Answer to Question #278191 in Differential Geometry | Topology for Rabia qadeer

Question

Find

the Curvature and

torsion of the Curves r = (u, (1 + u)/u, (1 - u ^ 2)/u)

Answer 1

The curvature of the curve given by the vector function $\bm r$ is

$$k(u)=\dfrac{|\bm r'(u)\times\bm r''(u)|}{|\bm r'(u)|^3}$$

$$\bm r'(u)=\langle 1 , \dfrac{u-(1+u)}{u^2}, \dfrac{-2u^2-(1-u^2)}{u^2}\rangle$$

$$=\langle 1 , -\dfrac{1}{u^2}, -(1+\dfrac{1}{u^2}\rangle$$

$$\bm r''(u)=\langle 0 , \dfrac{2}{u^3}, \dfrac{2}{u^3}\rangle$$

$$\bm r'(u)\times\bm r''(u)=\begin{vmatrix} \bm i & \bm j & \bm k \\ 1 & -1/u^2 & -1-1/u^2 \\ 0 & 2/u^3 & 2/u^3 \\ \end{vmatrix}$$

$$=(-2/u^5+2/u^3+2/u^5)\bm i-(2/u^3)\bm j+()2/u^3\bm k$$

$$=(2/u^3)\bm i-(2/u^3)\bm j+(2/u^3)\bm k$$

$$|\bm r'(u)|=\sqrt{1+1/u^4+1+2/u^2+1/u^4}$$

$$=\dfrac{\sqrt{2}}{u^2}\sqrt{u^4+u^2+1}$$

$$|\bm r'(u)\times\bm r''(u)|=\dfrac{\sqrt{6}}{u^2|u|}$$

$$k(u)=\dfrac{\sqrt{6}}{u^2|u|}\cdot\dfrac{u^6}{2\sqrt{2}(u^4+u^2+1)^{3/2}}$$

$$k(u)=\dfrac{\sqrt{3}u^2|u|}{2(u^4+u^2+1)^{3/2}}$$

The torsion of the curve given by the vector function $\bm r$ is

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

$$\bm r'''(u)=\langle 0 , -\dfrac{6}{u^4}, -\dfrac{6}{u^4}\rangle$$

$$(\bm r'(u)\times\bm r''(u))\cdot\bm r'''(u)=0+12/u^7-12/u^7=0$$

$$\tau(u)=0$$