Rearranging equation

[SMOF]

Hello.

I am hoping someone can help me rearrange the following equation for t.

I'm getting better at these, but this one stumps me.

C(x,t) = Cserfc(x/2√(Dt))

I was starting with C(x,t)/Cs = erfc(x/2√(Dt))

But from here I am not sure how to deal with the square root and the t buried in the square root.

Thanks.

Seán

[SHScanuck]

If I remember right, this equation corresponds to Dopant diffusion

C(x,t) = Cs*erfc(x/2√Dt)

C(x,t)/Cs = erfc(x/2√Dt)

erfc^-1*(C(x,t)/Cs) = erfc^-1(erfc(x/2√Dt))

(2*erfc^-1*(C(x,t)/Cs))/x = 1/√Dt

Square both sides,

1/t = D*((2*erfc^-1*(C(x,t)/Cs))/x)^2

[Ray Vickson]

Hello.

I am hoping someone can help me rearrange the following equation for t.

I'm getting better at these, but this one stumps me.

C(x,t) = Cserfc(x/2√(Dt))

I was starting with C(x,t)/Cs = erfc(x/2√(Dt))

But from here I am not sure how to deal with the square root and the t buried in the square root.

Thanks.

Seán

It is not clear what you are asking. Is Cserfc(x/2√(Dt)) the _definition_ of the object C(x,t)? If so, you presumably have an equation of the form Cserfc(x/2√(Dt)) = v (where v is some value) and you want to solve this for t. You would have t on only one side of this equation. It wold be farily easy to solve: just find a value of z giving erfc(z) = v/Cs (which can be found numerically if v and C_s are known numbers); then solve z = x/2√(Dt) for t. However, if C(x,t) is some other function of x and t that you want to equate to Cserfc(x/2√(Dt)) , then you would have t on both sides of your equation, and without knowing more about the function C(x,t) we cannot say very much.

RGV

[SMOF]

If I remember right, this equation corresponds to Dopant diffusion

C(x,t) = Cs*erfc(x/2√Dt)

C(x,t)/Cs = erfc(x/2√Dt)

erfc^-1*(C(x,t)/Cs) = erfc^-1(erfc(x/2√Dt))

(2*erfc^-1*(C(x,t)/Cs))/x = 1/√Dt

Square both sides,

1/t = D*((2*erfc^-1*(C(x,t)/Cs))/x)^2

It does, yea ...and thanks for the work through. I will go over it a few times until I have it in my head.

Again, many thanks.

Seán

[Ray Vickson]

If C(x,t) is a function with t in it this will NOT isolate t, as you still have t on both sides of the equation. That is why I asked you in my first post EXACTLY what was your question. You did not answer.

RGV

[SMOF]

You did not answer.

I am sorry if this offended you.

I did not answer for a number of reasons.

1. I did not fully understand what you were asking, and
2. Someone else posted a reply which seemed to give me the answer.


The (x,t) part seem to be largely ignored in sample questions. Maybe I should have just written my question as C = .....

Anyway, sorry I did not respond to your question directly, and thank you for taking the time to reply.

Seán

[Ray Vickson]

I am not offended. I was trying to help you better understand your problem, and to do that I posed a question whose answer would, I think, be helpful to you.

RGV

[SMOF]

Hello.

Well, I will try to understand your question better, and see where that takes me :)

Seán

[Ray Vickson]

Let me give you an example. Isolate "t" in the following two problems.
(1) erfc(x/sqrt(t))= 0.25 (an "easy" problem); or
(2) erfc(x/sqrt(t)) = .5*sin(x^3 + 3t^2 + t) (a really, really difficult problem).

RGV