Solution 6: Envelopes.

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The velocity vector of the parametrization is given by . Then a point (x,y) on the plane lies on one of the tangent lines to if and only if

where s is a real number. You may check that F(x, y, t) = y - 3t2x + 2t3 = 0 holds for x = t + s, y = t3 + 3st2. In order to find the envelope of the set of tangent lines to , we must solve the system S:

Which has solutions

That is, the envelope is the curve itself.


The analogous problem for the parabola is even simpler.


Exercises

    Exercise 5.7(2) from Bruce and Giblin: Let . Show that 0 is a regular value of each Ft and that F = 0 is the equation of the chord of the unit circle joining to . We have:

    (allowing t = 0 we find x = 1, y arbitrary is also a solution). The calculations for this are straightforward but trigonometrically messy - you may skip the messy part. Show that the envelope is a regular curve except for . Our embroidery will be the nephroid pictured below. (We can get even prettier embroidery by replacting 2t by mt).

    [Maple Plot]


Problem 6

References: Klain & Rota.