Drawing plane curve of r(t) = <cos(t),sin(t)> have answer, confuseD

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SUMMARY

The discussion revolves around the vector equation r(t) = and its implications for sketching a plane curve. The curve represents a circle in the Cartesian plane, traced counterclockwise as t varies from 0 to 2π. The user successfully computed the derivative r'(t) but sought clarification on how to derive the circular shape and the direction of traversal. Key calculations include determining the position vector r(π/4) and the tangent vector r'(π/4).

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mr_coffee
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Hello everyone, I'm confused on this problem. It says (a) sketch the plane curve with the given vector equatiion. (b) find r'(t) which is easy. (c)Sketch the position vector r(t) and the tagent vector r'(t) for the given value of t.

r(t) = <cos t, sin t>, t = PI/4;
I got part b of course. But I'm stuck on how they they got part a and c. How did they get a circle, and how did they know the circle is going counter clock wise? Thanks. Here is my work:
http://img134.imageshack.us/img134/8246/w00ta0qt.jpg
 
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As for a), what is [itex]x(t)^{2}+y(t)^{2}[/itex]?
As for c), what is [itex]\vec{r}(\frac{\pi}{4})[/itex] and [itex]\frac{d\vec{r}}{dt}(\frac{\pi}{4})[/tex]?[/itex]
 

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