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    How much work is done by a force= (2x)i N + (3)j N with x in meters, that moves a particle from a position (initial) = (2)i + (3)j to a position (final) = (-4)i + (-3)j?

    These are my workings (see post #3). Is my assumption about dr correct ?

    Also, the answer should be -6 and I dont get that....
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    (Original post by Ari Ben Canaan)
    How much work is done by a force= (2x)i N + (3)j N with x in meters, that moves a particle from a position (initial) = (2)i + (3)j to a position (final) = (-4)i + (-3)j?

    These are my workings (below). Is my assumption about dr correct ?

    Also, the answer should be -6 and I dont get that....
    Stay out of doing this with vectors if you can. If you have to do it vectors-wise, it's your parametrisation of the curve that seems dodgy to me (or even non-existent). To do a line integral, note that \int_C F \cdot dr = \int_{\alpha}^{\beta} F(r(u)) \cdot \frac{dr}{du} du for an appropriate parametrisation of the curve in terms of a function r(u). I can see how you've formed dr, but that's not really what dr means.
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    Do not consider the above workings..... These are my updated workings below.Name:  ImageUploadedByStudent Room1366362592.530247.jpg
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    (Original post by Smaug123)
    Stay out of doing this with vectors if you can. If you have to do it vectors-wise, it's your parametrisation of the curve that seems dodgy to me (or even non-existent). To do a line integral, note that \int_C F \cdot dr = \int_{\alpha}^{\beta} F(r(u)) \cdot \frac{dr}{du} du for an appropriate parametrisation of the curve in terms of a function r(u). I can see how you've formed dr, but that's not really what dr means.
    I dont understand your mathematical notation or its significance but lets ignore that for now.

    I HAVE to do this using a path integral so please walk me through the general method solving such questions which always seem to have vectors in them.
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    (Original post by Ari Ben Canaan)
    I dont understand your mathematical notation or its significance but lets ignore that for now.

    I HAVE to do this using a path integral so please walk me through the general method solving such questions which always seem to have vectors in them.
    OK, your second working is a bit closer, but I think you've still misunderstood what dr means.
    Because it would take ages in LaTeX, here's my written out working and explanation. Shout if something's illegible or you don't understand something.
    If you don't understand the notation in the first line ("A line integral is something that looks like…"), that's the normal notation for this, as far as I am aware - could you post what your notation for a path integral is?
    I parametrised in the way that seemed most obvious to me - you can indeed parametrise by arc length (as it looks like you were doing), and I can show you that if you like. (Sorry, I started out intending to use arc length, which is why I've used s instead of u - s is the conventional letter for arc length - but it was just much easier not to.)
    Sorry for the slower-than-intended reply - I used Dropbox to get the scan onto this computer, and it took longer than usual to copy.
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    (Original post by Smaug123)
    OK, your second working is a bit closer, but I think you've still misunderstood what dr means.
    Because it would take ages in LaTeX, here's my written out working and explanation. Shout if something's illegible or you don't understand something.
    If you don't understand the notation in the first line ("A line integral is something that looks like…"), that's the normal notation for this, as far as I am aware - could you post what your notation for a path integral is?
    I parametrised in the way that seemed most obvious to me - you can indeed parametrise by arc length (as it looks like you were doing), and I can show you that if you like. (Sorry, I started out intending to use arc length, which is why I've used s instead of u - s is the conventional letter for arc length - but it was just much easier not to.)
    Sorry for the slower-than-intended reply - I used Dropbox to get the scan onto this computer, and it took longer than usual to copy.
    I used your explanation and a few videos on the net to clarify some things and I've figured it out now.

    Thank you very much ! +REP
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    (Original post by Ari Ben Canaan)
    I used your explanation and a few videos on the net to clarify some things and I've figured it out now.

    Thank you very much ! +REP
    Thanks It took me a little while to get my head around line integrals, but once you've done a few they're fairly routine.
 
 
 
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