Particles traveling along space curves? collision?

Question

Two particles are traveling along the space curves defined by ( r_1(t) = (t, t^2, t^3) ) and ( r_2(t) = (1 + 4t, 1 + 16t, 1 + 52t) ). 
Please find the points at which their paths intersect. If there is no intersection, please indicate this by entering "DNE."
a) Provide the coordinates ((x, y, z)) corresponding to the smaller value of (t).
b) Provide the coordinates ((x, y, z)) corresponding to the larger value of (t).
Additionally, determine the points where the particles collide. Enter your answers as a comma-separated list. If there is no collision, please indicate this by entering "DNE."
c) (t = \ldots)

Anonymous · Accepted Answer

first, I assume you meant: r1(t) = Since only the path matters (and not the time they intersect), set the two paths equal to each other with different parameters = <1 + 4s, 1 + 16s, 1 + 52s>t = 1 + 4s s = (t – 1)/4t^2 = 1 + 16s = 1 + 4t – 4t^2 – 4t + 3 = 0(t – 1)(t – 3) = 0t = 1 or 3solving for s:t = 1, s = 0t = 3, s = 1/2Both of these are possible points of intersection but to be sure you must check that both values of t and s satisfy the last directiont^3 = 1 + 52sIt is clear that both do. So the points corresponding to t = 1 and t = 3 are solutionst = 1 —–> (1, 1, 1)t = 3 —–> (2, 9, 27)To find the points of collision, follow the same procedure as you did for finding points of intersection except use the same parameters rather than different ones. Since all the work is done above and we already saw that when t = 1, s = 1 (so it would have been a solution if instead we used t for both), the point of collision occurs at t = 1, point (1, 1, 1) = <1, 1, 1> + s<4, 16, 52>

Anonymous · Answer

every thing is correct except t=3 —–> (3,9,27)

Anonymous

Particles traveling along space curves? collision?

2 Answers

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