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Homework Help: Length of the curve

  1. Mar 17, 2005 #1
    Find the length of the curve

    [tex] x = 3y^{4/3} - 3/32y^{2/3} [/tex]

    from [tex] -8 \leq y \leq 64 [/tex]

    dy/dx = [tex] 4*y^{1/3}-\frac{1}{16y^{1/3}}[/tex] <-- dont need to check this, i used a math tool to check it.

    if i plug all of this into a integral, its going to be crazy. so i was wondering if there is a different approach to this problem

    [tex] \int_{-8}^{64} \sqrt{1+{(4y^{1/3}-\frac{1}{16y^{1/3}})}^2}[/tex]
     
    Last edited: Mar 17, 2005
  2. jcsd
  3. Mar 17, 2005 #2

    cepheid

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    DO it. Why would it be "crazy?" You've just got some power functions.
     
  4. Mar 17, 2005 #3
    is my integral setup correctly?
     
  5. Mar 17, 2005 #4

    HallsofIvy

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    So far, yes.

    Actually MOST functions would give "crazy" arclength integrals- so teachers choose functions that simplify. Go ahead, square the formula and add the "1"- and see what happens (hint: the result is a perfect square).
     
  6. Mar 17, 2005 #5
    your integral is correct, you just forgot the dy at the end.

    [tex] \int_{-8}^{64} \sqrt{1+{(4y^{1/3}-\frac{1}{16y^{1/3}})}^2} dy[/tex]
     
  7. Mar 17, 2005 #6

    ZapperZ

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    I don't want to be picky, but what you differentiated was dx/dy, not dy/dx. That is how you could dx = {garbage}*dy.

    Zz.
     
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