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Homework Help: Length of a curve

  1. Jan 20, 2004 #1
    Find the length of [tex] y = \frac{1}{3} (x^2 + 2)^{\frac{3}{2}} [/tex] from x=0 to x=3.

    I used the formula [tex] s = \int_{a}^{b} \sqrt{1 + \frac{\,dy}{\,dx}} \,dx [/tex].
    After plugging everything in, I got
    [tex] s = \int_{0}^{3} (1 + \frac{1}{4} (x^2 + 2))}^\frac{1}{2} \,dx [/tex]
    Now, this isn't an integral I've learned how to do, so
    1) Did I do anything wrong, and
    2) If yes, what?
     
    Last edited: Jan 20, 2004
  2. jcsd
  3. Jan 20, 2004 #2
    I realized I forgot to use the chain rule when doing dy/dx. However when I do that, I get even a messier integral,
    [tex] s = \int_{0}^{3} (1 + x^2(x^2 + 2))^\frac{1}{2} \,dx [/tex]

    ?
     
  4. Jan 20, 2004 #3
    It should be

    [tex] s = \int_{a}^{b} \sqrt{1 + (\frac{\,dy}{\,dx})^2} \,dx [/tex]

    But thats not the pro u just wrote the wrong one and applied the right

    u also have
    [tex]1+x^2(x^2+2) = 1+x^4+2x^2=x^4+2x^2+1=(x^2+1)^2[/tex]
    Is it enough
     
  5. Jan 21, 2004 #4
    Woah yeah, thanks, overlooked a simple thing there.
     
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