To what extent can you manipulate differentials like dy and dt like actual values?

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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

Answer 4

You can manipulate differentials in the following manner:

• You can add differentials: if $z=x+y,$ then $\mathrm{d}z=\mathrm{d}x+\mathrm{d}y.$
• You can multiply differentials by functions: if $y=f(x),$ then $\mathrm{d}y=f'(x)\mathrm{d}x$
• You are allowed to "divide" differentials, but only in the context of appliying the chain rule. So if $y=f\circ{g},$ where $\circ$ denotes composition, then $\frac{\mathrm{d}y}{\mathrm{d}s}=\frac{\mathrm{d}(f\circ{g})}{\mathrm{d}g}\frac{\mathrm{d}g}{\mathrm{d}s},$ which looks like the differential $\mathrm{d}g$ is being divided out, at least visually (but there genuinely is no actual division happening here).

Everything else is not permitted.