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### Problem Recap
- **Given:**
  - The height \( h \) of the funnel ranges from \( 0 \) to \( 10 \) inches.
  - The radius \( r \) at a height \( h \) is given by the function:
    \[
    r = \frac{1}{20}(3 + h^2)
    \]
  - At a specific instance, when \( h = 3 \) inches, the radius \( r \) is decreasing at the rate of \( \frac{1}{5} \) inches per second. 
- **Find:**
  - The rate of change of the height \( \frac{dh}{dt} \) at that instance.

### Solution

1. **Differentiate \( r(h) \) with respect to \( h \):**
   \[
   r = \frac{1}{20}(3 + h^2)
   \]
   Differentiating both sides with respect to \( h \), we get:
   \[
   \frac{dr}{dh} = \frac{1}{20} \cdot 2h = \frac{h}{10}
   \]

2. **Relate \( \frac{dr}{dt} \) and \( \frac{dh}{dt} \):**
   Using the chain rule:
   \[
   \frac{dr}{dt} = \frac{dr}{dh} \cdot \frac{dh}{dt}
   \]
   Substituting \( \frac{dr}{dh} = \frac{h}{10} \):
   \[
   \frac{dr}{dt} = \frac{h}{10} \cdot \frac{dh}{dt}
   \]

3. **Plug in the given values:**
   - At \( h = 3 \) inches, \( \frac{dr}{dt} = -\frac{1}{5} \) inches per second (since the radius is decreasing, we take the negative sign).
   \[
   -\frac{1}{5} = \frac{3}{10} \cdot \frac{dh}{dt}
   \]

4. **Solve for \( \frac{dh}{dt} \):**
   \[
   \frac{dh}{dt} = -\frac{1}{5} \times \frac{10}{3} = -\frac{2}{3} \text{ inches per second}
   \]

### Answer
The rate of change of the height \( \frac{dh}{dt} \) when \( h = 3 \) inches is \( -\frac{2}{3} \) inches per second.

Would you like more details or have any further questions? Here are some related questions:

1. How does the rate of change of height vary with different values of \( h \)?
2. What would the rate of change of height be if the radius decreased at a different rate?
3. How would the problem change if the funnel's shape were different?
4. Can you derive a general formula for \( \frac{dh}{dt} \) in terms of \( h \) and \( \frac{dr}{dt} \)?
5. What is the significance of the negative sign in \( \frac{dh}{dt} \)?

**Tip:** When dealing with related rates problems, always carefully apply the chain rule, keeping track of how variables change with respect to time.

At the instant when the height of the liquid is h = 3 inches, the radius of the surface of the liquid is decreasing at a rate of 1/5 inch per second. What is the rate of change of the height of the liquid with respect to time?

This problem involves calculating the rate of change of the height of a liquid in a funnel using related rates and the chain rule. Given the radius function r = (1/20)(3 + h^2) and the rate of change of the radius, we determine the rate at which the height changes when h = 3 inches.

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