Difference between revisions of "Page 536 Problem 1 MA181Fall2008bell" - Rhea

Revision as of 16:15, 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)