Answer to Question #265929 in Differential Geometry | Topology for Khizar

Question

Prove that

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

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

Answer 1

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

$\tau=\frac{r\gamma}{k^2}$

torsion:

$\tau=\frac{(\gamma'\times \gamma'')\cdot\gamma'''}{|\gamma'\times \gamma''|^2}$

curvature:

$k=\frac{|\gamma'\times \gamma''|}{|\gamma'|^3}$

then:

$(\gamma'\times \gamma'')\cdot\gamma'''=|\gamma'|^6r\gamma$

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

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

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

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

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

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

$|\gamma'|^6=((3a+8au^ 7 )^2+36a^2u ^ 2+(9a + 9au ^ 2))^3$