Single Variable Calculus, Chapter 3, 3.8, Section 3.8, Problem 33

At what rate is $R$ changing when $R_1 = 80 \Omega$ and $R_2 = 100 \Omega$?
$\displaystyle \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}$

$
\begin{equation}
\begin{aligned}
\text{Given: } \frac{dR_1}{dt} &= 0.30 \frac{\Omega}{s} \\
\\
\frac{dR_1}{dt} &= 0.20 \frac{\Omega}{s}\\
\\

\end{aligned}
\end{equation}
$


Required: $\displaystyle \frac{dR}{dt}$ when $R_1 = 80 \Omega$ and $R_2 = 100 \Omega$



$\displaystyle \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} \qquad \Longleftarrow \text{ Equation 1}$
By getting the derivative with respect to time


$
\begin{equation}
\begin{aligned}
\frac{-\frac{dR}{dt}}{R^2} &= \frac{-\frac{dR_1}{dt}}{R_1^2} + \frac{-\frac{dR_2}{dt}}{R_2^2}\\
\\
- \frac{dR}{dt} &= R^2 \left[\frac{-\frac{dR_1}{dt}}{R_1^2} + \frac{-\frac{dR_2}{dt}}{R_2^2} \right]
\end{aligned}
\end{equation}
$


We will use Equation 1 to get the value of $R$

$
\begin{equation}
\begin{aligned}
\frac{1}{R} &= \frac{1}{80}+ \frac{1}{100}\\
\\
R &= \frac{400}{9} \Omega
\end{aligned}
\end{equation}
$


Now, to solve for the required

$
\begin{equation}
\begin{aligned}
-\frac{dR}{dt} &= \left( \frac{400}{9}\right)^2 \left[ \frac{-0.30}{(80)^2} + \frac{-0.20}{(100)^2}\right]
\end{aligned}
\end{equation}\\
\boxed{\displaystyle \frac{dR}{dt} = \frac{107}{810} \frac{\Omega}{s}}
$