1. ## 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 $\bold{u}_\theta = \frac{(t \sqrt{2t-t^2}, t^2-t)}{\sqrt{(t^2-t)^{2} + t^{2}(2t-t^2)}}$?

2. Originally Posted by shilz222
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?

3. $\bold{u}_r = \frac{\bold{r}(t)}{r(t)}$ in the direction of $\bold{r}$ and $\bold{u}_\theta$ is a perpendicular unit vector.

4. Originally Posted by shilz222
$\bold{u}_r = \frac{\bold{r}(t)}{r(t)}$ 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 $$ two perpendicular vectors are $<-b,a>$ or $$, they point in opposite directions. you put the vector $$

5. yes, that was a typo. yeah in the book, $r(t) = ||\bold{r}(t)||$.