Answer to Question #265916 in Differential Geometry | Topology for Khizar

Question

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

Answer 1

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

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

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

$$=i\begin{vmatrix} -\cos t & (3\sin t)/2 \\ \sin t & (3\cos t)/2 \end{vmatrix}$$

$$-j\begin{vmatrix} -(\sin t)/2 & (3\sin t)/2 \\ -(\cos t)/2 & (3\cos t)/2 \end{vmatrix}$$

$$+k\begin{vmatrix} -(\sin t)/2 &-\cos t \\ -(\cos t)/2 & \sin t \end{vmatrix}$$

$$=-\dfrac{3}{2}i+\dfrac{3}{4}\sin t\cos tj-\dfrac{1}{2}k$$

$$|r'(t) \times r''(t)|=\sqrt{\dfrac{9}{4}+\dfrac{9}{16}\sin^2t\cos^2t+\dfrac{1}{4}}$$

$$=\dfrac{1}{4}\sqrt{40+9\sin^2t\cos^2t}$$

$$|r'(t)|=\sqrt{\dfrac{1}{4}\sin^2t+\cos^2t+\dfrac{9}{4}\sin^2t}$$

$$=\dfrac{1}{2}\sqrt{4+6\sin^2t}$$

$$k=\dfrac{|r'(t) \times r''(t)|}{|r'(t)|^3}=\dfrac{\dfrac{1}{4}\sqrt{40+9\sin^2t\cos^2t}}{\dfrac{1}{8}(4+6\sin^2t)\sqrt{4+6\sin^2t}}$$

$$=\dfrac{\sqrt{40+9\sin^2t\cos^2t}}{(2+3\sin^2t)\sqrt{4+6\sin^2t}}$$