# Help on partial derivative

## Main Question or Discussion Point

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

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HallsofIvy
Homework Helper
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
Gold Member
2019 Award
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
Homework Helper
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.