x(t) = 3cos(t) - cos(3t)This bisymmetric curve is traced out in a counterclockwise direction around the origin, starting from the cusp at (2,0). Its maximum width occurs at the point pairs which determine that the nephroid is non-convex by the multiple local extrema test. The values of t at these point pairs are ±π/4 (right pair) and ±3π/4 (left pair). The maximum height of the nephroid is between the point pair (0,±4), so the width x height of its bounding rectangle is
y(t) = 3sin(t) - sin(3t), 0 ≤ t < 2π
x'(t) = -3sin(t) + 3sin(3t)so by (L1), the perimeter of the convex hull is
y'(t) = 3cos(t) - 3cos(3t)
which is about 6% shorter than that of the nephroid.
To compute the area of the convex hull, we can see by
(A1) that the area A1 of the
region in green in the left diagram below is the integral of y dx from
Therefore, the area of the convex hull is
A + 4A1 = 12π + 4(5 - 3π/2) = 20 + 6π = 38.849556,which is about 3% more than the area of the nephroid.
x2(t) + y2(t) = 10 - 6cos(2t)has its extrema on the coordinate axes at (±2,0) and (0,±4), so the inradius is 2 and the circumradius is 4. A candidate for the inellipse is one enclosed by the annulus between these two circles, with
x(t)y(t) = [9sin(2t) - 6sin(4t) + sin(6t)]/2is maximized in the first quadrant when
The circumellipse dimensions are
For verification, we have
Figure Parameters Perimeter Area Centroid Incircle R = 2 12.566371 12.566371 Inellipse a = 2
b = 4
19.376896 25.132741 Nephroid 24 37.699112 Convex hull 22.627417 38.849556 Circumellipse 23.181126 41.848410 Circumcircle R = 4 25.132741 50.265482
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