(→(C)) |
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[[User:Idryg|Idryg]] 21:14, 6 October 2008 (UTC) | [[User:Idryg|Idryg]] 21:14, 6 October 2008 (UTC) | ||
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| + | Hmmm. I bet that works. But maybe L'Hopital's might useful here. --[[User:Bell|Bell]] 21:39, 6 October 2008 (UTC) | ||
Revision as of 16:39, 6 October 2008
(A)
So you know:
A(t) = the integral of e^(-x) dx from 0 to t
and
V(t) = the integral of Pi*[e^(-x)]^2 dx from 0 to t
Just evaluate the integrals:
A(t) = -e^-t + 1
and
V(t) = -(1/2)*Pi*e^-2x + Pi/2
and then take the limits as t approaches infinity.
(B)
Just put V(t) over A(t) and take the limits.
(C)
I'm not sure about this part..
Idryg 21:03, 6 October 2008 (UTC)
On part C, since e^a is valid for all real a, and since V(0) and A(0) are valid functions (i.e. V(0) does not give a no solution), the limit as t approaches zero from the right is the same as if t approaches infinity from the left. This means that you can just take the limit as t approaches 0 and ignore the 0+ aspect of the problem.
I am not %100 sure about this, but this is how I understood the problem, maybe if someone graphs this, we can see what V(t)/A(t) is approaching when t = 0. --Ctuchek 21:13, 6 October 2008 (UTC)
Just factor the top of the equation like this: (e^(-2t) - 1) into (e^(-t) + 1)*(e^(-t) - 1)
Then you can cancel some terms and there you go.
Idryg 21:14, 6 October 2008 (UTC)
Hmmm. I bet that works. But maybe L'Hopital's might useful here. --Bell 21:39, 6 October 2008 (UTC)
