# Gradient vector $\nabla F = (z_x, z_y, -1)$ is normal to the integral surface?

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).

• Ahh, you used different variables $z$ and $Z$, so it makes sense now. But the author used $z$ for both, but treated $z(x, y)$ as a function of 2 variables and $z$ as a constant in the same expression, which makes it confusing, because the author also just said that $z = z(x, y)$. Seems paradoxical? As if $z$ is both a function of $x$ and $y$ and a constant! – Wyuw Jul 29 '18 at 10:23
• @Wyuw yes well, this is the way of things unfortunately and you may need to get used to it – Calvin Khor Jul 29 '18 at 10:24
• Yes, thank you for your help! – Wyuw Jul 29 '18 at 10:25
• @Wyuw You're welcome. good luck :) – Calvin Khor Jul 29 '18 at 10:26

$$z=x^2+y^2$$ $$F(x,y,z)=x^2+y^2-z$$ $$\nabla F = (z_x, z_y, -1)=< 2x,2y,-1>$$
If a point is given, for example $P(1,2,5)$ Then at that point you have two normal vector to the surface.
Upward normal $$< -2x,-2y,1> = <-2,-4,1>$$ Downward normal $$< 2x,2y,-1> = <2,4,-1>$$
• Ahh, thank you for this. I understand your example, but in the case of $F(x, y, z) = z(x, y) - z = 0$, we have that $z = z(x, y)$, so it is not a constant? So how can we say that $\nabla F = (z_x, z_y, -1)$? That is why I am confused about where the $-1$ comes from? $z$ is still a function of $x$ and $y$, no? But it is being treated as a constant here, and then it is treated as a variable of $x$ and $y$ in the same expression when they calculate $z_x$ and $z_y$! – Wyuw Jul 29 '18 at 10:19
• $z=z(x,y)$is the explicit equation of the surface surface and $F(x,y,z)=z(x,y,)-z =0$ is the equation of the same surface implicitly. The vector of partial derivatives of $F(x,y,z)$ gets its $-1$ from the $-z$ in $z(x,y,)-z$ – Mohammad Riazi-Kermani Jul 29 '18 at 10:27