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# L'Hopitals Rule in Action - Help! watch

1. This is all to do with decay, @ and b are decay constants, etc...

B(t) = @A_0{[e^([email protected]) - e^(-bt)]/(b - @)}
Use L'Hopitals rule to show that the solution for b -> @ is:
B(t) = @A_0te^([email protected])

Help appreciated.
2. B(t) = aA_0 (e^(-at) - e^(-bt))/(b - a)

Now:
d/db [aA_0 (e^(-at) - e^(-bt))] = aA_0 * t e^(-bt)
d/db (b - a) = 1

Hence:
lim(b->a) B(t) = lim(b->a) (aA_0 * t e^(-bt))/1 = aA_0 t e^(-at), as required.
3. One little query: We treat t as a constant here, but it's a variable (time), so why can we just drop the t down?
4. Isn't a parameter instead of a constant or something odd... I dunno (I'm only an AS mathmo!)
5. your limiting b -> infinity. All your doing is taking a limit of a 2 variable function as one goes to infinity. L'hopitals rule just shows what this limit is the diffentiating has no actual effect on the problem being considered its just a method for finding the limit.
6. It's just kind of weird thinking about it lol.

Anyways, heres some more...

A decays into B at rate alpha, B decays into C at rate Beta.
I got my DE's as: dA/dt = -alpha.A, dB/dt = alpha.A - Beta.B, dC/dt = Beta.B

Well so far I get (which are correct):

A(t) = A_0.e^(-alpha.t)
B(t) = [alpha.A_0/(Beta - alpha)]{e^(-alpha.t) - e^(-Beta.t)}

Then it says find the solution for C using the solution for B, both for Beta =/= Alpha and Beta = Alpha.

But I get a solution:
C(t) = A_0/[Beta(Beta - Alpha]{Alpha[e^(-Beta.t) - 1] + Beta[1 - e^(-alpha.t)]}

So, is this correct, and if so, surely I don't get a solution for C in the case alpha = beta?

Help appreciated.
7. I'm not familiar with the physics involved here, but maybe you should also use the limiting case for B(t).
8. (Original post by dvs)
I'm not familiar with the physics involved here, but maybe you should also use the limiting case for B(t).
Yes I questioned that, but surely it would just say to solve for Beta -> alpha and not Beta = Alpha, which gives no solution. Meh, thanks anyways.
9. (Original post by Nima)
C(t) = A_0/[Beta(Beta - Alpha]{Alpha[e^(-Beta.t) - 1] + Beta[1 - e^(-alpha.t)]}
I agree with your expression for C(t) (actually Mathematica agrees with it ). After correcting the typo,

C(t) = [A_0/(Beta - Alpha)] {Alpha[e^(-Beta.t) - 1] + Beta[1 - e^(-alpha.t)]}

[Check: C(t) -> A0 as t -> infinity]

To deal with the Alpha = Beta case, replace Beta by Alpha in the differential equations and solve to find B(t) and then C(t).

Using L'Hopital's rule should give the same answers.

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Updated: April 12, 2006
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