# Tag Info

0

Treating $dx$ as a quantity you can multiply/divide both sides by is really just a shorthand for the chain rule, or equivalently u-substitution. Here's what's going on. You have $$v=\frac{dx}{dt}$$ Thus, if you integrate both sides against time, they should (up to a constant) be equal $$\int v dt = \int \frac{dx}{dt}dt$$ Doing a u-substitution on ...

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The term ${\rm d}t$ means an infinitesimal quantity of time. Like a slice of time so small, but still not zero. When something is moving with speed $v$ the infinitesimal distance it travels during the infitesimal time is designated ${\rm d}x = v \, {\rm d}t$. Speed just linearly scales time into distance. Now to get the total time passed, you sum of all ...

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v=dx/dt. It simply means that v is the slope of the x-t graph. v=dx/dt. dx=vdt. This is simple cross multiplication. Although I am not too sure, v=dx/dt has anything to do with deriving the equations of motion as it is acceleration that is the main factor in the equations of motion not velocity.

12

The equations are entirely equivalent, as can be proven using Gauss' and Stokes' theorems. The integral forms are most useful when dealing with macroscopic problems with high degrees of symmetry (e.g. spherical or axial symmetry; or, following on from comments below, a line/surface integrals where the field is either parallel or perpendicular to the ...

5

Your confusion lies in failing to recognize that they are exactly the same equations. Take for example Gauss's law $$\vec \nabla \cdot \vec E = \dfrac{\rho}{\epsilon_0}$$ You can see that there $\rho$ is the charge distribution, and in general can be a funcion of the position. Now consider a volume $V$, you can just integrate the density to obtain the ...

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All this tells you is that the fields satisfies both the inategral and the differential equations. The two are related by the mathematical identities called the divergence theorem and Stokes' theorem. So which do you apply? Well, which ever one you want! If you run into an integral, you use the integral form, and if you're ever asked for the divergence or ...

3

What an interesting question! It depends. In modern calculus, $\frac{df}{dt}$ is just a symbol for the derivative $$\lim_{h \to 0} \frac{f(t+h)-f(t)}{h}$$ As a matter of fact, mathematicians prefer different notations for the derivative of a function $f$, as $D f$ or $f'$. But the above definition of derivative became rigorous only when the concept of ...

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Using the word fraction is a more elementary way of naming the set of rational numbers : numbers that are represented by a ratio of integers. In physics when we model physical systems with differential equations we generally work in the domain of real numbers and sometimes complex numbers since we tend to think of real physical systems as existing in a ...

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In physics there are no infinitesimals, so dx is always treated as a "small but finite interval" during discussions, and only when the actual calculation is being done do we switch to mathematical mode, and "take the limit." During the 17th and 18th centuries, mathematicians and physicists both did this all the time. As they say in sports "no harm, no ...

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Yes: $$\partial_\mu \mathrm e^{ikx}=ik_\mu\mathrm e^{ikx}+\mathrm e^{ikx}\partial_\mu$$ because the differential operator acts on a test function (i.e., product rule).

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