the problem is:

$\displaystyle dx \cdot y=(A-dx)dy$

it yields:

$\displaystyle \frac{dx}{A-dx}=\frac{dy}{y}\rightarrow \frac{A}{dx}-1=\frac{y}{dy}$

I know to integrate only: f(x)dx=f(y)dy

Here the differentials aren't in the denominator

- Jan 12th 2013, 03:05 AMKarolHow to integrate when the dx, the differential, is in the denominator?
the problem is:

$\displaystyle dx \cdot y=(A-dx)dy$

it yields:

$\displaystyle \frac{dx}{A-dx}=\frac{dy}{y}\rightarrow \frac{A}{dx}-1=\frac{y}{dy}$

I know to integrate only: f(x)dx=f(y)dy

Here the differentials aren't in the denominator - Jan 12th 2013, 04:09 AMJJacquelinRe: How to integrate when the dx, the differential, is in the denominator?
- Jan 12th 2013, 09:25 AMKarolRe: How to integrate when the dx, the differential, is in the denominator?
Why is it absurd, mathematically? Can't it be?

I have a small correction, but i am sure it will not help:

$\displaystyle dx\cdot By=(A-dx)dy$ - Jan 12th 2013, 10:24 AMJJacquelinRe: How to integrate when the dx, the differential, is in the denominator?
They are two possibilities

First : It comes from a physical modelisation. Then (A-dx) should be replaced by A, because dx is infitinitely smaller than A. So, the modelisation was not finished. In this case B*y*dx=A*dy is a correct equation, easy to solve.

Second : It comes from a purely mathematical problem. Then show the initial wording and it could be seen where is the mistake. - Jan 12th 2013, 11:33 AMKarolRe: How to integrate when the dx, the differential, is in the denominator?
I cannot give up the dx in: (A-dx), since it is from a physical problem, but i am not sure i made it right.

A minimizes with time, until it gets to zero, so, i think, i need (A-dx).

But, tell me what's wrong with the original equation? why is it absurd, from a mathematical point of view? - Jan 12th 2013, 11:53 AMHallsofIvyRe: How to integrate when the dx, the differential, is in the denominator?
Then you are going to have to tell us what "A" and "dx" mean. If "dx" really is a differential (in one sense an infinitesmal compared with A) then either A- dx makes no sense or A- dx= A. If this is from a "physical problem" then it looks like there is some thing wrong with the mathematical model for the physical situation.

- Jan 12th 2013, 12:45 PMJJacquelinRe: How to integrate when the dx, the differential, is in the denominator?
An infinitesimal is some quantity that is explicitly nonzero and yet smaller in absolute value than any real quantity.

Roughly, if you supposes that dx have a value so that (A-dx) must be distinguished from (A), you are not acting in the field of pur mathematics.

Of course, in physics, while modelizing a phenomena, the symbol dx is commonly used in a less rigourous sens. But, at the end of modelization, the Physicians know what to do to bring their equation to a correct mathematical form, so that the equation is not mathematically absurd and is usable for further analytical transformations (may be "absurd" is not the right word ? ). If this final step is not made or uncompletely made, the resulting "absurd" equation cannot be correctly treated with the rules of mathematics : The equation might be ambiguous and even not understood by Mathematicians.

To be aware of that, one have to learn the founding principles of the infinitesimal calculus. Moreover, learning the Nonsandard Analysis gives a better understanding of the relationships with calculus in Physics. - Jan 13th 2013, 01:19 AMhollywoodRe: How to integrate when the dx, the differential, is in the denominator?
Hi Karol,

If you want us to help you, you're going to have to tell us about the first part of the problem - the physics, that is - since the problem seems to be in the conversion from the physical problem to a mathematical problem. Until then, we are all stuck just like you are....

- Hollywood - Jan 13th 2013, 01:44 AMJJacquelinRe: How to integrate when the dx, the differential, is in the denominator?
I fully agree with the post of hollywood. Your difficulty is in the modelization part, before going to the mathematical resolution. If you need help, the wording of the physical problem has to be posted on the forum as well as what you have done. May be, a better place should be on the physical forum instead of the mathematical forum.

- Jan 13th 2013, 09:59 AMKarolRe: How to integrate when the dx, the differential, is in the denominator?
To purify a long bar of semiconductor they place a small furnace of width l round it and move it, slowly, from one end to the other.

The bar has in it uniform impurity C_{0}which is in atoms per unit volume.

Where the furnace is the bar is molten liquid, and as it advances material freezes behind it, with impurity concentration K*C_{L}, K<1, and new material is molten in front, adding impurity of concentration C_{0}to the liquid zone. The concentration in the liquid zone is C_{L}.

So, as the furnace advances, the impurity concentration in the liquid zone rises till it reaches the constant value C_{0}, but at the end, before the furnace leaves the bar, the concentration rises, theoretically, to infinity.

I have to show it.

So, i wrote the equation:

$\displaystyle -(\delta x)k \cdot C_L=(l-\delta x)(\delta C_L)$

The left side shows the amount of impurity leaving the zone as the furnace advances dx, and the right side should be the impurity rise in the ever shortening length l. - Jan 13th 2013, 10:38 AMtopsquarkRe: How to integrate when the dx, the differential, is in the denominator?
- Jan 14th 2013, 10:41 AMKarolRe: How to integrate when the dx, the differential, is in the denominator?
This makes sense, but i don't get C

_{L}=f(x):

$\displaystyle \frac{dC_L}{dx}=\frac{kC_L}{L-x}$

$\displaystyle \Rightarrow C_L=kC_L \int \frac{1}{L-x}dx$

The C_{L}reduces on both sides. - Jan 14th 2013, 11:27 AMJJacquelinRe: How to integrate when the dx, the differential, is in the denominator?
You have not completely separed the variables. It must be:

$\displaystyle \frac{dC_L}{C_L}=\frac{kdx}{L-x}$ - Jan 14th 2013, 11:28 AMILikeSerenaRe: How to integrate when the dx, the differential, is in the denominator?
Try integrating the following:

$\displaystyle {dC_L \over C_L} = {k dx \over L - x}$

$\displaystyle \ln C_L = -k \ln(L - x) + C$ - Jan 16th 2013, 11:13 AMKarolRe: How to integrate when the dx, the differential, is in the denominator?
I continue:

$\displaystyle \ln C_L=-k\ln (L-x)+C \Rightarrow \ln \left( C_L (L-x)^k \right)=C$

$\displaystyle e^C=C_L (L-x)^k$

We know k=0.15, and that at x=0 the concentration in the zone was C_{L}=C_{0}/k, so the constant C must be:

$\displaystyle C=\ln \left( \frac {C_0}{k} L^k \right)$

And when i insert this C into:

$\displaystyle e^C=C_L (L-x)^k$

I get:

$\displaystyle C_L=\frac{L\sqrt[k]{\frac{C_0}{k}}}{L-x}$

But the answer should be:

$\displaystyle C_L=\frac{C_0}{0.15}\left( \frac {L}{L-x} \right)^{0.85}$