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Topic: tangent lines and parametric curves
Replies: 2   Last Post: Mar 22, 2001 5:59 PM

 Messages: [ Previous | Next ]
 BETH HENTGES Posts: 16 Registered: 12/6/04
tangent lines and parametric curves
Posted: Mar 22, 2001 2:02 PM

Here's a problem my calc. II class has been working on.
x=t^3-3t
y=t^2-2t
-inf. < t < inf.

Note that
dy/dt = 2t-2 = 2(t-1)
and
dx/dt = 3t^2-3 = 3(t+1)(t-1)

Both dy/dt and dx/dt are zero when t=1.

The question we have been discussing is whether or not there is a tangent line when t=1. When t=1, the curve has a cusp. If we approach t=1 from either side, the limit of the slopes of the secant lines through the point when t=1 is 1/3. So, do we say that there is a tangent line, and its slope is 1/3, or do we say that the derivative dy/dx does not exist and so there is no tangent line?

When t=1, (x,y)=(2,-1).

Thinking of moving forward in "time" toward t=1, from t=0.9, at the moment we arrive at (2,-1) we are facing down toward the left along the line y-(-1) = (1/3)(x-2).

Thinking of moving backward in "time" toward t=1, from t=1.1, at the moment we arrive at (2,-1) we are facing down toward the left along that same line.

However, thinking of moving forward in time from t=1 to t=1.1, the moment we leave (2,-1) we are facing up toward the right along that same line.

So, at t=1, which way are we going? Of course, we're not "going" anywhere since the speed is zero. So, did we instantaneously turn around?
Maybe there's no "front" and "back" to a particle. If the question is "which way is the particle moving at t=1?" is the answer "It's not moving." or is the answer "It is simultaneously moving up and down along that line."?

Is there an agreed upon definition that will answer our questions, or are there a couple of definitions that are not (quite) equivalent?

Maybe only the theoretical mathematicians care, and the engineering types don't. The engineering types should worry about the following, though.

I graphed this curve and used the CALC menu to ask imy TI-83 what dy/dx is when t=1. My TI-83 thinks the slope is zero because it returns a value of zero for dy/dt, but it returns a value of 1X10^ - 6 for dx/dt.

Some other interesting TI-83 results follow.
t=0.999 dy/dx=0.333557 (close to 1/3)
t=0.9999 dy/dx=0.3339990654 (FARTHER AWAY from 1/3)
t=0.99999 dy/dx=0.33898477
t=0.999999 dy/dx=0.4

It's interesting to see that the error is getting worse.

I hear my numerical analysis teacher's voice now. The most important (and only!?) thing I remember from my year-lone course is the following.

"If a number equal to zero is a problem mathematically,
then a number close to zero is a problem numerically."

This reminds me of teaching four sections of high school geometry, in a not too long ago previous life, from two different texts. (10th grade "regular" and 9th grade "honors") In one of the books, the definition of parallel lines requires that the lines be distinct whereas in the other book two lines with diferent "names" that are happen to be coincidental ARE considered parallel.

What do you think?

Beth Hentges
Century College
White Bear Lake, MN

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Date Subject Author
3/22/01 BETH HENTGES