
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. :(

The magnitude of a vector $\displaystyle \vec{v}=\sqrt{\vec{v}\cdot\vec{v}}$.

I got down to $\displaystyle \frac{root{2}}{3e^t}$
That look right? :/

Can you clean up the LaTeX code a bit?

Tried to make it more legible.

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!