If two resistors with resistances R1 and R2 are connected in parallel, as in the figure below, then the total resistance R, measured in ohms (Ω), is given by 1R = 1R1 + 1R2.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 = 110 Ω? (Round your answer to three decimal places.)

To solve this problem, we need to use the concept of related rates in calculus.

Step 1: Write down the given equation and differentiate it with respect to time (t).

The given equation is 1/R = 1/R1 + 1/R2.

Differentiating both sides with respect to time, we get -1/R^2 * dR/dt = -1/R1^2 * dR1/dt - 1/R2^2 * dR2/dt.

Step 2: Substitute the given values into the differentiated equation.

We know that R1 = 90 Ω, R2 = 110 Ω, dR1/dt = 0.3 Ω/s, and dR2/dt = 0.2 Ω/s.

We also need to find the value of R when R1 = 90 Ω and R2 = 110 Ω. We can find this by substituting R1 and R2 into the original equation: 1/R = 1/90 + 1/110. Solving this equation gives R ≈ 47.297 Ω.

Substituting all these values into the differentiated equation, we get -1/(47.297)^2 * dR/dt = -1/(90)^2 * 0.3 - 1/(110)^2 * 0.2.

Step 3: Solve for dR/dt.

Solving the above equation for dR/dt gives dR/dt ≈ -0.001 Ω/s.

So, the total resistance R is decreasing at a rate of approximately 0.001 Ω/s when R1 = 90 Ω and R2 = 110 Ω.