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

Question

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 = 110 Ω?

Answer 1

R^-1 = R1^-1 + R2^-1
-R^-2 dR/dt = -R1^-2 dR1/dt - R2^-2 dR2/dt

** 1/R^2 dR/dt = 1/R1^2 dR1/dt + 1/R2^2 dR2/dt **

when R1 = 80, R2 = 110
1/R = 1/80+1/110 = 19/880
R = 880/19

now you have the values in ** , and you can solve for dR/dt

Answer 2

this is very good. thanks for help