Results 1 to 6 of 6

Math Help - Find the length of curve

  1. #1
    Member
    Joined
    Sep 2007
    Posts
    198

    Find the length of curve

    The equations x = t + sin t, y = 1 cos t, 0≤t≤2pi, define a curve C parametrically from the origin O to A(2pi, 0), which is rotated completely about Ox to form a surface S.

    Find the length of the curve from O to A.



    As I did so far,

    dx/dt = 1 + cos t and dy/dt = sin t

    Curve length OA =  \int_{0}^{2\pi} [(1+cos t)^2 + sin^2t]^ \frac {1}{2}dt
    =\int_{0}^{2\pi} (2+2cos t)^\frac{1}{2}dt
    =\int_{0}^{2\pi}2 cos \frac{t}{2}dt

    My answer is 0. Where I did wrong?

    But its correct answer is 8. Please help me.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by geton View Post
    The equations x = t + sin t, y = 1 cos t, 0≤t≤2pi, define a curve C parametrically from the origin O to A(2pi, 0), which is rotated completely about Ox to form a surface S.

    Find the length of the curve from O to A.



    As I did so far,

    dx/dt = 1 + cos t and dy/dt = sin t

    Curve length OA =  \int_{0}^{2\pi} [(1+cos t)^2 + sin^2t]^ \frac {1}{2}dt
    =\int_{0}^{2\pi} (2+2cos t)^\frac{1}{2}dt
    =\int_{0}^{2\pi}2 cos \frac{t}{2}dt

    My answer is 0. Where I did wrong?

    But its correct answer is 8. Please help me.
    AL=\int_a^{b}\sqrt{\left(\frac{dx}{dt}\right)^2+\l  eft(\frac{dy}{dt}\right)^2}dt

    So seeing thta

    \frac{dx}{dt}=1+\cos(t)

    and

    \frac{dy}{dt}=\sin(t)

    We have

    \int_0^{2\pi}\sqrt{1+\cos^2(t)+2\cos(t)+<br />
\sin^2(t)}dt=\sqrt{2}\int_0^{2\pi}\sqrt{1+1\cos(t)  }dt

    Which is nasty...but

    I just realized

    \cos\left(\frac{x}{2}\right)=\pm\sqrt{\frac{1+\cos  (x)}{2}}

    So we see that

    in this quadrant

    \sqrt{2}\cos\left(\frac{x}{2}\right)=\sqrt{1+\cos(  x)}

    now make your sub

    (o wait...you had that haha)

    after your sub we get 8

    how did you get 0?

    Show your work so as that we can better help you
    Last edited by Mathstud28; June 15th 2008 at 05:10 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by geton View Post
    The equations x = t + sin t, y = 1 – cos t, 0≤t≤2pi, define a curve C parametrically from the origin O to A(2pi, 0), which is rotated completely about Ox to form a surface S.

    Find the length of the curve from O to A.



    As I did so far,

    dx/dt = 1 + cos t and dy/dt = sin t

    Curve length OA =  \int_{0}^{2\pi} [(1+cos t)^2 + sin^2t]^ \frac {1}{2}dt
    =\int_{0}^{2\pi} (2+2cos t)^\frac{1}{2}dt
    =\int_{0}^{2\pi}2 cos \frac{t}{2}dt

    My answer is 0. Where I did wrong?

    But its correct answer is 8. Please help me.
    Length is positive. So the length is the magnitude of the area enclosed by 2 \cos \left( \frac{t}{2} \right) and the t-axis between t = 0 and t = 2 pi. How would you find that area ......? You'd draw a graph of 2 \cos \left( \frac{t}{2} \right) versus t and break it up, right?:

    \int_{0}^{\pi}2 \cos \left( \frac{t}{2}\right) \, dt {\color{red}-} \int_{\pi}^{2 \pi}2 \cos \left( \frac{t}{2}\right)  \, dt since the signed area is negative from pi to 2pi ......

    Capisce?
    Last edited by mr fantastic; June 15th 2008 at 05:36 PM. Reason: Fixed an obvious typo
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by geton View Post
    Curve length OA =  \int_{0}^{2\pi} [(1+cos t)^2 + sin^2t]^ \frac {1}{2}dt
    =\int_{0}^{2\pi} (2+2cos t)^\frac{1}{2}dt
    =\int_{0}^{2\pi}2 cos \frac{t}{2}dt
    = 2*2 [sin \frac{t}{2}]_{0}^{2\pi}
    =4(sin\pi - sin 0)

    Here is my problem. sin\pi = 0
    Oh! We both made a stupid mistake

    There is a sign change! With the plus or minus, it isn't plus over 0 to 2pi, it changes signs at pi

    EDIT: Mr. F beat me to it
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Sep 2007
    Posts
    198
    Quote Originally Posted by mr fantastic View Post

    \int_{0}^{\pi}2 \cos \left( \frac{t}{2}\right) \, dt {\color{red}-} \int_{\pi}^{\pi}2 \cos \left( \frac{t}{2}\right)  \, dt since the signed area is negative from pi to 2pi ......
    But isn't

    <br /> <br />
\int_{0}^{\pi}2 \cos \left( \frac{t}{2}\right) \, dt - \int_{\pi}^{2\pi}2 \cos \left( \frac{t}{2}\right)  \, dt<br />
?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by geton View Post
    But isn't

    <br /> <br />
\int_{0}^{\pi}2 \cos \left( \frac{t}{2}\right) \, dt - \int_{\pi}^{2\pi}2 \cos \left( \frac{t}{2}\right) \, dt<br />
?
    Clearly I made an obvious typo. Fixed.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Find the length of the curve
    Posted in the Calculus Forum
    Replies: 4
    Last Post: April 29th 2010, 11:21 AM
  2. Trying to find the length of a curve.
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: February 13th 2010, 06:20 AM
  3. Find the length of the curve
    Posted in the Calculus Forum
    Replies: 1
    Last Post: December 6th 2009, 09:56 AM
  4. find length of curve
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 15th 2009, 05:55 PM
  5. Find length of the curve help
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 4th 2008, 10:15 PM

Search Tags


/mathhelpforum @mathhelpforum