# Intersection of two parametric equations at the origin

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

• $tcos(t)=tsin(t)$ so that $t=0$ or $cos(t)=sin(t)$ possibly? Note that you should use different parameters for each curve not use t twice.
– Paul
Commented Apr 30, 2020 at 10:27
• When setting $x_1 = x_2$ and $y_1 = y_2$ did you divide the $t$ out ? Commented Apr 30, 2020 at 10:27
• @Moeee yes, sorry, I've added my working. Commented Apr 30, 2020 at 10:34
• @Paul What do you mean? And yeah I've done that, sorry, I added the working. Commented Apr 30, 2020 at 10:34
• Not sure what you mean by "What do you mean?". You had no working, so I said "possibly", if that is what you meant by "What do you mean?", if you see what I mean..
– Paul
Commented Apr 30, 2020 at 10:39

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.