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Grade: 12
        
find radius of curvature at x=y,y=3a/2 to the curve x^3+y^3=3axy
one year ago

Answers : (1)

Arun
24510 Points
							
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| 


 
Regards
Arun (askIITians forum expert)
one year ago
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