Finding curvature

**Negativ** · June 10th 2010, 08:34 AM

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:

$ <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.

**Ackbeet** · June 10th 2010, 08:48 AM

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

**Negativ** · June 10th 2010, 09:51 AM

I got down to $\frac{root{2}}{3e^t}$

That look right? :/

**Ackbeet** · June 10th 2010, 10:01 AM

Can you clean up the LaTeX code a bit?

**Negativ** · June 10th 2010, 10:07 AM

Tried to make it more legible.

**Ackbeet** · June 10th 2010, 10:13 AM

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!

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