The parametric curve $r=(-2t^2+8t-2,cos({\pi}t),t^3-28t)$ crosses itself at one and only one point. The point is $(x,y,z)$. I found $t=-2$ to be the answer and $(x,y,z)=(-26,-1,48)$ to be the correct answer.
However it asks for the acute angle between the two tangent lines to the curve at the crossing point $cos({\Theta})=$ as well. I know what the formula will look like, but the formula implies that there will be another number, like $t=-2$, but I haven't been able to find that number? I feel like I"m done the majority of the question, but am not sure how to seal it all up.
Any help is greatly appreciated.
Best Answer
suppose the trajectory crosses itself at $t = a, t = b \neq a,$ the we have $r(a)= r(b).$ there are three simultaneous equation to solve: the first one $$-2a^2 + 8a - 2 = -2b^2 +8b -2$$ implies $b + a = 4.$ if i substitute in the $z$ component i get $$a^3 - 28a = (4-a)^3 - 28(4-a).$$ this equation has thre roots $a = -2, 2, 6$ and the corresponding $b = 6, 2, -2$. which shows that trajectory crosses itself at the point $(-26, 1, 48)$ at time $t = -2$ and again at $t = 6.$
now the tangent vector at $t$ is $(-4t + 8, -\pi\sin \pi t, 3t^2 - 28)$ evaluating this at $t = -2,$ we get $16(1,0,-1 )$ and at $t = 6$ we have $16(-1,0,5).$ the angle between these vector is $$\cos^{-1} \left(\frac{-6}{\sqrt 2\sqrt 26}\right) = 2.553 = 146.31^\circ $$
the acute angel is $33.69^\circ .$