# Finding curvature

• Jun 10th 2010, 08:34 AM
Negativ
Finding curvature
Find the curvature K of the curve.
$r(t) = < e^t cos(t) , e^t sin(t) , e^t >$

I have:
$
r(t) = < e^t cos(t) , e^t sin(t) , e^t >$

$r'(t) = < e^t cos(t) - e^t sin(t), e^t sin(t) + e^t cost(t), e^t >
$

$r''(t) = < - e^t cos(t) - e^t sin(t) + e^t cos(t) - e^t sin(t) ,$ $
e^t cos(t) + e^t sin(t) + e^t cos(t) - e^t sin(t), e^t >
$

Combine like terms:
$
r''(t) = <- 2e^t sin(t) , 2e^t cos(t) , e^t >
$

I have that curvature is

$
\kappa = \frac{|r' \times r''|}{|r'|^3}
$

For the numerator I crossed r' and r'' and got:

$

$

I don't know how to convert this into a magnitude; same for denominator. :(
• Jun 10th 2010, 08:48 AM
Ackbeet
The magnitude of a vector $|\vec{v}|=\sqrt{\vec{v}\cdot\vec{v}}$.
• Jun 10th 2010, 09:51 AM
Negativ
I got down to $\frac{root{2}}{3e^t}$

That look right? :/
• Jun 10th 2010, 10:01 AM
Ackbeet
Can you clean up the LaTeX code a bit?
• Jun 10th 2010, 10:07 AM
Negativ
Tried to make it more legible.
• Jun 10th 2010, 10:13 AM
Ackbeet
Try this: $\frac{\sqrt{2}}{3e^{t}}$. You can click on the equation I just wrote down to see what LaTeX code I wrote to obtain it.

That's what I got as well. Looks good!