# Curvature When t=0

1. Jun 1, 2014

### mill

1. The problem statement, all variables and given/known data

Find the curvature of

$r(t) = \frac 4 9 (1+t)^ \frac 3 2 i + \frac 4 9 (1-t)^ \frac 3 2 j + \frac 1 3 t \hat k$ at t=0

2. Relevant equations

K=1/|v| * |dT/dt|

3. The attempt at a solution

Found v.

$v= \frac 2 3 (1+t)^ \frac 1 2 i - \frac 2 3 (1-t)^ \frac 1 2 j + 1/3 \hat k$
|v|=1

v/|v|= <(2/3)(1+t)^(1/2), -(2/3)(1-t)^(3/2), 1/3>

$\vec T = \frac v (|v|)$

Take derivative of T.

K(0)=1/1 * |dT/dt|

$= \sqrt( \frac 1 3 ^2 + \frac 1 3 ^2 )= \sqrt \frac 2 3$

The answer is $(\sqrt 2) / 3$. Where did I go wrong?

Last edited: Jun 1, 2014
2. Jun 1, 2014

### Simon Bridge

Did you misplace a minus sign?

Aside:
* you used the letter k for two different things: avoid. Curvature is K or \kappa
The way you wrote it, it looks like r(t) depends on the curvature.
* it can help to indicate vectors: i,j,k are unit vectors?

\vec r, \hat\imath, \hat\jmath, \hat k gets you $\vec r, \hat\imath, \hat\jmath, \hat k$, or you can bold-face them.

3. Jun 1, 2014

### mill

Ok. I tried to use capital K to denote curvature and changed the $\hat k$

The problem was copied as is. I fixed the minus signs but since they were squared anyway, it doesn't seem to make a difference.

Last edited: Jun 1, 2014
4. Jun 1, 2014

### mill

My question is

1. was my method wrong?

2. if no to 1 then is there something obviously wrong in any part?

Last edited: Jun 1, 2014
5. Jun 1, 2014

### ehild

You did not square the denominators.

K(0)=1/1 * |dT/dt|

$= \sqrt{ \left(\frac {1} {3}\right) ^2 + \left(\frac {1} {3}\right) ^2 }=\frac{ \sqrt { 2}} {3}$

ehild

6. Jun 2, 2014

### mill

Thanks so much!