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Continuity and Differentiability
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Solution:
Let x=acos3θ,y=asin3θ. Then, x32+y32=(acos3θ)32+(asin3θ)32 =a32[(cos2θ+(sin2θ)]=a32
Hence, x=acos3θ,y=cosin3θ is parametric equation of x32+y32=a32
Now, dθdx=−3acos2θsinθ and dθdy=3asin2θcosθ dxdy=dθdxdθdy=−3acos2θsinθ3asin2θcosθ=−tanθ=−3xy