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Math Help - Local Coords

  1. #1
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    Local Coords

    If  \bold{r}(t) = (t^2-t, t \sqrt{2t-t^2}), 0 \leq t \leq 2 is  \bold{u}_r = \frac{(t^2-t, t \sqrt{2t-t^2})}{\sqrt{(t^2-t)^{2} + t^{2}(2t-t^2)}} and  \bold{u}_\theta = \frac{(t \sqrt{2t-t^2}, t^2-t)}{\sqrt{(t^2-t)^{2} + t^{2}(2t-t^2)}} ?
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by shilz222 View Post
    If  \bold{r}(t) = (t^2-t, t \sqrt{2t-t^2}), 0 \leq t \leq 2 is  \bold{u}_r = \frac{(t^2-t, t \sqrt{2t-t^2})}{\sqrt{(t^2-t)^{2} + t^{2}(2t-t^2)}} and  \bold{u}_\theta = \frac{(t \sqrt{2t-t^2}, t^2-t)}{\sqrt{(t^2-t)^{2} + t^{2}(2t-t^2)}} ?
    what are \bold{u}_r \mbox { and } \bold{u}_{\theta} ?

    they seem to be unit vectors, but unit vectors of what? or in what direction?
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  3. #3
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     \bold{u}_r = \frac{\bold{r}(t)}{r(t)} in the direction of  \bold{r} and  \bold{u}_\theta is a perpendicular unit vector.
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by shilz222 View Post
     \bold{u}_r = \frac{\bold{r}(t)}{r(t)} in the direction of  \bold{r} and  \bold{u}_\theta is a perpendicular unit vector.
    you mean \bold{u}_r = \frac {\bold{r}(t)}{|\bold{r}(t)|} ?

    note that: for a vector <a,b> two perpendicular vectors are <-b,a> or <b,-a>, they point in opposite directions. you put the vector <b,a>
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  5. #5
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    yes, that was a typo. yeah in the book,  r(t) = ||\bold{r}(t)|| .
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