Answer to Question #281786 in Differential Geometry | Topology for chaitu

Question #281786


1. Find the radius of curvature (ROC) at (a cos3 θ, a sin3


θ) on the curve x

2/3 +


y

2/3 = a

2/3



2. Find the radius of curvature (ROC) for the curve x = a(cos θ + θ sin θ), y =

a(sin θ − θ cos θ).


1
Expert's answer
2021-12-22T05:27:55-0500

1.

The parametric equations are


"x=a\\cos^3 \\theta=\\dfrac{a}{4}(3\\cos \\theta+\\cos 3\\theta),"

"y=a\\sin^3 \\theta=\\dfrac{a}{4}(3\\sin \\theta-\\sin 3\\theta),"

"x'=\\dfrac{a}{4}(-3\\sin \\theta-3\\sin 3\\theta)"

"x''=\\dfrac{a}{4}(-3\\cos \\theta-9\\cos 3\\theta)"

"y'=\\dfrac{a}{4}(3\\cos \\theta-3\\cos 3\\theta)"

"y''=\\dfrac{a}{4}(-3\\sin \\theta+9\\sin 3\\theta)"

"R=\\dfrac{((x')^2+(y')^2)^{3\/2}}{|x'y''-x''y'|}"

"(x')^2+(y')^2"

"=\\dfrac{a^2}{16}(9\\sin^2 \\theta+9\\sin^2 3\\theta+18\\sin\\theta \\sin 3\\theta)"

"+\\dfrac{a^2}{16}(9\\cos^2 \\theta+9\\cos^2 3\\theta-18\\cos\\theta \\cos 3\\theta)"

"=\\dfrac{9a^2}{8}(1-\\cos4\\theta)=\\dfrac{9a^2}{4}\\sin^22\\theta"


"((x')^2+(y')^2)^{3\/2}=\\dfrac{27a^3}{8}\\sin^32\\theta"

"x'y''-x''y'"

"=\\dfrac{a^2}{16}(9\\sin^2 \\theta-27\\sin^2 3\\theta-18\\sin\\theta \\sin 3\\theta)"

"-\\dfrac{a^2}{16}(-9\\cos^2 \\theta+27\\cos^2 3\\theta-18\\cos\\theta \\cos 3\\theta)"

"=-\\dfrac{9a^2}{8}(1-\\cos4\\theta)=-\\dfrac{9a^2}{4}\\sin^22\\theta"

"R=\\dfrac{\\dfrac{27a^3}{8}\\sin^32\\theta}{|-\\dfrac{9a^2}{4}\\sin^22\\theta|}"

"=\\dfrac{3a}{2}\\sin2\\theta=3a\\sin\\theta \\cos\\theta"

"R=3a\\sin\\theta \\cos\\theta"

2.


"x=a(\\cos \\theta+\\theta \\sin \\theta)"

"y=a(\\sin \\theta-\\theta \\cos \\theta)"

"x'=a(-\\sin \\theta+\\sin \\theta+\\theta\\cos \\theta)=a(\\theta \\cos \\theta)"

"x''=a(\\cos \\theta-\\theta \\sin \\theta )"

"y'=a(\\cos \\theta-\\cos \\theta+\\theta \\sin \\theta)=a(\\theta \\sin \\theta)"

"y''=a(\\sin \\theta+\\theta \\cos \\theta )"

"R=\\dfrac{((x')^2+(y')^2)^{3\/2}}{|x'y''-x''y'|}"

"(x')^2+(y')^2=a^2\\theta^2(\\cos62\\theta+\\sin^2\\theta)=a^2\\theta^2"

"((x')^2+(y')^2)^{3\/2}=a^3\\theta^3"

"x'y''-x''y'=a^2(\\theta\\sin\\theta \\cos\\theta+\\theta^2\\cos^2\\theta)"

"-a^2(\\theta\\sin\\theta \\cos\\theta-\\theta^2\\sin^2\\theta)=a^2\\theta^2"

"R=\\dfrac{a^3\\theta^3}{|a^2\\theta^2|}=a\\theta"


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