Answer to Question #265938 in Differential Geometry | Topology for Khizar

Question

find curvature r = (4/5 Cost, 1-sint, -3/2 Cost)

Answer 1

$r = [\frac{4}{5} cos(t), (1-sin(t)), -\frac{3}{2} cos(t)]$

Since f(t), g(t), and h(t) are the components of the vector r(t), then f, g, and h are real-valued functions called the component functions of r and we can write

$r(t)=\frac{4}{5} cos(t) i + (1-sin(t)) j - \frac{3}{2} cos(t) k$

$r'(t)=-\frac{4}{5} sin(t) i - cos(t) j + \frac{3}{2} sin(t) k$

$r''(t)=-\frac{4}{5} cos(t) i + sin(t) j + \frac{3}{2} cos(t) k$

So the curvature is

$κ(t)=\frac{\vert r'(t)× r''(t)\vert}{{\vert r'(t)\vert}^{3}}$

$r'(t)× r''(t)=\begin{vmatrix} i & j & k \\ -\frac{4}{5} sint & -cost & \frac{3}{2} sint\\ -\frac{4}{5} cost & sint & \frac{3}{2} cost \end{vmatrix}$

$r'(t)× r''(t)=(-\frac{3}{2}cos^{2}t -\frac{3}{2}sin^{2}t)i -(-\frac{12}{10}cost.sint +\frac{12}{10}cost.sint)j +(-\frac{4}{5}sin^{2}t -\frac{4}{5}cos^{2}t)$

$r'(t)× r''(t)=-\frac{3}{2}(cos^{2}t+sin^{2}t)i -0j -\frac{4}{5}(sin^{2}t +cos^{2}t)k$

$r'(t)× r''(t)=-\frac{3}{2}i +0j -\frac{4}{5}k$

So Now

$\vert r'(t)× r''(t)\vert ={[(-\frac{3}{2})^2+(-\frac{4}{5})^2]}^{1/2}$

$={[\tfrac{9}{4}+\tfrac{16}{25}]}^{1/2}$ $={[\frac{289}{100}]}^{1/2}$ $=\frac{17}{10}$

$$\vert r'(t)× r''(t)\vert =\frac{17}{10}$$

${\vert r'(t)\vert}^3={[(-\frac{4}{5}sint)^2+(cost)^{2}+(-\frac{3}{2}sint)^2]}^{3/2}$

${\vert r'(t)\vert}^3={[\frac{16}{25}sin^2t+cos^2t+\frac{9}{4}sin^2t]}^{3/2}$

${\vert r'(t)\vert}^3={[\frac{189}{100}sin^2t+cos^2t]}^{3/2}$

${\vert r'(t)\vert}^3={[\frac{189}{100}sin^2t+1-sin^2t]}^{3/2}={[1+\frac{189-100}{100}sin^2t]}^{3/2}$

${\vert r'(t)\vert}^3={[1+\frac{89}{100}sin^2t]}^{3/2}$

$κ(t)=\frac{\vert r'(t)× r''(t)\vert}{{\vert r'(t)\vert}^{3}} =\frac{\frac{17}{10}}{{[\frac{189}{100}sin^2t+cos^2t]}^{3/2}}$

So the curvature is ...

$$κ(t)=\frac{\frac{17}{10}}{{[\frac{189}{100}sin^2t+cos^2t]}^{3/2}} =\frac{\frac{17}{10}}{{[1+\frac{89}{100}sin^2t]}^{3/2}}$$