- Feb 2010

- 148

- 7

Hello all,

I have a regular parametrized curve \(\displaystyle \gamma: \mathbb{R} \rightarrow \mathbb{R}^{3}\) such that \(\displaystyle \Vert\gamma''(t)\Vert =1 \) for all \(\displaystyle t \in \mathbb{R}\).

Assume that \(\displaystyle \gamma(t)\) has constant curvature \(\displaystyle k \neq 0\) and constant torsio \(\displaystyle \tau=\frac{1}{\sqrt{2}}\). We also assume that:

\(\displaystyle \gamma(0)=\left( \frac{1}{\sqrt{2}},0,0 \right)\)

\(\displaystyle \gamma'(0)=\left( 0,\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}} \right)\)

\(\displaystyle \mathbf{b}(t)=\frac{1}{\sqrt{2}} \left( \sin(t),-\cos(t),1 \right)\)

where \(\displaystyle \mathbf{b}(t)\) is the binormal of \(\displaystyle \gamma(t)\)

I am asked to:

1) find the curvature K

2) find the \(\displaystyle \gamma(t)\) explicitely

I have done the following:

1)

Using the Frenet equations I have that \(\displaystyle \mathbf{n'}=-k\mathbf{t}+\tau\mathbf{b}\).

Differentiating on both sides yields \(\displaystyle \mathbf{n''}=-k\mathbf{t'}+\tau\mathbf{b'}\) (*).

Using \(\displaystyle \mathbf{b'}=-\tau\mathbf{n}\) and \(\displaystyle \mathbf{t'}=k\mathbf{n}\) and substituting in (*) gives \(\displaystyle k=\frac{\sqrt{2}}{2}\).

2)

I am still not done with this one but my suggestion is to use that \(\displaystyle \mathbf{n}=\frac{\gamma''(t)}{\Vert \gamma''(t) \Vert}\) and integrate together with the fact that \(\displaystyle \Vert\gamma''(t)t\Vert =1 \).

Could someone verify 1) and 2)?

Thanks.

I have a regular parametrized curve \(\displaystyle \gamma: \mathbb{R} \rightarrow \mathbb{R}^{3}\) such that \(\displaystyle \Vert\gamma''(t)\Vert =1 \) for all \(\displaystyle t \in \mathbb{R}\).

Assume that \(\displaystyle \gamma(t)\) has constant curvature \(\displaystyle k \neq 0\) and constant torsio \(\displaystyle \tau=\frac{1}{\sqrt{2}}\). We also assume that:

\(\displaystyle \gamma(0)=\left( \frac{1}{\sqrt{2}},0,0 \right)\)

\(\displaystyle \gamma'(0)=\left( 0,\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}} \right)\)

\(\displaystyle \mathbf{b}(t)=\frac{1}{\sqrt{2}} \left( \sin(t),-\cos(t),1 \right)\)

where \(\displaystyle \mathbf{b}(t)\) is the binormal of \(\displaystyle \gamma(t)\)

I am asked to:

1) find the curvature K

2) find the \(\displaystyle \gamma(t)\) explicitely

I have done the following:

1)

Using the Frenet equations I have that \(\displaystyle \mathbf{n'}=-k\mathbf{t}+\tau\mathbf{b}\).

Differentiating on both sides yields \(\displaystyle \mathbf{n''}=-k\mathbf{t'}+\tau\mathbf{b'}\) (*).

Using \(\displaystyle \mathbf{b'}=-\tau\mathbf{n}\) and \(\displaystyle \mathbf{t'}=k\mathbf{n}\) and substituting in (*) gives \(\displaystyle k=\frac{\sqrt{2}}{2}\).

2)

I am still not done with this one but my suggestion is to use that \(\displaystyle \mathbf{n}=\frac{\gamma''(t)}{\Vert \gamma''(t) \Vert}\) and integrate together with the fact that \(\displaystyle \Vert\gamma''(t)t\Vert =1 \).

Could someone verify 1) and 2)?

Thanks.

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