Question

If L=√(x^2+y^2), dx/dt =-4, dy/dt=3, find dL/dt when x=4 and y=3

Answer 1

Hello,

$L=\sqrt{x^2+y^2} \\\\\dfrac{dx}{dt}=-4\\\\\dfrac{dy}{dt}=3\\\\x=4, y=3\\\\\dfrac{\partial L} {\partial x} =\dfrac{x}{\sqrt{x^2+y^2}} \\\\\dfrac{\partial L} {\partial y} =\dfrac{y}{\sqrt{x^2+y^2}} \\\\\dfrac{dL}{dt} =\dfrac{\partial L} {\partial x}*\dfrac{dx}{dt}+\dfrac{\partial L} {\partial y}*\dfrac{dy}{dt}\\\\=-4*\frac{-4}{\sqrt{4^2+3^2}} +3*\frac{3}{\sqrt{4^2+3^2}}\\\\=\dfrac{16}{5} +\dfrac{9}{5} \\\\=5$