Question 1193543
Suppose that we have two resistors connected in parallel with resistances R1 and R2 measured in ohms (Ω). The total resistance, R, is then given by:

1/R=1/R1 + 1/R2

Suppose that R1 is increasing at a rate of 0.4Ω/min and R2 is decreasing at a rate of 0.7Ωmin. At what rate is R changing when R1 = 80Ω and R2=105Ω?
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At t = 0, R = R1*R2/(R1+R2) = 80*105/185 = 2160/37 ohms
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Call R1 a and R2 b
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1/r = 1/a + 1/b
r^-1 = a^-1 + b^-1
Differentiate
-(r^-2)dr/dt = -(a^-2)da/dt -(b^-2)db/dt
dr/dt = r^2*(da/dt/a^2 + db/dt/b^2)
dr/dt = (2160/37)^2*(0.4/6400 - 0.7/11025)
dr/dt = 0.123002191 ohms/minute
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At t=0, R = 1680/37 ohms, not 2160/37.  Oooops.
You can redo the arithmetic.