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I have a homework where I am supposed to find the involute of two curves:

  • $\alpha(t):=(t\cos(t), t\sin(t), \frac{2\sqrt{2}}{3} + \frac{3}{2})$
  • $\beta(t):=(\cos^3t, \sin^3t, \cos2t), t \in [0, \frac{\pi}{2}]$

I tried using the standard formula for involute of a cruve $\beta(t)$: $I(t)=\beta(t) - s(t) \frac{\beta'(t)}{|\beta'(t)|}$ where $s(t)=\int_{0}^{t} |\beta'(t)|dt$, but in both cases calculations get prohibitively complex.

Am I doing something wrong, or there is some "trick" in there I am not aware of?

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1 Answer 1

The only difficulty I see here is in the computation of the arclength $s(t)$. Well, for (a) we have $|\alpha'(s)|=\sqrt{s^2+1}$, and although $\int_0^t\sqrt{s^2+1}\,ds$ is not the nicest integral in the world, it's certainly doable (or findable in a calculus book).

$|\beta'(s)|^2=9\cos^4 s \sin^2 s+9\sin^4 s\cos^2 s+4 \sin^2 2s =9\cos^2 s\sin^2 s+4\sin^2 2s$, and then there's the double angle formula which puts everything in terms of $\sin 2s$.

Moral: know thy trig identities.

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