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Homework Help: Newton's Method

  1. Oct 10, 2007 #1
    1. The problem statement, all variables and given/known data
    Problem:
    A (area) = [tex]\int^{e}_{1}[/tex] ln(x) dx = 1

    Now we let k be such that 0 [tex]\leq[/tex] k [tex]\leq[/tex]1

    Consider the line y = k.

    Find k so that area computed by A is exactly one half.

    2. Relevant equations

    So, first, I found point of intersection:

    k = ln (x)

    e[tex]^{k}[/tex] = x​


    Now I have:

    1/2 = [tex]\int^{e}_{e^{k}}[/tex] ln (x) dx​




    3. The attempt at a solution

    I'm having a hard time grasping the big picture of this, that's pretty much why I'm stuck. I know that Newton's Method is used to find the roots of a function, but this area twist is really giving me a hard time. I'm supposed to use N.M. to solve for k after setting up integral to compute A/2. THEN I have to experiment to find 2A/3 and A/1000. Can anyone enlighten me? Where is this zero happening?? Am I supposed to use [tex]\int[/tex] ln (x) -k dx??

    HELP!!
     
  2. jcsd
  3. Oct 10, 2007 #2

    HallsofIvy

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    The problem, as stated, doesn't make a whole lot of sense. The "area computed by A" does not depend on k. Do you mean to say that y= k divides the area computed by A in half? That's different from saying that [itex]\int_{e^k}^e ln(x)dx= 1/2[/itex] which has nothing to do with A.

    In any case, you will have to integrate [itex]\int ln(x) dx[/itex]. Can you do that?
     
  4. Oct 10, 2007 #3
    I copied the problem verbatim from the handout I was given, so I'm sorry that it doesn't make a whole lot of sense to you...

    Yes, I mean to say that the line y = k splits the area in half. And yes, I know how to integrate ln(x), but it doesn't do me a whole lot if I don't know what I'm doing. :eek:(


    So I have:

    [tex]\int^{e}_{e^{k}}[/tex] ln (x) dx = x ln x - x ][tex]^{e}_{e^{k}}[/tex]

    = ( e ln(e) - e ) - (e[tex]^{k}[/tex] ln(e[tex]^{k}[/tex] )

    = (e*1 -e) - ( e[tex]^{k}[/tex]*k)

    = - ( e[tex]^{k}[/tex]*k)
    Since that integral is equal to 1/2 (since the line cuts it in half)

    1/2 = -k * e[tex]^{k}[/tex]

    0 = -k * e[tex]^{k}[/tex] - 1/2 <--- Is this supposed to be the equation I use N.M. on?
     
  5. Oct 10, 2007 #4

    HallsofIvy

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    You forgot the "-x". the integral is
    [tex]-ke^k+ e^k[/tex]
    and so this is [itex]e^k(1-k)[/itex]
    No, because you neglected the "-x" your equation is
    [tex] e^k (1-k)= 1/2[/tex]
     
    Last edited by a moderator: Oct 10, 2007
  6. Oct 10, 2007 #5
    Ok, yeah, that makes sense. So now,

    [tex]k - ( e^k (1-k) ) / d/dk ( e^k (1-k) ) [/tex] ​

    should give me what k converges to, meaning the value of k that when 1/2 is subtracted, equals to zero, right??

    So for 2/3, I'd use this equation to solve for k using Newton's method,
    [tex] e^k (1-k)= 2/3 [/tex] ​

    and for 1/1000,
    [tex] e^k (1-k)= 1/1000 [/tex]​

    Am I on the right track?
     
  7. Oct 10, 2007 #6

    Gib Z

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    Are you 100% sure that your original integral wasn't [tex]\int^e_1 \frac{1}{x} dx = 1[/tex] instead? Because [tex]\int^e_1 \log x dx = -1[/tex] in case you didn't check with the anti derivative... Also the integral of 1/x from t to 1, wrt t, is a common definition of the natural logarithm, so you might have just typed down something you were thinking about a few lines ahead. Just check up on the question.
     
  8. Oct 11, 2007 #7

    HallsofIvy

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    Since log(x) (ln(x)) is positive for x> 1, how do you get -1 as its integral? I get 1 for the integral!
     
  9. Oct 11, 2007 #8

    Gib Z

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    That's a good point...

    [tex]\int^e_1 \log x dx = x(\log x -1) \right|^e_1 = e(1-1) - 1(\log 1 -1)= 0 - (\log 1 -1) = 1[/tex]..

    Great, just great, i forgot the brackets on the log 1 -1 when I was doing it in my head, now i've made an ass of myself =]. Please ignore any thing i have said.
     
  10. Oct 11, 2007 #9
    So I am on the right track?
     
  11. Oct 11, 2007 #10

    Gib Z

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    Yes probably, seeing as Halls is helping you
     
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