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
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 = 80 Ω and R2 = 90 Ω? (Round your answer to three decimal places.)
well, just do the math
1/R = 1/R1 + 1/R2
plug in R1 and R2 to find R
1/R^2 dR/dt = 1/R1^2 dR1/dt + 1/R2^2 dR2/dt
Now just plug in your numbers to find dR/dt
To find the rate at which the total resistance R is changing, we need to differentiate the equation
1/R = 1/R1 + 1/R2
with respect to time and then solve for the rate of change of R.
Let's start by differentiating both sides of the equation:
d/dt(1/R) = d/dt(1/R1) + d/dt(1/R2)
To differentiate 1/R, we can use the chain rule:
d/dt(1/R) = -1/R^2 * dR/dt
Differentiating 1/R1 and 1/R2 with respect to time gives:
d/dt(1/R1) = -1/R1^2 * dR1/dt
d/dt(1/R2) = -1/R2^2 * dR2/dt
Now we can substitute these derivatives back into the equation:
-1/R^2 * dR/dt = -1/R1^2 * dR1/dt - 1/R2^2 * dR2/dt
To find the rate at which R is changing (dR/dt), we need to solve for it. Multiply both sides of the equation by -R^2 to isolate dR/dt:
dR/dt = -R^2 * (-1/R1^2 * dR1/dt - 1/R2^2 * dR2/dt)
Now let's plug in the given values: R1 = 80 Ω, dR1/dt = 0.3 Ω/s, R2 = 90 Ω, and dR2/dt = 0.2 Ω/s:
dR/dt = -R^2 * (-1/80^2 * 0.3 - 1/90^2 * 0.2)
To find R, we can use the equation 1/R = 1/R1 + 1/R2:
1/R = 1/80 + 1/90
1/R = (90 + 80) / (80 * 90)
1/R = 170/7200
R = 7200/170
Now we can substitute this value back into the equation for dR/dt:
dR/dt = -(7200/170)^2 * (-1/80^2 * 0.3 - 1/90^2 * 0.2)
Calculating this expression will give us the rate at which R is changing when R1 = 80 Ω and R2 = 90 Ω.