Find Radius of Curvature for x2y=a(x2+y2) at (-2a,2a)

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SUMMARY

The radius of curvature for the curve defined by the equation x2y = a(x2 + y2) at the point (-2a, 2a) cannot be determined due to the presence of a cusp, indicated by an undefined first derivative. The correct formula for calculating the radius of curvature is R = (1 + y'2)3/2 / y'', which accounts for both the first and second derivatives. At the specified point, both derivatives are undefined, confirming the existence of a vertical tangent and the inability to compute a radius of curvature.

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Homework Statement


how to find the radius of curvature for following curve-:

x^2y=a(x^2+y^2) at the point (-2a,2a)



Homework Equations



radius of curvature= {(1+y1)^3/2}/y2

where y1 and y2 are the first and second order derivatives



The Attempt at a Solution




to find the derivative at -2a,2a it is coming infinity , so how to find the radius of curvature>
 
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You can't. If the derivative does not exist at that point, then there is a "cusp" there so the surface is not smooth and there is no radius of curvature.
 
If there is infinite gradient at a point, it doesn't necessarily mean it's a cusp. In this case, there is a vertical tangent present.

I had to search this up because your formula wasn't working for another question I tested it on. It is instead R=\frac{(1+y' ^2)^{3/2}}{y''}

And now it's giving answers like it should.

While the first derivative at that point is undefined, so is its second derivative and you should find that they will cancel each other out.
 

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