A cone having fixed volume has semi - vertical angle of `(pi)/(4)`. At an instant when its height it decreasing at the rate of 2m/s, its radius increases at a rate equal to

Options

A. 2m/s
B. 4m/s
C. 1m/s
D. 8m/s

Answer

To solve the problem, we need to find the rate at which the radius of the cone is increasing when the height is decreasing at a rate of 2 m/s. We know that the cone has a fixed volume and a semi-vertical angle of \( \frac{\pi}{4} \) (or 45 degrees).

### Step-by-Step Solution:

1. **Understand the Geometry of the Cone**:
- The semi-vertical angle \( \theta = \frac{\pi}{4} \) implies that \( \tan(\theta) = 1 \). Therefore, the relationship between the radius \( r \) and height \( h \) of the cone is given by:
\[ r = h \]

2. **Volume of the Cone**:
- The volume \( V \) of the cone is given by the formula:
\[ V = \frac{1}{3} \pi r^2 h \]
- Since the volume is constant, we can differentiate this equation with respect to time \( t \).

3. **Differentiate the Volume with Respect to Time**:
- Differentiating both sides with respect to time \( t \):
\[ \frac{dV}{dt} = 0 = \frac{1}{3} \pi \left( 2r \frac{dr}{dt} h + r^2 \frac{dh}{dt} \right) \]
- This gives us the equation:
\[ 0 = \frac{2}{3} \pi r h \frac{dr}{dt} + \frac{1}{3} \pi r^2 \frac{dh}{dt} \]

4. **Substituting Known Values**:
- We know \( \frac{dh}{dt} = -2 \) m/s (since the height is decreasing).
- Substitute this into the equation:
\[ 0 = \frac{2}{3} \pi r h \frac{dr}{dt} + \frac{1}{3} \pi r^2 (-2) \]
- Simplifying, we get:
\[ 0 = \frac{2}{3} \pi r h \frac{dr}{dt} - \frac{2}{3} \pi r^2 \]

5. **Rearranging the Equation**:
- Rearranging gives:
\[ \frac{2}{3} \pi r h \frac{dr}{dt} = \frac{2}{3} \pi r^2 \]
- Dividing both sides by \( \frac{2}{3} \pi r \) (assuming \( r \neq 0 \)):
\[ h \frac{dr}{dt} = r \]

6. **Using the Relationship Between \( r \) and \( h \)**:
- Since \( r = h \) (from Step 1), we can substitute \( h \) with \( r \):
\[ r \frac{dr}{dt} = r \]
- Dividing both sides by \( r \) (assuming \( r \neq 0 \)):
\[ \frac{dr}{dt} = 1 \text{ m/s} \]

### Final Answer:
The radius of the cone is increasing at a rate of \( 1 \) m/s.