Answer to Question #271941 in Differential Geometry | Topology for Angel Nodado

Question

Find the curvature, the radius and the center of curvature at a point.

x=t, y=1/t ; t=1

Answer 1

curvature:

$K=\frac{|x'y''-y'x''|}{((x')^2+(y')^2)^{3/2}}$

$x'=1,x''=0$

$y'=-1/t^2,y''=2/t^3$

$K=\frac{2}{(1+1)^{3/2}}=\frac{1}{\sqrt 2}$

radius of curvature:

$R=1/K=\sqrt 2$

 center of curvature:

$x_c=x_0+Rsin|\theta|$

$y_c=y_0+Rcos|\theta|$

where $tan\theta$ is slope of tangent

$tan\theta=f'(x_0)$

$x_0=1,y_0=1$

$f'(x)=y'/x'=-1/t^2$

$f'(x_0)=-1$

$|\theta|=45\degree$

$x_c=1+1=2$

$y_c=1+1=2$