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Finding whether line integrals are independent of path. Watch

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    I have a question;

    "Determine whether of not the line integral  \int \vec{f} . d\vec{r} along the given curve C is independent of the path and evaluate the integral.

    (a)  \vec{f} = (xsin(y),y) along the parabola  y = x^2 from (-1,1) to (2,4)"




    I know you can work out if it's independent if  \int \vec{f} . d\vec{r} = 0 but I'm not exactly sure how to do it. My lecture notes don't explain this very well, what is  d\vec{r} ? It also says I should try to parametrise C, i.e  \vec{r} = \vec{r}(t) so that I can use  \int^b_a \vec{f}(\vec{r}(t)).\vec{r}'(t)d  t instead.

    It's all in another language, this section has been crammed in to a one page lecture note and I have no idea how I'm meant to approach this question. Can anyone help?
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    (Original post by Sasukekun)
    I know you can work out if it's independent if  \int \vec{f} . d\vec{r} = 0
    I think you might mean 'if the integral of every closed path is zero'.

    There's a couple of equivalent conditions for a line integral being independent of its path (and only depending on the start and end points):

    The curl of the integrand is zero.

    The integrand is a gradient (i.e. f=\nabla \phi for some scalar function \phi).

    The line integral along any closed path (start and end points are the same) is zero.

    what is  d\vec{r} ?
    Talking in a very non-rigourous way,  d\vec{r} is analogous to dx in the calculus you are used to: we can think of the integral as taking the scalar product of the integrand f with the line element  d\vec{r} at every point along the path and summing along the path, in the same way that a "normal" integral can be thought of as multiplying f by dx and summing along the range of integration.

    It also says I should try to parametrise C, i.e  \vec{r} = \vec{r}(t) so that I can use  \int^b_a \vec{f}(\vec{r}(t)).\vec{r}'(t)d  t instead.
    The standard way to evaluate a line integral like this is to parametrise the curve as your notes suggest, which turns it into a standard scalar integral. One way of parametrising the curve C in your question would be r(t)=(t,t^2) , with t ranging form -1 to 2.
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    (Original post by Mark13)
    Stuff
    Thanks, I think I've done that question now due to your help. Although I'm now stuck on another. It's asking for the same thing but for  \vec{f} = (sin(y), xcos(y)) for any closed circle.

    Now I realise for any closed circle it essentially means  (x - C)^2 + (y - D)^2 = r^2 where C and D are just random constants. Also  x = rcos(t) + C and  y = rsin(t) + D where t is the angle ranging from 0 to 2pi.

     \vec{r} = (x,y) = (rcos(t) + C,rsin(t) + D)

     d\vec{r} = (x,y) = (-rsin(t),rcos(t)) dt

    However how do I get  \vec{f} in terms of the parameter t? From instance the  sin(y) in  \vec{f} becomes  sin(rsin(t) +D) which surely can't be right?

    Am I doing the correct thing so far, what can I do?
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    (Original post by Sasukekun)
    Am I doing the correct thing so far, what can I do?
    I haven't checked your working thoroughly, but it all looks correct. However, you've probably noticed the algebra is pretty horrific.

    But from my first post, note that if the integrand is a gradient, then the integral along a closed path is zero. Can you show that the integrand is a gradient in this case?
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    (Original post by Mark13)
    Can you show that the integrand is a gradient in this case?
    Sorry man, I'm not exactly sure. What is the integrand in this case, and how would I know if it's a gradient?
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    (Original post by Sasukekun)
    Sorry man, I'm not exactly sure. What is the integrand in this case, and how would I know if it's a gradient?
    No worries. The integrand just means the bit inside the integral sign, which in this case is (sin(y),xcos(y)). If it is a gradient, then we can write it as:

    (\sin(y),x\cos(y))=\nabla \phi for some function \phi.

    If you haven't come across the \nabla notation before, all it means in this context is that ( \sin(y),x \cos(y))=( \frac{ \partial  \phi}{ \partial x}, \frac{ \partial \phi}{ \partial y}). So basically, if there's a function \phi that satisfies:

    \frac{\partial \phi}{\partial x}=\sin(y)
    and
    \frac{\partial \phi}{\partial y}=x\cos(y)

    then (sin(y),xcos(y)) is a gradient, and therefore any line integral of this function along a closed path is zero. So, you need to find \phi. If this all seems completely unfamiliar, then your lecturer might have wanted you to use some other method.
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    (Original post by Mark13)
    No worries. The integrand just means the bit inside the integral sign, which in this case is (sin(y),xcos(y)). If it is a gradient, then we can write it as:

    (\sin(y),x\cos(y))=\nabla \phi for some function \phi.

    If you haven't come across the \nabla notation before, all it means in this context is that ( \sin(y),x \cos(y))=( \frac{ \partial  \phi}{ \partial x}, \frac{ \partial \phi}{ \partial y}). So basically, if there's a function \phi that satisfies:

    \frac{\partial \phi}{\partial x}=\sin(y)
    and
    \frac{\partial \phi}{\partial y}=x\cos(y)

    then (sin(y),xcos(y)) is a gradient, and therefore any line integral of this function along a closed path is zero. So, you need to find \phi. If this all seems completely unfamiliar, then your lecturer might have wanted you to use some other method.
    I wish you were the one writing my lecture notes instead. Clear and concise explanation, thanks a lot for your help man.
 
 
 
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