Section 1.3 Intersection Points versus Collision Points

A parametrization carries more information than the curve it traces: it also records when the moving point visits each place on the curve. Because of this, there are two different questions we can ask about two parametric curves

\begin{equation*} \mathbf r_1(t) = \bigl(f_1(t),\, g_1(t)\bigr), \qquad \mathbf r_2(t) = \bigl(f_2(t),\, g_2(t)\bigr), \end{equation*}

thought of as the trajectories of two particles.

Definition 1.17. Intersection Points and Collision Points.

A point \(P\) is an intersection point of the two curves if \(P\) lies on both curves; that is, if there are parameter values \(t\) and \(s\text{,}\) not necessarily equal, with

\begin{equation*} \mathbf r_1(t) = \mathbf{OP} = \mathbf r_2(s). \end{equation*}

A point \(P\) is a collision point of the two particles if both particles are at \(P\) at the same time; that is, if there is a single value of \(t\) with

\begin{equation*} \mathbf r_1(t) = \mathbf r_2(t) = \mathbf{OP}. \end{equation*}

An intersection point is a statement about the two paths: they cross, like two roads on a map. A collision point is a statement about the two motions: the cars are at the crossroads at the same instant. Every collision point is automatically an intersection point, but, as the next two examples show, an intersection point need not be a collision point.

In practice this dictates how we set up the equations.

To find intersection points, give the two curves different parameter names and solve \(f_1(t) = f_2(s)\text{,}\) \(g_1(t) = g_2(s)\) for the pair \((t,s)\text{.}\) Equivalently, eliminate the parameters and intersect the two Cartesian curves.
To find collision points, use the same parameter \(t\) in both curves and solve \(f_1(t) = f_2(t)\text{,}\) \(g_1(t) = g_2(t)\) for \(t\text{.}\)

In the next two examples the two paths are always the same: the parabola \(y = x^2\) and the line \(y = x+2\text{.}\) Setting \(x^2 = x+2\) gives \(x^2 - x - 2 = (x-2)(x+1) = 0\text{,}\) so the paths meet at the two points

\begin{equation*} (-1,1) \qquad \text{and} \qquad (2,4). \end{equation*}

Only the schedule of the second particle will change.

Example 1.18. The paths cross, but the particles never meet.

Two particles move in the plane with position vectors

\begin{equation*} \mathbf r_1(t) = \bigl(t,\; t^2\bigr), \qquad \mathbf r_2(t) = \bigl(-t,\; 2-t\bigr), \qquad t \in \mathbb R . \end{equation*}

Find all intersection points of their paths, and all collision points of the particles.

Solution.

Paths. Eliminating the parameter, the first particle traces the parabola \(y = x^2\text{.}\) For the second, \(x = -t\) and \(y = 2-t = 2+x\text{,}\) so it traces the line \(y = x+2\text{.}\) As computed above, the two paths intersect at \((-1,1)\) and \((2,4)\text{.}\)

Collisions. A collision requires a single \(t\) with

\begin{equation*} t = -t \qquad \text{and} \qquad t^2 = 2 - t . \end{equation*}

The first equation forces \(t = 0\text{,}\) but then \(t^2 = 0\) while \(2 - t = 2\text{.}\) There is no solution, so the particles never collide.

We can see exactly how they miss each other. The first particle is at \((-1,1)\) when \(t=-1\text{,}\) while the second particle arrives there only at \(t=1\text{.}\) Similarly, the first particle reaches \((2,4)\) at \(t=2\text{,}\) long after the second particle passed through it at \(t=-2\text{.}\) Each intersection point is visited twice, but never simultaneously. See Figure 1.19 and Figure 1.20.

Figure 1.19. The particles \(\mathbf r_1(t) = (t,t^2)\) and \(\mathbf r_2(t) = (-t,2-t)\) move simultaneously. Their paths cross at \((-1,1)\) and \((2,4)\text{,}\) but the clock shows that the two particles are never at either crossing at the same time.

