Answer
$-\frac{\sqrt 5}{5}\approx-0.45$ ohms/sec.
Work Step by Step
Step 1. Identify the given conditions: $Z=\sqrt {R^2+X^2}, \frac{dR}{dt}=3ohms/sec, \frac{dX}{dt}=-2ohms/sec, R=10ohms, X=20ohms$
Step 2. Calculate: $Z=\sqrt {R^2+X^2}=\sqrt {10^2+20^2}=\sqrt {500}=10\sqrt 5$
Step 3. Take derivatives: $Z^2=R^2+X^2$, $2ZZ'=2RR'+2XX'$, thus $\frac{dZ}{dt}=Z'=\frac{RR'+XX'}{Z}=\frac{10(3)+20(-2)}{10\sqrt 5}=\frac{-10}{10\sqrt 5}=-\frac{\sqrt 5}{5}\approx-0.45$ ohms/sec.
