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 = 100 Ω and R2 = 110 Ω?

Answer 1

Since the connection is in parallel, i presume that it is given by the equation:
\[\frac{ 1 }{ R } = \frac{ 1 }{ R1 } + \frac{ 1 }{ R2 }\]

Answer 2

oh yea, that's what i meant

Answer 3

since this involves the change in total resistance per unit of time then you'll have it as:
\[\frac{ 1 }{ Rt } = \frac{ 1 }{ R_1 } + \frac{ 1 }{ R_2 }\]
expressing it as numerators
\[Rt^{-1} = R_{1} ^{-1} + R_{2} ^{-1}\]
taking the derivative of the equation you'll have:
\[-1(R ^{-2})\frac{ dR }{ dt } = -1(R _{1}^{-2})\frac{ dR _{1} }{ dt } + (-1)(R_{2}^{-2})\frac{ dR _{2} }{ dt }\]
dividing the whole equation with -1
\[R ^{-2}\frac{ dR }{ dt } = R _{1}^{-2}\frac{ dR _{1} }{ dt } + R_{2}^{-2}\frac{ dR _{2} }{ dt }\]
find R using the formula given and you are now ready to substitute the values and you will get dR/dt