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Space curve tangent vector

  1. Jun 15, 2010 #1
    1. The problem statement, all variables and given/known data

    Here's a worked problem, I can't understand how they have evaluated T at the given point (in part c):

    [PLAIN]http://img31.imageshack.us/img31/3725/97856984.gif [Broken]

    3. The attempt at a solution

    I just substituted [tex](0,1, \pi/2)[/tex] into r'(s) but

    [tex]\frac{1}{\sqrt{2}} cos \left(\frac{1}{\sqrt{2}}\right) = \frac{1}{\sqrt{2}} \neq 0[/tex]

    [tex]\frac{1}{\sqrt{2}}. -sin \left(\frac{0}{\sqrt{2}}\right) = 0 \neq \frac{-1}{\sqrt{2}}[/tex]

    Why is it that I'm not getting the right answer? Is there something else I need to do here?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jun 15, 2010 #2

    lanedance

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    Homework Helper

    to susbtitute into r'(s) you need to find s at that point
     
  4. Jun 17, 2010 #3
    Okay, but still it doesn't work:

    Since [tex]s= \sqrt{2}[/tex] , so at point 0 for example s=0. Then

    [tex]\frac{1}{\sqrt{2}} . -sin \left( \frac{0}{\sqrt{2}} \right)=0[/tex]

    You see, it should equal zero. But how did they get "[tex]-\frac{1}{\sqrt{2}}[/tex]"?? :rolleyes:
     
  5. Jun 17, 2010 #4

    HallsofIvy

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    "at [itex](0, 1, \pi/2)[/itex]" does NOT mean s= 0! It is referring to
    [tex]r(t)= \begin{pmatrix}cos(t)\\ sin(t) \\ t\end{pmatrix}[/tex]
    so t= [/itex]\pi/2[/itex].
     
    Last edited by a moderator: May 4, 2017
  6. Jun 18, 2010 #5
    How did you get [tex]t=\frac{\pi}{2}[/tex] out of that? Because by substituting these values into r(t) I got

    [tex]r(t)= \begin{pmatrix}cos(0)\\ sin(1) \\ \pi/2\end{pmatrix} = \begin{pmatrix}1\\ 0.84 \\ \pi/2\end{pmatrix}[/tex]

    And even if I set [tex]t=\frac{\pi}{2}[/tex] (therefore [tex]s= \frac{\pi}{\sqrt{2}}[/tex]) in r'(s), I still don't end up with [tex]-1/\sqrt{2}[/tex] in the first row like they have! :(
     
  7. Jun 18, 2010 #6
    You should get that,
    [tex]-\sin\left(\frac{\pi/\sqrt{2}}{\sqrt{2}}\right) = -\sin(\pi/2) = -1[/tex].

    t = pi/2 comes from r(t) = (0, 1, pi/2) = (cos t, sin t, t) and looking at the last entry.
     
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