by HallsofIvy » Mon Aug 24, 2020 11:18 am
The "osculating circle" to a curve, at a given point, is the circle that passes through that point, has the same tangent line as the curve at that point and has the same curvature. From the first two conditions, the osculating circle to [tex]x= y^2[/tex], at (0,0), must have center on the x- axis- the center is at (R, 0) and the radius is R for some number, R. The curvature of a circle of radius R is [tex]\frac{1}{R}[/tex]. What is the curvature of [tex]x= y^2[/tex] at (0, 0)?