Answer to Question #271917 in Calculus for Angel Nodado

Question

Find the curvature, the radius of curvature and the center of curvature having a parametric equations at the given point:

x = sin y, (1/2,1/6 π)

Answer 1

Solution;

Radius of curvature;

$R=\frac{(1+f'(y_0)^2)^{\frac32}}{f''(y_0)}$

$f(y)=sin y$

$f'(y)=cosy$

$f''(y)=-sin y$

Hence;

$R=\frac{(1+cos(30)^2)^{\frac32}}{-sin(30)}=-4.63$

$R=4.63$

$Curvature;$

$k=\frac{1}{R}$

$k=\frac{1}{4.63}=0.216$

Center of curvature;

$P=(x_0+R),(y_0+f'(y_0))R$

$P=(0.5+4.63),(\fracπ6+\frac{\sqrt3}{2})4.63$

$P=(5.13,6.4)$