Jhannybean
  • Jhannybean
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.
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
triciaal
  • triciaal
|dw:1441257093574:dw|
Jhannybean
  • Jhannybean
Sorry, havent posted questions for a long time.
Jhannybean
  • Jhannybean
SoI know you set them equal to eachother. \[mx = \frac{x}{x^2+1}\] Right?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

dan815
  • dan815
find asymtotes and turning points of x/x^2+1
triciaal
  • triciaal
|dw:1441257180743:dw|
Jhannybean
  • Jhannybean
And why are we using the derivative here?
dan815
  • dan815
http://prntscr.com/8c0uw1
ganeshie8
  • ganeshie8
|dw:1441257413232:dw|
ganeshie8
  • ganeshie8
\[mx = \frac{x}{x^2+1}\implies x = 0\lor x = \pm \sqrt{\frac{1}{m}-1}\]
Jhannybean
  • Jhannybean
Oh, this gives the points of intersection,I see
ganeshie8
  • ganeshie8
you need at least two points of intersection to get an "enclosed" region
Jhannybean
  • Jhannybean
Someone asked me about this question and I wasnt sure how to explain it to them lol.
ganeshie8
  • ganeshie8
\(x=0\) is always an intersection, no restrictions on this however look at the other intersection, it sure has restrictions because of that square root
Jhannybean
  • Jhannybean
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?
Jhannybean
  • Jhannybean
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....
anonymous
  • anonymous
It's symmetric, so you can just double it. Also, you'd want to note that the first part of the answer is \(0
Jhannybean
  • Jhannybean
Oh! I forgot about the doubling.
Jhannybean
  • Jhannybean
Yeah, realized the \(0 < m<1\) part. The area would then be :\[A= 2\int_0^{\sqrt{\frac{1}{m}-1}} \left(\frac{x}{x^2+1}-mx\right)dx\]
Jhannybean
  • Jhannybean
Not really sure how to go about solving this though :\
ganeshie8
  • ganeshie8
u substitution will do
ganeshie8
  • ganeshie8
let \(u=x^2+1\)
Jhannybean
  • Jhannybean
\[u=x^2+1 ~,~ du = 2xdx\]\[A=2\left[\frac{1}{2}\left(\frac{du}{u}\right) \right]_0^{\sqrt{\frac{1}{m}-1}} \]
Jhannybean
  • Jhannybean
Part of it, so incorrect...
Jhannybean
  • Jhannybean
If I let \(u=x^2+1\), then what will happen to the \(mx\)?
Jhannybean
  • Jhannybean
\[A=2\left[\left( \int_0^{\sqrt{\frac{1}{m}-1}} \frac{1}{2}\frac{du}{u} \right) -\left( \int_0^{\sqrt{\frac{1}{m}-1}} mx \right)\right]\]
Jhannybean
  • Jhannybean
eventually im supposed to get: \[\left[\ln \left(\frac{1}{x^2+1}\right) -mx^2\right]_0^{\sqrt{\frac{1}{m}-1}}\]
DanJS
  • DanJS
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
DanJS
  • DanJS
The integral of (1/u) * du = ln(u) why the answer 1/u
DanJS
  • DanJS
so ya need the Area as a function of the slope then right
DanJS
  • DanJS
I figure \[A(m) =m+\ln(\frac{ 1 }{ m })-1\] same answer if you use the negative root or the positive root
DanJS
  • DanJS
i have to run, i was just looking through this site... i forgot about it.. it is pretty nice http://tutorial.math.lamar.edu/
DanJS
  • DanJS
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
Jhannybean
  • Jhannybean
\[A = 2\int_0^{\sqrt{\frac{1}{m}-1}} \left(\frac{x}{x^2+1}-mx\right)dx\]\[A=2\int_0^{\sqrt{\frac{1}{m}-1}}\left[x\left(\frac{1}{x^2+1}-m\right)\right] dx\] \[u=x^2+1~,~ du = 2xdx \iff xdx = \frac{du}{2}\]\[x=\sqrt{\frac{1}{m}-1} ~\therefore~ u =\frac{1}{m}-1+1 = \frac{1}{m} ~,~ x=0 ~\therefore ~u=1 \] \[A=2\int_1^{\frac{1}{m}} \left(\frac{1}{u}-m\right)\frac{du}{2}\]\[A=2 \cdot \frac{1}{2} \int_1^{\frac{1}{m}} \left(\frac{1}{u}-m\right)du\] \[A = \left[\ln(u) -mu\right]_1^{1/m}\]\[A= \left(\ln\left(\frac{1}{m}\right)-m \cdot \frac{1}{m} \right) -\left(\ln(1)-(1)m\right)\]\[\boxed{A=\ln(m) - 1+m~~ \text{sq units}} \]
Jhannybean
  • Jhannybean
That's awesome @DanJS and really helpful!

Looking for something else?

Not the answer you are looking for? Search for more explanations.