Vector Calc: Can you verify my answer?

  • Thread starter rsq_a
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  • #1
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Just looking for a yes/no. I worked out the following question:

(A) Calculate the arc length of the curve [tex]r(t) = (\log t, 2t, t^2)[/tex] where [tex]1 \leq t \leq e[/tex]

(B) Let C be the ellipsed form by intersecting the cylinder [tex]x^2 + y^2 = 1[/tex] and the plane [tex]z = 2y + 1[/tex] and let [tex]\textbf{f}(x,y,z) = (y,z,x)[/tex]. What is [tex]\int_C \textbf{f} d\textbf{r}[/tex].

(C) Let C be the hyperbola formed by intersecting the cone [tex]x^2 + y^2 = z^2[/tex] and the plane [tex]x+y+z=1[/tex] and let [tex]\textbf{f}(x,y,z) = \textbf{k}/z^2[/tex]. What is [tex]\int_C \textbf{f} \times d\textbf{r}[/tex]
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:

Answers and Replies

  • #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.
 
  • #3
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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.
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?
 
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  • #4
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In (B), if you say that y=sin(t), then you should find z=2*y+1=2*sin(t)+1, right?
 
  • #5
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In (B), if you say that y=sin(t), then you should find z=2*y+1=2*sin(t)+1, right?
Doh! But I think the answer is the same. I've corrected it in the post above.
 

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