# Work done as change of potential, how total derivative is converted to partial derivative

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

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

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?

• Hint: $V=V(s) \implies dV(s)=\frac {\partial V(s)}{\partial s}ds.\;$ $V=V(s,t)\implies dV(s,t)=\frac {\partial V(s,t)}{\partial s}ds+\frac {\partial V(s,t)}{\partial t}dt.$ – Cinaed Simson Dec 8 '19 at 23:13

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 $$\vec{F}\cdot\vec{\mathrm{d}s}=-\frac{\partial V}{\partial s}\mathrm{d}s$$).