Asked by JDH
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)
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)
Answers
Answered by
Damon
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
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
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