How to integrate when the dx, the differential, is in the denominator?

**Karol** · January 12th 2013, 03:05 AM

the problem is:
$dx \cdot y=(A-dx)dy$
it yields:
$\frac{dx}{A-dx}=\frac{dy}{y}\rightarrow \frac{A}{dx}-1=\frac{y}{dy}$
I know to integrate only: f(x)dx=f(y)dy
Here the differentials aren't in the denominator

**JJacquelin** · January 12th 2013, 04:09 AM

Originally Posted by Karol

the problem is:
$dx \cdot y=(A-dx)dy$
it yields:
$\frac{dx}{A-dx}=\frac{dy}{y}\rightarrow \frac{A}{dx}-1=\frac{y}{dy}$
I know to integrate only: f(x)dx=f(y)dy
Here the differentials aren't in the denominator

In this case, we do not integrate. We look for the mistake which yields to the absurd equation $dx \cdot y=(A-dx)dy$

**Karol** · January 12th 2013, 09:25 AM

Why is it absurd, mathematically? Can't it be?
I have a small correction, but i am sure it will not help:
$dx\cdot By=(A-dx)dy$

**JJacquelin** · January 12th 2013, 10:24 AM

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.

**Karol** · January 12th 2013, 11:33 AM

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?

**HallsofIvy** · January 12th 2013, 11:53 AM

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.

**JJacquelin** · January 12th 2013, 12:45 PM

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.

**hollywood** · January 13th 2013, 01:19 AM

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

**JJacquelin** · January 13th 2013, 01:44 AM

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.

**Karol** · January 13th 2013, 09:59 AM

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 C₀ 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 C₀ 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 C₀, 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:
$-(\delta x)k \cdot C_L=(l-\delta x)(\delta C_L)$
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.

**topsquark** · January 13th 2013, 10:38 AM

Originally Posted by Karol

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 C₀ 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 C₀ 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 C₀, 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:
$-(\delta x)k \cdot C_L=(l-\delta x)(\delta C_L)$
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.

If I'm not mistaken your equation should be
$kC_L~dx = (L - x) dC_L$

-Dan

**Karol** · January 14th 2013, 10:41 AM

This makes sense, but i don't get C_L=f(x):
$\frac{dC_L}{dx}=\frac{kC_L}{L-x}$
$\Rightarrow C_L=kC_L \int \frac{1}{L-x}dx$
The C_L reduces on both sides.

**JJacquelin** · January 14th 2013, 11:27 AM

You have not completely separed the variables. It must be:
$\frac{dC_L}{C_L}=\frac{kdx}{L-x}$

**ILikeSerena** · January 14th 2013, 11:28 AM

Try integrating the following:

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

$\ln C_L = -k \ln(L - x) + C$

**Karol** · January 16th 2013, 11:13 AM

I continue:
$\ln C_L=-k\ln (L-x)+C \Rightarrow \ln \left( C_L (L-x)^k \right)=C$
$e^C=C_L (L-x)^k$
We know k=0.15, and that at x=0 the concentration in the zone was C_L=C₀/k, so the constant C must be:
$C=\ln \left( \frac {C_0}{k} L^k \right)$
And when i insert this C into:
$e^C=C_L (L-x)^k$
I get:
$C_L=\frac{L\sqrt[k]{\frac{C_0}{k}}}{L-x}$
But the answer should be:
$C_L=\frac{C_0}{0.15}\left( \frac {L}{L-x} \right)^{0.85}$

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