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The parabola \(y=x^2\) and the line \(y=x+2\) drawn on the same axes, crossing at the points \((-1,1)\) and \((2,4)\text{.}\) Each crossing is labeled with the two different parameter values at which the two particles pass through it.

Figure 1.20. The parabola \(C_1\) and the line \(C_2\) intersect at \((-1,1)\) and \((2,4)\text{.}\) At each of these points the two particles arrive at different times, so neither point is a collision point.

Example 1.21. The same paths, a different schedule—now they collide.

Keep the first particle, but re-schedule the second one:

\begin{equation*} \mathbf r_1(t) = \bigl(t,\; t^2\bigr), \qquad \mathbf r_2(t) = \bigl(2t-2,\; 2t\bigr), \qquad t \in \mathbb R . \end{equation*}

Find the intersection points of the paths and the collision points of the particles.

Solution.

Paths. For the second particle, \(x = 2t-2\) and \(y = 2t = x+2\text{,}\) so it travels along the very same line \(y = x+2\) as before, only faster and in the opposite direction. The paths are therefore unchanged, and they still intersect at \((-1,1)\) and \((2,4)\text{.}\)

Collisions. Now a collision requires a single \(t\) with

\begin{equation*} t = 2t - 2 \qquad \text{and} \qquad t^2 = 2t . \end{equation*}

The first equation gives \(t = 2\text{,}\) and this value does satisfy the second equation, since \(2^2 = 4 = 2\cdot 2\text{.}\) Hence the particles do collide, at

\begin{equation*} \mathbf r_1(2) = \mathbf r_2(2) = (2,4), \qquad \text{at time } t = 2 . \end{equation*}

The other intersection point is not a collision point: the first particle passes through \((-1,1)\) at \(t=-1\text{,}\) while the second particle passes through it at \(t = \tfrac12\text{.}\) Two intersection points, but only one collision. See Figure 1.22 and Figure 1.23.

Figure 1.22. With \(\mathbf r_2(t) = (2t-2,2t)\) the second particle follows the same line as before, but on a different schedule. It misses the first particle at \((-1,1)\) and collides with it at \((2,4)\) when \(t = 2\text{.}\)

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B	Activate menu driven exploration
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Cursor down	Explore next lower level
Cursor up	Explore next upper level
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Comma	Activate direct magnification
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The parabola \(y=x^2\) and the line \(y=x+2\) crossing at \((-1,1)\) and \((2,4)\text{.}\) The point \((2,4)\) is highlighted as a collision point, reached by both particles at time \(t=2\text{,}\) while the particles pass through \((-1,1)\) at the different times \(t=-1\) and \(t=\tfrac12\text{.}\)

Figure 1.23. The same two paths as in Figure 1.20. The point \((2,4)\) is now a collision point, reached by both particles at \(t=2\text{,}\) while \((-1,1)\) remains an intersection point that is not a collision point.

Insight 1.24. Every Collision Is an Intersection, but Not Conversely.

If two particles collide at \(P\text{,}\) then certainly \(P\) lies on both paths, so every collision point is an intersection point. The converse fails: the paths may cross at a point that the two particles visit at different times. Changing the parametrization of a curve does not move the curve, but it can create or destroy collisions.

Checkpoint 1.25.

Keep \(\mathbf r_1(t) = (t,t^2)\text{.}\) Find a parametrization \(\mathbf r_2\) of the line \(y = x+2\) for which the two particles collide at both intersection points.

Hint.

Try to arrange that the second particle has the same \(x\)-coordinate as the first one at every time \(t\text{.}\)

Answer.

\(\mathbf r_2(t) = (t,\; t+2)\text{.}\)

Solution.

With \(\mathbf r_2(t) = (t, t+2)\) the collision equations become \(t = t\) and \(t^2 = t+2\text{,}\) i.e. \((t-2)(t+1) = 0\text{.}\) Thus \(t = -1\) gives the collision at \((-1,1)\) and \(t = 2\) gives the collision at \((2,4)\text{:}\) both intersection points are now collision points.

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