(→(C)) |
(→(C)) |
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== (C) == | == (C) == | ||
| + | I'm not sure about this part.. | ||
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| + | |||
| + | 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. --[[User:Ctuchek|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) | Just factor the top of the equation like this: (e^(-2t) - 1) into (e^(-t) + 1)*(e^(-t) - 1) | ||
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Then you can cancel some terms and there you go. | Then you can cancel some terms and there you go. | ||
| − | [[User:Idryg|Idryg]] 21: | + | [[User:Idryg|Idryg]] 21:14, 6 October 2008 (UTC) |
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| − | + | ||
| − | + | ||
Revision as of 17:14, 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..
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)
