Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $\Gamma \subset \mathbb{R}^2$ be a curve. Define for a smooth function $f$, $$\nabla_\Gamma f = \nabla f - (\nabla f \cdot N)N$$ where $N$ is the unit normal.

Let $X:S \to \Gamma$ be a smooth regular parameterisation with $|\partial_s X(s) | > 0$.

Let $\tilde{f}(s) = f(X(s))$. How do I show that $$\nabla_\Gamma f = \frac{1}{|\partial_s X|} \partial_s \tilde{f}\frac{\partial_s X}{|\partial_s X(s)|}$$ ?

I don't know where to start. The notation is confusing..

share|cite|improve this question
This is called "covariant derivative" along the curve. If all else fails, try looking on Do Carmo's Differential geometry of curves and surfaces. – Giuseppe Negro Jun 29 '12 at 12:04
up vote 2 down vote accepted

We don't have to look far: It's the chain rule in disguise.

Let $$T:={\partial_s X\over|\partial_s X|}$$ be the unit tangent vector. Then the definition of $\nabla_\Gamma f$ amounts to $$\nabla_\Gamma f=(\nabla f\cdot T)T\ .$$ This is just two dimensional vector algebra: For any two orthogonal unit vectors $T$, $N$ and an arbitrary vector $V$ one has $V=(V\cdot T)T+(V\cdot N)N$. It follows that $$\nabla_\Gamma f=(\nabla f\cdot\partial_s X){\partial_s X\over|\partial_s X|^2}\ .$$ Now by the chain rule $\partial_s\tilde f=\nabla f\cdot\partial_s X$, and plugging this into the last formula you get the claim.

share|cite|improve this answer
Thanks a lot... – blahb Jun 29 '12 at 14:28

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.