Proving the total curvature of the trefoil curve is $4\pi $.

Question

I'm looking at the trefoil curve $\vec x (t) = (\cos (3t) \cos (t), \cos (3t)\sin (t)) $ defined along the interval $[0,\pi] $. First I tried calculating this by integrating curvature along the length of the curve, but it seemed to be getting extremely complicated. Then I thought of trying to count the number of times the tangent points parallel to one of the coordinate axes, but finding the zeros of the derivatives seemed to end up with really ugly expressions. I don't really know what to do now short of numerical integration but the problem is asking me to $prove$ the total curvature is $4\pi$.

Answer 1

The second component of the derivative of $x(t)$ is $2 \cos (4 t)-\cos (2 t)$. This is equal to $$2 \sin ^4(t)+\sin ^2(t)+2 \cos ^4(t)-\cos ^2(t)-12 \sin ^2(t) \cos ^2(t)$$ which, in turn, is $$16 \cos ^4(t)-18 \cos ^2(t)+3.$$ We want to know for what values of $t$ this is zero. The roots of $16 q^2-18 q+3=0$ are $\frac{1}{16} \left(9\pm\sqrt{33}\right)$, so the roots of out function are the numbers $t$ sich that $\cos^2(t)=\frac{1}{16} \left(9\pm\sqrt{33}\right)$. The numbers in the right hand side are $0.203465$ and $0.921535$. Clearly, this means that the equation has $4$ solutions in $[0,\pi]$, two for each right hand side.

This means that the tangent vector is horizontal at 4 points. Since it is closed, this means that it points in the direction of $(1,0)$ exactly two times. The integral of the curvature along the curve is equal to the total variation of the angle and equal to $2\pi$ the degree of the tangent, this means that the integral is $4\pi$.