The problem involves finding the rate of change of the combined electrical resistance $R$ of two resistors $R_1$ and $R_2$ connected in parallel. The equation for the combined resistance is given as:

$\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}$

Given:

• $R_1 = 49$ ohms and $R_2 = 87$ ohms,
• $\frac{dR_1}{dt} = 0.6$ ohms per second (rate of change of $R_1$),
• $\frac{dR_2}{dt} = 1.3$ ohms per second (rate of change of $R_2$).

We need to find the rate of change of $R$, i.e., $\frac{dR}{dt}$.

Steps:

1. Differentiate the equation implicitly with respect to time $t$: $\frac{d}{dt}\left(\frac{1}{R}\right) = \frac{d}{dt}\left(\frac{1}{R_1} + \frac{1}{R_2}\right)$ Using the chain rule: $-\frac{1}{R^2} \frac{dR}{dt} = -\frac{1}{R_1^2} \frac{dR_1}{dt} - \frac{1}{R_2^2} \frac{dR_2}{dt}$

2. Solve for $\frac{dR}{dt}$: $\frac{dR}{dt} = R^2 \left( \frac{1}{R_1^2} \frac{dR_1}{dt} + \frac{1}{R_2^2} \frac{dR_2}{dt} \right)$

Now, let's calculate $R$ using the formula for parallel resistors and then substitute all known values to find $\frac{dR}{dt}$. I'll compute that now.The combined resistance $R$ is approximately $31.346$ ohms.

The rate at which $R$ is changing, $\frac{dR}{dt}$, is approximately $0.414$ ohms per second.

Let me know if you would like further details or have any questions!

Here are five related questions to explore further:

1. How would the rate of change of $R$ be affected if $R_1$ and $R_2$ were decreasing instead of increasing?
2. How is the formula for resistors in series different from the formula for resistors in parallel?
3. What happens to the overall resistance $R$ if both resistances $R_1$ and $R_2$ increase indefinitely?
4. How can the rate of change formula be extended to more than two resistors connected in parallel?
5. In practical circuits, how are changing resistances modeled in time-dependent systems?

Tip: In physics and electronics, when working with rates of change, implicit differentiation is a powerful tool for tracking how multiple variables interact over time.