User:Sentriclecub/differentials help

What are the rules of treating differentials?
I learned calculus from a Stewart book which looks down on using dummy variables as actual terms like  dv = a dt 

I am wondering if someone can point me to a page (here or on the internet) that outlines the rules for how engineers and physicists treat the differentials as meaningful terms. My Stewart book said that the dt that appears after the integrand is only to remind us how Leibniz developed them from the concept of Reimann sums.

I'm specifically looking at

$$ dv = -v_{rel} \frac{dM}{M}$$

$$ \int_{v_i}^{v_f} dv = -v_{rel} \int_{M_i}^{M_f} \frac{dM}{M} $$

I don't understand how in the top equation, I can integrate both sides between different limits of integration? In my newbie understanding, that is like dividing one side of an equation by 12 candelas, but dividing the other side by 13 pascals.

My second question, is changing the limits of integration from v_i and v_f to M_i and M_f. I understand Integration by substitution clearly as it is taught in my Stewart book. I could understand if both sides were integrated between limits of integration if the terms had the same dimensions (velocity and velocity) but then one of the integrals changed the integrand from a velocity to a mass (and in complying with the FTC-2 of the subsitution rule, also changing the limits v_i to M_i and v_f to M_f).

I guess my apprehension is worrying about if the left side and right side will balance.

Third question is 5 second easy one. When you evaluate a definite integral, does the answer always have the same dimension as the integrand multiplied by the dummy variable? This is a pattern I've noticed.

Thanks, I'm a contributor to the Differential article and other pages where I can help out with clearity and links and navigability. Also, I've read 7 articles trying to find my answer, and have tried looking online. Thanks in advance. My only 3 questions at the ref desk have all been about differentials. Sentriclecub (talk) 14:56, 10 September 2008 (UTC)

Here's my work so far...

$$ \int_{a}^{b} dv = (b-a)*y $$

and I also have this worked out

$$ -v_{rel} \int_a^b \frac{dM}{M} = -v_{rel}*(ln b - ln a) = -v_{rel}*ln \frac{b}{a}$$

So as you can see, I'm stuck on only seeing the dM and the dv as a dummy variable only, and want to use y = f(x) lingo that is in my engrams. —Preceding unsigned comment added by Sentriclecub (talk • contribs) 15:23, 10 September 2008 (UTC)


 * You can think (extremely unrigorously) of the "dv" or "dM" or whatever as being an infinitesimal change in v or M, or whatever, so dM is probably the mass of an infinitesimal amount of the item in question. You can manipulate them in much the same way as any other variable and it usually works just fine, but just don't let a mathematician see you doing it! (And sanity check your answers just in case you've found one of the few cases where it doesn't work.) As for adding limits on both sides, I think they need to be the same limit, just in terms of the appropriate variable (so your limits are probably the values at t=0 and the values at t=T, or something, you just use the velocity at those times on one side and the mass at those times on the other). And yes, the dimension of d(Something) is the same as the dimension of the something and integration just means adding lots of them together, so that doesn't change the dimension. --Tango (talk) 15:31, 10 September 2008 (UTC)

If you like to be rigorous, just treat dx and dy as new variables. The rules of algebra are the addition rule: d(x+y) = dx+dy, the multiplication rule: d(x&middot;y) = x&middot;dy+dx&middot;y, the power rule: d(xy) = y&middot;xy&minus;1&middot;dx+xy&middot;log(x)&middot;dy, and the constant rule: dk = 0 if k is constant. Refer to differential algebra in case mathematicians object, as they sometimes do because they got scared by the unrigorous interpretation of differentials as infinitesimals. Note that the limits refer to the dummy variable behind the d. Explicitely:
 * $$ \int_{a}^{b} dv = \int_{v=a}^{b} dv = \int_{x=a}^{b} dx =b-a $$
 * $$ \int_{v=v_i}^{v_f} dv = -v_{rel} \int_{M=M_i}^{M_f} \frac{dM}{M} $$

Note that Mi=M(vi). Bo Jacoby (talk) 16:52, 10 September 2008 (UTC).


 * You've certainly got your work cut out to distinguish clearly between the various forms of differentials and derivatives. Offhand I can think of things like partial derivative, exterior differential, covariant derivative, never mind the boundary operator looking like a partial derivative. Using differentials is good I think but there is some problems especially with twice differentiating. In differential forms d2 is zero, and for ordinary differentials y"dx2 isn't the same as d(dy) which if one works it out would come to y"dx2+y'd(dx). Dmcq (talk) 18:04, 10 September 2008 (UTC)


 * Ulp I thought I say the quote marks in what I wrote but obviously I didn't and they have turned into the thing to make italic writing. I'll fix that last line and write it here again. For ordinary differentials y"dx2 isn't the same as d(dy) which if one works it out would come to y"dx2+y'd(dx). One has to resist the urge to use d2x here. There's also Lie derivatives if you're interested. Dmcq (talk) 18:52, 10 September 2008 (UTC)