Hola
Si dices que el determinante es 216, entonces tampoco me coincidiría con la curvatura, o me equivoco? podrías revisar si mi curvatura está bien obtenida?
Si. Está bien:
\( r(t)=(3t-t^3,3t^2,3t+t^3) \)
\( r'(t)=(3-3t^2,6t,3+3t^2)=3(1-t^2,2t,1+t^2) \)
\( r''(t)=3(-2t,2,2t)=6(-t,1,t) \)
\( r'''(t)=3(-2t,2,2t)=6(-1,0,1) \)
\( \|r'(t)\|=3\sqrt{(1-t^2)^2+(2t)^2+(1+t^2)^2}=3\sqrt{(2t^2+4t^2+2)}=3\sqrt{2}(t^2+1) \)
\( r'(t)\times r''(t)=18\left|\matrix{e_1&e_2&e_3\\1-t^2&2t&1+t^2\\-t&1&t}\right|=18(t^2-1,-2t,t^2+1) \)
\( \|r'(t)\times r''(t)\|=18\sqrt{(1+t^2)^2+(-2t)^2+(t^2+1)^2}=18\sqrt{(2t^2+4t^2+2)}=18\sqrt{2}(t^2+1) \)
\( \kappa(t)=\dfrac{\|r'(t)\times r''(t)\|}{\|r'(t)\|^3}=\dfrac{18\sqrt{2}(t^2+1)}{54\sqrt{2}(t^2+1)^3}=\dfrac{1}{3(t^2+1)^2} \)
\( det(r'(t),r''(t),r'''(t))=3\cdot 6^2\cdot \left|\matrix{1-t^2&2t&1+t^2\\-t&1&t\\-1&0&1}\right|=
3\cdot 6^2\cdot \left|\matrix{2&2t&1+t^2\\0&1&t\\0&0&1}\right|=6^3 \)
\( \tau(t)=\dfrac{det(r'(t),r''(t),r'''(t))}{\|r'(t)\times r''(t)\|^2}=\dfrac{6^3}{2\cdot 18^2(t^2+1)^2}=\dfrac{1}{3(t^2+1)^2} \)
Saludos.
