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To solve this problem, we’ll differentiate the formula for the total resistance \( R \) with respect to time \( t \) and then substitute the values given for \( R_1 \), \( R_2 \), and their rates of change.

### Problem Analysis
1. The formula for total resistance \( R \) in parallel is:
   \[
   \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}
   \]
   
2. Given:
   - \( \frac{dR_1}{dt} = 0.3 \, \Omega/s \)
   - \( \frac{dR_2}{dt} = 0.2 \, \Omega/s \)
   - \( R_1 = 80 \, \Omega \)
   - \( R_2 = 100 \, \Omega \)

We need to find \( \frac{dR}{dt} \), the rate at which \( R \) is changing.

### Step-by-Step Solution

1. **Differentiate both sides with respect to \( t \):**
   \[
   \frac{d}{dt} \left( \frac{1}{R} \right) = \frac{d}{dt} \left( \frac{1}{R_1} + \frac{1}{R_2} \right)
   \]
   
2. Using the chain rule:
   \[
   -\frac{1}{R^2} \frac{dR}{dt} = -\frac{1}{R_1^2} \frac{dR_1}{dt} - \frac{1}{R_2^2} \frac{dR_2}{dt}
   \]
   
3. Solving for \( \frac{dR}{dt} \):
   \[
   \frac{dR}{dt} = R^2 \left( \frac{1}{R_1^2} \frac{dR_1}{dt} + \frac{1}{R_2^2} \frac{dR_2}{dt} \right)
   \]

4. **Calculate \( R \) when \( R_1 = 80 \) and \( R_2 = 100 \):**
   \[
   \frac{1}{R} = \frac{1}{80} + \frac{1}{100} = \frac{5 + 4}{400} = \frac{9}{400}
   \]
   So,
   \[
   R = \frac{400}{9} \approx 44.44 \, \Omega
   \]

5. **Substitute values into the derivative expression:**
   - \( R = \frac{400}{9} \)
   - \( R_1 = 80 \), \( R_2 = 100 \)
   - \( \frac{dR_1}{dt} = 0.3 \), \( \frac{dR_2}{dt} = 0.2 \)

   \[
   \frac{dR}{dt} = \left( \frac{400}{9} \right)^2 \left( \frac{1}{80^2} \cdot 0.3 + \frac{1}{100^2} \cdot 0.2 \right)
   \]

6. **Calculate each term:**
   - \( \left( \frac{400}{9} \right)^2 = \frac{160000}{81} \)
   - \( \frac{1}{80^2} = \frac{1}{6400} \)
   - \( \frac{1}{100^2} = \frac{1}{10000} \)

   So,
   \[
   \frac{dR}{dt} = \frac{160000}{81} \left( \frac{0.3}{6400} + \frac{0.2}{10000} \right)
   \]

7. **Evaluate inside the parentheses:**
   \[
   \frac{0.3}{6400} = 4.6875 \times 10^{-5}
   \]
   \[
   \frac{0.2}{10000} = 2 \times 10^{-5}
   \]
   \[
   \frac{0.3}{6400} + \frac{0.2}{10000} = 6.6875 \times 10^{-5}
   \]

8. **Final calculation:**
   \[
   \frac{dR}{dt} = \frac{160000}{81} \times 6.6875 \times 10^{-5}
   \]
   \[
   \frac{dR}{dt} \approx 0.132 \, \Omega/s
   \]

### Answer:
The rate at which \( R \) is increasing is approximately \( 0.132 \, \Omega/s \).

Would you like a more detailed breakdown of any step, or have any questions?

### Related Questions:
1. How would the rate of change of \( R \) vary if \( R_1 \) and \( R_2 \) were decreasing?
2. What happens to the total resistance \( R \) if \( R_1 \) and \( R_2 \) approach infinity?
3. How would this calculation differ if there were three resistors in parallel?
4. Can this method be applied to resistors in series? Why or why not?
5. What would the result be if \( R_1 \) and \( R_2 \) were equal?

#### Tip:
In problems with implicit differentiation, be careful with signs when differentiating expressions with \( \frac{1}{R} \) or similar inverse terms.

If two resistors with resistances R1 and R2 are connected in parallel, as in the figure, then the total resistance R, measured in Ohms (Ω), is given by: 1/R = 1/R1 + 1/R2. If R1 and R2 are increasing at rates of 0.3 Ω/s and 0.2 Ω/s, respectively, how fast is R increasing when R1 = 80 Ω and R2 = 100 Ω?

Learn how to calculate the rate at which the total resistance of two resistors in parallel increases when their individual resistances are changing. This example covers differentiation and uses the parallel resistance formula, suitable for physics or engineering students.

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