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

Question #271943

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


r = sin 3 theta , theta = 0


1
Expert's answer
2021-12-06T16:35:42-0500
K=r2+2(r)2rr[r2+(r)2]3/2K=\dfrac{|r^2+2(r')^2-rr''|}{[r^2+(r')^2]^{3/2}}

r=3cos3θ,r=9sin3θr'=3\cos 3\theta, r''=-9\sin3\thetaK=(sin3θ)2+2(3cos3θ)2+9(sin3θ)2[(sin3θ)2+(3cos3θ)2]3/2K=\dfrac{|(\sin3\theta)^2+2(3\cos3\theta)^2+9(\sin3\theta)^2|}{[(\sin3\theta)^2+(3\cos 3\theta)^2]^{3/2}}

=10+8(cos3θ)2[1+8(cos3θ)2]3/2=\dfrac{10+8(\cos3\theta)^2}{[1+8(\cos 3\theta)^2]^{3/2}}

θ=0\theta=0

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

K=(0)2+2(3)20(0)[(0)2+(3)2]3/2=23K=\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 KK of the curve at this point:


R=1K=32R=\dfrac{1}{K}=\dfrac{3}{2}

xC=xy(1+(y)2)yx_C=x-\dfrac{y'(1+(y')^2)}{y''}


yC=y+1+(y)2yy_C=y+\dfrac{1+(y')^2}{y''}

x=rcosθ=sin(3θ)cosθx=r\cos\theta=\sin(3\theta)\cos \thetay=rsinθ=sin(3θ)sinθy=r\sin\theta=\sin(3\theta)\sin \theta

yx=yθxθ=rsinθ+rcosθrcosθrsinθy'_x=\dfrac{y'_\theta}{x'_\theta}=\dfrac{r'\sin \theta+r\cos \theta}{r'\cos\theta-r\sin \theta}


(yx)θ=(rsinθ+rcosθ)(rcosθrsinθ)(rcosθrsinθ)2(y'_x)'_{\theta}=\dfrac{(r''\sin \theta+r'\cos\theta)(r'\cos\theta-r\sin \theta)}{(r'\cos\theta-r\sin \theta)^2}

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

(rcosθrsinθ)(rsinθ+rcosθ)(rcosθrsinθ)2-\dfrac{(r''\cos \theta-r'\sin\theta)(r'\sin \theta+r\cos \theta)}{(r'\cos\theta-r\sin \theta)^2}

(rsinθrcosθ)(rsinθ+rcosθ)(rcosθrsinθ)2-\dfrac{(-r'\sin\theta-r\cos \theta)(r'\sin \theta+r\cos \theta)}{(r'\cos\theta-r\sin \theta)^2}

yxx=(rsinθ+rcosθ)(rcosθrsinθ)(rcosθrsinθ)3y''_{xx}=\dfrac{(r''\sin \theta+r'\cos\theta)(r'\cos\theta-r\sin \theta)}{(r'\cos\theta-r\sin \theta)^3}

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

(rcosθrsinθ)(rsinθ+rcosθ)(rcosθrsinθ)3-\dfrac{(r''\cos \theta-r'\sin\theta)(r'\sin \theta+r\cos \theta)}{(r'\cos\theta-r\sin \theta)^3}

(rsinθrcosθ)(rsinθ+rcosθ)(rcosθrsinθ)3-\dfrac{(-r'\sin\theta-r\cos \theta)(r'\sin \theta+r\cos \theta)}{(r'\cos\theta-r\sin \theta)^3}

θ=0\theta=0


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

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

yxx(0)=(0+3)(30)(3)3y''_{xx}(0)=\dfrac{(0+3)(3-0)}{(3)^3}

+(30)(30)(3)3+\dfrac{(3-0)(3-0)}{(3)^3}

(00)(0+0)(3)3-\dfrac{(0-0)(0+0)}{(3)^3}

(00)(0+0)(3)3=23-\dfrac{(0-0)(0+0)}{(3)^3}=\dfrac{2}{3}

xC=00(1+(0)2)23=0x_C=0-\dfrac{0(1+(0)^2)}{\dfrac{2}{3}}=0

yC=0+1+(0)223=32y_C=0+\dfrac{1+(0)^2}{\dfrac{2}{3}}=\dfrac{3}{2}

C(0,32)C(0, \dfrac{3}{2})


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