# Showing $\textbf{F} \cdot d\textbf{s} = -dV$ is equivalent to $F_s = -\frac{\partial V}{\partial s}$

Can someone please explain how the following

$$\textbf{F} \cdot d\textbf{s} = -dV$$

is equivalent to

$$F_s = -\frac{\partial V}{\partial s}$$

using some intermediate steps. I don't follow this in Goldstein's text. Thanks.

-

$$\text d f(x)=f'(x)\text d x\equiv\frac{\partial f}{\partial x}\text d x.$$
The second equation you posted is of course only a component representation, the vectors in the first equation are most definately higher dimensional (3 dimensional). And so the derivative in the second equation is chosen such that it goes in the direction of the path, denoted y s. I'm afraid $F_s$ is supposed to denote the "s-component".
Thanks, I'm beginning to get it. But why does $ds$ become a partial derivative in the second equation? –  Joebevo Jul 20 '12 at 12:11
@Joebevo: It doesn't. As I said, the second equation is only the component representation of the first one: The expression $V(\vec x)$ gets evaluated to a scalar at every point $\vec x$ and $\text d V$ is considere to be the change of $V$ in the direction of the path $s$. So by the equation I posted you have $\text d V = V'(s)\text d s\equiv \frac{\partial V}{\partial s}\text d s$. Plug that into the first equation and you get the second equation - apart from the $\text d s$ (what's left is "only the component expression") and a complication that they dropped the vector notation for $F$. –  NikolajK Jul 20 '12 at 12:34
Shouldn't the equation you posted be $\text d f(x)=f'(x)\text d x\equiv\frac{df}{dx}\text d x$, since f is a function only of x? According to my knowledge, the partial derivative comes in when the function is one of multiple variables. –  Joebevo Jul 21 '12 at 7:54