Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Finding an arc length, and why isn't latex working for me?

  1. Oct 16, 2005 #1
    Finding an arc length

    I am attempting to find the arc length of y = cuberoot[x] between (1,1) and (8,2).

    I solved the integral from 1 to 2 of sqrt[1+(3y^2)^2]dy. I used a formula from a table of integrals in my text to solve this integral. The solution I get is 68.19. I can see that this is not a reasonable answer. Is my setup incorrect or am I solving the integral incorrectly?

    Thanks for any help.

    Steve
     
    Last edited: Oct 16, 2005
  2. jcsd
  3. Oct 16, 2005 #2
    Error vv
     
    Last edited: Oct 16, 2005
  4. Oct 16, 2005 #3

    Gokul43201

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    The set-up is okay. Looks like you're solving the integral incorrectly. Perhaps, if you show the working, someone will spot the error.
     
  5. Oct 16, 2005 #4

    Fermat

    User Avatar
    Homework Helper

    I thought it wasn't working for me either. But it seems to be not working properly only in the preview pane, when you are composing your thread/response.

    When posted, the latex code works properly (at least for me it did :))

    You can hit the edit button on your post and edit it there as need be.
     
  6. Oct 16, 2005 #5
    You're right Fermat, I was only trying it in the preview page, but it's working now.

    Thanks Gokul, this is how I got to my answer.

    This is the formula I used from my text:

    [tex]\int \sqrt {a^2 + u^2} du = \frac u 2 \sqrt {a^2 + u^2} + \frac {a^2} {2} \ln{(u + \sqrt {a^2 + u^2})} + C[/tex]

    I used a = 1 and u = 3y^2, and that is how I came up with 68.19.

    I'm trying to find a method to solve this integral without using that formula but I am having trouble. I can't see any logical u substitution that would work. I have tried a trigonometric substitution, with [tex]3y^2 = \tan{\theta}[/tex], but when I replace [tex]dy[/tex] with [tex]d\theta[/tex] I'm left with a harder integral than I started with. Thanks again for any help.

    Steve
     
    Last edited: Oct 16, 2005
  7. Oct 16, 2005 #6

    Fermat

    User Avatar
    Homework Helper

    have you tried,

    u = a.sinht ?
     
  8. Oct 16, 2005 #7

    Fermat

    User Avatar
    Homework Helper

    I think the integal you derived

    [tex]\int \sqrt {a^2 + u^2}\ du[/tex]

    is wrong.

    If the integral is,

    [tex]\int \sqrt {1 + (3y^2)^2}\ dy[/tex]

    then the substitution u = 3y² gives,

    [tex]\int \sqrt {1 + u^2}\ dy[/tex]

    but

    du = 6y dy

    or

    dy = du/6y = √3.du/(6√u)

    which givves the integral as,

    [tex]\int \sqrt {1 + u^2}\ \sqrt{3}du/(6\sqrt{u})[/tex]

    Also, I put your origianl integral,

    [tex]\int \sqrt {1 + (3y^2)^2}\ dy[/tex]

    into this http://integrals.wolfram.com/" [Broken] and got back exotic formulas (I think they're called)

    I don't know how to evaluate those, sorry :frown:
     
    Last edited by a moderator: May 2, 2017
  9. Oct 16, 2005 #8
    I tried it at a similar site and got back some crazy stuff that didn't mean much to me. :frown:

    I'm still not getting anywhere with this, I'm really curious about what I'm doing wrong. I'm sure it's just a simple mistake.

    I guess I'll start working on another method and see if I get anywhere trying to find:

    [tex]\int_{1}^{8} \sqrt{1 + \left[\frac {1}{3\sqrt[3]{x^2}}\right]^2} dx[/tex]

    Steve
     
    Last edited by a moderator: May 2, 2017
  10. Oct 16, 2005 #9
    Is this integral not elementary? Do I need to use approximate integration?

    Steve
     
  11. Oct 17, 2005 #10
    btt.......
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook