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Question

Show that the total length of the astroid x2/3+y2/3=a2/3x^{2 / 3}+y^{2 / 3}=a^{2 / 3}, which can be parameterised as x=acos3θ,y=asin3θx=a \cos ^3 \theta, y=a \sin ^3 \theta, is 6a6 a.

Solution

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Answered 2 years ago
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Total length is given as:

x23+y23=a23x^{\frac{2}{3}}+y^{\frac{2}{3}} =a^{\frac{2}{3}}

Parametrized form is given as:\text{Parametrized form is given as}:

x=acos3tx= a\cos^{3}t

y=asin3ty= a\sin^{3}t

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