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Homework Help: 2 Finding length of curve problems

  1. Jan 24, 2007 #1
    1. Find the exact length of the curve analytically by antidifferentiation. You will need to simplify the integrand algebraically before finding an antiderivative.

    y = the integral from -2 to x of the SQUARE ROOT (3t^4-1)dt, -2 < x < -1

    note that the "<" is actually less than or equal to, don't know how to post that.

    For this one, do I just plug the x in? x_x I'm really clueless on how to start




    2. Find the length of the curve.

    y = the integral of 0 to x of SQUARE ROOT (cos(2t))dt from x = 0 to x = pi/4


    The problem with me is I know how to do it in terms of y and x, but I am terrible at parametrics.

    If you can, please help me with any of the above =) thanks
     
    Last edited: Jan 24, 2007
  2. jcsd
  3. Jan 25, 2007 #2

    Gib Z

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    Heres the arc length formula:

    [tex]\int_a^b \sqrt{1+\frac{dy}{dx}} dx [/tex]. Sub in the requirements, easy enough to get.
     
    Last edited: Jan 25, 2007
  4. Jan 25, 2007 #3

    Gib Z

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    O sorry didnt actually read it through well. For the first one,

    [tex]\int^{-2}_{x} \sqrt{3t^4 -1} dt [/tex]. If there wasn't an X there, but instead a normal number like you normally see, you would find the integral and then sub in b into it, and - the integral with a subbed in. In this case just sub in X.
     
  5. Jan 25, 2007 #4

    Gib Z

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    For the second one, there should be a second parametric equation >.<
     
  6. Jan 25, 2007 #5
    how do you actually type in the integral sign and stuff?
     
  7. Jan 25, 2007 #6
    Here's one source:
    http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuideCommands.php [Broken]
    which will help with formatting the integrals

    And another source for starters: (crash course in LaTeX at these forums:
    https://www.physicsforums.com/misc/howtolatex.pdf
     
    Last edited by a moderator: May 2, 2017
  8. Jan 25, 2007 #7
    Oh, and to see specifically how Gib Z did it, click to quote him, and take a look at what he has.

    However, the formula has a small mistake in it...
    (So, I copied and pasted from the quote so I could change it more simply)
    [tex]\int_a^b \sqrt{1+(\frac{dy}{dx})^2} dx [/tex].
    There's supposed to be a squared in there...
     
  9. Jan 25, 2007 #8

    Gib Z

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    Yes, of course, drpizza's correct...I forget the squared, my bad :p
     
  10. Jan 25, 2007 #9

    Gib Z

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    Btw, rather than actually having to quote me, just click on my latex pictures, that'll show up what I typed to show that code.
     
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