Answer
$R$ is increasing at a rate of $~~0.132~\Omega/s$
Work Step by Step
We can find the value of $R$:
$\frac{1}{R} = \frac{1}{R_1}+\frac{1}{R_2}$
$\frac{1}{R} = \frac{1}{80~\Omega}+\frac{1}{100~\Omega}$
$\frac{1}{R} = \frac{5}{400~\Omega}+\frac{4}{400~\Omega}$
$\frac{1}{R} = \frac{9}{400~\Omega}$
$R = \frac{400}{9}~\Omega$
We can find $\frac{dR}{dt}$:
$-\frac{1}{R^2}~\frac{dR}{dt} = -\frac{1}{R_1^2}~\frac{dR_1}{dt}-\frac{1}{R_2^2}~\frac{dR_2}{dt}$
$\frac{dR}{dt} = R^2~(\frac{1}{R_1^2}~\frac{dR_1}{dt}+\frac{1}{R_2^2}~\frac{dR_2}{dt})$
$\frac{dR}{dt} = (\frac{400}{9})^2~[(\frac{1}{80^2})~(0.3)+(\frac{1}{100^2})~(0.2)]$
$\frac{dR}{dt} = 0.132~\Omega/s$
$R$ is increasing at a rate of $~~0.132~\Omega/s$
