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I have been thinking about the differentials that we use in derivatives and integrals.

For example, I have an equation:

$${\int{w}{dr}} = \text{other stuff}$$

The context for this strange equation was: $$\begin{align*} q'(t) &= a - \frac{b}{r(t)},\\ r'(t) &= c - d r(t) - e q(t)\\ r''(t)&= d r'(t) - e q'(t)\\ r''(t)&= d r' - e a + \frac{eb}{r}\\ w&=r'\\ w' &= d w - ea + \frac{eb}{r}\\ \frac{dw}{dr} \frac{dr}{dt}&= d w - ea + \frac{eb}{r}\\ w \frac{dw}{dr} &= dw - ea + \frac{eb}{r} \end{align*}$$

Integrate both sides with respect to $r$: $$\frac{w^2}{2} = \int{d*w dr} - ea r + eb log(r)$$ Shift the integral to the left hand side: $$\int d * w dr = - \frac{w^2}{2} - ea * r - eb * log(r)$$

Divide by $d$: $$\int w dr = \frac{w^2}{2d} - \frac{ear}{d}-\frac{eb}{d} log(r)$$ And I get to that $w\, dr$.

So if I can replace $w$ with ${\frac{dr}{dt}}$, and differentials can be multiplied like normal, I would get ${\frac{dr^2}{dt}}$.

Then I have the idea of multiplying by ${\frac{dt}{dt}}$. Can I go ahead and do this? That would give me $${\int{\frac{dr^2}{dt^2} dt}}$$ Which looks like ${w^2}$ to me. But can I do that?

  • 3
    Look up "non-standard analysis" on Wikipedia. Basically, the answer is, provided you stick to a certain set of rules, yes you can manipulate them like a type of variable.2010-11-04
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    @Ryan, provided you stick to a certain set of rules, you can manipulate them in non-non-standard analysis, too :)2010-11-04
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    @Mariano, ja ja. Non-standard analysis somehow makes it "feel" more like an actual variable. But either way is fine, I suppose.2010-11-04
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    One intriguing application of this is a simple derivation of the arc length formula integral(sqrt(1+(dy/dx)^2) dx) from integral(sqrt(dx^2 + dy^2)) (no dx at the end!), rather than having to go through all that [summation notation nonsense](http://en.wikipedia.org/wiki/Arc_length#Another_way_to_obtain_the_integral_formula).2010-11-04
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    Another difficultly is $\mathrm{d}^2 t$ vs $\mathrm{d}t^2$.2010-11-04
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    @Joey: "nonsense" is a bit harsh, but I otherwise agree.2010-11-04

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