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 step by step, we will analyze the relationship between the radius and height of the cone, and how their rates of change relate to each other given the fixed volume of the cone.

### Step 1: Understand the Geometry of the Cone
The semi-vertical angle of the cone is given as \(\frac{\pi}{4}\). This means that the tangent of the angle is:
\[ \tan\left(\frac{\pi}{4}\right) = 1 \]
This implies that the radius \(r\) and height \(h\) of the cone are equal:
\[ \frac{r}{h} = 1 \quad \Rightarrow \quad r = h \]

**Hint:** Recall that the tangent of an angle in a right triangle is the ratio of the opposite side to the adjacent side.

### Step 2: Volume of the Cone
The volume \(V\) of a cone is given by the formula:
\[ V = \frac{1}{3} \pi r^2 h \]
Since the volume is fixed, we can express this as:
\[ \frac{dV}{dt} = 0 \]

**Hint:** Remember that if the volume is constant, the derivative with respect to time will be zero.

### Step 3: Differentiate the Volume with Respect to Time
Using the product rule, we differentiate the volume with respect to time:
\[ \frac{dV}{dt} = \frac{1}{3} \pi \left(2r \frac{dr}{dt} h + r^2 \frac{dh}{dt}\right) = 0 \]
We can simplify this to:
\[ 2r h \frac{dr}{dt} + r^2 \frac{dh}{dt} = 0 \]

**Hint:** Apply the product rule carefully to each term involving \(r\) and \(h\).

### Step 4: Substitute Known Values
We know that the height is decreasing at a rate of \(2 \, \text{m/s}\), so:
\[ \frac{dh}{dt} = -2 \, \text{m/s} \]
Substituting this into the equation gives:
\[ 2r h \frac{dr}{dt} + r^2 (-2) = 0 \]
This simplifies to:
\[ 2r h \frac{dr}{dt} = 2r^2 \]

**Hint:** Pay attention to the signs when substituting rates of change.

### Step 5: Solve for \(\frac{dr}{dt}\)
Dividing both sides by \(2r\) (assuming \(r \neq 0\)):
\[ h \frac{dr}{dt} = r \]
Thus, we can express \(\frac{dr}{dt}\) as:
\[ \frac{dr}{dt} = \frac{r}{h} \]

**Hint:** Isolate the term you want to solve for by dividing both sides by the appropriate quantities.

### Step 6: Substitute \(h\) in Terms of \(r\)
Since we established earlier that \(r = h\), we can substitute \(h\) with \(r\):
\[ \frac{dr}{dt} = \frac{r}{r} = 1 \, \text{m/s} \]

**Hint:** Use the relationship between \(r\) and \(h\) to simplify your final expression.

### Conclusion
The radius \(r\) of the cone is increasing at a rate of:
\[ \frac{dr}{dt} = 1 \, \text{m/s} \]