Geometric interpretation of ${\partial f\over \partial x}= {\partial f \over \partial y}$

Question

I know that $${\partial f\over \partial x}= {\partial f \over \partial y}$$ iff there exists a differentiable function $g$ (of one variable) such that $g(x+y)=f(x,y)$ (where $f : D\subseteq \mathbb R^2 \to \mathbb R$ and $D$ is an open set), but I don´t know what is the geometric interpretation of this fact. So I would really appreciate if you can help me with this

Answer 1

Consider $f(x,y)$ as a sort of contour map, with the height $z$ in 3 dimensions equal to $f(x,y)$. The function defines a continuous surface in that space.

Now stare at that function from far off in the "South-East" direction, looking North-West.

Then if $\frac{\partial f}{\partial x} = \frac{\partial f}{\partial y}$, the projection of that surface that you would see is just a one-dimensional curve. And the shape of that curve is $g(x+y)$

Answer 2

I would like to add to the good (old) answer by @Mark Fischler a graphical representation of the thing and, as well, the way to program it (see figure and Matlab program below)

Fig. 1 : In black the graphical representation of the "profile" function $g$ with triangles materializing a same (negative) slope at the same point, one of them being the variation wrt $x$, the other one the variation wrt to $y$... Please note that in the program below, one has to use $g(2t)$ instead of $g(t)$.

clear all;close all;hold on;
g=@(t)(t.*sin(0.1*t))
[X,Y] = meshgrid(0:0.1:100);
surf(X,Y,g(X+Y),'edgecolor','none');view([32,42]);
t=0:0.1:100;plot3(t,t,g(2*t),'k');
patch([20,20,20],[20,40,20],[-30,-30,0],'r');
patch([20,40,20],[20,20,20],[-30,-30,0],'r');