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anonymous

  • 5 years ago

ok, so now how would you integrate (5/2) i t ^ (3/2)? Same way? What happens to the i?

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  1. anonymous
    • 5 years ago
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    The i stays put (it's a constant), assuming i is the imaginary number and you're integrating with respect to t. So you integrate normally here.

  2. anonymous
    • 5 years ago
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    and 5/2 is a constant too :)

  3. anonymous
    • 5 years ago
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    that's what I thought....okkkkk, so can you check my work out on this one? I had a really long integral problem and think I did all the simplifying corectly but I end up with (1/3) t^3 + (2/3) i t^(5/2) + i t^(3/2) between 0 and 1...do i just plug in the 1 and then the 0 for t and the i remains in the answer?

  4. anonymous
    • 5 years ago
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    Can you write out the question using the equation editor?

  5. anonymous
    • 5 years ago
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    oh, how do I pull that up?

  6. anonymous
    • 5 years ago
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    this is a new laptop and not sure what is on it exactly!!

  7. anonymous
    • 5 years ago
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    (1/3) t^3 + (2/3) i t^(5/2) + i t^(3/2)

  8. anonymous
    • 5 years ago
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    The equation editor is on the bottom of the message box. It says, "Equation"

  9. anonymous
    • 5 years ago
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    \[\frac{1}{3}t^3+\frac{2}{3} i t^{5/2}+i t ^{3/2}\]...is that the thing you have to integrate or is that an answer to something?

  10. anonymous
    • 5 years ago
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    actually, I have to intergrate the curve from (0,0) to (1,1) where y = sq rt of x using m(z) dz where m(z) is x^2 +y^2 + 2ixy

  11. anonymous
    • 5 years ago
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    ok, I hit the eqn button but my tyoibng didnt come out like yours!

  12. anonymous
    • 5 years ago
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    actually, the "contour" ...this is in Complex Analysis.

  13. anonymous
    • 5 years ago
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    Yeah, I guessed. The path you have to integrate over is \[y=\sqrt{x}\]right?

  14. anonymous
    • 5 years ago
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    and x = t

  15. anonymous
    • 5 years ago
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    Yes, that's what I was going to suggest

  16. anonymous
    • 5 years ago
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    and I THINK I did the distributive correctly...the dz/dt at the end of the original integral changes to (dx/dt) + i (dy/dt) and and i came up with integral of [t^2 + t +2i t^(3/2)] [ 1 + i (1/(2sqrt t)] dt

  17. anonymous
    • 5 years ago
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    I double distributed and before simplifyling, came up with integral of t^2 + t + 2i t^(3/2) + i t^2 (1/(2sqrt t)) + i t (1/(2sqrt t) + 2i t^(3/2) (i) ((t^(1/2)/2)

  18. anonymous
    • 5 years ago
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    could that be right for the distributive part??

  19. anonymous
    • 5 years ago
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    just let me see

  20. anonymous
    • 5 years ago
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    it's past my bedtime

  21. anonymous
    • 5 years ago
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    oh, sorry...it's only past 10am here (east coast)

  22. anonymous
    • 5 years ago
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    all of these fractional exponents and fractional constants are givng me a headache and making my eyes cross!

  23. anonymous
    • 5 years ago
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    Were you actually given the function to integrate as\[\mu (z)=x^2+y^2+2i x y \]?

  24. anonymous
    • 5 years ago
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    yes! :-)

  25. anonymous
    • 5 years ago
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    Okay, you have \[z=x+iy\]as your variable and over this path, x and y will take values,\[x=t, y=t^{1/2}\]so\[z=z(t)=t+i t^{1/2}\]Your integral is\[\int\limits_{\gamma}^{}\mu (z) dz=\int\limits_{0}^{1}\mu(z(t))z'(t)dt\]where gamma is our defined path. Agree so far?

  26. anonymous
    • 5 years ago
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    yep! that's sort of what the prof wrote...

  27. anonymous
    • 5 years ago
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    Now, since mu(z) is as you defined it above in terms of x and y, and now that we have those x and y in terms of t along this contour, mu as a function of t is,\[\mu (t)=t^2+t+2i t^{3/2} \rightarrow \mu'(t)=2t+1+3 i t^{1/2} \]

  28. anonymous
    • 5 years ago
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    oh forget mu'(t)...

  29. anonymous
    • 5 years ago
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    late

  30. anonymous
    • 5 years ago
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    \[z'(t)=1+i \frac{1}{2}t\]

  31. anonymous
    • 5 years ago
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    woah....where did the 2 i t^ (3/2) come from?

  32. anonymous
    • 5 years ago
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    So your integral becomes,\[\int\limits_{0}^{1}\left( t^2+t+2\it^{3/2} \right)\left( 1+i \frac{1}{2}t^{-1/2} \right) dt\]

  33. anonymous
    • 5 years ago
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    yes, have that...

  34. anonymous
    • 5 years ago
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    2it^(3/2) from subbing in x=t, y=t^(1/2) into 2ixy

  35. anonymous
    • 5 years ago
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    Agreed?

  36. anonymous
    • 5 years ago
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    got you, you were just wrtiting the original problem for me...thought htat had come from all the distributing...

  37. anonymous
    • 5 years ago
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    The "i t" didn't show up for some reason in the integral above in the first bracket.

