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

Hmm, when I do the same insertion, I get:

$\displaystyle {C_0 \over k} L^k = C_L(L-x)^k$

$\displaystyle C_L = {{C_0 \over k} L^k \over (L-x)^k}$

$\displaystyle C_L = {C_0 \over k} \left({L \over L-x}\right)^k$

Substituting k = 0.15 gives:

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

Yes, there is a difference, which suggests the original formulation is incorrect.

Apparently, there should be a (1-k) somewhere instead of a k.

I haven't tried to understand your process yet, but is it possible it should have been:

$\displaystyle (1-k)C_L dx=(L-x) dC_L$

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

It makes sense, on left is the quantity of the impurity that is left in the zone after dx has solidified, (1-x), and it is equal to the increase, dC_{L}, in the concentration in the shortening zone (L-x). then the result is as it should be.

But let me ask a mathematical question:

Why don't i have dx on the right side also, on the side of (L-x)?

L-x is shortening with the amount dx also.

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

Hmm, I still don't understand your process.

But okay, let's take a look at your shortening zone.

Then apparently at some point in time the zone is (L-x).

Then it is shortened by an infinitesimal amount dx, so it becomes (L-x-dx).

Since dx is very small, it can be neglected relative to (L-x).

So it is the same as just (L-x), especially if we let dx approach zero.

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

I don't get it.

How should L be shortening?

Isn't L the fixed width of the furnace?

Oh wait, are we looking at the last part of the process only? When the last part of the bar is leaving the furnace?

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

yes, but i think i am getting the answer to my question, thanks to you all

Karol