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MathSofiya Group Title

Evaluate the line integral \[\int_c F\cdot dr\] where c is the vector function \[\hat{r}(t)=t^3\hat i-t^2 \hat j+ t \hat k\] and \[F(x,y,z)=sinx\hat i + cosy \hat j+zx \hat k\] Here is how far I have gotten so far: \[\hat{r}'(t)=3t^2\hat i-2t \hat j+ \hat k\]

  • 2 years ago
  • 2 years ago

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  1. lgbasallote Group Title
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    you might want to use \cdot rather than \bullet hehe

    • 2 years ago
  2. MathSofiya Group Title
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    I'll change the Function F in terms of t's in just a second

    • 2 years ago
  3. MathSofiya Group Title
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    thanks @lgbasallote. Better? :P

    • 2 years ago
  4. lgbasallote Group Title
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    yes. yes it is.

    • 2 years ago
  5. MathSofiya Group Title
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    oh my, I did I similar problem 28 days ago http://openstudy.com/study#/updates/50060b97e4b06241806745b8 and I'm also looking at http://tutorial.math.lamar.edu/Classes/CalcIII/LineIntegralsVectorFields.aspx but I can't seem to remember... What do i substitute in for x and y, do I pick which one I want t to be?

    • 2 years ago
  6. MathSofiya Group Title
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    Hey i think I got it! \[x=t^3;y=-t^2;z=t\] YES?

    • 2 years ago
  7. MathSofiya Group Title
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    That would give me: \[F(r(t))=sin(t^3)\hat i +cos(-t^2)\hat j+t^4 \hat k \]

    • 2 years ago
  8. MathSofiya Group Title
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    \[\int_{t=0}^{t=1}\left(sin(t^3)\hat i +cos(-t^2)\hat j+t^4 \hat k\right)\cdot\left(3t^2\hat i -2t\hat j +\hat k \right) dt\]

    • 2 years ago
  9. MathSofiya Group Title
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    and now I would just do the dot product correct?

    • 2 years ago
  10. mukushla Group Title
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    exactly

    • 2 years ago
  11. MathSofiya Group Title
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    \[\int_0^1(3t^2sin(t^3)-2tcos(-t^2)+t^4)dt\]

    • 2 years ago
  12. mukushla Group Title
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    well done

    • 2 years ago
  13. MathSofiya Group Title
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    wait, I have to take the integral of several products?

    • 2 years ago
  14. mukushla Group Title
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    separate them\[\int_{0}^{1} 3t^2 \sin t^3 dt+...+...\]

    • 2 years ago
  15. MathSofiya Group Title
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    oh ok \[\int_0^1 3t^2sin(t^3)dt-\int_0^12tcos(-t^2)dt+\int_0^1t^4dt\] like this?

    • 2 years ago
  16. mukushla Group Title
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    yep

    • 2 years ago
  17. MathSofiya Group Title
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    oh integration by parts! duh =D

    • 2 years ago
  18. mukushla Group Title
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    \[-\cos t^3\] :-)

    • 2 years ago
  19. TuringTest Group Title
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    no integration by parts you can do it all with u-subs

    • 2 years ago
  20. TuringTest Group Title
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    do you get that?

    • 2 years ago
  21. MathSofiya Group Title
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    Oh I see it now!!!!

    • 2 years ago
  22. TuringTest Group Title
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    cool :)

    • 2 years ago
  23. MathSofiya Group Title
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    u=t^2 du=2t dt for the second integral

    • 2 years ago
  24. TuringTest Group Title
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    yep

    • 2 years ago
  25. MathSofiya Group Title
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    u=t^3 and du=3t^2 dt for the first integral...LOL that took me a while!

    • 2 years ago
  26. TuringTest Group Title
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    yeah, it's well set-up for the u-sub thing :)

    • 2 years ago
  27. mukushla Group Title
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    u cooked the problem

    • 2 years ago
  28. MathSofiya Group Title
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    \[-cos(1)+sin(1)+2\] My algebra is probably wrong...but Yeah @mukushla that was one long recipe!

    • 2 years ago
  29. mukushla Group Title
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    \[\large -\cos t^3]_{0}^{1}+\sin -t^2]_{0}^{1}+t^5/5]_{0}^{1}\]

    • 2 years ago
  30. MathSofiya Group Title
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    \[(-cos(1)+1)+(-sin(1)+1)+\frac 1 5\] \[-cos(1)-sin(1)+\frac{11}{5}\]

    • 2 years ago
  31. MathSofiya Group Title
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    oops sine of 0 is zero

    • 2 years ago
  32. TuringTest Group Title
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    you have an extra +1 in there - -yep

    • 2 years ago
  33. MathSofiya Group Title
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    -cos(1)-sin(1)+ 6/5

    • 2 years ago
  34. TuringTest Group Title
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    looks good to me :)

    • 2 years ago
  35. MathSofiya Group Title
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    sigh...finally! Thanks guys!

    • 2 years ago
  36. TuringTest Group Title
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    welcome !

    • 2 years ago
  37. mukushla Group Title
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    :)

    • 2 years ago
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