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Line Integrals

  1. Nov 1, 2014 #1
    Hi everyone. I ran into a minor problem while trying to solve a problem on line integral. I suspect this question to be very straight-forward as it gave the parametric equations of the curve C. However, I am still unable to get the answer for some reason. May I have someone to point out where I went wrong? Thanks!

    Question:

    Ah, the image didn't get uploaded.. Please refer to the attached image. Thanks! The question in question is Q1(a).

    My attempt:

    Since x = t2 + 1 and y = t3+t , we have x(t) and y(t). Following on, I went one to apply the formula, ∫ F⋅ dr = ∫ f(x(t), y(t))||r'(t)|| dt.

    I had r(t) as (t2 + 1) i + (t3+t) j. And I have r'(t) = (2t) i + (3t2 + 1) j, and ||r'(t)|| = √ (9t2 + 1)(t2 + 1).

    Since the range of t is 0 ≤ t ≤ 1, I integrated over this range and my calculator returned 3.73 as the answer, but the answer in my answer key wrote 6. May I know where I had gone wrong? Thanks! :D

    <In case the picture cannot be seen, I have uploaded it here.> Thanks again!
     

    Attached Files:

  2. jcsd
  3. Nov 1, 2014 #2

    Mark44

    Staff: Mentor

    Your r'(t) looks fine, but |r'(t)| doesn't.
     
  4. Nov 1, 2014 #3

    td21

    User Avatar
    Gold Member

    Can you show me your work on the finding the gradient of f?
     
  5. Nov 1, 2014 #4
    To Mark44: Oh dear. I must have been hallucinating when doing Math at nearly 3am. T_T Sorry. ||r'(t)|| is √(t^2 + (1/9))(t^2 + 1).

    To td21: How do I find the gradient of f? Do I just use the formula Gradient = (y2-y1)/(x2-x1)?
     
  6. Nov 1, 2014 #5
    However, my calculator returned 1.24 as the answer.. :/
     
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