SOLUTION: If two resistors with resistance R1 and R2 are connected in parallel, the total resistance R in ohms is given by 1/R = 1/R1 + 1/R2. If R1 and R2 are increasing at 0.4 ohms/s and 0.

Question 1117081: If two resistors with resistance R1 and R2 are connected in parallel, the total resistance R in ohms is given by 1/R = 1/R1 + 1/R2. If R1 and R2 are increasing at 0.4 ohms/s and 0.25 ohms/s, respectively, how fast is R changing when R1 = 600 ohms and R2 = 400 ohms?
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
If two resistors with resistance R1 and R2 are connected in parallel, the total resistance R in ohms is given by 1/R = 1/R1 + 1/R2. If R1 and R2 are increasing at 0.4 ohms/s and 0.25 ohms/s, respectively, how fast is R changing when R1 = 600 ohms and R2 = 400 ohms?
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I'll use x or R1 and y for R2 for clarity and to save typing.
1/R = 1/x + 1/y
R^-1 = x^-1 + y^-1
R = 600*400/1000 = 240 ohms
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Differentiate implicitly
-R^-2*dR/dt = -x^-2*dx/dt - y^-2*dy/dt
(dR/dt)/R^2 = (dx/dt)/x^2 + (dy/dt)/y^2
(dR/dt)/240^2 = 0.4/600^2 + 0.25/400^2
dR/dt = 0.4*240^2/600^2 + 0.25*240^2/400^2
dR/dt = 0.4*2^2/5^2 + 0.25*6^2/10^2
dR/dt = 0.4*4/25 + 0.25*36/100
dR/dt = 6.4/100 + 9/100 = 15.4/100
dR/dt = 0.154 ohms/sec
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I've done resistor problems since the 1950's, but have never seen one like this.

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