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I am reading Goldsetein's Classic Mechanics 3rd edition in Chapter 1 it says,

If work done in moving form point 1 to 2 denoted by $W_{12}$, is independent of the path it should be possible to express it as a change in quantity that depends only on the positions of end points. This quantity may be designates as $-V$, so that for a differential length we have the relation

$$ \begin{equation} \label{eq:1} \mathbf{F}\cdot\text d\mathbf{s} = -\text dV \tag{1} \end{equation} $$

or

$$ F_{\mathrm{s}}=-\frac{\partial V}{\partial s} \tag{2} $$

I am not sure how (2) comes from (1)? How does the dot product goes away and a partial derivative is introduced?

Also, it goes on to say, if the force applied on particle is given by a gradient of a scalar function that depends both on position and time. The work done when it travels a distance $\text ds$ is,

$$\mathbf{F} \cdot \text d \mathbf{s}=-\frac{\partial V}{\partial s} \text ds \tag{3}$$

I am clear as to how (1) comes but how does that change to (3) if $V$ is function of both position and time?

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  • $\begingroup$ Hint: $V=V(s) \implies dV(s)=\frac {\partial V(s)}{\partial s}ds.\;$ $V=V(s,t)\implies dV(s,t)=\frac {\partial V(s,t)}{\partial s}ds+\frac {\partial V(s,t)}{\partial t}dt. $ $\endgroup$ – Cinaed Simson Dec 8 '19 at 23:13
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We know that $F_s=\vec{F}\cdot \vec{u}_s$ moreover $\vec{F}\cdot \vec{\mathrm{d}s}=\vec{F}(\mathrm{d}s\cdot \vec{u}_s)=(\vec{F}\cdot \vec{u}_s)\mathrm{d}s=F_s\mathrm{d}s$ and according to your third equation: $\vec{F}\cdot\vec{\mathrm{d}s}=-\dfrac{\partial V}{\partial s}\mathrm{d}s$.

So at the end we get: $F_s\mathrm{d}s=-\dfrac{\partial V}{\partial s}\mathrm{d}s$ and so: $F_s=-\dfrac{\partial V}{\partial s}$.

Now if $V$ is time dependent, we have : $F_s=-\dfrac{\partial V}{\partial s}+\dfrac{\partial V}{\partial t}\dfrac{dt}{ds}=-\dfrac{\partial V}{\partial s}+\dfrac{\partial V}{\partial t}\dfrac{1}{v_s}$ (because $dV=\frac{\partial V}{\partial s}\mathrm{d}s+\frac{\partial V}{\partial t}\mathrm{d}t$ and $\vec{F}\cdot\vec{\mathrm{d}s}=-\frac{\partial V}{\partial s}\mathrm{d}s$).

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F.ds is the product of the component of F in the direction described by ds and ds itself so it is Fsds. Now dividing both sides by ds gives equation (2). I'm not sure of the mathematical rigor, but I think the use of the partial derivative is because, as you say, V depends on more than just s.

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