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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.
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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triciaal
 one year ago
Best ResponseYou've already chosen the best response.0dw:1441257093574:dw

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.3Sorry, havent posted questions for a long time.

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.3SoI know you set them equal to eachother. \[mx = \frac{x}{x^2+1}\] Right?

dan815
 one year ago
Best ResponseYou've already chosen the best response.0find asymtotes and turning points of x/x^2+1

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0dw:1441257180743:dw

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.3And why are we using the derivative here?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2dw:1441257413232:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2\[mx = \frac{x}{x^2+1}\implies x = 0\lor x = \pm \sqrt{\frac{1}{m}1}\]

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.3Oh, this gives the points of intersection,I see

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2you need at least two points of intersection to get an "enclosed" region

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.3Someone asked me about this question and I wasnt sure how to explain it to them lol.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2\(x=0\) is always an intersection, no restrictions on this however look at the other intersection, it sure has restrictions because of that square root

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.3So 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
 one year ago
Best ResponseYou've already chosen the best response.3Oh 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
 one year ago
Best ResponseYou've already chosen the best response.0It's symmetric, so you can just double it. Also, you'd want to note that the first part of the answer is \(0 <m < 1 \).

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.3Oh! I forgot about the doubling.

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.3Yeah, 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
 one year ago
Best ResponseYou've already chosen the best response.3Not really sure how to go about solving this though :\

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2u substitution will do

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.3\[u=x^2+1 ~,~ du = 2xdx\]\[A=2\left[\frac{1}{2}\left(\frac{du}{u}\right) \right]_0^{\sqrt{\frac{1}{m}1}} \]

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.3Part of it, so incorrect...

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.3If I let \(u=x^2+1\), then what will happen to the \(mx\)?

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.3\[A=2\left[\left( \int_0^{\sqrt{\frac{1}{m}1}} \frac{1}{2}\frac{du}{u} \right) \left( \int_0^{\sqrt{\frac{1}{m}1}} mx \right)\right]\]

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.3eventually im supposed to get: \[\left[\ln \left(\frac{1}{x^2+1}\right) mx^2\right]_0^{\sqrt{\frac{1}{m}1}}\]

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1just 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
 one year ago
Best ResponseYou've already chosen the best response.1The integral of (1/u) * du = ln(u) why the answer 1/u

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1so ya need the Area as a function of the slope then right

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1I figure \[A(m) =m+\ln(\frac{ 1 }{ m })1\] same answer if you use the negative root or the positive root

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1i have to run, i was just looking through this site... i forgot about it.. it is pretty nice http://tutorial.math.lamar.edu/

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1oh 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
 one year ago
Best ResponseYou've already chosen the best response.3\[A = 2\int_0^{\sqrt{\frac{1}{m}1}} \left(\frac{x}{x^2+1}mx\right)dx\]\[A=2\int_0^{\sqrt{\frac{1}{m}1}}\left[x\left(\frac{1}{x^2+1}m\right)\right] dx\] \[u=x^2+1~,~ du = 2xdx \iff xdx = \frac{du}{2}\]\[x=\sqrt{\frac{1}{m}1} ~\therefore~ u =\frac{1}{m}1+1 = \frac{1}{m} ~,~ x=0 ~\therefore ~u=1 \] \[A=2\int_1^{\frac{1}{m}} \left(\frac{1}{u}m\right)\frac{du}{2}\]\[A=2 \cdot \frac{1}{2} \int_1^{\frac{1}{m}} \left(\frac{1}{u}m\right)du\] \[A = \left[\ln(u) mu\right]_1^{1/m}\]\[A= \left(\ln\left(\frac{1}{m}\right)m \cdot \frac{1}{m} \right) \left(\ln(1)(1)m\right)\]\[\boxed{A=\ln(m)  1+m~~ \text{sq units}} \]

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.3That's awesome @DanJS and really helpful!
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