Question #281785

find the radius of the curvature at (acostheta^3,asintheta^3) on the curve x^2/3+y^2/3=a^2/3


1
Expert's answer
2021-12-22T13:11:02-0500

radius of the curvature:

R=(1+(dy/dx)2)3/2d2y/dx2R=\frac{(1+(dy/dx)^2)^{3/2}}{|d^2y/dx^2|}


2x/3+2yy/3=02x/3+2yy'/3=0

y=xyy'=-\frac{x}{y}


y=yxyy2y''=-\frac{y-xy'}{y^2}


at point (acosθ3,asinθ3):(acos \theta^3 ,asin \theta ^3):

y=cotθ3y'=-cot\theta^3


y=sinθ3cos2θ3/sinθ3sin2θ3y''=-\frac{sin\theta^3-cos^2\theta^3/sin\theta^3}{sin^2\theta ^3}


R=sin2θ3(1+cot2θ3)3/2sinθ3cos2θ3/sinθ3R=\frac{sin^2\theta ^3(1+cot^2\theta^3)^{3/2}}{sin\theta^3-cos^2\theta^3/sin\theta^3}


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