# How can I find the curvature of $r(t) = ( \sqrt{2}t, e^t, e^{-t} )$ using the unit tangent vector?

I've been given the function $r(t) = ( \sqrt{2}t, e^t, e^{-t} )$ and asked to find:

(a) the unit tangent and unit normal vectors, and

(b) the curvature using the formula $|T'(t)| / |r'(t)|$ only (i.e. other formulae for K disallowed)

My answers for (a) match the textbook answer, but my answer for (b) does not. Unfortunately, the textbook only gives the final expected answer, no intermediate values or working, and I am unable to locate my error.

For reference I obtained:

$r'(t) = (\sqrt{2}, e^t, -e^{-t} )$ (unconfirmed)

$T(t) = \frac{1}{e^{2t} + 1}( \sqrt{2}e^t, e^{2t}, -1 )$ (apparently correct)

$T'(t) = 2e^{2t} ( \frac{1 - e^{2t}}{\sqrt{2}e^t}, 1, 1 )$ (unconfirmed)

$N(t) = \frac{1}{e^{2t} + 1} ( (1 - e^{2t}, \sqrt{2}e^t, \sqrt{2}e^t)$ (correct)

and $K = |T'(t)| / |r'(t)| = \frac{(\sqrt{2}e^t)(1 + e^{2t})}{(e^{2t} + 1)/e^t} = \sqrt{2}e^{2t}$ (apparently incorrect)

However the text book obtains $K = \frac{\sqrt{2}e^{2t}}{(e^{2t} + 1)^2}$

Can anyone tell me where I've gone wrong?

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It is a bit hard to read but it seems that the last coordinate on the 3rd equation should be $0$. – azarel Feb 20 '12 at 18:33
Could you please write e^{-t} and \sqrt{2} instead of e^(-t) and sqrt(2)? If you add also \$at the beginning and end of each formula then it would look much nicer: \$ e^{-1} \$->$e^{-t}$, \$ \sqrt{2} \$->$\sqrt{2}$. – savick01 Feb 20 '12 at 18:43 I edited your question, please check, for any introduce errors. There are some links posted here to learn Latex, or you can use MathJax help on the right, when you ask a question, or click here. – draks ... Feb 20 '12 at 18:54 I don't want to be rude, but I can see that it's not your first question and every time someone has to come and correct your non-LaTeX-question. It isn't really about learning the whole$\LaTeX$, in this case changing () into {} and adding some \$s and \s would be absolutely enough (@draks did a little bit more because he used \frac{a}{b} to make a fraction a/b, but that's all). – savick01 Feb 20 '12 at 19:35
That's a good start to do it your own. Good luck! – draks ... Feb 20 '12 at 19:40