The radius of a cylinder increases at the rate of 1 cm/s and its height decreases at the rate of 1 cm/s. Find the rate of change of its volume when the radius is 5 cm and the height is 5 cm.
If the volume should not change even when the radius and height are changed, what is the relation between the radius and height?

Answer

To solve the problem step by step, we will follow these instructions:

### Step 1: Identify the given rates and the formula for the volume of a cylinder.
- The radius \( r \) of the cylinder is increasing at the rate of \( \frac{dr}{dt} = 1 \) cm/s.
- The height \( h \) of the cylinder is decreasing at the rate of \( \frac{dh}{dt} = -1 \) cm/s.
- The formula for the volume \( V \) of a cylinder is given by:
\[ V = \pi r^2 h \]

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

### Step 3: Substitute the known values.
Now, we will substitute \( r = 5 \) cm, \( h = 5 \) cm, \( \frac{dr}{dt} = 1 \) cm/s, and \( \frac{dh}{dt} = -1 \) cm/s into the differentiated equation:
\[ \frac{dV}{dt} = \pi \left( 5^2 \cdot (-1) + 5 \cdot 2 \cdot 5 \cdot 1 \right) \]
Calculating each term:
- \( 5^2 \cdot (-1) = -25 \)
- \( 5 \cdot 2 \cdot 5 \cdot 1 = 50 \)

Thus, we have:
\[ \frac{dV}{dt} = \pi \left( -25 + 50 \right) = \pi \cdot 25 = 25\pi \text{ cm}^3/\text{s} \]

### Step 4: Conclusion for the first part.
The rate of change of the volume when the radius is 5 cm and the height is 5 cm is:
\[ \frac{dV}{dt} = 25\pi \text{ cm}^3/\text{s} \approx 78.5 \text{ cm}^3/\text{s} \]

### Step 5: Determine the condition for constant volume.
For the volume to remain constant, we set \( \frac{dV}{dt} = 0 \):
\[ 0 = \pi \left( r^2 \frac{dh}{dt} + h \cdot 2r \frac{dr}{dt} \right) \]
This implies:
\[ r^2 \frac{dh}{dt} + 2hr \frac{dr}{dt} = 0 \]
Rearranging gives:
\[ r^2 \frac{dh}{dt} = -2hr \frac{dr}{dt} \]
Dividing both sides by \( hr \) (assuming \( h \neq 0 \) and \( r \neq 0 \)):
\[ \frac{dh}{dt} = -\frac{2h}{r} \frac{dr}{dt} \]

### Step 6: Conclusion for the second part.
If \( \frac{dr}{dt} = 1 \) cm/s, then:
\[ \frac{dh}{dt} = -\frac{2h}{r} \]
This means that for the volume to remain constant, the height \( h \) must be half of the radius \( r \):
\[ r = 2h \]

### Final Answers:
1. The rate of change of volume when the radius is 5 cm and height is 5 cm is \( 25\pi \) cm³/s.
2. The relation between the radius and height for constant volume is \( r = 2h \).