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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?

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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. – Ryan Budney Nov 4 '10 at 20:38
@Ryan, provided you stick to a certain set of rules, you can manipulate them in non-non-standard analysis, too :) – Mariano Suárez-Alvarez Nov 4 '10 at 20:51
@Mariano, ja ja. Non-standard analysis somehow makes it "feel" more like an actual variable. But either way is fine, I suppose. – Ryan Budney Nov 4 '10 at 21:13
Thanks for the tips. – Tony Peterson Nov 4 '10 at 22:28
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. – Joey Adams Nov 4 '10 at 22:38

None of your examples make any sense to me. The manipulations with differentials which are allowed in calculus are basically only those that follow from the chain rule. In particular, an integral can be transformed using the change-of-variables formula, but that's it. (OK, maybe I'm over-simplifying a little here, but the basic advice is that if you're unsure whether something is allowed or not, don't do it.)

I think much of the confusion surrounding this comes from the fact that the notation $dy/dx$ doesn't show at which point the derivative is supposed to be evaluated. For example, the derivative of an inverse function is given by the rule $dx/dy=1/(dy/dx)$ in this notation, which is suggestive and useful, but not very precise. What this really says is that if $(a,b)$ is a point on the graph $y=f(x)$ and if the slope at that point is $f'(a)=s$, then the point $(b,a)$ lies on the graph of the inverse function, $y=f^{-1}(x)$, and the slope of that graph at that point is $1/s$: $$ (f^{-1})'(b)=1/s=1/f'(a).$$ There's no "magic" here, just a simple geometric fact which can be illustrated in a picture.

For another example of a similar flavour, see this question.

EDIT: I was a bit too quick. There is something behind your manipulations after all.

If I understand you correctly, you are looking at a second order ODE of the form $d^2 r/dt^2 = f(r,dr/dt)$; let's call this equation (A). This is equivalent to the first order system $dr/dt=w$, $dw/dt=f(r,w)$, which I'll call (B). Any solution $r=R(t)$ of (A) corresponds to a solution $(r,w)=(R(t),R'(t))$ of (B).

Fix one such solution for the moment. (We're trying to solve the equation, so we probably don't know such a solution yet, but we can pretend that we have one and see where that leads.)

Now $(r,w)=(R(t),R'(t))$ is a parametrized curve in the $(r,w)$-plane (the phase plane of the system). Provided this curve doesn't have vertical slope, it looks (at least locally) like the graph of a function, $w=H(r)$ say. This function $H$ satisfies a first-order ODE, $$H'(r)=\frac{f(r,H(r))}{H(r)},$$ which is the useful fact that you learned from Mariano in that other thread. [It follows from "$dw/dr=(dw/dt)/(dr/dt)=f(r,w)/w$" as explained in the answer that I linked to above.]

The integral you asked about is $$ \int_{r_0}^{r_1} H(r) dr.$$ (I've attached some limits $r_0$ and $r_1$ just because that makes it easier for me to think.) Making the change of variables $r=R(t)$, where $R$ is the solution of (A) that we used in order to define the function $H$, we get $$ \int_{t_0}^{t_1} H(R(t)) R'(t) dt$$ where $t_0$ and $t_1$ are suitable time values corresponding to $r_0$ and $r_1$ [that is, $r_0=R(t_0)$ and $r_1=R(t_1)$]. By the definition of $H$ we have $H(R(t))=R'(t)$, since $w=H(r)$ defines a point on the curve $(w,r)=(R(t),R'(t))$. Thus the integral becomes $$ \int_{t_0}^{t_1} (R'(t))^2 dt,$$ just like you (and Dan in his answer) said, although expressed in a more precise way.

Now, whether this leads to something useful, I don't know. But at least it should be a bit clearer what the manipulations mean, I hope.

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I thought that might be the case. But separable differential equations made me think it might work, because you break apart a y' into dy/dt and multiply dt to the right side of the equation. – Tony Peterson Nov 4 '10 at 22:27
@Tony: The separation of variables technique for differential equations is also justified by using the chain rule (or a substitution - the integral version of the chain rule). See, for example, page 26 of the third edition of Blanchard, Devaney, and Hall' s Differential Equations. – Mike Spivey Nov 5 '10 at 0:00
@Tony: Maybe we could try to make sense of your ideas if you provide a little more context. What I found strange was "but $w$ is $dr/dt$" when $w$ appear in an integral with respect to $r$. How do you get into that situation in the first place? What kind of quantities do $r$, $w$ and $t$ represent here? – Hans Lundmark Nov 5 '10 at 7:26
I had an equation with second derivatives but no t component. In an earlier thread here, I learned some ways to simplify that kind of equation. So I set w = r'. And then I had an equation that was w' = f(r,w). So I configured derivatives in the form of dw/dr instead of dw/dt. w' was dw / dt, which is dw / dr * dr / dt. Which is w * dw / dr. Then I integrated both sides wrt r. Giving me w^2 / 2, and on the other side, there was a term n w, which became $\int{nw dr}$. Since everything else integrated fine, I brought that to the other side of the equation and divided by n. – Tony Peterson Nov 5 '10 at 11:59
@Tony: Thanks. Now I understand better, and I've updated the answer accordingly. – Hans Lundmark Nov 5 '10 at 13:18

Your calculation looks to me like it produces the right answer provided one interprets it correctly. The only sensible meaning one can attach to $dr^2/dt^2$ should be $(dr/dt)^2$. So it seems your calculation shows that $\int w dr = \int w^2 dt$.

My experience from my own undergraduate calculus is that: (i) any calculation I did by multiplying or dividing differentials like this would produce a meaningful and correct answer in the end; (ii) whatever answer was found in the end could also be justified by only using the normal chain rule. In this case, for instance, the change-of-variables formula tells you that $dr = (dr/dt)dt = wdt$, which gives the above formula.

Edit: I realised afterwards that point (i) above might be misleading when one takes a course on calculus in several variables, since $dx \neq \frac{\partial x}{\partial y} dy$ if $x$ depends on more variables than $y$. In principle I would guess that multiplication and division of differentials is still possible as a heuristic but one needs to be very careful to distinguish between d's and $\partial$'s.

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I had a similar question a while back which collected some great answers. It might be helpful here.

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