University Calculus: Early Transcendentals (3rd Edition)

$\frac{dR}{dt}=\frac{1}{50}$
We are given: $R_{1}=75$ and $R_{2}=50$ $\frac{dR_{1}}{dt}=-1$ and $\frac{dR_{2}}{dt}=0.5$ $\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}$ diffrentiating above with respect to t: $\frac{1}{R^2}\frac{dR}{dt}=\frac{1}{R_{1}^2}\frac{dR_{1}}{dt}+\frac{1}{R_{2}^2}\frac{dR_{2}}{dt}$ $(\frac{1}{R_{1}}+\frac{1}{R_{2}})^2\frac{dR}{dt}=\frac{1}{R_{1}^2}\frac{dR_{1}}{dt}+\frac{1}{R_{2}^2}\frac{dR_{2}}{dt}$ on putting in the values: $(\frac{1}{75}+\frac{1}{50})^2\frac{dR}{dt}=\frac{1}{75^2}(-1)+\frac{1}{50^2}(0.5)$ $(\frac{5}{150})^2\frac{dR}{dt}=\frac{1}{150^2}(0.5)$ $\frac{dR}{dt}=\frac{1}{50}$ thus,$\frac{dR}{dt}=\frac{1}{50}$