(→(C)) |
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V(t) = the integral of Pi*[e^(-x)]^2 dx from 0 to t | V(t) = the integral of Pi*[e^(-x)]^2 dx from 0 to t | ||
| − | Just evaluate the integrals: | + | Just evaluate the integrals to get: |
| − | A(t) = -e^-t + 1 | + | <math>A(t)=-e^{-t}+1</math> |
and | and | ||
| − | V(t) = -( | + | <math>V(t) = -(Pi*e^{-2x})/2 + Pi/2</math> |
and then take the limits as t approaches infinity. | and then take the limits as t approaches infinity. | ||
| Line 32: | Line 32: | ||
| − | Just factor the top of the equation like this: (e^ | + | Just factor the top of the equation like this: <math>(e^{-2t} - 1)</math> into <math>(e^{-t} + 1)(e^{-t} - 1)</math> |
Then you can cancel some terms and there you go. | Then you can cancel some terms and there you go. | ||
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[[User:Idryg|Idryg]] 21:14, 6 October 2008 (UTC) | [[User:Idryg|Idryg]] 21:14, 6 October 2008 (UTC) | ||
| − | Hmmm. I bet that works. But maybe L'Hopital's | + | Hmmm. I bet that works. But maybe L'Hopital's could be useful here. --[[User:Bell|Bell]] 21:39, 6 October 2008 (UTC) |
Latest revision as of 18:31, 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 to get:
$ A(t)=-e^{-t}+1 $
and
$ V(t) = -(Pi*e^{-2x})/2 + 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 could be useful here. --Bell 21:39, 6 October 2008 (UTC)
