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

[mr_coffee]

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

 

[arildno]

As for a), what is $x(t)^{2}+y(t)^{2}$?
As for c), what is $\vec{r}(\frac{\pi}{4})$ and [itex]\frac{d\vec{r}}{dt}(\frac{\pi}{4})[/tex]?

 

[quicknote]

Hi there,

It seems like you are struggling with understanding how to sketch a plane curve given a vector equation. Don't worry, it can be a bit confusing at first. Let me break it down for you.

Firstly, let's look at the given vector equation: r(t) = <cos t, sin t>. This equation represents a parametric curve in the xy-plane, where the x-coordinate is given by cos t and the y-coordinate is given by sin t. This means that as t varies, the point (x,y) on the curve also varies.

For part (a), you are asked to sketch the plane curve. This means you need to draw the path that the point (x,y) takes as t varies. To do this, you can plug in different values of t and plot the corresponding points on a coordinate plane. For example, when t = 0, we have (x,y) = (cos 0, sin 0) = (1,0). This means that the point (1,0) is on the curve. Similarly, when t = π/2, we have (x,y) = (cos(π/2), sin(π/2)) = (0,1). This means that the point (0,1) is also on the curve. If you continue to plug in different values of t and plot the corresponding points, you will notice that they form a circle centered at the origin with a radius of 1. This is how we know that the curve is a circle.

For part (c), you are asked to sketch the position vector r(t) and the tangent vector r'(t) for a given value of t. The position vector r(t) represents the point (x,y) on the curve at a specific value of t. In this case, t = π/4, so we need to find the point (x,y) on the curve when t = π/4. Plugging this value into the vector equation, we have r(π/4) = <cos(π/4), sin(π/4)> = <√2/2, √2/2>. This means that the point (√2/2, √2/2) is on the curve when t = π/4. To sketch the tangent vector r'(t), we need to find the derivative of r(t) with respect to t. In