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    Calculate the work done by moving along the straight line from (2,1,1) to (3,2,2) in a force field:

    F=\mathbf{r}/|\mathbf{r^2}|  = [ x\mathbf{i}+y\mathbf{j}+ z\mathbf{k} ]/  [x^2+y^2+z^2]
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    anyone?
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    (Original post by djpailo)
    anyone?
    Were you ever unsure about what you were doing but it worked out so nicely you thought it must be right?

    Well that's what just happened to me with this question.

    I did it as \int \vec{F} \frac{d\vec{r}}{dt}dt
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    (Original post by Get me off the £\?%!^@ computer)
    Were you ever unsure about what you were doing but it worked out so nicely you thought it must be right?

    Well that's what just happened to me with this question.

    I did it as \int \vec{F} \frac{d\vec{r}}{dt}dt
    cool, what answer did you get, i got 119/3
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    (Original post by djpailo)
    cool, what answer did you get, i got 119/3
    I got \frac{1}{2}\ln \frac{17}{6} so it seems that at least one of us is wrong.
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    (Original post by Get me off the £\?%!^@ computer)
    I got \frac{1}{2}\ln \frac{17}{6} so it seems that at least one of us is wrong.
    how did you get that?
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    The straight line from (2,1,1) to (3,2,2) is x(t)=2+t, y(t)=1+t, z(t)=1+t, 0<t<1.

    dx/dt=dy/dt=dz/dt=1.

    Subbing in you get

    \int_0^1 \frac{2+t+1+t+1+t}{(2+t)^2+(1+t)  ^2+(1+t)^2}dt
 
 
 
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