The semi-vertical angle of a cone remains constant. If its height increases by 2%, then find the approximate increase in its volume.

Answer

To find the approximate increase in the volume of a cone when its height increases by 2%, while keeping the semi-vertical angle constant, we can follow these steps:

### Step 1: Understand the relationship between the dimensions of the cone
Let:
- \( h \) = height of the cone
- \( r \) = radius of the base of the cone
- \( \alpha \) = semi-vertical angle of the cone

Since the semi-vertical angle \( \alpha \) remains constant, we can express the radius \( r \) in terms of height \( h \):
\[ \tan \alpha = \frac{r}{h} \implies r = h \tan \alpha \]

### Step 2: Write the formula for the volume of the cone
The volume \( V \) of the cone is given by:
\[ V = \frac{1}{3} \pi r^2 h \]
Substituting \( r = h \tan \alpha \) into the volume formula:
\[ V = \frac{1}{3} \pi (h \tan \alpha)^2 h = \frac{1}{3} \pi h^2 \tan^2 \alpha \cdot h = \frac{1}{3} \pi h^3 \tan^2 \alpha \]

### Step 3: Differentiate the volume with respect to height
To find the change in volume \( dV \) when height changes, we differentiate \( V \):
\[ dV = \frac{d}{dh} \left( \frac{1}{3} \pi h^3 \tan^2 \alpha \right) = \frac{1}{3} \pi \tan^2 \alpha \cdot 3h^2 \, dh = \pi h^2 \tan^2 \alpha \, dh \]

### Step 4: Calculate the relative change in volume
The relative change in volume \( \frac{dV}{V} \) can be expressed as:
\[ \frac{dV}{V} = \frac{\pi h^2 \tan^2 \alpha \, dh}{\frac{1}{3} \pi h^3 \tan^2 \alpha} \]
Simplifying this expression:
\[ \frac{dV}{V} = \frac{3h^2 \tan^2 \alpha \, dh}{h^3 \tan^2 \alpha} = \frac{3 \, dh}{h} \]

### Step 5: Substitute the percentage increase in height
Given that the height increases by 2%, we have:
\[ \frac{dh}{h} = \frac{2}{100} = 0.02 \]
Now substituting this into the relative change in volume:
\[ \frac{dV}{V} = 3 \cdot 0.02 = 0.06 \]

### Step 6: Convert to percentage
To express this as a percentage increase in volume:
\[ \text{Percentage increase in volume} = 0.06 \times 100 = 6\% \]

### Final Answer
The approximate increase in the volume of the cone is **6%**.

---