Let $\mathbf c(t)$ be a flow line of a gradient field $\mathbf F = - \nabla V$. Prove that $V(\mathbf c(t))$ is a decreasing function of t.

Not sure where to begin here, although it might have to do with the gradient chain rule?

My attempt:

$$\mathbf c'(t) = \mathbf F(\mathbf c(t)) = -\nabla V(\mathbf c(t))$$

So for $\mathbf c'(t) > 0, \nabla V(\mathbf c(t)) < 0$ indicating that $V$ is decreasing. Is that right?

  • $\begingroup$ What does this "$\mathbf c'(t) > 0, \nabla V < 0$" mean? $\endgroup$ – Raskolnikov Mar 7 '13 at 22:09
  • $\begingroup$ I edited my post, is it clearer now? $\endgroup$ – Sean Haugh Mar 7 '13 at 22:13
  • 1
    $\begingroup$ No. Let me be more clear: an inequality for vectors does not make any sense. There's no "natural" order on a vector space like there is on the reals. $\endgroup$ – Raskolnikov Mar 8 '13 at 22:19

I would start by picking $t_1$ and $t_2$ which define a segment of curve $c_{12}$ which starts at $c(t_1)$ and ends at $c(t_2)$. The integral

$$ \int_{c_{12}} \textbf{F}\cdot dr$$

will tell you the change in the function $-V(c(t)$ from $t_1$ to $t_2$. Prove that this is always positive, and you will prove that $V(c(t))$ is decreasing in t.

Specifically, when you expand the integral, you will be able to find the directional derivative of $-V$ in the direction of $c'(t)$. Argue that this must always be positive, so your integral must be positive.

If this isn't enough help, let me know in comments and I can expand my answer. I don't want to give it away if you are close.

  • $\begingroup$ Why does the integral tell you the change in the function $-V(c(t))$ and not $V(c(t))$? $\endgroup$ – Joe Z. Feb 26 '14 at 6:27
  • $\begingroup$ (I'm doing this exact homework question for calculus and while your hint helped quite a lot, I'm stuck on just that last portion.) $\endgroup$ – Joe Z. Feb 26 '14 at 13:40

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