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Homework Help: Area - Riemann Sum/Integration problem

  1. Jan 13, 2006 #1
    This problem is in the section of my book called "The Natural Logarithmic Function" so I am guessing that I would have to use a natural log somewhere...but anyways...the problem says:

    For what values of [tex]m[/tex] do the line [tex]y=mx[/tex] and the curve [tex]y=\frac{x}{(x^2+1)}[/tex] enclose a region? Find the area of the region.

    For some reason...I related this problem to an optimization problem because I thought if I could find the line with the greatest slope that is tangent to the curve, then I would know the values for which [tex]m[/tex] can be used with...but that didn't seem to use any natural logs or anything..
     
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  3. Jan 13, 2006 #2

    TD

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    It may be a good idea to find the intersection points. If there aren't any intersection points, there's no enclosed region.
    It can also be useful to graph the problem for a few values of m, this gives you a better insight in the question.
     
  4. Jan 13, 2006 #3
    Well, from graphing the problem...it seems like [tex]y=x[/tex] may be the right bound of the interval for which [tex]y=mx[/tex] and the curve enclose a region...but I would need to show a calculation for this...

    I tried: [tex]mx= \frac{x}{(x^2+1)} m=\frac{1}{x^2+1}[/tex] but that only gives me the same equation of the curve or perhaps it gives me the equation for the bounds of the interval for which [tex]m[/tex] can be used with..

    Seems like [tex]m[/tex] would belong to [tex][0,1)[/tex]...
     
  5. Jan 13, 2006 #4

    TD

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    I think you're on the right track:

    [tex]m = \frac{1}
    {{x^2 + 1}} \Leftrightarrow m\left( {x^2 + 1} \right) = 1 \Leftrightarrow mx^2 + m - 1 = 0[/tex]

    Now, solving this would give the intersection points (which you'll be needing for you integration limits, I assume), but for the moment we're only interested in when there are solutions (i.e. when they intersect and thus enclose a region). You can express this by saying that the discriminant of this quadratic equation in x has to be positive, this will give a condition on m.
     
  6. Jan 14, 2006 #5
    Ahh....well, sorry to say this but I am still somewhat confused as to how that equation would be solved...two variables...need to solve for [tex]m[/tex] but that would just give us [tex]\frac{1}{x^2+1}[/tex] ... hmmm...
     
  7. Jan 14, 2006 #6

    siddharth

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    For the the straight line [tex] y=mx [/tex] and the curve [tex] y=\frac{x}{x^2+1} [/tex] to enclose some finite region, there should be atleast 2 points of intersection.
    One point is (0,0). To find the other points, you need to solve
    [tex] mx^2 + m -1 =0 [/tex].
    So for what values of 'm' does a solution to the above equation exist? That will give you the values of 'm' for which a region is enclosed.
    To find the area, find the points of intersection in terms of 'm'. You don't need to solve for both the variables, you only need to find x in terms of 'm'.
     
  8. Jan 14, 2006 #7
    Ohhhhhh, now I understand. I had understood the part about the 2 points of intersection but was trying to figure out which variable we solve that equation for and I now realize that we find the point on the curve using the intersecting line...which is why we solve for the x in terms of m. I got it, hehe.

    so to continue...I would then continue to do:
    [tex]\int_{0}^{\sqrt{\frac{1-m}{m}}} (\frac{x}{x^2+1}-mx)dx[/tex]

    Is this right?..m would also be a constant, right?
     
    Last edited: Jan 14, 2006
  9. Jan 14, 2006 #8

    siddharth

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    Yeah, that is almost right. What about the other point of intersection (ie, [itex]-\sqrt\frac{1-m}{m} [/itex]? so the area will be twice your integral.
     
  10. Apr 10, 2008 #9
    With:
    [tex] mx^2 + m -1 =0 [/tex]

    How can I solve for m?
     
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