space that an object occupies. Thus, we can say that the space available in the cone can be determined by its volume. We need to keep in mind that if we change the height and radius of the cone, the volume of the cone will also change.