Vector-valued function is smooth over an interval

[A.Magnus]

I am reading Larson's Calculus textbook and come across this paragraph about vector-valued function:

The parametrization of the curve represented by the vector-valued function

$$\textbf{r}(t) = f(t)\textbf{i} + g(t)\textbf{j} + h(t)k$$

is smooth on an open interval $I$ when $f'$, $g'$ and $h'$ are continuous on $I$ and $\textbf{r}'(t) \neq \textbf{0}$ for any value of $t$ in the interval $I$.

Can somebody please tell me why is that $\textbf{r}'(t)$ has to be not equal to zero? Thank you beforehand for your time and gracious helping hand. ~MA

 

[I like Serena]

I am reading Larson's Calculus textbook and come across this paragraph about vector-valued function:

The parametrization of the curve represented by the vector-valued function

$$\textbf{r}(t) = f(t)\textbf{i} + g(t)\textbf{j} + h(t)k$$

is smooth on an open interval $I$ when $f'$, $g'$ and $h'$ are continuous on $I$ and $\textbf{r}'(t) \neq \textbf{0}$ for any value of $t$ in the interval $I$.

Can somebody please tell me why is that $\textbf{r}'(t)$ has to be not equal to zero? Thank you beforehand for your time and gracious helping hand. ~MA

Hey MaryAnn! (Smile)

The vector $\mathbf{r}'(t)$ is the tangent of the curve.
If it is zero somewhere, it means that the curve slows to a stop or even backtracks on itself.

The $\mathbf{r}'(t) \ne \mathbf 0$ restriction is not generally part of the definition of smooth though.
Instead it's part of the definition of a regular curve (of order $1$).
So we're really talking about a regular smooth curve (regular of order $1$ and smooth of class $C^1$).