8. Properties of Curves

f. Curvature and Torsion

2.a. Torsion Formula Proof

By definition, \[ \tau=-\,\dfrac{d\hat{B}}{ds}\cdot\hat{N} \] We want to prove \[ \tau=\dfrac{\vec{v}\times\vec{a}\cdot\vec{j}}{|\vec{v}\times\vec{a}|^2} \]

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We start by converting \(\dfrac{d\hat{B}}{ds}\) from an \(s\) derivative to a \(t\) derivative, and then apply the quotient rule to \(\hat{B}=\dfrac{\vec{v}\times\vec{a}}{|\vec{v}\times\vec{a}|}\). \[ \dfrac{d\hat{B}}{ds} =\dfrac{1}{|\vec{v}|}\dfrac{d\hat{B}}{dt} =\dfrac{1}{|\vec{v}|}\dfrac {|\vec{v}\times\vec{a}|\dfrac{d}{dt}(\vec{v}\times\vec{a}) -\,\dfrac{d}{dt}|\vec{v}\times\vec{a}|(\vec{v}\times\vec{a})} {|\vec{v}\times\vec{a}|^2} \] This quotient simplifies because \(\vec{v}\times\vec{a}\) cancels in the first term and \(\dfrac{\vec{v}\times\vec{a}}{|\vec{v}\times\vec{a}|}=\hat{B}\) in the second term: \[ \dfrac{d\hat{B}}{ds} =\dfrac{1}{|\vec{v}|\,|\vec{v}\times\vec{a}|} \left(\dfrac{d}{dt}(\vec{v}\times\vec{a}) -\,\dfrac{d}{dt}|\vec{v}\times\vec{a}|\hat{B}\right) \] In the first term, we apply the product rule. (Note when you apply the product rule to a cross product, you cannot reverse the order of the factors, or else you get a minus sign error.) Then we use the definitions of acceleration and jerk and the fact that the cross product of a vector with itself is \(0\). \[\begin{aligned} \dfrac{d}{dt}(\vec{v}\times\vec{a}) &=\dfrac{d\vec{v}}{dt}\times\vec{a}+\vec{v}\times\dfrac{d\vec{a}}{dt} \\ &=\vec{a}\times\vec{a}+\vec{v}\times\vec{j} =\vec{v}\times\vec{j} \end{aligned}\] So \[ \dfrac{d\hat{B}}{ds }=\dfrac{1}{|\vec{v}|\,|\vec{v}\times\vec{a}|} \left(\vec{v}\times\vec{j}-\,\dfrac{d}{dt}|\vec{v}\times\vec{a}|\hat{B}\right) \] We now use the definition of torsion and the fact that \(\hat{B}\) and \(\hat{N}\) are orthogonal. \[\begin{aligned} \tau&=-\,\dfrac{d\hat{B}}{ds}\cdot\hat{N} =\dfrac{-1}{|\vec{v}|\,|\vec{v}\times\vec{a}|} \left(\vec{v}\times\vec{j}-\,\dfrac{d}{dt}|\vec{v}\times\vec{a}|\hat{B}\right)\cdot\hat{N} \\ &=\dfrac{-1}{|\vec{v}|\,|\vec{v}\times\vec{a}|}\vec{v}\times\vec{j}\cdot\hat{N} \end{aligned}\] Finally, we make use of one of the formulas for \(\hat{N}\): \[ \hat{N} =\dfrac{\text{proj}_{\bot \vec{v}}\vec{a}} {\left|\text{proj}_{\bot \vec{v}}\vec{a}\right|} =\dfrac{|\vec{v}|}{|\vec{v}\times\vec{a}|} \left(\vec{a}-\,\dfrac{\vec{a}\cdot\vec{v}}{|\vec{v}|^2}\vec{v}\right) \] So the torsion is: \[\begin{aligned} \tau&=\dfrac{-1}{|\vec{v}\times\vec{a}|^2}\vec{v}\times\vec{j} \cdot\left(\vec{a}-\,\dfrac{\vec{a}\cdot\vec{v}}{|\vec{v}|^2}\vec{v}\right) \\ &=\dfrac{-\vec{v}\times\vec{j}\cdot\vec{a}}{|\vec{v}\times\vec{a}|^2} =\dfrac{\vec{v}\times\vec{a}\cdot\vec{j}}{|\vec{v}\times\vec{a}|^2} \end{aligned}\] since \(\vec{v}\times\vec{j}\) is perpendicular to \(\vec{v}\) and the triple product changes sign when two vectors are interchanged.