Chapter 13 - Section 13.6 - Tangent Planes and Differentials - Exercises - Page 728: 21

$0$

Here, $u=\dfrac{v}{|v|}=\dfrac{\lt 2,2,-2 \gt}{\sqrt{2^2+2^2+(-2)^2}}$ or, $u=\lt \dfrac{1}{\sqrt 3},\dfrac{1}{\sqrt 3},\dfrac{-1}{\sqrt 3} \gt$ Now $dg=(\nabla g \cdot u) ds$ or, $=[(\lt0\gt)(\lt \dfrac{1}{\sqrt 3},\dfrac{1}{\sqrt 3},\dfrac{-1}{\sqrt 3} \gt)](0.2)$ or, $=0$