Parametric Plotting Intersection not working

[alcoonslists]

It seems that the intersection command when applied to two parametric functions does not work.  Call TI-CARES and they do not have a quick solution.  Anyone have any info on this?

Thanks,

Al

Al Coons

Department of Mathematics

Buckingham Browne & Nichols

Cambridge, MA

AP Statistics Site: http://www.bbn-school.org/us/math/ap_stats/

[alcoonslists]

FYI.

TI just called back and said that the intersection command has not been implemented for parametrics.  Wow.  This one really is fundamental.  This, along with lack of dynamic tracing without a kludge.  Ugh.

I just checked.  Concurrent tracing and intersections for polar functions does not work either.

Ugh! Ugh!

Al

[John Hanna]

The reason that parametrics and polars do not have intersection tools is because the paths may ‘cross’ at different values of t or theta. Is that or is that not an ‘intersection’?

 

     John Hanna

     jeh...@optonline.net

     www.johnhanna.us

     T³ - Teachers Teaching with Technology

     "the future isn't what it used to be."

 

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[John Losse]

I think this will be hard for TI to implement.  

Thinking of the parametric equations as representing the paths of two particles in time, are we looking just for a place where the paths cross, or a time when the particles will be in the same place simultaneously?

By doing graph trace, you can get good estimates of where the graphs cross and see whether the t-values are the same, or at least close.

If you are looking for values of t where the particles are in the same spot simultaneously, one approach is to define f1(x) = (x2(x)-x1(x))*(y2(x)-x2(x)) and look at its zeros.  These will be candidates for the t-values you want, but they have to be checked.

[alcoonslists]

Thanks both Johns,

I knew that.  Me getting old:).  There is no such command on the TI-84 either.

In fact, that is why we want concurrent tracing in both parametric and polar modes.  So that we can watch "particles" move along both curves maintaining identical "t-values".  That way we can see that some intersection points are solutions to both and some are not.  Your suggestion below is a way to get at the same thing, but it sure is not intuitive to a student new to this material.  

Thanks,

Al

On the other hand