vector-valued function

**cinder** · August 6th 2007, 04:34 PM

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

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

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

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

**ThePerfectHacker** · August 6th 2007, 07:24 PM

Originally Posted by cinder

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

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

A vector function is smooth iff it is differenciable and $\bold{r}'(t)\not = 0$ .

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

**topsquark** · August 7th 2007, 07:42 AM

Originally Posted by ThePerfectHacker

A vector function is smooth iff it is differenciable and $\bold{r}'(t)\not = 0$ .

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

Why would smoothness require that $\bold{r}'(t)\not = 0$ ?

-Dan

**ThePerfectHacker** · August 7th 2007, 08:29 AM

Originally Posted by topsquark

Why would smoothness require that $\bold{r}'(t)\not = 0$ ?

-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 $\bold{r}'$ within the computation. And to keep it valid we require vectors being smooth for certain theorerms.

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