The radius of a right circular cylinder increases at the rate of 0.1 cm/min, and the height decreases at the rate of 0.2 cm/min. The rate of change of the volume of the cylinder, in `c m^2//m in` , when the radius is `2c m` and the height is 3cm is (a) `-2p` (b) `-(8pi)/5` `-(3pi)/5` (d) `(2pi)/5`

Options

A. `-2 pi`
B. `-(8pi)/(5)`
C. `16//6`
D. `-8//15`

Answer

To find the rate of change of the volume of a right circular cylinder, we can follow these steps:

### Step 1: Write the formula for the volume of a cylinder
The volume \( V \) of a right circular cylinder is given by the formula:
\[ V = \pi r^2 h \]
where \( r \) is the radius and \( h \) is the height.

### Step 2: Differentiate the volume with respect to time
To find the rate of change of volume with respect to time, we differentiate both sides of the volume formula with respect to time \( t \):
\[ \frac{dV}{dt} = \pi \left(2r \frac{dr}{dt} h + r^2 \frac{dh}{dt}\right) \]

### Step 3: Substitute the known values
We are given:
- The radius \( r = 2 \) cm
- The height \( h = 3 \) cm
- The rate of change of the radius \( \frac{dr}{dt} = 0.1 \) cm/min
- The rate of change of the height \( \frac{dh}{dt} = -0.2 \) cm/min (negative because the height is decreasing)

Now, substituting these values into the differentiated volume formula:
\[ \frac{dV}{dt} = \pi \left(2 \cdot 2 \cdot 0.1 \cdot 3 + 2^2 \cdot (-0.2)\right) \]

### Step 4: Calculate the terms
Calculating each term:
1. \( 2 \cdot 2 \cdot 0.1 \cdot 3 = 1.2 \)
2. \( 2^2 \cdot (-0.2) = -0.8 \)

Now substitute these back into the equation:
\[ \frac{dV}{dt} = \pi (1.2 - 0.8) = \pi \cdot 0.4 \]

### Step 5: Final result
Thus, the rate of change of the volume of the cylinder is:
\[ \frac{dV}{dt} = 0.4\pi \text{ cm}^3/\text{min} \]

To express this in terms of \( \frac{cm^3}{min} \):
\[ \frac{dV}{dt} = \frac{2\pi}{5} \text{ cm}^3/\text{min} \]

### Conclusion
The answer is:
\[ \frac{dV}{dt} = \frac{2\pi}{5} \text{ cm}^3/\text{min} \]