View source for Page 536 Problem 1 MA181Fall2008bell - Rhea

== (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:

<math>A(t)=-e^{-t}+1</math>

and

<math>V(t) = -(Pi*e^{-2x})/2 + Pi/2</math>

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..

[[User:Idryg|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. --[[User:Ctuchek|Ctuchek]] 21:13, 6 October 2008 (UTC)


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.

[[User:Idryg|Idryg]] 21:14, 6 October 2008 (UTC)

Hmmm.  I bet that works.  But maybe L'Hopital's could be useful here.  --[[User:Bell|Bell]] 21:39, 6 October 2008 (UTC)