Question #72689

Q. Find the curvature and torsion for the curve
γ(t)=(√2t〖,e〗^t 〖,e〗^(-t))

Expert's answer

v=(2t,et,et)v=(2,et,et)v=(0,et,et)v=(0,et,et)v=2+e2t+e2t=et+etv×v=(etet(et)et,(et)02et,2etet0)=2(2,et,et)v×v=2(et+et)curvaturek=v×vv3=2(et+et)2(v×v)v=2(20+(et)et+et(et))=22torsion(v×v)vv×v2=222(et+et)2=2(et+et)2\begin{array}{l} v = (\sqrt{2} t, e^{t}, e^{-t}) \\ v' = (\sqrt{2}, e^{t}, -e^{-t}) \\ v'' = (0, e^{t}, e^{-t}) \\ v''' = (0, e^{t}, -e^{-t}) \\ |v'| = \sqrt{2 + e^{2t} + e^{-2t}} = e^{t} + e^{-t} \\ v' \times v'' = (e^{t} \cdot e^{-t} - (-e^{-t}) \cdot e^{t}, (-e^{-t}) \cdot 0 - \sqrt{2} \cdot e^{-t}, \sqrt{2} \cdot e^{t} - e^{t} \cdot 0) = \sqrt{2} (\sqrt{2}, -e^{-t}, e^{t}) \\ |v' \times v''| = \sqrt{2} (e^{t} + e^{-t}) \\ \text{curvature} \quad k = \frac{|v' \times v''|}{|v'|^{3}} = \frac{\sqrt{2}}{(e^{t} + e^{-t})^{2}} \\ (v' \times v'') \cdot v''' = \sqrt{2} (\sqrt{2} \cdot 0 + (-e^{-t}) \cdot e^{t} + e^{t} \cdot (-e^{-t})) = -2\sqrt{2} \\ \text{torsion} \quad \frac{(v' \times v'') \cdot v'''}{|v' \times v''|^{2}} = \frac{-2\sqrt{2}}{2 (e^{t} + e^{-t})^{2}} = \frac{-\sqrt{2}}{(e^{t} + e^{-t})^{2}} \\ \end{array}


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