Flow lines of a gradient field 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?
 A: 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.
A: I'd like to share a trick to solve this problem.
Let $f:M\subset\mathbf{R}^n\longrightarrow\mathbf{R}$ be a function defined on $M$, let $x\in M, X$ be an arbitrary vector on the tangent space $T_xM$, then the gradient field will induce a vector field, namely
$$\bar{X}(x)=(df)_xX=\left<\mathrm{grad}_xf,X\right>$$
Then, let $\phi^s(x)$ be the flow line induced by $-\mathrm{grad}_xf$, i.e.
$$-\mathrm{grad}_{\phi}f=\phi'$$
Now if we take $X=-\mathrm{grad}_{\phi^s(x)}f$, by the definition of the tangent map:
$$\frac{d(f(\phi^s))}{ds}=-\Vert \mathrm{grad}_{\phi^s}f \Vert^2$$
Thus it is decreasing along the flow line.
Actually, the result holds for any Riemannian Manifold $M$.
