Answer to Question #289748 in Differential Geometry | Topology for Abi

Question

find the curvature and torsion of the curve z=u,y=1+u/u,z=1-u^2/u

Answer 1

a)

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

$$r'(u)=\langle1, -\dfrac{1}{u^2},-\dfrac{1}{u^2}-1\rangle$$

$$r''(u)=\langle0, \dfrac{2}{u^3}, \dfrac{2}{u^3}\rangle$$

$$|r'(u)|=\sqrt{(1)^2+(-\dfrac{1}{u^2})^2+(-\dfrac{1}{u^2}-1)^2}$$

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

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

$$=i\begin{vmatrix} -\dfrac{1}{u^2} & -\dfrac{1}{u^2}-1 \\ \\ \dfrac{2}{u^3} & \dfrac{2}{u^3} \end{vmatrix}-j\begin{vmatrix} 1 & -\dfrac{1}{u^2}-1 \\ \\ 0& \dfrac{2}{u^3} \end{vmatrix}$$

$$+k\begin{vmatrix} 1 & -\dfrac{1}{u^2} \\ \\ 0 & \dfrac{2}{u^3} \end{vmatrix}=-\dfrac{2}{u^3}i-\dfrac{2}{u^3}j+\dfrac{2}{u^3}k$$

$$|r'(t)\times r''(t)|=\sqrt{(-\dfrac{2}{u^3})^2+(-\dfrac{2}{u^3})^2+(\dfrac{2}{u^3})^2}$$

$$=\dfrac{2\sqrt{3}}{u^2|u|}$$

Find curvature

$$\kappa(t)=\dfrac{|r'(t)\times r''(t)|}{(|r'(t)|)^{3}}$$

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

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

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

b)

$$r'''(u)=\langle0, -\dfrac{6}{u^4}, -\dfrac{6}{u^4}\rangle$$

$$(r'(u)\times r''(u))\cdot r'''(u)=0-\dfrac{2}{u^3}(-\dfrac{6}{u^4})+\dfrac{2}{u^3}(-\dfrac{6}{u^4})$$

$$=0$$

$$\tau(u)=\dfrac{(r'(u)\times r''(u))\cdot r'''(u)}{(|r'(t)\times r''(t)|)^2}=0$$