Question

Hello,

Please help!!

Show that the parabola y=ax^2 has its largest curvature at its vertex and no minimum curvature. (Note: since the curvature remains the same if the curve is translated or rotated, this result is true for any parabola)

Answer 1

The curvature of a function is the inverse of the radius of curvature, R

R = [ 1 + (dy/dx)^2 ]^1.5 / d^2y/dx^2

here
R = [ 1+ (2ax)^2 ]^1.5 / 2a

So the curvature, 1/R


=2a/ [1+(2ax)^2]^1.5

now (2ax)^2 is always positive and will be 0 when x = 0
That will give the smallest denominator and therefore the biggest curvature
as x becomes huge (+ or -), the curvature becomes small, but never gets bigger again, so there is no minimum