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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:

ara

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

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  • $\begingroup$ $tcos(t)=tsin(t)$ so that $t=0$ or $cos(t)=sin(t)$ possibly? Note that you should use different parameters for each curve not use t twice. $\endgroup$
    – Paul
    Commented Apr 30, 2020 at 10:27
  • $\begingroup$ When setting $x_1 = x_2$ and $y_1 = y_2$ did you divide the $t$ out ? $\endgroup$
    – Moeee
    Commented Apr 30, 2020 at 10:27
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    $\begingroup$ @Moeee yes, sorry, I've added my working. $\endgroup$
    – BigData
    Commented Apr 30, 2020 at 10:34
  • $\begingroup$ @Paul What do you mean? And yeah I've done that, sorry, I added the working. $\endgroup$
    – BigData
    Commented Apr 30, 2020 at 10:34
  • $\begingroup$ 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.. $\endgroup$
    – Paul
    Commented Apr 30, 2020 at 10:39

1 Answer 1

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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.

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