I have an integral surface $z = z(x, y)$.
Writing this integral surface in implicit form, we get
$$F(x, y, z) = z(x, y) - z = 0$$
I am then told that the gradient vector $\nabla F = (z_x, z_y, -1)$ is normal to the integral surface $F(x, y, z) = 0$.
First of all, how was this calculated? I understand how the gradient is calculated, but I don't understand how it was calculated in this case?
And lastly, where did the $-1$ come from and why? Couldn't they also have had $\nabla F = (z_x, z_y, 1)$, where this would just be the normal vector in the other direction? Why and how did they pick the $-1$ direction instead?
I apologise. My vector calculus understanding is not particularly strong, and I strive to improve it.
Thank you for any help.
maybe its helpful to give the surface equation a different symbol $$ S = \{ (x,y,z) : z=Z(x,y) \}$$ then with $F(x,y,z):= Z(x,y) - z$,
$$∇ F (x,y,z) = \begin{pmatrix}\partial_x( Z(x,y) - z)\\\partial_y( Z(x,y) - z)\\\partial_z( Z(x,y) - z)\end{pmatrix}= \begin{pmatrix}\partial_x Z(x,y)\\\partial_yZ(x,y)\\\ - 1\end{pmatrix}$$ The $-1$ came from the definition of $F$. If you forced $+1$ instead of $-1$ you would be talking about $Z(x,y) + z$ which is not related to $S = \{F = 0 \}$.
If you take any curve $\mathbf x=\mathbf x(t)$ such that $\mathbf x(t) ∈ S$ for every $t$, then
$$ F(\mathbf x(t)) = 0 $$ taking derivatives, $$∇ F(\mathbf x)· \mathbf x' = 0$$ but the collection of all such $\mathbf x'$s as you consider different curves $\mathbf x$ form the tangent vectors at each point of $S$. Therefore, $∇ F$ is perpendicular to all tangent vectors of $S$ at that point. In dimension 3, there are 2 linearly independent tangent vectors and you get a unique normal vector (up to scaling and sign).