SOLUTION: Please help me with this related rates question if two resistors with resistances R1 and R2 are connected in parallel, as in the figure below, then the total resistance R, measure

Question 1156187: Please help me with this related rates question
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 = 100 Ω? (Round your answer to three decimal places.)
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
1/R=1/80+1/100; multiply by LCD of 400R and 400=5R+4R=9R, so R=400/9=44.44 ohms
R^(-1)=(R1)^(-1)+R2^(-1)
-1/R^2 dR/ds=-1/R1^2 dR1/ds-1/R2^2 dR2/ds
-1/1975.31 dR/ds=-1/6400*0.3-1/10000*0.2
so
-0.000506 dR/ds=-0.0000469-0.00002=-0.0000669. Change all signs and divide by 0.000506
dR/ds=0.132 ohms/sec.
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