I have the two parametric curves defined by $r_1(t) = tcos(t)i + tsin(t)j$ and $r_2(t)=\frac{t}{\sqrt{2}} i+ \frac{t}{\sqrt{2}} j$.
I am asked to find the time values at which these curves "collide", and I have found the set of these points by solving the systems $x_1=x_2$ and $y_1=y_2$.
$ \Rightarrow t_1cos(t_1) = \frac{t_2}{\sqrt{2}} $ and $ t_1sin(t_1) = \frac{t_2}{\sqrt{2}} $
$ \Rightarrow t_1cos(t_1) = \frac{t_2}{\sqrt{2}} \Rightarrow t_2 = \sqrt{2}t_1cos(t_1) $
Sub $t_2$ into $y_2$ and solve for $y$:
$ \frac{\sqrt{2}t_1cos(t_1)}{\sqrt{2}} = t_1sin(t_1) $
$ cos(t_1) = sin(t_1) $
This gives the set $A$ of $t$-values $A = \lbrace \frac{\pi}{4} + 2k\pi \: \vert \: k \in \mathbb{N} \rbrace$.
Graphing these two parametric curves, however, it seems like there is an intersection at the origin when t = 0, shown here:
Is this counted as a collision/intersection? If so, what have I done wrong to not have a solution for this in the set $A$?
Thanks
Best Answer
Note that the equation $$\frac{\sqrt{2}t_1\cos(t_1)}{\sqrt{2}}=t_1\sin(t_1)$$ has two solutions: either $t_1=0$ or $t_1$ such that $\cos(t_1)=\sin(t_1)$. You are missing the first one.