## anonymous 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.

1. triciaal

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2. anonymous

Sorry, havent posted questions for a long time.

3. anonymous

SoI know you set them equal to eachother. $mx = \frac{x}{x^2+1}$ Right?

4. dan815

find asymtotes and turning points of x/x^2+1

5. triciaal

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6. anonymous

And why are we using the derivative here?

7. dan815
8. ganeshie8

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9. ganeshie8

$mx = \frac{x}{x^2+1}\implies x = 0\lor x = \pm \sqrt{\frac{1}{m}-1}$

10. anonymous

Oh, this gives the points of intersection,I see

11. ganeshie8

you need at least two points of intersection to get an "enclosed" region

12. anonymous

13. 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

14. anonymous

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. anonymous

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

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. anonymous

Oh! I forgot about the doubling.

18. anonymous

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. anonymous

Not really sure how to go about solving this though :\

20. ganeshie8

u substitution will do

21. ganeshie8

let $$u=x^2+1$$

22. anonymous

$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. anonymous

Part of it, so incorrect...

24. anonymous

If I let $$u=x^2+1$$, then what will happen to the $$mx$$?

25. anonymous

$A=2\left[\left( \int_0^{\sqrt{\frac{1}{m}-1}} \frac{1}{2}\frac{du}{u} \right) -\left( \int_0^{\sqrt{\frac{1}{m}-1}} mx \right)\right]$

26. anonymous

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

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

The integral of (1/u) * du = ln(u) why the answer 1/u

29. DanJS

so ya need the Area as a function of the slope then right

30. DanJS

I figure $A(m) =m+\ln(\frac{ 1 }{ m })-1$ same answer if you use the negative root or the positive root

31. 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/

32. 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

33. anonymous

$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. anonymous

That's awesome @DanJS and really helpful!