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We are given the formula for the total resistance \( R \) when two resistors \( R_1 \) and \( R_2 \) are connected in parallel:

\[
\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}
\]

The problem also asks us to find how fast \( R \) is changing when \( R_1 \) and \( R_2 \) are increasing at rates of \( 0.3 \, \Omega/s \) and \( 0.2 \, \Omega/s \), respectively. We are given that \( R_1 = 90 \, \Omega \) and \( R_2 = 100 \, \Omega \).

### Step 1: Differentiate the formula

We need to find \( \frac{dR}{dt} \), the rate of change of \( R \), using implicit differentiation. First, differentiate both sides of the equation:

\[
\frac{d}{dt} \left( \frac{1}{R} \right) = \frac{d}{dt} \left( \frac{1}{R_1} + \frac{1}{R_2} \right)
\]

Using the chain rule, this becomes:

\[
-\frac{1}{R^2} \cdot \frac{dR}{dt} = -\frac{1}{R_1^2} \cdot \frac{dR_1}{dt} - \frac{1}{R_2^2} \cdot \frac{dR_2}{dt}
\]

### Step 2: Substitute known values

We know that:

- \( \frac{dR_1}{dt} = 0.3 \, \Omega/s \)
- \( \frac{dR_2}{dt} = 0.2 \, \Omega/s \)
- \( R_1 = 90 \, \Omega \)
- \( R_2 = 100 \, \Omega \)

First, calculate \( R \) using the given values for \( R_1 \) and \( R_2 \):

\[
\frac{1}{R} = \frac{1}{90} + \frac{1}{100}
\]

Finding the common denominator:

\[
\frac{1}{R} = \frac{100 + 90}{9000} = \frac{190}{9000}
\]

So:

\[
R = \frac{9000}{190} \approx 47.368 \, \Omega
\]

### Step 3: Solve for \( \frac{dR}{dt} \)

Now, substitute these values into the differentiated equation:

\[
-\frac{1}{(47.368)^2} \cdot \frac{dR}{dt} = -\frac{1}{90^2} \cdot 0.3 - \frac{1}{100^2} \cdot 0.2
\]

Simplifying each term:

\[
-\frac{1}{(47.368)^2} \cdot \frac{dR}{dt} = -\frac{1}{8100} \cdot 0.3 - \frac{1}{10000} \cdot 0.2
\]

\[
-\frac{1}{(47.368)^2} \cdot \frac{dR}{dt} = -0.00003704 - 0.00002
\]

\[
-\frac{1}{(47.368)^2} \cdot \frac{dR}{dt} = -0.00005704
\]

Finally, solving for \( \frac{dR}{dt} \):

\[
\frac{dR}{dt} = \frac{0.00005704 \times (47.368)^2}{1}
\]

Calculating \( (47.368)^2 \approx 2243.79 \):

\[
\frac{dR}{dt} = 0.00005704 \times 2243.79 \approx 0.128 \, \Omega/s
\]

### Final Answer:
The rate at which \( R \) is changing is approximately \( \boxed{0.128} \, \Omega/s \).

---

Would you like more details or have any questions?

Here are 5 related questions:
1. How does the resistance change if only one of the resistors is increasing?
2. How does the formula for total resistance change if the resistors were in series instead of parallel?
3. What happens to the total resistance when \( R_1 \) and \( R_2 \) are both equal?
4. Can you derive a formula for the rate of change of \( R \) as a function of \( R_1 \), \( R_2 \), and their rates of change?
5. What would be the effect of decreasing \( R_1 \) or \( R_2 \) on the total resistance?

**Tip**: Implicit differentiation is a powerful tool when dealing with relationships between multiple changing variables!

If two resistors with resistances R₁ and R₂ are connected in parallel, the total resistance R is given by the formula 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 changing when R₁ = 90 Ω and R₂ = 100 Ω?

Learn how to calculate the rate of change in total resistance when two resistors in parallel, R₁ and R₂, are increasing at different rates. In this example, R₁ = 90 Ω and R₂ = 100 Ω, with rates of 0.3 Ω/s and 0.2 Ω/s. Step-by-step solution using implicit differentiation.

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