# Smooth curves

1. Sep 21, 2009

### 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.

2. Sep 21, 2009

### 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.

3. Sep 21, 2009

### bodensee9

Thanks!

4. Sep 21, 2009

### LCKurtz

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:

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

and

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

and look at the curve

$$\vec{r}(t) = < f(t), g(t) >$$

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.