Answer
$\triangle z=0.9225$ and $dz=0.9$
Work Step by Step
Given the function $z=5x^2+y^2$
The differential form can be evaluated as follows:
$dz=10x dx+2y dy$
Here, $\triangle x=1.05-1=0.05$ and $\triangle y=2.1-2=0.1$
Now, we have
$dz=10 \cdot (1) (0.05)+2 \cdot (2) (0.1)=0.9$
Thus,
$\triangle z=f(1.05,2.1)-f(1, 2)=5x^2+y^2=5(1.05)^2+(2.1)^2-(5+4)$
or, $\triangle z=0.9225$
Hence, $\triangle z=0.9225$ and $dz=0.9$
