## University Calculus: Early Transcendentals (3rd Edition)

$\frac{dZ}{dt}=\frac{-1}{\sqrt{5}}$
We are given: $R=10$ and $X=20$ $\frac{dR}{dt}=3$ and $\frac{dX}{dt}=2$ $Z=\sqrt{R^2+X^2}$ diffrentiating above with respect to t: $\frac{dZ}{dt}=\frac{1}{2\sqrt{(R^2+X^2)}}(2R\frac{dR}{dt}+2X\frac{dX}{dt})$ $\frac{dZ}{dt}=\frac{1}{\sqrt{(R^2+X^2)}}(R\frac{dR}{dt}+X\frac{dX}{dt})$ $\frac{dZ}{dt}=\frac{1}{\sqrt{(10^2+20^2)}}(10\times3+20\times-2)$ $\frac{dZ}{dt}=\frac{1}{10\sqrt{(5)}}(-10)$ $\frac{dZ}{dt}=\frac{-1}{\sqrt{5}}$ thus: $\frac{dZ}{dt}=\frac{-1}{\sqrt{5}}$