Answer
\begin{aligned} \text { (a) } f(2,1) & =5 \\ f(2.1,1.05) & =5.5125\\ \Delta z &=0.5125 \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \text {( b) }
dz &=0.5 \end{aligned}
Work Step by Step
Given $$f(x, y)=x^2+ y^2$$
so, we have
\begin{aligned} \text { (a) } f(2,1) &=2^2+1^2=5 \\ f(2.1,1.05) &=2.1^2+1.05^2=4.41+1.1025\\
&=5.5125\\
\Delta z &=f(2.1,1.05)-f(2,1)=5.5125-5\\
&=0.5125\\ \text {( b) } \text { since }d z &=\frac{\partial f}{\partial x} dx+\frac{\partial f}{\partial y} dy,\\
&\frac{\partial f}{\partial x}= 2x=2(2)=4,\\
&\frac{\partial f}{\partial y} =2y=2(1)=2,\\
& dx\approx\Delta x=2.1-2=0.1 \ \text{and} \\& dy\approx\Delta y=1.05-1=0.05 \\& \text{So we get},\\
&dz=4 d x+2 d y=4(0.1)+2(0.05) \\
&\ \ \ \ =0.4+0.1=0.5 \end{aligned}
