The radius of a cyclinder is increasing 2 cm/ sec and its altitude is decreasing at the rate of 3 cm/ sec Find the rate of change of volume when radius is 3 cm and altitude is 5cm.

Answer

To solve the problem, we need to find the rate of change of the volume of a cylinder when the radius and height are changing. Let's break it down step by step.

### Step 1: Understand 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 (altitude) of the cylinder.

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

### Step 3: Identify the given values
From the problem, we have:
- \( \frac{dr}{dt} = 2 \) cm/sec (the radius is increasing)
- \( \frac{dh}{dt} = -3 \) cm/sec (the altitude is decreasing)
- \( r = 3 \) cm
- \( h = 5 \) cm

### Step 4: Substitute the values into the differentiated equation
Now, we substitute the values into the differentiated equation:
\[ \frac{dV}{dt} = \pi \left(2(3)(2)(5) + (3^2)(-3)\right) \]
Calculating each term:
1. \( 2(3)(2)(5) = 60 \)
2. \( (3^2)(-3) = 9(-3) = -27 \)

So, we have:
\[ \frac{dV}{dt} = \pi (60 - 27) \]
\[ \frac{dV}{dt} = \pi (33) \]

### Step 5: Final answer
Thus, the rate of change of volume when the radius is 3 cm and the altitude is 5 cm is:
\[ \frac{dV}{dt} = 33\pi \text{ cm}^3/\text{sec} \]