Answer to Question #271943 in Differential Geometry | Topology for Angel Nodado

Question

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

r = sin 3 theta , theta = 0

Answer 1

$$K=\dfrac{|r^2+2(r')^2-rr''|}{[r^2+(r')^2]^{3/2}}$$

$$r'=3\cos 3\theta, r''=-9\sin3\theta$$ $$K=\dfrac{|(\sin3\theta)^2+2(3\cos3\theta)^2+9(\sin3\theta)^2|}{[(\sin3\theta)^2+(3\cos 3\theta)^2]^{3/2}}$$

$$=\dfrac{10+8(\cos3\theta)^2}{[1+8(\cos 3\theta)^2]^{3/2}}$$

$\theta=0$

$$r(0)=0, r'(0)=3, r''(0)=0$$

$$K=\dfrac{|(0)^2+2(3)^2-0(0)|}{[(0)^2+(3)^2]^{3/2}}=\dfrac{2}{3}$$

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

$$R=\dfrac{1}{K}=\dfrac{3}{2}$$

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

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

$$x=r\cos\theta=\sin(3\theta)\cos \theta$$ $$y=r\sin\theta=\sin(3\theta)\sin \theta$$

$$y'_x=\dfrac{y'_\theta}{x'_\theta}=\dfrac{r'\sin \theta+r\cos \theta}{r'\cos\theta-r\sin \theta}$$

$$(y'_x)'_{\theta}=\dfrac{(r''\sin \theta+r'\cos\theta)(r'\cos\theta-r\sin \theta)}{(r'\cos\theta-r\sin \theta)^2}$$

$$+\dfrac{(r'\cos \theta-r\sin \theta)(r'\cos\theta-r\sin \theta)}{(r'\cos\theta-r\sin \theta)^2}$$

$$-\dfrac{(r''\cos \theta-r'\sin\theta)(r'\sin \theta+r\cos \theta)}{(r'\cos\theta-r\sin \theta)^2}$$

$$-\dfrac{(-r'\sin\theta-r\cos \theta)(r'\sin \theta+r\cos \theta)}{(r'\cos\theta-r\sin \theta)^2}$$

$$y''_{xx}=\dfrac{(r''\sin \theta+r'\cos\theta)(r'\cos\theta-r\sin \theta)}{(r'\cos\theta-r\sin \theta)^3}$$

$$+\dfrac{(r'\cos \theta-r\sin \theta)(r'\cos\theta-r\sin \theta)}{(r'\cos\theta-r\sin \theta)^3}$$

$$-\dfrac{(r''\cos \theta-r'\sin\theta)(r'\sin \theta+r\cos \theta)}{(r'\cos\theta-r\sin \theta)^3}$$

$$-\dfrac{(-r'\sin\theta-r\cos \theta)(r'\sin \theta+r\cos \theta)}{(r'\cos\theta-r\sin \theta)^3}$$

$\theta=0$

$$x(0)=0, y(0)=0$$

$$y'_{x}(0)=\dfrac{3(0)+0(1)}{3(1)-0(0)}=0$$

$$y''_{xx}(0)=\dfrac{(0+3)(3-0)}{(3)^3}$$

$$+\dfrac{(3-0)(3-0)}{(3)^3}$$

$$-\dfrac{(0-0)(0+0)}{(3)^3}$$

$$-\dfrac{(0-0)(0+0)}{(3)^3}=\dfrac{2}{3}$$

$$x_C=0-\dfrac{0(1+(0)^2)}{\dfrac{2}{3}}=0$$

$$y_C=0+\dfrac{1+(0)^2}{\dfrac{2}{3}}=\dfrac{3}{2}$$

$C(0, \dfrac{3}{2})$