Answer to Question #265924 in Differential Geometry | Topology for Khizar

Question

Find curvature r(t) = ( ½ Cost, 1-sint, -3/2 cost)

Answer 1

$$r(t)=\langle\dfrac{1}{2}\cos t, 1-\sin t, -\dfrac{3}{2}\cos t\rangle$$

$$r'(t)=\langle-\dfrac{1}{2}\sin t, -\cos t, \dfrac{3}{2}\sin t\rangle$$

$$r''(t)=\langle-\dfrac{1}{2}\cos t, \sin t, \dfrac{3}{2}\cos t\rangle$$

$$|r'(t)|=\sqrt{(-\dfrac{1}{2}\sin t)^2+( -\cos t)^2+(\dfrac{3}{2}\sin t)^2}$$

$$=\dfrac{\sqrt{10-6\cos ^2t}}{2}$$

$$r'(t)\times r''(t)=\begin{vmatrix} i & j & k \\ \\ -\dfrac{1}{2}\sin t & -\cos t &\dfrac{3}{2}\sin t \\ \\ -\dfrac{1}{2}\cos t & \sin t &\dfrac{3}{2}\cos t \end{vmatrix}$$

$$=i\begin{vmatrix} -\cos t & \dfrac{3}{2}\sin t \\ \\ \sin t & \dfrac{3}{2}\cos t \end{vmatrix}-j\begin{vmatrix} -\dfrac{1}{2}\sin t & \dfrac{3}{2}\sin t \\ \\ -\dfrac{1}{2}\cos t& \dfrac{3}{2}\cos t \end{vmatrix}$$

$$+k\begin{vmatrix} -\dfrac{1}{2}\sin t & -\cos t \\ \\ -\dfrac{1}{2}\cos t & \sin t \end{vmatrix}=-\dfrac{3}{2}i-\dfrac{1}{2}k$$

$$|r'(t)\times r''(t)|=\sqrt{(-\dfrac{3}{2})^2+(0)^2+(-\dfrac{1}{2})^2}=\dfrac{\sqrt{10}}{2}$$

Find curvature

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

$$=\dfrac{\dfrac{\sqrt{10}}{2}}{(\dfrac{\sqrt{10-6\cos ^2t}}{2})^{3}}$$

$$=\dfrac{4\sqrt{10}}{(10-6\cos ^2t)^{3/2}}$$

$$=\dfrac{2\sqrt{5}}{(5-3\cos ^2t)^{3/2}}$$

$$\kappa(t)=\dfrac{2\sqrt{5}}{(5-3\cos ^2t)^{3/2}}$$