Graphing Parametric Equations on TI-Nspire CAS

[Mike Bevelacqua]

I have tried to graph a circle with x(t)=4cos(t) & y(t)=4sin(t) for t
from 0 to 2π. If the step is small enough you get a circle, but if
you
play around with the step it does not do what you would intuitively
think. For example a step of π /2 should be 4 steps and hence graph a
square.
When in fact is looks to have 15 steps. A step of π should be a line
segment, but graphs an octagon. However, when I "trace" the graph in
after using "Zoom Fit" the trace will jump to the correct angle.

What's happing here?

[-TJ]

Dear Mike,

Hmm... I'm not sure, and I might be wrong (correct me if I am), but
this sounds like a bug!
Let me fiddle around with the step settings. I'll try to find a
conclusion to this.

Sincerely,
-TJ

[-TJ]

Dear Mike,

Haha! One portion of the circle is severely flattened! =P
At the instance of graphing that function and tracing it, the "x"
cursor doesn't even touch the flattened part.
But when I apply "zoom fit", it still does not correct it.

I don't know if this is called a "bug", but it seems like the
calculator loses accuracy when graphing tstep is larger than its
default number.

I'm still a student; although I am starting pre-calculus, I haven't
learned anything about parametric graphing. =(

Sincerely,
-TJ

[Mike Bevelacqua]

Figured this out...

Basically you just need to change one of the "attributes" and you are
set.

On Oct 2, 8:48 am, Mike Bevelacqua <mbevelac...@aea10.k12.ia.us>
wrote:

[Sean Bird]

Ah, the discrete points connected instead of continuous. Continuous with a step size of pi still 'tries' to be continues. Was it something like 4 extra points that continuous gives between each requested discrete point.

Nice find. I wonder if it some kind of smoothing algorithm.

- Sean Bird

2009/10/5 Mike Bevelacqua <mbeve...@aea10.k12.ia.us>