| Line 8: | Line 8: | ||
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. | ||
Revision as of 17:29, 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)
