# Flow line in conservative vector field

## Homework Statement

Recall that a flow line, c(t), of a vector field F has c'(t)=F(c(t)) at all times t. Show all work below.
a.) Let c(t) be the flow line of a particle moving in a conservative force field F=-grad(f), where f:R^3->R, f(x,y,z) >=0 for all (x,y,z), represents the potential energy at each point in space. Prove that the particle will always move towards a point with lower potential energy. What is the limit of F(c(t)) as t goes to infinity?

c'(t)=F(c(t))

## The Attempt at a Solution

We are trying to show that f(c(t)) is a decreasing function of t. So far, I have that

But I am not sure if this really helps at all. Is this the right direction to go in, or am I completely off base?

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Zondrina
Homework Helper
Recall that vectors in a vector field are tangent to the flow lines. Since $\vec c(t)$ is a flow line, you know that $\vec F(\vec c(t)) = \vec c'(t)$. That is, the vector field of the flow line is equal to the tangent to the flow line for all times $t$.

You know $\vec F$ is conservative, and so $\vec F = - \vec{\nabla f} = -(f_x \hat i + f_y \hat j + f_z \hat k)$ for some potential function $f(x, y, z) \geq 0$. The gradient of a function gives the direction of fastest increase. So minus the gradient gives the direction of fastest decrease.

Hence $\vec F(\vec c(t)) = - \vec{\nabla f}(\vec c(t)) = \vec c'(t)$. Notice the tangent to the flow line is equal to the negative of the gradient of the potential function (which is decreasing). So it must be the case $\vec c'(t)$ is also decreasing. So as the particle moves along the flow line $\vec c(t)$, its potential energy is decreasing.

Now what about $\displaystyle \lim_{t \rightarrow \infty} \vec F(\vec c(t))$?