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Curvature, More of a cross product question really.

  1. Oct 30, 2008 #1
    I've been using various methods of finding curvature, and using the forumla

    [tex]

    K=\frac{||\vec{r}'X \vec{r}''||}{||\vec{r}'||^3}

    [/tex]

    I took the cross product of my two vectors and came up with (just assume this is correct, if my question isn't answered i'll post all related work)

    [tex]

    128 \pi^3 sin(2 \pi t)^2 - -128 \pi^3 cos(2 \pi t)^2

    [/tex]

    My question is: Can I factor out the 128[itex] \pi^3[/itex] and then use a trig identity to make sin^2 + cos^2 =1?

    I assumed that since these were components of vectors I couldn't do this. However, if I do, I get the correct answer. I left them as is, and then took the magnitude of the resulting cross product.

    Which produced

    [tex]

    128 \pi^3 \sqrt{sin(2 \pi t)^4+cos(2 \pi t)^4}

    [/tex]

    This would also produce the correct answer if sin(x)^n +cos(x)^n =1, but I'm fairly certain it doesn't.

    Any help would be greatly appreciated. Thanks
     
  2. jcsd
  3. Oct 30, 2008 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Well, sin(x)^4+cos(x)^4 is NOT 1. Yes, I think you'll need to post more work.
     
  4. Oct 31, 2008 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Do you mean you got this as one of the components of the cross product or do you mean, rather, something like
    [tex]128\pi^3 sin(2\pi t)^2\vec{i} - 128 \pi^2 cos(2\pi t)^2\vec{j}[/tex]?
    If the latter, you can certainly factor out the [itex]128\pi^3[/itex] but not combine the different components.

     
  5. Oct 31, 2008 #4
    The original question is: Find the curvature of

    [tex]

    \vec{r}(t) = 4cos(2 \pi t)\vec{i} + 4sin(2 \pi t)\vec{j}

    [/tex]

    I have a solutions guide and found the curvature using

    [tex]

    K = \frac{||\vec{T}'(t)||}{||\vec{r}'(t)||}

    [/tex]

    but I wanted to try the cross product method to see if it is easier, and just for some review.

    [tex]

    \vec{r}'(t) = -8 \pi sin(2 \pi t)\vec{i} + 8 \pi cos(2 \pi t)\vec{j}

    [/tex]

    [tex]

    \vec{r}''(t) = -16 \pi^2 cos(2 \pi t)\vec{i} - 16 \pi^2 sin(2 \pi t)\vec{j}

    [/tex]

    [tex]

    \vec{r}'(t) X \vec{r}''(t) = 128 \pi^3 sin(2 \pi t )^2\vec{i} - - 128 \pi^3 cos(2 \pi t)^2\vec{j}

    [/tex]

    [tex]

    || \vec{r}'(t) \crcoss \vec{r}''(t) || = \sqrt{(128 \pi^3 sin(2 \pi t )^2)^2+(28 \pi^3 cos(2 \pi t)^2)^2}

    [/tex]

    [tex]

    ||\vec{r}'(t)|| = \sqrt{(-8 \pi sin(2 \pi t))^2+(8 \pi cos(2 \pi t))^2} = 8 \pi

    [/tex]

    [tex]

    K=\frac{||\vec{r}'(t)X \vec{r}'(t)||}{||\vec{r}'(t)||^3}=\frac{128 \pi^3 \sqrt{sin(2 \pi t)^4+cos(2 \pi t)^4}}{(8 \pi)^3}

    [/tex]

    As I typed all this out, I realized that I took the cross product of a 2 X 2 matrix, and probably should have taken a 3 X 3 cofactor expansion with vectors [itex] \vec{i},\vec{j} [/itex] and [itex] \vec{k} [/itex]

    so i'll try that and see if it works, until then this is the problem in its entirety.
     
  6. Oct 31, 2008 #5
    Actually, looking back over the problem, I think I assumed my 2 X 2 cross product was a cofactor expansion, where it should have been a constant. This would have allowed me to simplify that result and get the [itex] 128 \pi^3 [/itex] I needed in the numerator.

    Cofactor expansion results in [itex] 0 \vec{i} + 0\vec{j} + 128 \pi^3 \vec{k} [/itex] which if I take the magnitude of that vector I get the result I need.

    Do I have this somewhat straightened out not?
     
  7. Oct 31, 2008 #6

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    That looks much better, yes.
     
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