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Homework Help: Need Help Integrating This Tricky Integrand

  1. Sep 17, 2011 #1
    Find the length of the curve:

    r(t) = <2t, t2, (1/3)t3>
    r'(t) = <2, 2t, t2>

    From bounds of t: 0 to 1.

    So length = integral of the modulus of r'(t):

    Integral of sqrt(t4+4t2+4)

    I'm just dead stuck on how to attack it. I tried to make it integral of sqrt((t2+2)2), and then just getting rid of the square, but I'm feeling intrinsically unsure that that way will work.

    Would setting a u = t4 help at all?

    Any help is appreciated!
     
  2. jcsd
  3. Sep 17, 2011 #2
    Yeah, sure, try it that way!!
     
  4. Sep 17, 2011 #3
    I tried it and got a numerical value. So, I guess it's alright then.

    Is it too much to ask for help on a different one? It's in the same form.

    r(t) = <12t, 8t3/2, 3t2>
    r'(t) = <12, 12t1/2, 9t>

    Bounds of t = 0 to 1

    Integral of sqrt(122 +(12t1/2)2 + (9t)2)dt

    = integral sqrt(144 + 144t +81t2)dt.

    I'm really just baffled; these are problems 1-10, which are technically supposed to be pretty simple. Am I just missing a glaringly easy way to integrate these things?
     
  5. Sep 17, 2011 #4
    Try to complete the square here. Prepare for a trigonometric substitution.
     
  6. Sep 17, 2011 #5
    I'm having trouble after completing the square. I don't remember too much about trigonometric substitution... :(

    I factored out the 81, to move a constant out:

    9 * (integral of sqrt(t2 + (16/9)t +16/9))

    So I took the square of half of 16/9:

    9 * (integral of sqrt([t2 + (16/9)t + (64/81)] + 80/81))

    9 * (integral of sqrt((t+(8/9))2 + 80/81)

    EDIT: And just for questioning's sake; the bounds are from 0 to 1, so I'm guessing the answer itself shouldn't be hard to compute???
     
  7. Sep 17, 2011 #6
    Factor out 80/81 and set

    [tex]\tan^2{u}=\frac{81}{80}(t+\frac{8}{9})^2[/tex]
     
  8. Sep 17, 2011 #7
    It might be the energy drink crash I'm getting here, but I don't see how the 80/81 can be factored out...

    Sorry :/
     
  9. Sep 17, 2011 #8
    [tex]\sqrt{x-a}=\sqrt{a}\sqrt{\frac{x}{a}-1}[/tex]

    Do something like this.
     
  10. Sep 17, 2011 #9
    So I get

    a = 80/81

    sqrt(a) * (sqrt((t + 8/9)2)/a + 1)

    I can't see how the sqrt goes away. Was I supposed to change the sign in b/t the binomal squared term and the 1?
     
  11. Sep 17, 2011 #10
    Now make an appropriate trigonometric substitution. Substitute tan in.
     
  12. Sep 17, 2011 #11
    Do you mean in the sense of the derivative of arctan? (1/(1+x2))?

    If I substitute tan in,

    u = tan(t+(8/9))2,

    Is this the correct substitution? It seems different from yours.
     
  13. Sep 21, 2011 #12
    Ah ha! I found it. By factoring out scalars from the vectors, I made my life a ton easier.

    I found it out after consulting my textbook.

    Thanks though :)
     
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