Answer to Question #271939 in Differential Geometry | Topology for Angel Nodado

Question

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

y=sin x, x = pi/2

Answer 1

$$K=\dfrac{|y''|}{[1+(y')^2]^{3/2}}$$

$$y'=\cos x, y''=-\sin x$$

$$K=\dfrac{|-\sin x|}{[1+\cos ^2 x]^{3/2}}$$

$x=\pi/2$

$$K=\dfrac{|-\sin (\pi/2)|}{[1+\cos ^2 (\pi/2)]^{3/2}}=1$$

The radius of curvature of a curve at a point is called the inverse of the curvature $K$ of the curve at this point:

$$R=\dfrac{1}{K}=1$$

$$x_C=x-\dfrac{y'(1+(y')^2)}{y''}$$

$$y_C=y+\dfrac{1+(y')^2}{y''}$$

$x=\pi/2$

$$y(\pi/2)=\sin(\pi/2)=1$$

$$y'(\pi/2)=-\cos(\pi/2)=0$$

$$y''(\pi/2)=-\sin(\pi/2)=-1$$

$$x_C=\pi/2-\dfrac{0(1+(0)^2)}{-1}=\pi/2$$

$$y_C=1+\dfrac{1+(0)^2}{-1}=0$$

$C(\pi/2, 0)$