Answer
$\frac{dL}{dt}=\frac{31}{13}$
Work Step by Step
given $L=\sqrt{x^2+y^2}$,$\frac{dx}{dt}=-1$and $\frac{dy}{dt}=3$
on differentiating L with respect to t:
$\frac{dL}{dt}=\frac{1}{2}\frac{1}{\sqrt{(x^2+y^2)}}(2x\frac{dx}{dt}+2y\frac{dy}{dt})$
from the above data:
$\frac{dL}{dt}=\frac{1}{\sqrt{(x^2+y^2)}}(x(-1)+y(3))$
on putting value of $x=5$ and $y=12$
$\frac{dL}{dt}=\frac{1}{\sqrt{(5^2+12^2)}}(5(-1)+12(3))=\frac{31}{\sqrt{169}}$
$\frac{dL}{dt}=\frac{31}{13}$
