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

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.

## 1 Answer

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.