The height of a cone increases by k%, its semi-vertical angle remains the same. What is the approximate percentage increase in volume, assuming that k is small?

Options

A. a) `2K%`
B. b) `k^(2)%`
C. c)
`3k%`
D. d) None of these

Answer

To solve the problem, we need to find the approximate percentage increase in the volume of a cone when its height increases by \( k\% \) while keeping the semi-vertical angle constant.

### Step-by-Step Solution:

1. **Understanding the Cone's Volume Formula**:
The volume \( V \) of a cone is given by the formula:
\[ V = \frac{1}{3} \pi r^2 h \]
where \( r \) is the radius and \( h \) is the height of the cone.

2. **Relating Radius and Height**:
Given that the semi-vertical angle \( \alpha \) remains constant, we can express the radius \( r \) in terms of the height \( h \):
\[ r = h \tan(\alpha) \]
Substituting this into the volume formula gives:
\[ V = \frac{1}{3} \pi (h \tan(\alpha))^2 h = \frac{1}{3} \pi h^3 \tan^2(\alpha) \]

3. **Differentiating the Volume**:
To find the change in volume with respect to height, we differentiate \( V \) with respect to \( h \):
\[ \frac{dV}{dh} = \frac{d}{dh} \left( \frac{1}{3} \pi h^3 \tan^2(\alpha) \right) = \pi h^2 \tan^2(\alpha) \]

4. **Finding the Change in Height**:
If the height increases by \( k\% \), then the change in height \( dh \) can be expressed as:
\[ dh = \frac{k}{100} h \]

5. **Calculating the Change in Volume**:
Substituting \( dh \) into the derivative of the volume:
\[ dV = \frac{dV}{dh} \cdot dh = \pi h^2 \tan^2(\alpha) \cdot \frac{k}{100} h = \frac{\pi k}{100} h^3 \tan^2(\alpha) \]

6. **Expressing the Change in Volume as a Percentage of the Original Volume**:
The original volume \( V \) is:
\[ V = \frac{1}{3} \pi h^3 \tan^2(\alpha) \]
Now, we can express \( dV \) in terms of \( V \):
\[ dV = \frac{3k}{100} V \]

7. **Calculating the Percentage Increase**:
The percentage increase in volume is given by:
\[ \text{Percentage Increase} = \left( \frac{dV}{V} \right) \times 100 = \frac{3k}{100} \times 100 = 3k\% \]

### Final Answer:
The approximate percentage increase in volume is \( 3k\% \).