Results 1 to 4 of 4

Math Help - Length Of the curve

  1. #1
    Newbie
    Joined
    May 2007
    Posts
    3

    Length Of the curve

    Hey all, I cant seem to figure out this last question..I have the answer but I cant get the full step by step process. Please help

    Find the length of the curve: Y = ln x on 1 less than and equal to x less than or equal to square root 3.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by sikhest View Post
    Hey all, I cant seem to figure out this last question..I have the answer but I cant get the full step by step process. Please help

    Find the length of the curve: Y = ln x on 1 less than and equal to x less than or equal to square root 3.
    use the formula: s = \int_{1}^{ \sqrt {3}} \sqrt {1 + (y')^2}dx

    where s is the length of the curve on the desired interval and y' is the derivative of y with respect to x
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    May 2007
    Posts
    3
    Hey thanks, All that is left to do is change the bounds and solve it..any ideas on how the bounds are changed?

    Thanks
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,904
    Thanks
    765
    Hello, sikhest!

    Find the length of the curve: . y = \ln x .on 1 \leq x \leq \sqrt{3}

    If you followed Jhevon's advice: . y' = \frac{1}{x}

    Then: . S \;=\;\int^{\sqrt{3}}_1\sqrt{1 + \frac{1}{x^2}}\:dx \;=\; \int^{\sqrt{3}}_1\frac{\sqrt{x^2+1}}{x}\:dx


    Let x = \tan\theta\quad\Rightarrow\quad dx = \sec^2\!\theta\ d\theta

    . . and we have: . S \;=\;\int\frac{\sec\theta}{\tan\theta}(\sec^2\!\th  eta\ d\theta) \;=\;\int\frac{\sec^3\!\theta}{\tan\theta}\ d\theta


    To change the limits, we have: . \tan\theta = x
    . . When x = 1, we have: . \tan\theta = 1\quad\Rightarrow\quad \theta = \frac{\pi}{4}
    . . When x = \sqrt{3}, we have: . \tan\theta = \sqrt{3}\quad\Rightarrow\quad\theta = \frac{\pi}{3}


    Therefore, the integral is: . S \;=\;\int^{\frac{\pi}{3}}_{\frac{\pi}{4}} \frac{\sec^3\!\theta}{\tan\theta}\ d\theta

    Good luck!

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Length of a curve
    Posted in the Calculus Forum
    Replies: 6
    Last Post: December 6th 2010, 12:24 PM
  2. Length of a curve
    Posted in the Calculus Forum
    Replies: 13
    Last Post: March 29th 2009, 05:32 PM
  3. arc length and parameterized curve length
    Posted in the Calculus Forum
    Replies: 1
    Last Post: December 5th 2008, 03:33 AM
  4. length of curve
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 19th 2007, 06:21 PM
  5. length of curve
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 6th 2007, 09:54 PM

Search Tags


/mathhelpforum @mathhelpforum