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.