Results 1 to 9 of 9

Math Help - Fundamental Theorem for Line Integrals

  1. #1
    Junior Member
    Joined
    Sep 2007
    Posts
    56

    Fundamental Theorem for Line Integrals

    (a) Find a function f such that F=\nabla f, and (b) use part (a) to evaluate \oint_C \vec{F}\cdot d  \vec{r} along the given curve C.

    F(x,y) = <xy^2, x^2y>
    C: r(t) = <t + sin(\frac{\pi t}{2}), t + cos(\frac{\pi t}{2})>

    _____________________________

    MY work:

    (a)

    F(x,y)=<xy^2, x^2y>

    Check if it's conservative or not:
    P=xy^2 and Q=x^2y

    \frac{\partial P}{\partial y} = 2xy
    \frac{\partial Q}{\partial x} = 2xy

    It is conservative.

    f_x = xy^2
    f=\frac{1}{2}x^2y^2 +g(y)
    f_y=x^2y +g'(y) = x^2y. So  g'(y) = 0, which means  g(y) = K

    Therefore, f=\frac{1}{2}xy^2 +K

    Is that correct so far? I'm kind of lost here. I'm confused what I'm supposed to do with the whole F=\nabla f thing.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Feb 2010
    Posts
    422
    Do you know how to compute the gradient? The f you have doesn't have F as its gradient (but its really close).
    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 Cursed View Post
    (a) Find a function f such that F=\nabla f, and (b) use part (a) to evaluate \oint_C \vec{F}\cdot d \vec{r} along the given curve C.

    F(x,y) = <xy^2, x^2y>
    C: r(t) = <t + sin(\frac{\pi t}{2}), t + cos(\frac{\pi t}{2})>

    _____________________________

    MY work:

    (a)

    F(x,y)=<xy^2, x^2y>

    Check if it's conservative or not:
    P=xy^2 and Q=x^2y

    \frac{\partial P}{\partial y} = 2xy
    \frac{\partial Q}{\partial x} = 2xy

    It is conservative.

    f_x = xy^2
    f=\frac{1}{2}x^2y^2 +g(y)
    f_y=x^2y +g'(y) = x^2y. So  g'(y) = 0, which means  g(y) = K

    Therefore, f=\frac{1}{2}xy^2 +K

    Is that correct so far? I'm kind of lost here. I'm confused what I'm supposed to do with the whole F=\nabla f thing.
    You can make the arbitrary choice K = 0. So f = \frac{1}{2} x^2 y^2.

    Then the given line integral is simply equal to f(B) - f(A) where A and B are the start and endpoints of the curve you are integrating along.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Sep 2007
    Posts
    56
    whoops. I forgot the constraints:

    \vec{r}(t) = <t + sin(\frac{\pi t}{2}), t + cos(\frac{\pi t}{2})>, where 0 \leq t \leq 1

    I know how to plug in points when you're given them, but what do you do with the vector equation?

    If I do \vec{F}(\vec{r}(u,v)), I get something really ugly. o.O
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by Cursed View Post
    whoops. I forgot the constraints:

    \vec{r}(t) = <t + sin(\frac{\pi t}{2}), t + cos(\frac{\pi t}{2})>, where 0 \leq t \leq 1

    I know how to plug in points when you're given them, but what do you do with the vector equation?

    If I do \vec{F}(\vec{r}(u,v)), I get something really ugly. o.O
    I have told you what to do. Where are you stuck? Can you not get the coordinates of A and B and substitute into f?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Sep 2007
    Posts
    56
    How do I get the coordinates of A and B if the problem doesn't give them?

    The problem just gives a vector function \vec{F} and a curve C.

    That's what I'm confused about.

    If I'm given the coordinates, I can plug them in. I just don't have any.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by Cursed View Post
    How do I get the coordinates of A and B if the problem doesn't give them?

    The problem just gives a vector function \vec{F} and a curve C.

    That's what I'm confused about.

    If I'm given the coordinates, I can plug them in. I just don't have any.
    You are told 0 \leq t \leq 1. Therefore A has coordinates corresponding to t = 0 and B has coordinates corresponding to t = 1. And you are told x = t + \sin \left(\frac{\pi}{2} t\right) and y = t + \cos \left(\frac{\pi}{2} t\right). Therefore the coordinates of A and B are ....
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    Sep 2007
    Posts
    56
    Oh, Okay. I didn't think you could do that. Thanks.

    My book said to plug \vec{r} into \vec{f} symbolically and then evaluate the integral from A to B, in this case: from t=0 to t=1. o_o
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,373
    Thanks
    1314
    Quote Originally Posted by Cursed View Post
    (a) Find a function f such that F=\nabla f, and (b) use part (a) to evaluate \oint_C \vec{F}\cdot d  \vec{r} along the given curve C.

    F(x,y) = <xy^2, x^2y>
    C: r(t) = <t + sin(\frac{\pi t}{2}), t + cos(\frac{\pi t}{2})>

    _____________________________

    MY work:

    (a)

    F(x,y)=<xy^2, x^2y>

    Check if it's conservative or not:
    P=xy^2 and Q=x^2y

    \frac{\partial P}{\partial y} = 2xy
    \frac{\partial Q}{\partial x} = 2xy

    It is conservative.

    f_x = xy^2
    f=\frac{1}{2}x^2y^2 +g(y)
    f_y=x^2y +g'(y) = x^2y. So  g'(y) = 0, which means  g(y) = K

    Therefore, f=\frac{1}{2}xy^2 +K
    This is probably a typo- you dropped the ^2 on the x.

    Now, what exactly IS the "given curve c"? You give parametric equations for a curve but no beginning and ending values for t.

    If you have such values, calculate the corresponding beginning and ending values for x and y, calculate F(x,y) for those values and subtract, just like you do for integrals on the real numbers. Of course, the constant, K, will cancel out when you do that. That is why mr fantastic said "You can make an arbitrary choice of K= 0".

    Is that correct so far? I'm kind of lost here. I'm confused what I'm supposed to do with the whole F=\nabla f thing.[/QUOTE]
    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. Replies: 5
    Last Post: April 22nd 2010, 05:17 PM
  3. Fundamental Theorem of Line Integrals
    Posted in the Calculus Forum
    Replies: 4
    Last Post: April 13th 2009, 01:11 PM
  4. Fundamental Theorem for Line Integrals
    Posted in the Calculus Forum
    Replies: 9
    Last Post: November 16th 2008, 04:06 PM
  5. integrals by fundamental theorem
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 24th 2007, 10:16 AM

Search Tags


/mathhelpforum @mathhelpforum