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MathSofiya

  • 3 years ago

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: \[\hat{r}'(t)=3t^2\hat i-2t \hat j+ \hat k\]

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  1. lgbasallote
    • 3 years ago
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    you might want to use \cdot rather than \bullet hehe

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

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

  4. lgbasallote
    • 3 years ago
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    yes. yes it is.

  5. MathSofiya
    • 3 years ago
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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?

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

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

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

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

  10. mukushla
    • 3 years ago
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    exactly

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

  12. mukushla
    • 3 years ago
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    well done

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

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

  15. MathSofiya
    • 3 years ago
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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?

  16. mukushla
    • 3 years ago
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    yep

  17. MathSofiya
    • 3 years ago
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    oh integration by parts! duh =D

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

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

  20. TuringTest
    • 3 years ago
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    do you get that?

  21. MathSofiya
    • 3 years ago
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    Oh I see it now!!!!

  22. TuringTest
    • 3 years ago
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    cool :)

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

  24. TuringTest
    • 3 years ago
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    yep

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

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

  27. mukushla
    • 3 years ago
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    u cooked the problem

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

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

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

  31. MathSofiya
    • 3 years ago
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    oops sine of 0 is zero

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

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

  34. TuringTest
    • 3 years ago
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    looks good to me :)

  35. MathSofiya
    • 3 years ago
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    sigh...finally! Thanks guys!

  36. TuringTest
    • 3 years ago
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    welcome !

  37. mukushla
    • 3 years ago
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    :)

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