Gradients of two functions with the same level sets are parallel for all points in the intersection of the level sets

Question

Is the intuition behind this statement, based on : 1. The definition that a gradient is perpendicular to the level curves 2. Since the level sets are the same for both functions, the corresponding gradients are parallel to each other?

Answer 1

No this is not quite the case although it is tempting to believe. One such counter example is as follows:

Let $f(x,y,z)=x$ and $g(x,y,z)=z$. Then lets consider $f=1$ and $g=1$.

• The level surface for $f$, $x=1$, gives us a vertical plane with a constant gradient vector $<1,0,0>$.
• The level surface for $g$, $z=1$, gives us a horizontal plane with a constant gradient vector $<0,0,1>$.

We know that these two level surfaces intersect at a line, however their gradients are never parallel.

Answer 2

If we assume the level set are regular. That means, we can parametrize the level set as $\vec{x}(s_1, s_2, ..., s_{n-1})$. Also if we assume two functions having the same level set means if $f = g = t$, then we can use the same paramertization to represent them $$\vec{x}(s_1, s_2, ..., s_{n-1}, t)$$ such that $$f(\vec{x}(s_1, s_2, ..., s_{n-1}, t)) = g(\vec{x}(s_1, s_2, ..., s_{n-1}, t))= t$$ Then we have $$\nabla f \cdot \frac{\partial \vec{x}}{\partial t} = 1$$ $$\nabla f \cdot \frac{\partial \vec{x}}{\partial s_i} = 0$$

$$\nabla g \cdot \frac{\partial \vec{x}}{\partial t} = 1$$ $$\nabla g \cdot \frac{\partial \vec{x}}{\partial s_i} = 0$$

As long as the level is not degerated, $( \frac{\partial \vec{x}}{\partial t},\frac{\partial \vec{x}}{\partial s_1}, ..., \frac{\partial \vec{x}}{\partial s_{n-1}} )$ is invertible, it is true under my assumptions of the OP. If the level set is degenerated, then it could be wrong.