Differentiability and parametric curves

1. Oct 25, 2007

kingwinner

f(t)=(t^3, |t|^3) is a parametric representation of y=f(x)=|x|.

Consider y=|x|,
the left hand derivative f '-(0)=-1 and the right hand derivative f '+(0)=1, so f(x) is clearly not differentiable at 0.

But
f '(t)=(3t^2, 3t^2) for t>=0
f '(t)=(3t^2, -3t^2) for t<=0
f '(0)=(0,0) and f(t) is differentiable at 0 (my textbook says this explicitly)

These 2 are talking about the same point, how can one gives that it's differentiable at 0 and the other gives that it's not differentiable at 0?

==============================================

Secondly, my textbook says that f '(a)>0 (a is real number) doesn't imply that f is increasing in some neighbourhood of a, how come?

Thanks for explaining!

2. Oct 25, 2007

HallsofIvy

Staff Emeritus
In order to be a "valid" parameterization, the derivatives of all components must not be 0 for the same t. Since x'(0)= y'(0)= 0, this is not a valid parameterization.

3. Oct 26, 2007

kingwinner

But my textbook says that "f(t)=(t^3, |t|^3) is a parametric representation of y=f(x)=|x|." (direct quote)

4. Oct 28, 2007

kingwinner

any ideas?

How come the results are inconsistent? How come the parametrization shows that y=|x| is differentiable at x=0?

5. Oct 29, 2007

kingwinner

Please help me...I am confused...

6. Oct 29, 2007

HallsofIvy

Staff Emeritus
I have already answered this. Yes, what you give is a "parameterization" of the curve. But it is not a "valid" parameterization (some texts say "smooth" parameterization) because at t=0, f'(0) is the 0 vector. It is precisely because of problems like you have here that "valid" parameterizations are normally required.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?