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Wislar

Please Check! Line integral: \[\int\limits_{C}^{}x ^{2}dx+y ^{2}dy\] of the line segment from (0,2) to (4,3)

  • one year ago
  • one year ago

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  1. Wislar
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    Is it: \[\int\limits_{0}^{4}16t ^{2}(4)+(2+t)^{2}(1)dt\]

    • one year ago
  2. Wislar
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    The parameterized curve is r(t)=(4t, 2+t) right?

    • one year ago
  3. Wislar
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    @phi @amistre64 @Zarkon @TuringTest

    • one year ago
  4. amistre64
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    id have a little reading to do to catch up on this :)

    • one year ago
  5. Wislar
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    I think I have it down correctly, I'm just not sure if the limit is from 0 to 4

    • one year ago
  6. amistre64
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    we need to parametrize the line from 0,2 to 4,3 right?

    • one year ago
  7. Wislar
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    Yes

    • one year ago
  8. amistre64
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    4,3 -0-2 ----- <4,1> is the direction vector, and we can apply it to either point given so one parametric could be: x = 0 + 4t y = 2 + t

    • one year ago
  9. Wislar
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    That's what I got! I'm not sure about the limits of integration though

    • one year ago
  10. abb0t
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    sketch out the graph.

    • one year ago
  11. amistre64
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    the dx and dy parts in your post got me confuddled. im used to seeing a ds in it

    • one year ago
  12. amistre64
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    here we go http://tutorial.math.lamar.edu/Classes/CalcIII/LineIntegralsPtII.aspx

    • one year ago
  13. amistre64
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    \[\int_CPdx+Qdy=\int_CPdx+\int_CQdy\]

    • one year ago
  14. amistre64
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    x = 0 + 4t dx = 4 dt y = 2 + t dy = dt

    • one year ago
  15. Wislar
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    I have that down @amistre64 I'm just really not sure about the limits. I just used the x coordinates.

    • one year ago
  16. amistre64
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    what are the limits of t for the given line?

    • one year ago
  17. Wislar
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    1 and 4?

    • one year ago
  18. amistre64
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    when t=0 we are at (0,2) when t=? we are at (4,3)

    • one year ago
  19. Wislar
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    I'm not sure?

    • one year ago
  20. amistre64
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    0 = 0 + 4t 2 = 2 + t t=0 4 = 0 + 4t 3 = 2 + t t = 1 so the limit of our t is 0 to 1

    • one year ago
  21. Wislar
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    Wait how did you do that?

    • one year ago
  22. amistre64
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    we are changing x and y into functions of t correct?

    • one year ago
  23. Wislar
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    Yes

    • one year ago
  24. amistre64
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    then "with respect to t" the line is from t=0 to t=1 since (0,2) is t=0; and (4,3) is t=1

    • one year ago
  25. Wislar
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    I see!

    • one year ago
  26. amistre64
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    we are simply defining all the terms with respect to t; x,y and movement

    • one year ago
  27. Wislar
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    I have another question that asks for the line integral that consists of the arc of the circle x^2+y^2=4 from (2,0 to (0,2) I know how to find the line integral, but the limits are what I'm not sure about again

    • one year ago
  28. Wislar
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    The parameterized curve is (2cost,2sint)

    • one year ago
  29. amistre64
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    is it asking you to define it by the line that goes from 2,0 to 0,2? or from the arc that goes from 2,0 to 0,2 ?

    • one year ago
  30. Wislar
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    It says the arc of the circle from (2,0) to (0,2)

    • one year ago
  31. amistre64
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    x = r cos(t) y = r sin(t) and from the equation r = 2 x = 2 cos(t) dx = -2sin(t) y = 2 sin(t) dy = 2 cos(t) what is the value of t for the point x=2, y=0? what is the value of t for the point x=0, y=2?

    • one year ago
  32. Wislar
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    cos(t)=1 and sin(t)=1?

    • one year ago
  33. amistre64
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    2 = 2 cos(t) 0 = 2 sin(t) cos(t) = 1 and sin(t) = 0 when t = 0 0 = 2 cos(t) 2 = 2 sin(t) cos(t) = 0 and sin(t) = 1 when t = pi/2

    • one year ago
  34. Wislar
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    So it's from 0 to pi/2 right?

    • one year ago
  35. amistre64
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    yes

    • one year ago
  36. Wislar
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    Thank you so much!!

    • one year ago
  37. amistre64
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    youre welcome, and good luck :)

    • one year ago
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