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I'm slightly confused after reading the problem posed here: Points of intersection of two parametric curves

Why is it that the "t" result at the intersection point of the two curves is not the same? I.e. the solution involves the answer "t1" for the first curve and "t2" for the second curve, but if the two curves are being made equal, why is t1 != t2?

Also, is there any way we can "manufacture" two different curves (say two parametric straight lines) that, at a specific t value for BOTH curves, yield the SAME point in space? How would you go about doing that? (given, for example, 3 points in space)

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    $\begingroup$ An intersection only requires the $x$ and $y$ values to be equal, which may occur at different $t$ values. $\endgroup$ – Tavish Jan 21 at 11:27
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Maybe a simple example helps to see what's going on. Let's say you have two curves $$c_1:I \rightarrow \mathbb{R}^2, t \mapsto (\sin(t), \cos(t))$$ $$c_2:I \rightarrow \mathbb{R}^2, t \mapsto (\cos(t), \sin(t)),$$ where $I = [0, 2\pi]$.

Both these curves are the unit circle around the origin. So if you ask, where do these two curves intersect, the answer should basically be 'everywhere'. So that means that for any point $p$ on the unit circle, you can find $t_1$ and $t_2$ such that $c_1(t_1) = c_2(t_2) = p$. But only rarely, if ever, will you find a value $t$ such that $c_1(t) = c_2(t) = p$, that's just too much to ask. So in the exercise you link, the first question asks for $t_1$ and $t_2$, where the second one asks for a single $t$.

Hope this helps.

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  • $\begingroup$ Thanks! This was the intuitive explanation I was looking for $\endgroup$ – Gary Allen Jan 21 at 11:55
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Say you had two flies, fly A and fly B, which trace out respective parametric curves, but fly B is not as fast as fly A. We are looking for a shared point, i.e. the curves pass through the same coordinate in space, so fly A might get to the shared point quicker than fly B. They arrive at the shared point at different times. It’s not necessary that they arrive to the shared point at the same time, although it may happen in special cases as well.

When the intersection happens at the same time, that is when the two particles would collide. Two particles equidistant from a point travelling at the same speed in the direction of the that point would be an example when $t_1 = t_2$. Given lines $\mathcal{L}_1,\mathcal{L}_2, \mathcal{L’}_1$

$ \mathcal{L}_1 = \begin{cases} x=1-t \\ y=0 \end{cases} , \mathcal{L’}_1 = \begin{cases} x=1-\frac{t}{2} \\ y=0 \end{cases} $

$ \mathcal{L}_2 = \begin{cases} x=0\\ y=1-t \end{cases}$

$\mathcal{L_1,L_2}$ cross the origin $(0,0)$ at time $t_1=t_2=1$. $\mathcal{L’_1,L_2}$ cross the origin $(0,0)$ at times $t_1’=2$ and $t_2 =1$.

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  • $\begingroup$ Thanks a lot! This really helped $\endgroup$ – Gary Allen Jan 21 at 11:56
  • $\begingroup$ glad to help :) $\endgroup$ – no lemon no melon Jan 21 at 11:57
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As for your second question, given two parametric curves $$C_1 :\ (a(t),b(t)) \\ C_2 :\ (c(t), d(t)) $$

they can be made to intersect at $t=t_0$ by choosing functions $a,b,c,d$ such that $$a(t_0) = c(t_0) \\ b(t_0)= d(t_0) $$

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    $\begingroup$ Thanks a lot - this is ultimately what I'm trying to do so definitely helps to understand it first. $\endgroup$ – Gary Allen Jan 21 at 11:55
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Let's think of the two curves $c_1, c_2$ as the routes two different cars take. The parameter $t$ then represents time, and $c_1(t), c_2(t)$ describe the position of the cars at time $t$.

If both cars stop at the same gas station, does it follow that they have to do it at the same time? Of course not!

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