Vectors of a gradient field

Question

The scalar field function s is defined by s = (x + y + z)².

How do I show that the vectors of the associated gradient field are all parallel, have different lengths and do not always point in the same direction?

Answer 1

First note $\big(\vec{\nabla} s\big)(x,y,z)=2(x+y+z)\Big<1,1,1\Big>$ which means $\big(\vec{\nabla}s\big)(x,y,z)$ is parallel to the vector $\big<1,1,1\big>$ for every $(x,y,z)\in \mathbb{R}^3$.

The length of $\vec{\nabla}s$ at the point $(x,y,z)$ is $\|\vec{\nabla}s\|=2\sqrt{3}\cdot |x+y+z|$ which you should also note ultimately depends on your argument $(x,y,z)$.

Lastly, $\big(\vec{\nabla}s\big)(0,0,1/2)=\big<1,1,1\big>$ whereas $\big(\vec{\nabla}s\big)(0,0,-1/2)=-\big<1,1,1\big>$ so these vectors don't all point in the same direction.