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Homework Help: Length of a curve

  1. Oct 28, 2006 #1
    The Length of a Curve

    Hi, i'm stumped! :confused:

    I've been asked to calculate the length of a curve with the equation...

    [tex]\frac{x^2}{a^2}+\frac{y^2}{b^2}=1[/tex]

    for the range [tex]0 \leq x \leq a[/tex]


    I've been trying to apply the equation...

    [tex]\int\sqrt{1+\left(\frac{dy}{dx}\right)^2}\times dx[/tex]

    I think...

    [tex]y=b\times\sqrt{1-\frac{x^2}{a^2}}[/tex]

    [tex]\frac{dy}{dx}=\frac{-b\times x}{a^2\times\sqrt{1-\frac{x^2}{a^2}}}[/tex]

    [tex]\left(\frac{dy}{dx}\right)^2=\frac{b^2\times x^2}{a^2\times(a^2-x^2)}[/tex]

    [tex]Length=\int\sqrt{1+\frac{b^2\times x^2}{a^2\times(a^2-x^2)}}\times dx[/tex]

    and then if that is right i get stuck and don't know how to do that integral with the limits above.

    I tried to say that since...

    [tex]\frac{1}{a^2}[/tex] and [tex]\frac{1}{b^2}[/tex] are constants i can ignore them and follow the method here:

    http://www.maths.abdn.ac.uk/~igc/tch/ma1002/int/node21.html (example 2.8 the last one on the page)

    but instead of [tex]x^2[/tex] and [tex]y^2[/tex] i'd call my new variables [tex]c^2[/tex] and [tex]D^2[/tex] and then i thought if i did that i'd have to change the range and since [tex]\frac{x^2}{a^2}=c^2[/tex] then the new range would be [tex]0 \leq \sqrt{c^2 \times a^2}\leq a[/tex] which i thought

    might be the same as [tex]0 \leq c \leq 1[/tex] which i think gives an answer of [tex]\frac{\pi}{2}[/tex].

    But i think i'm probably wrong any help would be appreciated thanks :smile:
     
    Last edited by a moderator: Apr 22, 2017
  2. jcsd
  3. Oct 28, 2006 #2

    Office_Shredder

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    As a quick check to why your answer is wrong:

    The curve you're looking at is an ellipse. Surely you can agree that if a and b increase, the length will increase. So [tex]\frac{\pi}{2}[/tex] doesn't really make much sense
     
  4. Oct 28, 2006 #3

    arildno

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    sanitykey:
    Just forget about about trying to find the general expression for the length of the circumference of an ellipse.

    It can be done, but you'll only end up with a rather pointless series solution.
     
  5. Oct 28, 2006 #4
    Thanks for your replies, to Office_Shredder ~ yeah i was thinking [tex]\frac{\pi}{2}[/tex] didn't really make sense in fact errm oh yeah how can the length be a constant if i'm finding a general expression :redface: to arildno ~ would you suggest i should just leave my expression as that integral then if that is right?
     
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