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Math Forum / Mathematics / Recreational Math / July 2009Curvature
Find and prove the numerical value of the curvature at the point (x,y) of the ellipse (x^2/a^2) + (y^2/b^2) = 1. What is the ratio of the greatest to the least curvature of the ellipse ?. Hi Bill, Ellipse x^2/a^2 + y^2/b^2 = 1 Assume WLOG a>b dy/dx = -b^2.x / a^2.y d2y/dx2 = -b^2.(b^2.x + a^2.y) / a^4.y^3 curvature = K = -b^2.a^2.(b^2.x^2+a^2.y^2) / (a^4.y^2 + b^4.x^2)^ 3/2 dK/dx = 0 at x = 0, and ellipse is symmetrical in x and y At x=0 Kmin = b/a^2 At y=0 Kmax = a/b^2 Ratio Kmax/Kmin = a^3/b^3 Regards, Peter Scales. Hi Peter, For the curvature I have (a^4*b^4)/[(a^4*y^2 + b^4*x^2)^(3/2)] Regards Bill Hi Bill, I don't think so. Curvature must have a dimension of L^-1 Also please note a typo in my result for d2y/dx2 It should have been: d2y/dx2 = -b^2.(b^2.x^2+a^2.y^2) / a^4.y^3 Regards, Peter Scales. Sorry, Bill. Your result has the right dimension! I think the problem is elsewhere. Regards, Peter. |
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