Question

Find the curvature and torsion of the curve x=a(u-sinu),y=(1-cosu),z=bu

Answer 1

let

$r=[a(u-sinu),1-cosu,bu]$

so

$r^{'}$ (single deriverative of r) $=[a(1-cosu),sinu,0]$

$r^{''}=[asinu,cosu,0]$

$r^{'''}=acosu,-sinu,0$

Now :

Curvature $=\dfrac{|r^{'} \times r^{''}|}{|r^{'}|^3}$

Now: $r^{'} \times r^{''}=$ $i[-bcosu]-j[-absinu]+k[a(cosu-cos^2u)-asin^2u]$

$=(-bcosu)i +(absinu)j+(acosu-a)k$

$|r^{'} \times r^{''}|= \sqrt{b^2(cos^2u+a^2sin^2u)+a^2+a^2(cos^2u-2cosu)}$

$|r^{'}|= \sqrt{a^2(1+cos^2u)-2acosu+sin^2u+b}$

Curvature $= \dfrac{|r^{'} \times r^{''}|}{|r^{'}|^3} = \dfrac{\sqrt{b^2(cos^2u+a^2sin^2u)+a^2+a^2(cos^2u-2cosu)}}{(a^2(1+cos^2u)-2acosu+sin^2u+b)^{3/2}}$

Torsion $= \dfrac{[r^{'},r^{''},r^{'''}]}{|r^{'} \times r^{''}|^2}$

$[r^{'},r^{''},r^{'''}]=[a(1-cosu)(0)] -[sinu(0)] +b(-asin^2u-acos^2u)= -ab$

so Torsion $= \dfrac{-ab}{b^2(cos^2u+a^2sin^2u)+a^2+a^2(cos^2u-2cosu)}$