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Vector Calc: Can you verify my answer?

  1. Jun 7, 2009 #1
    Just looking for a yes/no. I worked out the following question:

    I got,

    (A) [tex]e^2[/tex]
    (B) [tex]\pi[/tex]
    (C) [tex]-(1+3\pi/2, 1+3\pi/2, 0)[/tex]

    Did I hit the mark?
     
    Last edited: Jun 8, 2009
  2. jcsd
  3. Jun 7, 2009 #2

    HallsofIvy

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    I haven't done (C) yet but I don't get anything like you have for either (A) or (B). Show how you got those answers.
     
  4. Jun 7, 2009 #3
    Okay. So for (A):

    [tex]\begin{align*}L &= \int_1^e |r'(t)| dt \\ &= \int_1^e |(1/t, 2, 2t)| dt \\
    &= \int_1^e \sqrt{\frac{(2t^2+1)^2}{t^2}} dt = \int_1^e 2t + \frac{1}{t} dt \\
    &= e^2 - 1 + 1 = e^2\end{align*}[/tex]

    For (B), the contour is,

    [tex]\gamma(t) = (\cos t, \sin t, 2\cos t + 1)[/tex]

    So the relevant integral is,

    [tex]\int_0^{2\pi} (\sin t, 2 \sin t + 1, \cos t) \cdot (-\sin t, \cos t, -2 \cos t) \ dt = \pi[/tex]

    Does that look wrong?
     
    Last edited: Jun 8, 2009
  5. Jun 8, 2009 #4
    In (B), if you say that y=sin(t), then you should find z=2*y+1=2*sin(t)+1, right?
     
  6. Jun 8, 2009 #5
    Doh! But I think the answer is the same. I've corrected it in the post above.
     
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