Answer

• Make a table, using values of t for which we can easily find the sine and cosine:

%%%%%%DATA ENTRY This data is in a table.

t & cost & sin t\hline0&1&0$\frac{π}{6}$&$\frac{\sqrt 3}{2}$&$\frac{1}{2}$$\frac{π}{4}$&$\frac{\sqrt 2}{2}$&$\frac{\sqrt2}{2}$$\frac{π}{3}$&$\frac{1}{2}$&$\frac{\sqrt3}{2}$$\frac{π}{2}$&0&1

0 ≤ t ≤ \frac{π}{2}These outline a quarter of a circle, which is what we find when we connect the dots:

PICTURE graph dots and the curve through themIf we replace t with θ, the coordinates (cos θ, sin θ) describe a point on the unit circle.Since $0 ≤ t ≤ \frac{π}{2}$, we get the portion of the unit circle where the angle goes from 0 to $\frac{π}{2}$.

• Since the coordinates are still describing the unit circle, this time we go around the whole unit circle:PICTURE: draw unit circle, label the point (1,0) with both t = 0 and t = 2π

• The coordinates are still describing the unit circle. The picture looks the same as the graph in part (B):PICTURE: vector funcs 6

However, since we're allowing ``all real values of <em>t" it's like we're traveling around the circle over and over. Every time <em>tem> is an integer multiple of 2π we pass through the point (1,0). We'll talk more about this when we get to parametric equations.

Answer

If we didn't have the third coordinate to worry about, this would be the unit circle again. Now it's like the unit circle, but each point is higher than the point before it:

It looks like a slinky or a blood-draining helical turn on a roller coaster.