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Hi everyone. I ran into a minor problem while trying to solve a problem on line integral. I suspect this question to be very straightforward 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 = t^{2} + 1 and y = t^{3}+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 (t^{2} + 1) i + (t^{3}+t) j. And I have r'(t) = (2t) i + (3t^{2} + 1) j, and r'(t) = √ (9t^{2} + 1)(t^{2} + 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!
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 = t^{2} + 1 and y = t^{3}+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 (t^{2} + 1) i + (t^{3}+t) j. And I have r'(t) = (2t) i + (3t^{2} + 1) j, and r'(t) = √ (9t^{2} + 1)(t^{2} + 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!
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