# Proving there exists a curve whose tangent vector $v$ satisfies $\nabla f \cdot v = 0$

Let $f:\mathbb{R}^3\to \mathbb{R}$ a $C^1$ function, $(x_0,y_0,z_0)\in \mathbb{R}^3$ such that $f(x_0,y_0,z_0)=0$ and $\nabla f(x_0,y_0,z_0)\neq 0$. Let $$S=\{(x,y,z) \ | \ f(x,y,z)=0\}$$ and $v=(v_1,v_2,v_3)\in \mathbb{R}^3$ such that $\nabla f(x_0,y_0,z_0) \cdot v =0$.

Prove that there exists $\delta>0$ and $C:(-\delta,\delta) \to \mathbb{R^3}$ of class $C^1$ such that $C(-\delta,\delta)\subseteq S$, $C(0)=(x_0,y_0,z_0)$ and $C'(0)=v$.

I'm dealing with this exercise. I've tried to use the Implicit function theorem, but I have not reached anything. Any hint you can give me would be very appreciated!

There may be a cleaner solution, but here's how I would approach the problem:

I want a curve whose tangent vector is orthogonal to $\nabla f$. As long as this is true, I know I will stay on the level set $f=0$. So I want

$$C'(t) \cdot \nabla f(C(t)) = 0$$ or $$C'(t) = \nabla f(C(t)) \times \textrm{something}$$ In space such a curve is not unique. I can pick more or less whatever function of $t$ I want for the "something", but I do need $C'(0) = v.$ Can you think of how to pick a "something" to make this true?

For a more explicit solution:

Since $\nabla f(x_0, y_0, z_0)$ and $v$ are perpendicular, $$\nabla f(x_0, y_0, z_0) \times (\nabla f(x_0, y_0, z_0) \times v) = -v \|\nabla f(x_0, y_0, z_0)\|^2$$

${}$

which suggests the natural $$C'(t) = \nabla f(C(t)) \times \frac{-(\nabla f(x_0, y_0, z_0) \times v)}{\|\nabla f(x_0, y_0, z_0)\|^2}.$$ I leave it to you to apply first-order ODE existence results to show that for the initial condition $C(0) = (x_0, y_0, z_0)$ I can integrate this ODE up over a small neighborhood of $t=0$.

• Excellent answer! Thank you very much. I gave a different solution, but yours is fascinating since it uses ODE (mine doesn't). Thanks again! – Daniel Dec 18 '14 at 18:21