Question
If two objects travel through space along two different curves, it is often important to know whether they will collide. The curves might intersect showing that objects will be in the same location at some time, but we need to know whether the objects are in the same locationat the same time. Suppose the trajectories of two objects are given by
r1(t)=<t,t^2,t^3+5(t-1) and r2(s)=<1+3s,1+9s,1+36s>
point = ( , , ) at t =
and at s=
point = ( , , ) at t =
and at s=
Explanation / Answer
f the two particles collide, they will need to be in the same location at the same time. That means the coordinates of their positions must be the same first of all... r_1(t) = r_2(t) But the positions need to be the same at the same time, so the t in the r_1 expression will be the same t as in the r_2 expression. So you can write: t1 = t2 = t And then from the first expression up there... r_1(t) = r_2(t) = < 11t -28, t^2, 15t -56 > These are vector equations. Envision them looking like this, with the commas separating x, y, and z, each of which represents an expression in t: = You'll need to set each component of r_1 equal to the corresponding component of r_2. That means x1(t) = x2(t), y1(t) = y2(t), and z1(t) = z2(t). Of course, if you look at the original vector equations for r_1 and r_2, you'll see that those x's, y's, and z's just correspond to the t expressions separated by commas. So set appropriate expressions equal to each other: t^2 = 11t - 28 9t - 14 = t^2 t^2 = 15t - 56 Now rearrange them so they can be factored conveniently... t^2 - 11t + 28 = 0 t^2 - 9t + 14 = 0 t^2 - 15t + 56 = 0 Now solve for t each time. Note that each expression is quadratic, so solving each one will yield two values for t. If the objects are to collide, at least one t result from each equation must be the same. If such a value exists, the two particles collide and they do so at this value of t. If no such value of t exists, the particles do not collide. You can determine this value of t by factoring and solving the equations: t^2 - 11t + 28 = 0 (t - 7)(t - 4) = 0 t = 7,4 t^2 - 9t + 14 = 0 (t - 7)(t - 2) = 0 t = 7,2 t^2 - 15t + 56 = 0 (t - 7)(t - 8) = 0 t = 7,8 As you can see, each time we get t = 7 and t = some number not a seven. But the important thing is that there exists a time t where the coordinates of the first particle match the coordinates of the second particle, and this time is t = 7. If you plug this value into your original vector equations, you will see that it yields the same coordinates for both r_1 and r_2... r_1(t) = r_1(7) = < 49, 63 -14, 49 > r_1(7) = < 49, 49, 49 > r_2(t) = < 11t -28, t^2, 15t -56 > r_2(7) = < 77 -28, 49, 15(7) -56 > r_2(7) = < 49, 49, 49 > So as you can see r_1 = r_2 at t = 7.