Answer to Question #265931 in Differential Geometry | Topology for Khizar

Question

Find Torison r=(a (3u, u³), 3au², 3a (3u+u³)

Answer 1

$$r=\langle a (3a-u^3), 3au^2, 3a (3u+u^3)\rangle$$

$$r'=\langle -3au^2, 6au, 9a (1+u^2)\rangle$$

$$r''=\langle -6au, 6a, 18au\rangle$$

$$r'''=\langle -6a, 0, 18a\rangle$$

$$r'\times r''=\begin{vmatrix} i & j & k \\ -3au^2 & 6au & 9a(1+u^2)\\ -6au & 6a & 18au \end{vmatrix}$$

$$=i(108a^2u^2-54a^2(1+u^2))$$

$$-j(-54a^2u^3+54a^2u(1+u^2))$$

$$+k(-18a^2u^2+36a^2u^2)$$

$$=54a^2(u^2-1)i-54a^2uj+18a^2u^2k$$

$$(r'\times r'')\cdot r'''=54a^2(u^2-1)(-6a)-0+18a^2u^2(18a)$$

$$=-324a^3u^2+324a^3+324a^3u^2=324a^3$$

$$||r'\times r''||^2=(54a^2(u^2-1))^2+(-54a^2u)^2+(18a^2u^2)^2$$

$$=324a^4(9u^4-18u^2+9+9u^2+u^4)$$

$$=324a^4(10u^4-9u^2+9)$$

$$\tau=\dfrac{(r'\times r'')\cdot r'''}{||r'\times r''||^2}$$

$$=\dfrac{324a^3}{324a^4(10u^4-9u^2+9)}=\dfrac{1}{a(10u^4-9u^2+9)}$$

$$\tau=\dfrac{1}{a(10u^4-9u^2+9)}$$