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June 11th, 2017, 03:38 PM   #1
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parameterization and Integral

One parameterization for the unit circle C is r(t) = <cos(2t), sin(2t)>, for 0 <= t <= pi.

Set up but do not evaluate the line integral of f(x, y) = x^2 + y^2 + 3 along C, using the parameterization above.

Answer


But.. according to my calculation, I got 4 dt only

My reasoning: x^2+y^2+3 = (sin(2t))^2 + (cos(2t))^2+3 = 1 + 3 = 4
Any idea where I got it wrong??
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June 11th, 2017, 03:51 PM   #2
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You got 4 dt only? With no integral limits?
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June 11th, 2017, 04:05 PM   #3
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Quote:
Originally Posted by skipjack View Post
You got 4 dt only? With no integral limits?

Here is the limit: 0 <= t <= pi
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June 11th, 2017, 05:18 PM   #4
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Okay. By default, a line integral is with respect to s, not t. For this problem, ds/dt = 2, so the answer given (in terms of t) is twice the answer you found.

Do you know the formula by which ds/dt can be found?
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June 11th, 2017, 05:20 PM   #5
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Quote:
Originally Posted by skipjack View Post
Okay. By default, a line integral is with respect to s, not t. For this problem, ds/dt = 2, so the answer given (in terms of t) is twice the answer you found.

Do you know the formula by which ds/dt can be found?
You think I may have missed putting some vital info about the question?
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June 11th, 2017, 10:12 PM   #6
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You should previously have been shown a comparable example of a line integral.

You have to obtain $\displaystyle \int_0^\pi \!\text{f}(x,y)\frac{ds}{dt}dt$, where $\displaystyle \frac{ds}{dt} = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}$, which implies that $ds/dt = 2$ for this problem, and where $\text{f}(x, y) = x^2 + y^2 + 3 = \cos^2(2t) + \sin^2(2t) + 3$, which equals 4.
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