Help on partial derivative

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

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  • #2
HallsofIvy
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You are trying to apply the "one variable" chain rule to a multivariable function. The chain rule for multivariable functions is
[tex]\frac{d\phi}{dz}= \frac{\partial \phi}{\partial x}\frac{dx}{dt}+ \frac{\partial \phi}{\partial y}\frac{dy}{dt}[/tex]
or in "differential form"
[tex]d\phi= \frac{\partial \phi}{\partial x}dx+ \frac{\partial \phi}{\partial y}dy[/tex]
 
  • #3
Erland
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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.
 
  • #4
vanhees71
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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
[tex]C: \quad \vec{x}=\vec{x}(t), \quad t \in [t_1,t_2].[/tex]
Let further be [itex]\vec{V}(\vec{x})[/itex] a vector field. Then by definition the line integral of this field along the curve is given by
[tex]\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)].[/tex]

Now suppose [itex]\vec{V}=-\vec{\nabla} \phi[/itex]. Now according to the chain rule for multi-variable functions we have
[tex]\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).[/tex]
Plugging this into the above integral gives
[tex]\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)],[/tex]
where [itex]\vec{x}_1=\vec{x}(t_1)[/itex] and [itex]\vec{x}_2=\vec{x}(t_2)[/itex] 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.
 
  • #5
verty
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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.
 

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