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The problem asks to find the rate of change of the current \( I \) with respect to time, denoted \( \frac{dI}{dt} \), using the given information and Ohm's law, \( V = IR \).

### Given dataThe problem asks to find the rate of change of the current \( I \) with respect to time, denoted \( \frac{dI}{dt} \), using the given information and Ohm's law, \( V = IR \).

### Given 
- \( R = 498 \, \Omega \)
- \( I = 0.08 \, A \)
- \( \frac{dV}{dt} = -0.04 \, V/s \)
- \( \frac{dR}{dt} = 0.06 \, \Omega/s \)

### Step-by-step solution:

1. **Start with Ohm's law**: 
   \[
   V = IR
   \]

2. **Differentiate both sides with respect to time \( t \)** using the product rule:
   \[
   \frac{dV}{dt} = \frac{dI}{dt} R + I \frac{dR}{dt}
   \]
   
3. **Substitute the known values**:
   \[
   -0.04 = \frac{dI}{dt} \times 498 + 0.08 \times 0.06
   \]

4. **Simplify the equation**:
   \[
   -0.04 = 498 \frac{dI}{dt} + 0.0048
   \]

5. **Isolate \( \frac{dI}{dt} \)**:
   \[
   -0.04 - 0.0048 = 498 \frac{dI}{dt}
   \]
   \[
   -0.0448 = 498 \frac{dI}{dt}
   \]
   \[
   \frac{dI}{dt} = \frac{-0.0448}{498}
   \]

6. **Calculate \( \frac{dI}{dt} \)**:
   \[
   \frac{dI}{dt} \approx -0.00008996 \, A/s
   \]

### Final answer:
The rate of change of the current is approximately \( -0.000090 \, A/s \) (rounded to six decimal places).

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

Here are some related questions for further practice:
1. What happens to the current if \( \frac{dV}{dt} \) is positive instead?
2. How does the product rule apply to functions of multiple variables in physics problems?
3. How would the answer change if \( R \) were constant?
4. Can Ohm's law be used if the resistor is non-linear?
5. How would the problem change if both voltage and resistance were increasing?

**Tip:** When applying the product rule to physics problems involving time rates, always double-check units to ensure consistency throughout the calculation.

The voltage V in a simple electrical circuit is slowly decreasing as the battery wears out. The resistance R is slowly increasing as the resistor heats up. Use Ohm's law, V = IR, to find how the current I is changing (in A/s) at the moment when R = 498 Ω, I = 0.08 A, dV/dt = -0.04 V/s, and dR/dt = 0.06 Ω/s. (Round your answer to six decimal places.)

This problem explores the rate of change of current in an electrical circuit as voltage decreases and resistance increases. Learn how to apply Ohm's Law and the product rule to differentiate the equation V = IR. Given values include R = 498 Ω, I = 0.08 A, dV/dt = -0.04 V/s, and dR/dt = 0.06 Ω/s.

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