Finding curvature

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

I have:
$\displaystyle r(t) = < e^t cos(t) , e^t sin(t) , e^t >$
$\displaystyle r'(t) = < e^t cos(t) - e^t sin(t), e^t sin(t) + e^t cost(t), e^t >$
$\displaystyle r''(t) = < - e^t cos(t) - e^t sin(t) + e^t cos(t) - e^t sin(t) ,$$\displaystyle e^t cos(t) + e^t sin(t) + e^t cos(t) - e^t sin(t), e^t >$

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

I have that curvature is

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

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

$\displaystyle <e^{2t} [sin(t)-cos(t)], -e^{2t} [sin(t)+cos(t)], 2e^{2t}>$

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 $\displaystyle |\vec{v}|=\sqrt{\vec{v}\cdot\vec{v}}$.
• Jun 10th 2010, 09:51 AM
Negativ
I got down to $\displaystyle \frac{root{2}}{3e^t}$

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: $\displaystyle \frac{\sqrt{2}}{3e^{t}}$. 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!