# Thread: vector-valued function

1. ## vector-valued function

Find the open interval(s) on which the curve given by the vector-valued function is smooth.

$\displaystyle r(t)=t^2i+t^3j$

If I follow the book correctly, I get the derivative which is $\displaystyle r'(t)=2ti+3t^2j$ and then after that I have no idea.

In the example problem it shows $\displaystyle r(t)=0i+0j$ and list some intervals, but I'm not sure what they're doing or what's going on.

2. Originally Posted by cinder
Find the open interval(s) on which the curve given by the vector-valued function is smooth.

$\displaystyle r(t)=t^2i+t^3j$
A vector function is smooth iff it is differenciable and $\displaystyle \bold{r}'(t)\not = 0$.

Thus,
$\displaystyle \bold{r}'(t) = 2t\bold{i}+3t^2\bold{j}$
It can only be be zero if $\displaystyle t=0$. Thus any open interval NOT containing zero makes this smooth.

3. Originally Posted by ThePerfectHacker
A vector function is smooth iff it is differenciable and $\displaystyle \bold{r}'(t)\not = 0$.

Thus,
$\displaystyle \bold{r}'(t) = 2t\bold{i}+3t^2\bold{j}$
It can only be be zero if $\displaystyle t=0$. Thus any open interval NOT containing zero makes this smooth.
Why would smoothness require that $\displaystyle \bold{r}'(t)\not = 0$?

-Dan

4. Originally Posted by topsquark
Why would smoothness require that $\displaystyle \bold{r}'(t)\not = 0$?

-Dan
Because the intuitive meaning is that we can draw a tangent vector at every point. Now what type of tangent is a zero vector?

I believe the mathematical meaning is that we end up dividing by $\displaystyle \bold{r}'$ within the computation. And to keep it valid we require vectors being smooth for certain theorerms.

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### vector valued function is smooth or not

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