Dorsa R.
asked • 04/13/20can you find the unit tangent vector, unit normal and unit binormal , curvature and torsion for this as t= 0 ?
x = cosht
y = sinht
z = t
and t = 0
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![{\displaystyle {\begin{aligned}&\color {black}{\text{wyzant.com/resources/answers/752486}}\\[3ex]&{\underline {\text{Problem:}}}\\&{\text{Find the Frenet frame }}{\bigl [}\mathbf {T} ,\mathbf {N} ,\mathbf {B} {\bigr ]},{\text{ the curvature, and the torsion for the curve }}\\&\qquad r(t)=(\cosh t,\sinh t,t)\,.\\[2ex]&{\underline {\text{Solution:}}}\\&\qquad {\begin{aligned}{\dot {r}}(t)&=(\sinh t,\cosh t,1)\\{\ddot {r}}(t)&=(\cosh t,\sinh t,0)\\{\overset {\cdot \cdot \cdot }{r}}(t)&=(\sinh t,\cosh t,0)\end{aligned}}\\&{\text{Using the relation }}\cosh ^{2}t-\sinh ^{2}t=1,{\text{ the arc length satisfies }}\\&\qquad {\dot {s}}=\vert {\dot {r}}\vert ={\sqrt {2}}\cosh t>0\,.\\&{\text{It follows }}\\&\qquad \mathbf {T} (t)={\frac {1}{{\sqrt {2}}\cosh t}}(\sinh t,\cosh t,1)\,.\\&{\text{The cross product }}{\dot {r}}\times {\ddot {r}}=(-\sinh t,\cosh t,-1){\text{ points in the direction of }}\mathbf {B} ,\\&\qquad \mathbf {B} (t)={\frac {1}{{\sqrt {2}}\cosh t}}(-\sinh t,\cosh t,-1)\,.\\&{\text{Finally, }}\mathbf {N} =\mathbf {B} \times \mathbf {T} ,\\&\qquad \mathbf {N} (t)={\frac {1}{\cosh t}}(1,0,-\sinh t)\,.\\&{\text{In particular, }}\mathbf {T} (0)={\tfrac {1}{\sqrt {2}}}(0,1,1),\mathbf {N} (0)=(1,0,0),\mathbf {B} (0)={\tfrac {1}{\sqrt {2}}}(0,1,-1).\\&{\text{Using the Frenet-Serret equations, and the chain rule }}{\dot {s}}\cdot {\tfrac {d}{ds}}={\tfrac {d}{dt}},{\text{ one can }}\\&{\text{derive the formulae}}\\&\qquad \kappa ={\frac {\vert {\dot {r}}\times {\ddot {r}}\vert }{{\dot {s}}^{3}}}\quad {\text{ and }}\quad \kappa ^{2}\tau ={\frac {\det({\dot {r}},{\ddot {r}},{\overset {\cdot \cdot \cdot }{r}})}{{\dot {s}}^{6}}}\,.\\&{\text{If the arc length is measured from }}r(0)=(1,0,0),{\text{i.e., }}s(t)={\sqrt {2}}\sinh t,\\&{\text{one finds }}\\&\qquad \kappa =\tau ={\frac {1}{2\cosh ^{2}t}}={\frac {1}{s^{2}+2}}\,.\\&{\text{In particular, }}\kappa (0)=\tau (0)={\tfrac {1}{2}}.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bfbf65d4f595da4c701c1c8087fc64146aeb0ccd)
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