parametric curve sketching question

[sigma]

Also had this on my test yesterday. If somebody could run it through some graphing program, I just want to see if I'm right.

\(\displaystyle \
\L\
x = 3t - t^3 ,y = 4 - t^2
\\)

When it was all done it looked something like this:

 

[stapel]

I get something similar.

Eliz.

 

[skeeter]

free graphing program

 

[sigma]

Is there a way to show where the endpoints start and end?

 

[tkhunny]

What end points?

"End points start"? What does that mean? Points are just points.

Do you mean what values are for various values of t?
t = 0 ==> (0,4)
t = 1 ==> (2,3)
t = -1 ==> (-2,3)

Where is x = y? t = -1.83117721

Where does it intersect itself? \(\displaystyle t = \sqrt{3}\) and \(\displaystyle t = -\sqrt{3}\)

Learning the language is part of the study.

 

[sigma]

Let me rephrase my question. Where does the graph start and end? Meaning what direction is the graph going because on my test I said it was moving from the positive x axis, negative y axis (that's where it started) and ended in the negative x and y axis but the limits to infinity didn't seem to agree with my concavity so thats why I'm not sure about that. I was wondering if the graph program shows that somehow.

 

[tkhunny]

"Start" and "End" don't mean anything. "Direction" is a useful term.

From the points I provided earlier, one can get a sense of direction. Adding the x-intercepts can also help the study of direction. Just order them by increasing t.

t = -2 ==> (2,0)
\(\displaystyle t = -\sqrt{3}\) ==> (0,1)
t = -1 ==> (-2,3)
t = 0 ==> (0,4)
t = 1 ==> (2,3)
\(\displaystyle t = \sqrt{3}\) ==> (0,1)
t = 2 ==> (-2,0)

So, t < -2 is in Quadrant IV and t > 2 is in Quadrant III. You supply the rest of the description.