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Calculate the Frenet apparatus $\kappa,\tau,T,N,B$ where $T,N,B$ are the tangent, normal and binormal vectors.

The curve i'm trying to calculate the Frenet apparatus for is given by:

$\gamma(t)=(e^tcos(t),e^tsin(t),e^t)$

In the example we did in class our curve was paramaterized by arc length, which meant $T$ would automatically be a unit vector. Is the definition of $T$ that it must be a unit vector? Or does $T=\gamma'(t)$ in general? Anyway, from here on out do I just start calculating? It looks like it's going to get very messy.

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  • $\begingroup$ $T$ must be a unit vector so just divide the derivative by its magnitude. "It looks like it's going to get very messy". That says it all. You haven't attempted it yet. I suggest you do. $\endgroup$ – John Douma Jan 18 at 15:31
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    $\begingroup$ I recommend you look at my differential geometry text, in particular, at pp. 12-14. $\endgroup$ – Ted Shifrin Jan 18 at 19:54
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$\newcommand{\cyclic}[1]{\langle #1\rangle}%cyclic group \newcommand{\ann}{\text{Ann\,}}%annihilator \newcommand{\aut}{\text{Aut}\,}%aut G.$ Use the formulas from Shiffrin's notes for a (possibly) non-ALP regularly parameterized curve.

$$\vec{\alpha'}(t)=e^t\cyclic{\cos(t)-\sin(t),\sin(t)+\cos(t),1}$$ $$\vec{\alpha'}(t)=e^t\cyclic{\cos(t)-\sin(t),\cos(t)+\sin(t),1}$$ $$\vec{\alpha}''(t)=e^t\cyclic{-2\sin(t),2\cos(t),1}$$ $$\vec{\alpha}'''(t)=e^t\cyclic{-2(\sin(t)+\cos(t)),2(\cos(t)-\sin(t)),1}$$ $$\vec{T}(t)=\frac{\vec{\alpha}'(t)}{|\vec{\alpha}'(t)|}=\frac{\cyclic{\cos(t)-\sin(t),\sin(t)+\cos(t),1}}{\sqrt{1+2\cos ^2\left(t\right)+2\sin ^2\left(t\right)}}=\frac{1}{\sqrt{3}}{\cyclic{\cos(t)-\sin(t),\sin(t)+\cos(t),1}}$$ $$\vec{T'}(t)=\frac1{\sqrt3}\cyclic{-\sin(t)-\cos(t),\cos(t)-\sin(t),0}~~~~~|\vec{T}'(t)|=\frac{1}{\sqrt3}\sqrt{2(\sin^2(t)+\cos^2(t))}=\sqrt{\frac23}$$ $$\vec{N}(t)=\frac{\vec{T}'(t)}{|\vec{T}'(t)|}=\frac1{\sqrt2}\cyclic{-\sin(t)-\cos(t),\cos(t)-\sin(t),0}$$ $$\vec{T}(t) \times \vec{N}(t)=\frac{1}{\sqrt6}\det \begin{pmatrix}\hat{i}&\hat{j}&\hat{k}\\ \:\cos\left(t\right)-\sin\left(t\right)&\sin\left(t\right)+\cos\left(t\right)&1\\ \:-\sin\left(t\right)-\cos\left(t\right)&\cos\left(t\right)-\sin\left(t\right)&0\end{pmatrix}$$ $$\vec{B}(t)=\frac{1}{\sqrt6}\cyclic{\sin(t)-\cos(t),\sin(t)+\cos(t),2}$$ $$\vec{\alpha}'(t)\times\vec{\alpha}''(t)=e^{2t}\det\begin{pmatrix}\hat{i}&\hat{j}&\hat{k}\\ \cos \left(t\right)-\sin \left(t\right)&\sin \left(t\right)+\cos \left(t\right)&1\\ -2\sin \left(t\right)&2\cos \left(t\right)&1\end{pmatrix}=e^{2t}\cyclic{\sin(t)-\cos(t),\cos(t)+\sin(t),2}$$ $$|\vec{\alpha}'(t)\times\vec{\alpha}''(t)|=\sqrt{6}e^{2t}~~~~~~|\alpha'|^3=(\sqrt3e^t)^3=3\sqrt3e^{3t}$$ $$\kappa(t)=\frac{|\vec{\alpha}'(t)\times\vec{\alpha}''(t)|}{|\vec{\alpha}'|^3}=\frac{\sqrt2}{3}e^{-t}$$ $$\tau(t)=\vec{\alpha}'(t)\cdot(\vec{\alpha}''(t)\times\vec{\alpha}'''(t))=e^{3t}\det \begin{pmatrix} \cos(t)-\sin(t) & \cos(t)+\sin(t) & 1\\ -2\sin(t) & 2\cos(t) & 1\\ -2(\sin(t)+\cos(t)) & 2(\cos(t)-\sin(t)) & 1\end{pmatrix}$$ $$=(2\cos \left(t\right)\sin \left(t\right)-2\sin ^2\left(t\right)) -(2\cos ^2\left(t\right)+2\sin \left(t\right)\cos \left(t\right))+4(\sin^2(t)+\cos^2(t))=2e^{3t} $$

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