Results 1 to 4 of 4

Math Help - Arc Length Trig substitution

  1. #1
    Senior Member mollymcf2009's Avatar
    Joined
    Jan 2009
    From
    Charleston, SC
    Posts
    490
    Awards
    1

    Arc Length Trig substitution

    Find the arc length:
    <br />
y = \frac{1}{2} x^2 from point P to point Q: P(-3,\frac{9}{2}) Q(3,-\frac{9}{2})

    Here is what I have so far:

    Length = 2 \int\limits^{3}_{0} \sqrt{1 + x^2} dx

    = 2 \int\limits^{\sqrt{10}}_{1} \sqrt{tan^2\theta + 1} \cdot sec^2\theta d\theta

    <br />
= 2 \int\limits^{\sqrt{10}}_{1} sec^3\theta d\theta

    Now what?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Feb 2009
    Posts
    36
    I=integration(sec^3 A dA)=INT.[secA*sec^2AdA]

    now,apply integration by parts,
    I=secA*INT[sec^2AdA]-int.[secA*tan^2AdA]


    =secA*tanA-int.(sec^3AdA)+log(secA+tanA)

    so,2I=secA*tanA+log(secA+tanA)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member mollymcf2009's Avatar
    Joined
    Jan 2009
    From
    Charleston, SC
    Posts
    490
    Awards
    1
    Quote Originally Posted by sbcd90 View Post
    I=integration(sec^3 A dA)=INT.[secA*sec^2AdA]

    now,apply integration by parts,
    I=secA*INT[sec^2AdA]-int.[secA*tan^2AdA]


    =secA*tanA-int.(sec^3AdA)+log(secA+tanA)

    so,2I=secA*tanA+log(secA+tanA)


    I am not getting what you typed. Under the "now apply int. by parts" you did two separate int by parts there? I can't tell what that is. Can you tell me what you used for your substitutions u & dv?
    Thanks!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Feb 2009
    Posts
    36
    u=secA

    v=sec^2 A.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Trig Substitution
    Posted in the Calculus Forum
    Replies: 2
    Last Post: February 23rd 2010, 01:12 PM
  2. Trig question help (Determining length)
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: November 13th 2009, 01:09 PM
  3. Trig substitution
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 4th 2009, 12:37 PM
  4. Trig Substitution Help
    Posted in the Calculus Forum
    Replies: 2
    Last Post: March 1st 2009, 11:29 PM
  5. Trig Substitution
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 1st 2009, 11:21 AM

Search Tags


/mathhelpforum @mathhelpforum