The radius of a right circular cylinder increases at the rate of 0.2 cm/sec and the height decreases at the rate of 0.1 cm/sec. The rate of change of the volume of the cylinder when the radius is 1 cm and the height is 2 cm is

Options

A. `(pi)/(10)cm^(3)//sec`
B. `(9pi)/(10)cm^(3)//sec`
C. `(7pi)/(10)cm^(3)//sec`
D. `(pi)/(5)cm^(3)//sec`

Answer

To find the rate of change of the volume of a right circular cylinder when the radius is 1 cm and the height is 2 cm, we will follow these steps:

### Step 1: Understand the problem
We know that:
- The radius \( r \) of the cylinder is increasing at a rate of \( \frac{dr}{dt} = 0.2 \) cm/sec.
- The height \( h \) of the cylinder is decreasing at a rate of \( \frac{dh}{dt} = -0.1 \) cm/sec (negative because it is decreasing).
- We need to find the rate of change of the volume \( \frac{dV}{dt} \) when \( r = 1 \) cm and \( h = 2 \) cm.

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

### Step 3: Differentiate the volume with respect to time
To find the rate of change of volume, we differentiate \( V \) with respect to time \( t \):
\[ \frac{dV}{dt} = \frac{d}{dt}(\pi r^2 h) \]
Using the product rule, we differentiate \( r^2 h \):
\[ \frac{dV}{dt} = \pi \left( \frac{d(r^2)}{dt} h + r^2 \frac{dh}{dt} \right) \]

### Step 4: Differentiate \( r^2 \)
Using the chain rule, we have:
\[ \frac{d(r^2)}{dt} = 2r \frac{dr}{dt} \]
Substituting this into the equation gives:
\[ \frac{dV}{dt} = \pi \left( 2r \frac{dr}{dt} h + r^2 \frac{dh}{dt} \right) \]

### Step 5: Substitute the known values
Now, substitute \( r = 1 \) cm, \( h = 2 \) cm, \( \frac{dr}{dt} = 0.2 \) cm/sec, and \( \frac{dh}{dt} = -0.1 \) cm/sec into the equation:
\[ \frac{dV}{dt} = \pi \left( 2(1)(0.2)(2) + (1^2)(-0.1) \right) \]

### Step 6: Calculate the terms
Calculating the first term:
\[ 2(1)(0.2)(2) = 0.8 \]
Calculating the second term:
\[ (1^2)(-0.1) = -0.1 \]
Now substituting these values back:
\[ \frac{dV}{dt} = \pi (0.8 - 0.1) = \pi (0.7) \]

### Step 7: Final result
Thus, we find:
\[ \frac{dV}{dt} = 0.7\pi \text{ cm}^3/\text{sec} \]

### Conclusion
The rate of change of the volume of the cylinder when the radius is 1 cm and the height is 2 cm is:
\[ \frac{dV}{dt} = \frac{7\pi}{10} \text{ cm}^3/\text{sec} \]