Question #265929

Prove that





gamma=(a(3u,u^ 8 ), 3a * u ^ 2, 3a * (3u + u ^ 3)





tau = 1/(k ^ 2) * (r*gamma)

1
Expert's answer
2021-11-15T17:13:33-0500

γ=(a(3u+u8),3au2,3a(3u+u3))\gamma=(a(3u+u^ 8 ), 3a u ^ 2, 3a (3u + u ^ 3))


τ=rγk2\tau=\frac{r\gamma}{k^2}


torsion:


τ=(γ×γ)γγ×γ2\tau=\frac{(\gamma'\times \gamma'')\cdot\gamma'''}{|\gamma'\times \gamma''|^2}


curvature:


k=γ×γγ3k=\frac{|\gamma'\times \gamma''|}{|\gamma'|^3}


then:

(γ×γ)γ=γ6rγ(\gamma'\times \gamma'')\cdot\gamma'''=|\gamma'|^6r\gamma



γ=(a(3+8u7),6au,3a(3+3u2))\gamma'=(a(3+8u^ 7 ), 6au , 3a (3 + 3u ^2))

γ=(56au6,6a,18au)\gamma''=(56au^ 6 , 6a , 18au)

γ=(336au5,0,18a)\gamma'''=(336au^ 5 , 0 , 18a)


(γ×γ)γ=a(3+8u7)6au3a(3+3u2)56au66a18au336au5018a=(\gamma'\times \gamma'')\cdot\gamma'''=\begin{vmatrix} a(3+8u^ 7 ) & 6au& 3a (3 + 3u ^2)\\ 56au^ 6 & 6a&18au\\ 336au^ 5&0&18a \end{vmatrix}=


=6au(1008a2u66048a2u6)+6a(54a2+144a2u73024a2u53024a2u7)==-6au(1008a^2u^6-6048a^2u^6)+6a(54a^2+144a^2u^7-3024a^2u^5-3024a^2u^7)=

=12960a3u718144a3u5+324a3=12960a^3u^7-18144a^3u^5+324a^3


γ6=((3a+8au7)2+36a2u2+(9a+9au2))3|\gamma'|^6=((3a+8au^ 7 )^2+36a^2u ^ 2+(9a + 9au ^ 2))^3


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