A tube has an inner radius of 2mm and a height of 20mm. The tube maintains a cylindrical shape as it gets stretched out. If the height increases at a rate of 3mm per second, and the radius decreases at a rate of 0.1mm per second, what is happening to the volume at this instant?

To determine what is happening to the volume of the tube at the given instant, we need to start by finding the formula for the volume of a cylinder and then apply the related rates to that formula.

### Step 1: Formula for the Volume of a Cylinder
The volume \( V \) of a cylinder is given by:
\[
V = \pi r^2 h
\]
where:
- \( r \) is the radius of the cylinder,
- \( h \) is the height of the cylinder.

### Step 2: Differentiating with Respect to Time
We need to find the rate of change of volume \( \frac{dV}{dt} \). To do this, we differentiate the volume formula with respect to time \( t \):
\[
\frac{dV}{dt} = \frac{d}{dt} \left(\pi r^2 h\right)
\]
Using the product rule:
\[
\frac{dV}{dt} = \pi \left(2r \frac{dr}{dt} \cdot h + r^2 \frac{dh}{dt}\right)
\]
Where:
- \( \frac{dr}{dt} \) is the rate of change of the radius,
- \( \frac{dh}{dt} \) is the rate of change of the height.

### Step 3: Substituting the Given Values
At the instant in question:
- The initial radius \( r = 2 \) mm,
- The height \( h = 20 \) mm,
- The rate of change of the radius \( \frac{dr}{dt} = -0.1 \) mm/s (negative because the radius is decreasing),
- The rate of change of the height \( \frac{dh}{dt} = 3 \) mm/s.

Substituting these values into the derivative formula:
\[
\frac{dV}{dt} = \pi \left(2(2)(-0.1)(20) + (2^2)(3)\right)
\]
\[
\frac{dV}{dt} = \pi \left(-8 + 12\right)
\]
\[
\frac{dV}{dt} = \pi \times 4
\]
\[
\frac{dV}{dt} = 4\pi \ \text{mm}^3/\text{s}
\]

### Conclusion
The volume of the tube is increasing at a rate of \( 4\pi \) mm\(^3\) per second at this instant.

Would you like more details or have any questions?

Here are five related questions to expand on this topic:

1. How would the rate of volume change differ if the radius and height were changing at different rates?
2. What would happen to the volume if both the radius and height were decreasing?
3. How does the relationship between radius and height affect the overall rate of volume change?
4. What if the tube was not cylindrical, but another shape—how would you calculate the volume change then?
5. How does the volume change if the tube's radius decreases at an increasing rate?

**Tip:** When solving related rates problems, carefully consider the sign of the rates of change (positive for increasing, negative for decreasing) to avoid errors.

Explore how to calculate the rate of volume change for a tube with a decreasing radius and increasing height. This advanced calculus problem involves differentiating the volume formula and applying related rates. Learn the step-by-step solution and key mathematical concepts.

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