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

[amaresh92]

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>

 

[HallsofIvy]

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.

 

[Mentallic]

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.