The radius of a cylinder is increasing at the rate of 3 cm /sec and its height is decreasing at the rate of 4 cm/sec. The rate of change of its volume when radius is 4 cm and height is 6 cm , is a

Options

A. `64 cm^(3) //sec `
B. `144 cm^(3) // sec`
C. `80 cm^(3)//sec`
D. `80 pi cm^(3)//sec`

Answer

To find the rate of change of the volume of a cylinder when the radius is increasing and the height is decreasing, we can follow these steps:

### Step 1: Write down the formula for the volume of a cylinder.
The volume \( V \) of a 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 \( t \).
To find the rate of change of volume with respect to time, we differentiate both sides of the volume formula using the product rule:
\[ \frac{dV}{dt} = \pi \left( 2r \frac{dr}{dt} h + r^2 \frac{dh}{dt} \right) \]

### Step 3: Substitute the known values.
We are given:
- \( \frac{dr}{dt} = 3 \) cm/sec (rate of increase of radius)
- \( \frac{dh}{dt} = -4 \) cm/sec (rate of decrease of height)
- \( r = 4 \) cm (radius at the moment of interest)
- \( h = 6 \) cm (height at the moment of interest)

Substituting these values into the differentiated volume equation:
\[ \frac{dV}{dt} = \pi \left( 2 \cdot 4 \cdot 3 \cdot 6 + 4^2 \cdot (-4) \right) \]

### Step 4: Calculate the individual components.
Calculating the first term:
\[ 2 \cdot 4 \cdot 3 \cdot 6 = 144 \]
Calculating the second term:
\[ 4^2 \cdot (-4) = 16 \cdot (-4) = -64 \]

### Step 5: Combine the results.
Now, substituting back into the equation:
\[ \frac{dV}{dt} = \pi (144 - 64) = \pi \cdot 80 \]

### Step 6: Final answer.
Thus, the rate of change of the volume of the cylinder is:
\[ \frac{dV}{dt} = 80\pi \text{ cm}^3/\text{sec} \]