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find radius of curvature at x=y,y=3a/2 to the curve x^3+y^3=3axy
Dear student Since this equation is given in cartesian coordinates instead of parametric, we can rewrite this as y = (x^3 + y^3) / (3ax) R = | ( (1 + (dy/dx)^2 ) ^ (3/2)) / d²y / dx² | So dy/dx = (x/a) - (1/3)(x^3+y^3)/(ax^2) (dy/dx)^2 = (4/9)(x^2/a^2) - (4/9)(y^3/(x)(a^2)) + (1/9) ( y^6) /((a^2)(x^4)) d²y / dx² = (2/3)(x^3+y^3)/(ax^3) Now that we have finished computing the first derivative, its square and the second derivative we are ready to use the formula -- R = | [(1 + (4/9)(x^2/a^2) - (4/9)(y^3/(x)(a^2)) + (1/9) ( y^6) /((a^2)(x^4)) ) ^(3/2) )] / ( 2/3)(x^3+y^3)/(ax^3) ) | Algebraic manipulation of the above expression leads to, R = | (3/2)(1 + (x/a - (1/3)(x^3+y^3) / (a)(x^2))^2)^ (3/2*(a) (x^3) /(x^3+y^3) | Plugging in 3a/2 for x and y it results in R = |(15/64)sqrt(5)sqrt(4) a| RegardsArun (askIITians forum expert)
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