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MathSofiya

  • 3 years ago

Evaluate the line integral \[\int_c F\bullet dr\] where C is given by the vector function r(t). \[F(x,y,z)=(x+y)\hat{i}+(y-z)\hat{j}+z^2\hat{k}\] \[r(t)=t^2\hat{i}+t^3\hat{j}+t^2\hat{k}\] \[0 \le t \le 1\] \[r'(t)=2t\hat{i}+3t^2\hat{j}+2t\hat{k}\] \[\int_c F \bullet dr=\int_a^b F(r(t)) \bullet r'(t)dt\] \[\int_0^1 (t^2\hat{i}+(t^3-t^2)\hat{j} + t^4\hat{k})\bullet (2t\hat{i} +3t^2\hat{j}+2t\hat{k})dt\] how am I doing?

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  1. MathSofiya
    • 3 years ago
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    Oops wrong question, it's \[F(x,y,z)=z\hat{i}+y\hat{j}-z\hat{k}\] \[r(t)=t\hat{i}+sint\hat{j}+cost\hat{k}\] \[0 \le t \le pi\]

  2. Outkast3r09
    • 3 years ago
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    for your i, x+y

  3. Outkast3r09
    • 3 years ago
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    alright guess we'll start over hahha continues

  4. MathSofiya
    • 3 years ago
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    would that make \[r'(t)=\hat{i}+cost\hat{j}-sint\hat{k}\] and

  5. MathSofiya
    • 3 years ago
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    \[\int_0^{\pi} (t\hat{i}+sint\hat{j}+cost\hat{k})\bullet (\hat{i}+cost\hat{j}-sint\hat{k})dt\]

  6. KingGeorge
    • 3 years ago
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    So far that looks right to me.

  7. MathSofiya
    • 3 years ago
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    thanks

  8. KingGeorge
    • 3 years ago
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    You're welcome. Once you do the dot product, it looks like it will be very easy to evaluate.

  9. KingGeorge
    • 3 years ago
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    So can you show me what you think the dot product will be?

  10. MathSofiya
    • 3 years ago
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    I can't just multiply the i j and k coefficients can I? \[t+sintcost-sintcost\]

  11. KingGeorge
    • 3 years ago
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    That's exactly what it is. You should end up with \[t+\sin(t)\cos(t)-\sin(t)\cos(t)=t\]

  12. MathSofiya
    • 3 years ago
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    Yaaaaayyyyy!!!!! I just need to build confidence i guess.... sigh

  13. KingGeorge
    • 3 years ago
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    You'll get used to it.

  14. MathSofiya
    • 3 years ago
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    here is the mistake I see: x=t y=sint z=cost that would give us \[\int_0^{\pi} cost dt+\int_0^{\pi}sintcostdt+\int_0^{\pi}tsint dt\]

  15. TuringTest
    • 3 years ago
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    \[\vec F(x,y,z)=z\hat{i}+y\hat{j}-z\hat{k}\]\[\vec r(t)=t\hat i+\sin t\hat j+\cos t\hat k\]\[\vec r'(t)=\hat i+\cos t\hat j-\sin t\hat k\]\[\int_0^\pi\vec F(\vec r(t))\cdot\vec r'(t)dt=\int_0^\pi\langle\cos t,\sin t,\cos t\rangle\cdot\langle1,\cos t,-\sin t\rangle dt\]do you see a problem with that?

  16. TuringTest
    • 3 years ago
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    \[=\int_0^\pi\cos t+\sin t\cos t-\sin t\cos tdt=\int_0^\pi\cos tdt=-2\]

  17. MathSofiya
    • 3 years ago
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    oh I see what I did! It's z y -z and not z y -x

  18. MathSofiya
    • 3 years ago
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    thanks!

  19. TuringTest
    • 3 years ago
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    welcome!

  20. TuringTest
    • 3 years ago
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    now I go to bed :)

  21. MathSofiya
    • 3 years ago
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    Have a good night's rest!

  22. TuringTest
    • 3 years ago
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    likewise!

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