# vector-valued function

• Aug 6th 2007, 04:34 PM
cinder
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.
• Aug 6th 2007, 07:24 PM
ThePerfectHacker
Quote:

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.
• Aug 7th 2007, 07:42 AM
topsquark
Quote:

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\$? :confused:

-Dan
• Aug 7th 2007, 08:29 AM
ThePerfectHacker
Quote:

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

-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? :confused:

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.