  38. anonymous
    • 5 years ago
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    \[\int\limits_{0}^{1}\left( t^2+t+2 i t^{2/3} \right)\left( 1+i \frac{1}{2}t^{-1/2} \right)dt\]

  39. anonymous
    • 5 years ago
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    That should be mistake-free

  40. anonymous
    • 5 years ago
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    You have to expand that out and group the terms into an integral for the real and imaginary components, and then do Riemann integration ('normal' integration) on each interval.

  41. anonymous
    • 5 years ago
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    \[\int\limits_{0}^{1}u(t)dt+i \int\limits_{0}^{1}v(t)dt\]

  42. anonymous
    • 5 years ago
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    Happy so far?

  43. anonymous
    • 5 years ago
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    yeah, I can finally type again...froze up or something

  44. anonymous
    • 5 years ago
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    ok

  45. anonymous
    • 5 years ago
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    ok, my question is, did I do the distributive math correctly?

  46. anonymous
    • 5 years ago
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    You can check that with wolframalpha - have you heard of that?

  47. anonymous
    • 5 years ago
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    way back up at 43 minutes ago... no, but I'll check it out...

  48. anonymous
    • 5 years ago
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    Don't be surprised/put off if you distribute and get something rotten.

  49. anonymous
    • 5 years ago
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    It happens all the time in complex analysis. It's only ever nice when the give you "Integrate z over the unit circle" or some pos.

  50. anonymous
    • 5 years ago
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    actually, not bad... now can't find it, just had it

  51. anonymous
    • 5 years ago
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    could I actually come up with something like t^2 + (3/2) i t ^(3/2) + (1/2 i t^ (1/2) before I integrate?

  52. anonymous
    • 5 years ago
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    I got some like terms when I distributed so combined them

  53. anonymous
    • 5 years ago
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    Before I answer that, again, there's a mistake in the last integral where I declared, "No mistakes"...it's late and this site is awkward. The power on the 2it in the first bracket should be 3/2, not 2/3.

  54. anonymous
    • 5 years ago
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    You could come up with something like that before you integrate.

  55. anonymous
    • 5 years ago
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    I'm double-checking my expansion.

  56. anonymous
    • 5 years ago
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    sorry, didn't catch that...I hacethe exponent as 3/2

  57. anonymous
    • 5 years ago
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    \[t^2+i \left( \frac{5}{2}t^{3/2}+\frac{1}{2}t^{1/2} \right)\]

  58. anonymous
    • 5 years ago
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    You get 3/2, where I get 5/2

  59. anonymous
    • 5 years ago
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    yeah! that's what I have !!!!!!! (just saw my mistake, yes the 3/2 should be a 5/2

  60. anonymous
    • 5 years ago
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    So it's all good to go...

  61. anonymous
    • 5 years ago
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    yes, so when integrating , I would get (1/3) t^3 for the real part and i [(t^5/2) + (1/3) t^(3/2) for the imaginary part?

  62. anonymous
    • 5 years ago
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    That looks right.

  63. anonymous
    • 5 years ago
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    Now sub your limits.

  64. anonymous
    • 5 years ago
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    wahoo!@! so just plug in the 1 and 0 and evaluate as usual?

  65. anonymous
    • 5 years ago
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    \[\frac{1}{3}+i \frac{4}{3}\]

  66. anonymous
    • 5 years ago
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    so (1/3) + 2i???

  67. anonymous
    • 5 years ago
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    Yes, you can evaluate as usual since you've converted a contour integral to one over an interval...so it's 'normal', for lack of a better term.

  68. anonymous
    • 5 years ago
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    oos! wherer did your i (4/3) come from

  69. anonymous
    • 5 years ago
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    Look at the imaginary part of your result. It's \[t^{5/2}+\frac{1}{2}t^{3/2}\]

  70. anonymous
    • 5 years ago
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    sorry ^ 1/3

  71. anonymous
    • 5 years ago
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    (1)^{5/2}+(1/3)*(1)^{3/2} = 3/3 + 1/3 = 4/3

  72. anonymous
    • 5 years ago
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    fool thing froze again!! I saw my mistake while you were typing! duh, forgot to multiply in the 1/3

  73. anonymous
    • 5 years ago
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    So you're fine now?

  74. anonymous
    • 5 years ago
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    so I had (1/3)+ i + (1/3)i to get the two terms (1/3) + (4/3) i

  75. anonymous
    • 5 years ago
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    yes

  76. anonymous
    • 5 years ago
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    and when t=0, the result for the other limit is 0.

  77. anonymous
    • 5 years ago
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    Sure am!! Now just have to do the same thing for contours y=x and y=x^2 and compare all 3 answers.

  78. anonymous
    • 5 years ago
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    Well, good luck. It's the same deal :P

  79. anonymous
    • 5 years ago
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    sorry, computer delayed again! THANK YOU soooooo MUCH FOR HELPING ME!! ;- )

  80. anonymous
    • 5 years ago
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    No probs. I'm hoping you fanned me ;)

  81. anonymous
    • 5 years ago
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    sure did!!!!! :-) and this site is free????

  82. anonymous
    • 5 years ago
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    Yes

  83. anonymous
    • 5 years ago
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    Okay, I'm off for sleep...have fun with complex analysis ;)

  84. anonymous
    • 5 years ago
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    thanks, I will! :-)

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