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### Problem Recap:
Given a funnel of height 10 inches, where the radius \( r \) at a height \( h \) is given by:
\[
r = \frac{1}{20}(3 + h^2)
\]
We need to determine the rate of change of the height \( h \) of the liquid with respect to time \( t \), at the instant when \( h = 3 \) inches, given that the radius \( r \) is decreasing at the rate of \(\frac{1}{5}\) inch per second.

### Solution Approach:
We need to find \(\frac{dh}{dt}\) given \(\frac{dr}{dt} = -\frac{1}{5}\) when \( h = 3 \).

1. **Given relationship:**
   \[
   r = \frac{1}{20}(3 + h^2)
   \]

2. **Differentiate \( r \) with respect to time \( t \):**
   \[
   \frac{dr}{dt} = \frac{d}{dt}\left(\frac{1}{20}(3 + h^2)\right)
   \]
   \[
   \frac{dr}{dt} = \frac{1}{20} \cdot 2h \cdot \frac{dh}{dt} = \frac{h}{10} \cdot \frac{dh}{dt}
   \]

3. **Substitute the given values:**
   When \( h = 3 \) inches, and \( \frac{dr}{dt} = -\frac{1}{5} \) inch/second:
   \[
   -\frac{1}{5} = \frac{3}{10} \cdot \frac{dh}{dt}
   \]

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

### Final Answer:
The rate of change of the height \( h \) of the liquid with respect to time \( t \) at the instant when \( h = 3 \) inches is \(-\frac{2}{3}\) inches per second.

Would you like further details on any part of the solution? 

Here are some related questions to expand on the topic:

1. How would the rate of change of the height change if the funnel had a different height?
2. What if the radius decreased at a different rate, say \(-\frac{1}{10}\) inch per second?
3. How does the equation for \( r \) change if the funnel's cross-sectional shape is altered?
4. Can you derive the volume of the liquid as a function of height and time?
5. What assumptions are made about the shape of the funnel in this problem?

**Tip:** When working with related rates problems, always carefully differentiate each variable with respect to time, keeping track of how each one changes!

Learn how to solve a related rates problem involving a funnel's height and radius. This problem covers the calculation of the rate of change of the height of liquid with respect to time, given a decreasing radius. Suitable for advanced high school students.

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