Norm of the gradient of function $f$ on manifold $g(x)=c$

Question

Let $g,f:\mathbb{R}^2 \to \mathbb{R}$, $M=g^{-1}(c)$. Let say that we manage to write $f(x,y)=f_{*}(x)$ for $x\in M$. When I was calculating the square of norm of $$\nabla f_M (x,y)=\nabla f(x,y)-\frac{\langle\nabla f(x,y),\nabla g(x,y)\rangle}{\|\nabla g(x,y)\|^{2}}\nabla g(x,y)$$ I get $\frac{(f_x g_y-f_y g_x)^2}{\|\nabla g\|^2}$, which is $\frac{{f_* }^2}{1+(\frac{g_x}{g_y})^2}$. I was expecting simply ${{f_* }^2}$. What is the geometrical meaning of $\frac{1}{1+(\frac{g_x}{g_y})^2}$? What kind of result can I expect in higher dimensions?

Answer 1

The result $$\|\nabla_Mf\|^2 = \frac{(f_x g_y-f_y g_x)^2}{\|\nabla g\|^2}$$ is correct and geometrically natural. Indeed, $(g_y,-g_x)$ is the vector $\nabla g$ rotated by $90$ degrees, that is, a tangent vector to the level set. The formula expresses the projection of $\nabla f$ onto the tangent space, which is what $\nabla_M f$ is.

In higher dimensions, it is probably easier to find $\|\nabla_M f\|^2$ by subtracting the square of the normal component from $\|\nabla f\|^2$. That is, $$\|\nabla_Mf\|^2 = \|\nabla f\|^2 - \frac{|\langle\nabla f,\nabla g\rangle|^2}{\|\nabla g\|^2}$$

The problem with $\nabla f_*$ is that it's coordinate-dependent unlike $\nabla f$ and $\nabla _M f$ themselves. So the formula in terms of $\nabla f_*$ is a bit ugly.