Answer
Thus, $\sqrt 4.96^{2}+12.06^{2} \approx 13 + 0.04= 13.04$
Calculator gives $\sqrt 4.96^{2}+12.06^{2}\approx13.0401$
The error is approximately 0.0001
Work Step by Step
We are given: $\sqrt 4.96^{2}+12.06^{2}$
Let $f(x,y)=\sqrt x^{2}+y^{2}$ with $x=5, dx=-0.04, y=12, dy=0.06$
then use dz to approximate $\Delta z$
$dz=f_{x}(x,y).dx+d_{y}(x,y).dy$
$dz=(\frac{1}{2\sqrt x^{2}+y^{2}}.2x)dx+(\frac{1}{2\sqrt x^{2}+y^{2}}.2y)dy$
$dz=(\frac{x}{\sqrt x^{2}+y^{2}})dx+(\frac{y}{\sqrt x^{2}+y^{2}})dy$
$dz=\frac{5}{13}(-0.04)+\frac{12}{13}(0.06)$
$dz=0.04$
Thus, $\sqrt 4.96^{2}+12.06^{2} \approx 13 + 0.04= 13.04$
Calculator gives $\sqrt 4.96^{2}+12.06^{2}\approx13.0401$
The error is approximately 0.0001
