Differentiability and parametric curves

[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?

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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!

 

[HallsofIvy]

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.

 

[kingwinner]

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.

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

 

[kingwinner]

any ideas?

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

 

[kingwinner]

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

 

[HallsofIvy]

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.