Substituting ds by dt in Frenet-Serret Formulas

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SUMMARY

This discussion focuses on substituting \( ds \) with \( dt \) in the Frenet-Serret Formulas, specifically addressing the curvature \( \kappa \) and torsion \( \tau \). The user aims to express the derivatives of the tangent vector \( \vec{T} \), normal vector \( \vec{N} \), and binormal vector \( \vec{B} \) in terms of \( t \) using the chain rule. The relationship \( \dot{s} = \frac{ds}{dt} = \left| \frac{d\vec{r}}{dt} \right| = \left| \dot{\vec{r}} \right| \) is crucial for this transformation, leading to the expression \( \vec{T} = \frac{\dot{\vec{r}}}{\dot{s}} \).

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Philosophaie
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I want to substitute ds by dt in the Frenet-Serret Formulas where κ is the curvature and is the torsion:
Tangential:\frac{d\vec{T}}{ds} = κ*\vec{N}
Normal:\frac{d\vec{N}}{ds} = -κ*\vec{T}+τ*\vec{B}
Binormal:\frac{d\vec{B}}{ds} =- τ*\vec{N}
I want to substitute \frac{d\vec{T}}{ds} → \frac{d}{dt} T(t) N(t), B(t) and solve for κ and τ.
 
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Just use the chain rule and the fact that
\dot{s}=\frac{\mathrm{d} s}{\mathrm{d} t}=\left | \frac{\mathrm{d} \vec{r}}{\mathrm{d} t} \right|=\left |\dot{\vec{r}} \right|,
where the function \vec{r}(t)[/itex] is the parametrization of the curve. Then you have
\vec{T}=\frac{\dot{\vec{r}}}{\dot{s}}
etc.
 

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