Why does the dx cancel out in partial derivatives for conservative fields?

[DunWorry]

Hi, I was reading something on conservative fields, in this example \phi is a scalar potential. (Please refer to the attatched thumbnail). It's partial derivatives, but I'm not sure why the d\phi/dx * dx, the dx should cancel out? and that should leave d\phi. So the integral should be -3∫d\phi. I know this is wrong, but I'm not sure why, can someone explain?

Thanks

 

[HallsofIvy]

You are trying to apply the "one variable" chain rule to a multivariable function. The chain rule for multivariable functions is
\frac{d\phi}{dz}= \frac{\partial \phi}{\partial x}\frac{dx}{dt}+ \frac{\partial \phi}{\partial y}\frac{dy}{dt}
or in "differential form"
d\phi= \frac{\partial \phi}{\partial x}dx+ \frac{\partial \phi}{\partial y}dy

 

[Erland]

It does not say $\frac{d\phi}{dx}dx$ etc, it says $\frac{\partial \phi}{\partial x}dx$ etc. and that is not the same thing. A function of three variables $\phi(x,y,z)$ changes if any of three variables changes, not just $x$, and if all the variables change, then these changes all contribute to the change in $\phi$.
To derive the formula, choose a path from $a$ to $b$ and parametrize it, and then evaluate the line integral using this parametrization.

 

[vanhees71]

To answer your question, it's better to go back to the definition of a line integral. To that end we give the curve in parameter representation
C: \quad \vec{x}=\vec{x}(t), \quad t \in [t_1,t_2].
Let further be \vec{V}(\vec{x}) a vector field. Then by definition the line integral of this field along the curve is given by
\int_C \mathrm{d} \vec{x} \cdot \vec{V}(\vec{x})=\int_{t_1}^{t_2} \mathrm{dt} \frac{\mathrm{d} \vec{x}}{\mathrm{d} t} \cdot \vec{V}[\vec{x}(t)].

Now suppose \vec{V}=-\vec{\nabla} \phi. Now according to the chain rule for multi-variable functions we have
\frac{\mathrm{d}}{\mathrm{d} t} \phi[\vec{x}(t)]=\frac{\mathrm{d} x}{\mathrm{d}t} \frac{\partial \phi}{\partial x}+\frac{\mathrm{d} y}{\mathrm{d}t} \frac{\partial \phi}{\partial y}+\frac{\mathrm{d} z}{\mathrm{d}t} \frac{\partial \phi}{\partial z}=\frac{\mathrm{d} \vec{x}}{\mathrm{d} t} \cdot \vec{\nabla} \phi[\vec{x}(t)]=-\frac{\mathrm{d} \vec{x}}{\mathrm{d} t} \cdot \vec{V}[\vec{x}(t).
Plugging this into the above integral gives
\int_{C} \mathrm{d} \vec{x} \cdot \vec{V}(x)=-\int_{t_1}^{t_2} \mathrm{d} t \frac{\mathrm{d}}{\mathrm{d} t} V[\vec{x}(t)]=-[V(\vec{x}_2)-V(\vec{x}_1)],
where \vec{x}_1=\vec{x}(t_1) and \vec{x}_2=\vec{x}(t_2) are the boundary points of the curve.

Note that this result implies that if the vector field is conservative, i.e., if it is the gradient of a scalar field, the line integral connecting two points is independent of the shape of the curve.

 

[verty]

For me, the most intuitive way to think about this is to pretend that $dx$ is a rate, so that makes $\phi_x dx = {\partial\phi \over \partial x} dx$ the related rate and $\phi_x$ the x-sensitivity of $\phi$. $\phi$'s rate of change is the dot product of $\phi$'s sensitivity vector (the gradient) and the variable rates of change.