Results 1 to 5 of 5

Math Help - Line integrals

  1. #1
    Member
    Joined
    Jun 2006
    Posts
    93

    Line integrals

    F, i, j, r, q, p are vectors

    F(x,y) = (2xy)i - (x^2)j

    div F(x,y) = 2y

    curl F(x,y) = -4x which is not equal to zero, allowing no potential.

    Now this is the confusing part. Work = 25/3

    Here is how they got it

    Row (or atleast I think that's what it is) is parameterized by r(t) = p + t(q-p)
    x(t) = 1+t
    y(t) = 3+t
    p=<1,3>
    q=<2,4>
    0 <= t <= 1

    INT(F dot dr) = INT(0 to 1) 2(1+t)(3+t)dt - (1+t)^2 dt
    u = 1+t
    du = dt
    then on from there to get 25/3

    My question is... what is all of that stuff for Work? I don't understand where the r(t) = p + t(q-p) comes from... same goes for all the values of x(t), y(t) etc...

    Someone help please! I had to postpone my finals due to military training and now I've forgotten so much.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by pakman View Post
    F, i, j, r, q, p are vectors

    F(x,y) = (2xy)i - (x^2)j

    div F(x,y) = 2y

    curl F(x,y) = -4x which is not equal to zero, allowing no potential.

    Now this is the confusing part. Work = 25/3
    .
    Since it is not path independent, you need to specify what path are you intergrating?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,839
    Thanks
    320
    Awards
    1
    Quote Originally Posted by ThePerfectHacker View Post
    Since it is not path independent, you need to specify what path are you intergrating?
    I don't really understand the parametrization part, but isn't he specifying a path there? (It's just a guess, it's too confusing for me to follow.)

    -Dan
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Jan 2007
    Posts
    42
    Quote Originally Posted by pakman View Post
    F, i, j, r, q, p are vectors

    F(x,y) = (2xy)i - (x^2)j

    div F(x,y) = 2y

    curl F(x,y) = -4x which is not equal to zero, allowing no potential.

    Now this is the confusing part. Work = 25/3

    Here is how they got it

    Row (or atleast I think that's what it is) is parameterized by r(t) = p + t(q-p)
    x(t) = 1+t
    y(t) = 3+t
    p=<1,3>
    q=<2,4>
    0 <= t <= 1

    INT(F dot dr) = INT(0 to 1) 2(1+t)(3+t)dt - (1+t)^2 dt
    u = 1+t
    du = dt
    then on from there to get 25/3

    My question is... what is all of that stuff for Work? I don't understand where the r(t) = p + t(q-p) comes from... same goes for all the values of x(t), y(t) etc...

    Someone help please! I had to postpone my finals due to military training and now I've forgotten so much.

    if you are asking about the integral

    INT(0 to 1) 2(1+t)(3+t)dt - (1+t)^2 dt

    just expand and integrate term by term you do not need a substitution

    if you are asking about r(t) = p + t(q-p), i think it is the equation of a line passing through points with position vector p and q where t is a parameter

    you might have been asked to evaluate the work done in moving from p=<1,3> to q=<2,4>

    substituting into r(t) = p + t(q-p) and simplifying,

    you get r(t) =<1+t , 3+t>

    if r(t) = <x(t),y(t)>

    <x(t),y(t)>=<1+t , 3+t> will give you x(t) = 1+t y(t) = 3+t

    work done is line integral of F.dr

    replace the x and y in F(x,y) = (2xy)i - (x^2)j with x(t) = 1+t y(t) = 3+t
    and take the dot product with r(t) =<1+t , 3+t>

    to get INT(0 to 1) 2(1+t)(3+t)dt - (1+t)^2 dt

    when t =0 and 1, you get p and q from r(t) =<1+t , 3+t> hence the limits
    Last edited by qpmathelp; March 29th 2007 at 07:07 AM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member ecMathGeek's Avatar
    Joined
    Mar 2007
    Posts
    436
    Quote Originally Posted by qpmathelp View Post
    if you are asking about the integral

    INT(0 to 1) 2(1+t)(3+t)dt - (1+t)^2 dt

    just expand and integrate term by term you do not need a substitution
    He doesn't understand the setup for the integration. And I don't blame him. Look at the steps he used to arrive at that integration. They make almost no sense.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: December 11th 2011, 11:30 PM
  2. Line Integrals
    Posted in the Calculus Forum
    Replies: 2
    Last Post: April 24th 2011, 06:52 AM
  3. Line Integrals
    Posted in the Calculus Forum
    Replies: 2
    Last Post: March 11th 2009, 08:01 PM
  4. Line Integrals Across Line Segment
    Posted in the Calculus Forum
    Replies: 1
    Last Post: December 3rd 2008, 12:19 PM
  5. Line integrals
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 14th 2007, 07:10 PM

Search Tags


/mathhelpforum @mathhelpforum