Finding Curvature and Torsion: Derivatives and Unit Vectors Explained

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SUMMARY

This discussion focuses on the calculation of curvature and torsion using derivatives and unit vectors in vector calculus. The user successfully computed the unit vector for dN/ds as {.21i + 0.91j - 0.42k} normalized to {.205i + 0.889j - 0.4102k}. They derived the expression for dN/ds using the formula dN/ds = (dB/ds) x T + B x (dT/ds) and simplified it to T⋅(dN/ds) = -τN. Ultimately, the user resolved their confusion independently.

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Homework Statement



http://www.mathhelpforum.com/math-help/attachments/f6/22423d1317129472-curvature-torsion-untitled.png



The Attempt at a Solution



What I did was I calculated the unit vector for dN/ds={.21i+0.91j-0.42k}/ sqrt(.21^2+.91^2+0.42^2)=.205i+0.889j-.4102k


then, I took a derivative of N

since N=BxT,

dN/ds=(dB/ds)xT+Bx(dT/ds)
then taking T and doing dot product on both sides:
T⋅(dN/ds)=T⋅(dB/ds)xT+Bx(dT/ds) which simplifies to:
T⋅(dN/ds)=dB/ds
T⋅(dN/ds)=-τN

From here I have no clue what I should be doing..

Any suggestions will be greatly appreciated
 
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nvm figured it out
 

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