Is a gradient perpendicular to the osculating plane of a regular curve?

Question

Prove the following or disprove with a counterexample:

Let $f$ be a differentiable function in an open set $U \subset \mathbb{R}^3$ and $(a, b, c)$ be a point in $U$ where the gradient of the function $f$ isn't zero. If $r : I \to U$ is a regular curve with a regular derivative on an open interval $I$, which contains the zero point, and satisfies the following conditions:

• $r(0) = (a, b, c)$
• $r(I)$ is contained in the contour surface of the function $f$ that goes through the point $(a, b, c)$

then its osculating plane in $t = 0$ is perpendicular to the gradient of the function $f$ in the point $(a, b, c)$.

Attempt at a solution: The osculating plane is given with $\{r(0) + ar^{\prime}(0) + br^{\prime}(0) : a, b \in \mathbb{R}\}$. Since $r$ is regular with with a regular derivative then $r^{\prime}\cdot r^{\prime\prime} = 0$. Since $r$ is contained in the contour surface then $\nabla f\cdot r^{\prime} = 0$.

If I can show that the gradient of $f \cdot r^{\prime\prime}$ is or isn't $0$ then this is done. I can't find any relevant theorems for this problem though.

Answer 1

It's good that you found no way to prove $\nabla f\cdot \vec r\,''=0$, because this is false. Counterexample: $f(x,y,z)=x^2+y^2+z^2$ and $\vec r(t)=(\cos t, \sin t, 1)$. Clearly, $f(\vec r)\equiv 2$. It is equally clear that the curve described by $\vec r$ lies in the plane $z=1$, which is its osculating plane. Yet, the plane $z=1$ meets the sphere $x^2+y^2+z^2=2$ at the 45 degree angle.