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Jhannybean

  • one year ago

Calculus 2: For what values of m do the line \(y=mx\) and the curve \(y=\dfrac{x}{x^2+1}\) enclose a region? Find the area of the region.

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  1. triciaal
    • one year ago
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    |dw:1441257093574:dw|

  2. Jhannybean
    • one year ago
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    Sorry, havent posted questions for a long time.

  3. Jhannybean
    • one year ago
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    SoI know you set them equal to eachother. \[mx = \frac{x}{x^2+1}\] Right?

  4. dan815
    • one year ago
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    find asymtotes and turning points of x/x^2+1

  5. triciaal
    • one year ago
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    |dw:1441257180743:dw|

  6. Jhannybean
    • one year ago
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    And why are we using the derivative here?

  7. dan815
    • one year ago
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    http://prntscr.com/8c0uw1

  8. ganeshie8
    • one year ago
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    |dw:1441257413232:dw|

  9. ganeshie8
    • one year ago
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    \[mx = \frac{x}{x^2+1}\implies x = 0\lor x = \pm \sqrt{\frac{1}{m}-1}\]

  10. Jhannybean
    • one year ago
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    Oh, this gives the points of intersection,I see

  11. ganeshie8
    • one year ago
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    you need at least two points of intersection to get an "enclosed" region

  12. Jhannybean
    • one year ago
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    Someone asked me about this question and I wasnt sure how to explain it to them lol.

  13. ganeshie8
    • one year ago
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    \(x=0\) is always an intersection, no restrictions on this however look at the other intersection, it sure has restrictions because of that square root

  14. Jhannybean
    • one year ago
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    So then our points on the enclosed region would be \(x=0,~ \pm \sqrt{\frac{1}{m}-1}\) and so therefore to find the enclosed region we could just take... \[A=\int_0^{\sqrt{\frac{1}{m}-1}} \left(\frac{x}{x^2+1}-mx\right)dx\] Am I missing anything?

  15. Jhannybean
    • one year ago
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    Oh wait... if I took the region from \(x=0 \rightarrow x=\sqrt{\frac{1}{m}-1}\) I'd be leaving out the other region wouldnt i....

  16. anonymous
    • one year ago
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    It's symmetric, so you can just double it. Also, you'd want to note that the first part of the answer is \(0 <m < 1 \).

  17. Jhannybean
    • one year ago
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    Oh! I forgot about the doubling.

  18. Jhannybean
    • one year ago
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    Yeah, realized the \(0 < m<1\) part. The area would then be :\[A= 2\int_0^{\sqrt{\frac{1}{m}-1}} \left(\frac{x}{x^2+1}-mx\right)dx\]

  19. Jhannybean
    • one year ago
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    Not really sure how to go about solving this though :\

  20. ganeshie8
    • one year ago
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    u substitution will do

  21. ganeshie8
    • one year ago
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    let \(u=x^2+1\)

  22. Jhannybean
    • one year ago
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    \[u=x^2+1 ~,~ du = 2xdx\]\[A=2\left[\frac{1}{2}\left(\frac{du}{u}\right) \right]_0^{\sqrt{\frac{1}{m}-1}} \]

  23. Jhannybean
    • one year ago
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    Part of it, so incorrect...

  24. Jhannybean
    • one year ago
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    If I let \(u=x^2+1\), then what will happen to the \(mx\)?

  25. Jhannybean
    • one year ago
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    \[A=2\left[\left( \int_0^{\sqrt{\frac{1}{m}-1}} \frac{1}{2}\frac{du}{u} \right) -\left( \int_0^{\sqrt{\frac{1}{m}-1}} mx \right)\right]\]

  26. Jhannybean
    • one year ago
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    eventually im supposed to get: \[\left[\ln \left(\frac{1}{x^2+1}\right) -mx^2\right]_0^{\sqrt{\frac{1}{m}-1}}\]

  27. DanJS
    • one year ago
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    just read through the thing, looks good, my first thought was the thing is an odd function , reflected over both axis... so you can worry about one side maybe... nice prob

  28. DanJS
    • one year ago
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    The integral of (1/u) * du = ln(u) why the answer 1/u

  29. DanJS
    • one year ago
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    so ya need the Area as a function of the slope then right

  30. DanJS
    • one year ago
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    I figure \[A(m) =m+\ln(\frac{ 1 }{ m })-1\] same answer if you use the negative root or the positive root

  31. DanJS
    • one year ago
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    i have to run, i was just looking through this site... i forgot about it.. it is pretty nice http://tutorial.math.lamar.edu/

  32. DanJS
    • one year ago
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    oh yeah, check this out too, there are a crap ton of all types of probs w/ the solutions from that patricjmt dude on youtube... https://drive.google.com/folderview?id=0B1YZD9uzvB5TfnkzNzg5LU54VC1zUENBNHZoWlUxYXk1OXQtNklXdVVtZnA1T3c4X3hMMWM&usp=drive_web

  33. Jhannybean
    • one year ago
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    \[A = 2\int_0^{\sqrt{\frac{1}{m}-1}} \left(\frac{x}{x^2+1}-mx\right)dx\]\[A=2\int_0^{\sqrt{\frac{1}{m}-1}}\left[x\left(\frac{1}{x^2+1}-m\right)\right] dx\] \[u=x^2+1~,~ du = 2xdx \iff xdx = \frac{du}{2}\]\[x=\sqrt{\frac{1}{m}-1} ~\therefore~ u =\frac{1}{m}-1+1 = \frac{1}{m} ~,~ x=0 ~\therefore ~u=1 \] \[A=2\int_1^{\frac{1}{m}} \left(\frac{1}{u}-m\right)\frac{du}{2}\]\[A=2 \cdot \frac{1}{2} \int_1^{\frac{1}{m}} \left(\frac{1}{u}-m\right)du\] \[A = \left[\ln(u) -mu\right]_1^{1/m}\]\[A= \left(\ln\left(\frac{1}{m}\right)-m \cdot \frac{1}{m} \right) -\left(\ln(1)-(1)m\right)\]\[\boxed{A=\ln(m) - 1+m~~ \text{sq units}} \]

  34. Jhannybean
    • one year ago
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    That's awesome @DanJS and really helpful!

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