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Homework Help: Arc Length (no calculators)?

  1. Jan 27, 2009 #1
    2ng5rg7.jpg

    How in gods name do I do that? I attempted that integral and... it just can't be integrated!

    What I tried:
    ih5obl.jpg

    That doesn't help one bit... How do I do this? NOTE: No graphing calculator is to be used.
     
  2. jcsd
  3. Jan 27, 2009 #2

    Mark44

    Staff: Mentor

    You have as the integrand
    [tex]\sqrt{1 + (x - 1/(4x))^2}[/tex]
    [tex]= \sqrt{1 + (x^2 - 1/2 + 1/16x^2)}[/tex]

    When you group together the only terms that can be grouped, you'll have a perfect square under the radical.
     
  4. Jan 27, 2009 #3
    when you squared out the (x-x/4)^2 you should have got 1/2 not 3/2. this should simplify things hopefully cause you should be able to factorise the numerator into
    sqrt[(4x^2+1)^2]
     
  5. Jan 27, 2009 #4
    Yes that gives me:
    2945imx.jpg

    However, what good is that, since there's still a 1 in there? I can't squareroot it to simplify...

    You do get 1/2, but I added a 1, therefore 3/2.
     
  6. Jan 27, 2009 #5

    Mark44

    Staff: Mentor

    Continuing from this point...
    [tex]= \sqrt{x^2 + 1/2 + 1/16x^2}[/tex]
    The part under the radical is a perfect square. Surely you can take it from here!
     
  7. Jan 27, 2009 #6
    OH... I added wrong... wow... Stupid mistakes FTL... Thanks all!
     
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