Smoothness of a Parametrized Curve: Understanding the Definition

[bodensee9]

Hi

I have a general question about the definition. I know that a curve parametrized by t is smooth if the derivative r'(t) is not 0. I assume this is the 0 vector?

So then does that mean that if we have r(t) = f(t)i + g(t)j + r(t)k then any of the two can be 0 simultaneously while the third isn't 0 and the curve is still smooth? For example, can f'(t) = 0 and g'(t) = 0 at some point t0 and with r'(t) not 0 then the curve is smooth? Or must f'(t), g'(t) and r'(t) all not be 0?

Thank you.

 

[Dick]

Yes to both. r'(t)=0 means the zero vector and f'(t), g'(t) and h'(t) not all zero is the correct condition.

 

[bodensee9]

Thanks!

 

[LCKurtz]

Hi

I have a general question about the definition. I know that a curve parametrized by t is smooth if the derivative r'(t) is not 0. I assume this is the 0 vector?

So then does that mean that if we have r(t) = f(t)i + g(t)j + r(t)k then any of the two can be 0 simultaneously while the third isn't 0 and the curve is still smooth? For example, can f'(t) = 0 and g'(t) = 0 at some point t0 and with r'(t) not 0 then the curve is smooth? Or must f'(t), g'(t) and r'(t) all not be 0?

Thank you.

It means the vector is not the zero vector. So as long as one of the components isn't zero, you are OK. Think of t as time and r(t) representing a moving point. You don't want the point to smoothly come to a stop and then go in a different direction. For example consider:

f(t) =<br /> <br /> \left \{<br /> \begin{array}{}<br /> t^2, t \ge 0\\<br /> 0, t \leq 0<br /> \end{array}<br /> \right.<br />

and

g(t) =<br /> <br /> \left \{<br /> \begin{array}{}<br /> 0, t \ge 0\\<br /> t^2, t \leq 0<br /> \end{array}<br /> \right.<br />

and look at the curve

\vec{r}(t) = &lt; f(t), g(t) &gt;

This \vec{r}(t) has a continuous derivative even at t = 0. The problem is that the point stops at the origin and makes a right angle turn. You don't want to call that "smooth", so you don't want the velocity, which is to say the derivative, to be zero.