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mikau
2006-02-08 15:58:42

This is the drawback of being self taught. When your stuck you can't ask teacher.

Anyway, the question is: Can I use a graphing calculator to find the undifferentiable points? Do you suspect thats how I'm supposed to solve this type of problem? If so, I may have to get a graphing calculator and take a day or two to get acquainted with it.

Please respond soon as my math study will be stalled until I know the answer to this question.

mikau
2006-02-08 15:34:38

So you can spot it on a graphing calculator if it has a corner point or cusp?

I mean you can see the point on the graph easy, but can you ask the calculator to give you the value of t that produced the x, y coordinates of that point?

Ricky
2006-02-08 15:31:42

If you can use a graphing calculator, you can spot non-differentiable parts at a point if:

It isn't continuous (you have to lift up your pencil when drawing it)
It has a corner point

You can't do this with the defintion of differentiable because the function goes from R⇒R², which means the graph of it is in R³.  But you're only used to dealing with functions in R².  So partial derivatives don't quite make sense yet.

mikau
2006-02-08 15:23:08

I'm not sure what you mean. That must be above my level. This is calculus 1. I have no idea what "matrix of partial derivative" is.

In any case, in my book they merely graphed it on the graphing calculator, and noted there were undifferentiable points in the graph, but then they simply stated where they were without explaing how they got them. Frusterating...

Ricky
2006-02-08 15:19:18

f(t) = ( 4 cos(t) - 4 cos(4t) , 4 sin (t) - 4 sin(4t) )

So f : R ⇒ R²

That function is differentiable at x0 if:

where Df(x0) is the matrix of partial derivatives.  In this case:

Df = ( 16sin(4t) - 4sin(t) , 4cos(t) - 16cos(4t) )

Since there is only one variable in this function, there are no partials.

Try using that definition to see if it works.

mikau
2006-02-08 15:04:56

Ok right. I got it backwards.

Anyways, what would you suggest? Is there a function in graphing calcs for finding undifferentiable points?

Ricky
2006-02-08 15:02:31

If a function is differentiable, then it is continuous.

That statement is true.

If a function is continuious, then it is differentiable.

That statement is not true.  Remember that the definition of continuous at x0 means:

But consider the function f(x) = |x|.  It is continuous at 0, but not differentiable.

mikau
2006-02-08 14:55:12

Doesn't continuity imply differentiability?

Anyway, its not this particular problem thats the question. I just learned these problems today of finding the arc length of a parametric curve, but before I can begin I need to know if there are discontinuous points and where they are. Can a graphing calculator give you these? I don't have one.. :-(

Ricky
2006-02-08 14:24:02

You want differentiability, not continuity.  I'm still working on this problem though.

mikau
2006-02-08 14:14:08

you can only integrate using the arc length formula to find the length of a curve if the curve is continuous on the given interval. If their is a sharp point at b between a and c, you need to integrate from a to b, then b to c. a to c won't work since the function must be continuous to integrate.

My problem is don't know how to find the points where function is not continuous.

Take for instance the parametric equaton:

x = 4 cos(t) - 4 cos(4t)
y = 4 sin (t) - 4 sin(4t)

the graph of this function looks similar to a 3 leaf clover, and at the meeting point of the leafs the function has a sharp corner (discontinuous) at t = 2/3 pi, t = 4/3 pi and t = 2 pi (or zero degree's).

But how am I supposed to find where the parametric curve is discontinuous? :-(

I had an idea that it should be at the values of t where dy/dt AND dx/dt = 0. These will be the points where both x and y change direction. If both change direction the curve should pretty much go back the way it came producing a sharp point. I think this might work assuming x(t) and y(t) are continuous in the first place. (which may be a bad assumption :-/ ) any other time x or y changes direction should be ok so long as the other does not change direction at the same instant.

Still, this would requre to you find all values of t for which dy/dt and dx/dt equal zero, which will not always be easy. This particular example has 3 values of t where the function is discontinuous, others may have more. Also solving for t when dy/dt = 0 is not always going to be easy. Even in this example its no cakewalk.

I don't know what to do... :-(