Results 1 to 3 of 3

Math Help - Find Arc Length

  1. #1
    Member FalconPUNCH!'s Avatar
    Joined
    Sep 2007
    From
    Daly City, CA
    Posts
    167

    Find Arc Length

    Find the Arc Length of the curve:

    1.  x = \frac{y^4}{8} + \frac{1}{4y^(2)}

    1 < x < 2

    2.  y = ln(cos(x))

     0 < x < \frac{\pi}{3}
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Mar 2008
    Posts
    148
    For problem 2:
    L = \int_0^\frac{\pi}{3} \sqrt{1+\left(\frac{d(\ln(\cos(x))}{dx}\right)^2}

    1 + \left(\frac{d(\ln(\cos(x))}{dx}\right)^2 = 1 + \frac{\sin^2(x)}{\cos^2(x)} = 1 + \tan^2(x) = \sec^2(x).

    Taking the square root gives:

    L = \int_0^\frac{\pi}{3}\sec(x) = \ln\left(\frac{\cos(x/2)+\sin(x/2)}{\cos(x/2)-\sin(x/2)}\right) evaluated at the limits.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,689
    Thanks
    617
    Hello, FalconPUNCH!

    Formula: . L \;=\;\int^b_a\sqrt{1 + \left(\frac{dy}{dx}\right)^2}\,dx \;\;\text{ or }\;\;\int^b_a\sqrt{1 + \left(\frac{dx}{dy}\right)^2}\,dy


    I hope there is a typo in #1 . . .


    Find the Arc Length of the curve:

    1)\;\;x \:= \:\frac{y^4}{8} + \frac{1}{4y^2}\qquad 1 < {\color{red}y} < 2
    We have: . x \;=\;\frac{1}{8}y^4 + \frac{1}{4}y^{-2}

    . . . . . . . \frac{dx}{dy} \;=\;\frac{1}{2}y^3 - \frac{1}{2}y^{-3}

    . . . . \left(\frac{dx}{dy}\right)^2 \;=\;\left(\frac{1}{2}y^3 - \frac{1}{2}y^{-3}\right)^2 \;=\;\frac{1}{4}y^6 - \frac{1}{2} + \frac{1}{4}y^{-6}

    . . 1 + \left(\frac{dx}{dy}\right)^2 \;=\;1 + \frac{1}{4}y^6 - \frac{1}{2} + \frac{1}{4}y^{-6} \;=\;\frac{1}{4}y^6 + \frac{1}{2}+ \frac{1}{4}y^{-6} \;=\;\left(\frac{1}{2}y^3 + \frac{1}{2}y^{-3}\right)^2

    \sqrt{1 + \left(\frac{dx}{dy}\right)^2} \;=\;\frac{1}{2}y^3 + \frac{1}{2}y^{-3} \;=\;\frac{1}{2}\left(y^3 + y^{-3}\right)


    Therefore: . L \;=\;\frac{1}{2}\int^2_1\left(y^3 + y^{-3}\right)\,dy\quad\hdots Got it?



    2)\;\;y \:= \:\ln(\cos x) \qquad 0 < x < \frac{\pi}{3}

    We have: . \frac{dy}{dx}\;=\;\frac{-\sin x}{\cos x} \;=\;-\tan x

    . . . 1 + \left(\frac{dy}{dx}\right)^2 \;=\;1 + \tan^2\!x \;=\;\sec^2\!x

    . \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \;=\;\sqrt{\sec^2\!x} \;=\;\sec x


    Therefore: . L \;=\;\int^{\frac{\pi}{3}}_0 \sec x\,dx\quad\hdots Okay?

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Find length of arc
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 11th 2010, 09:06 PM
  2. Find Length a
    Posted in the Geometry Forum
    Replies: 2
    Last Post: January 5th 2009, 09:57 PM
  3. Find the length for PR
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: September 20th 2008, 04:28 AM
  4. Find the Length
    Posted in the Calculus Forum
    Replies: 4
    Last Post: October 29th 2007, 12:18 AM
  5. find arc length
    Posted in the Calculus Forum
    Replies: 2
    Last Post: March 18th 2007, 12:39 PM

Search Tags


/mathhelpforum @mathhelpforum