Question

two particles travel along the space curves ( r_1(t)=langle t,t^{2},t^{3}
angle) ( r_2(t)=langle 1 + 2t,1 + 6t,1 + 14t
angle). find the points at which their paths intersect. (if an answer does not exist, enter dne.) ((x,y,z)=(quad)) (smaller (x)-value) ((x,y,z)=(quad)) (larger (x)-value) find the time(s) when the particles collide. (enter your answers as a comma - separated list. if an answer does not exist, enter dne.) (t=)

Explanation:

Step1: Set the components equal

Set \(t = 1 + 2s\), \(t^{2}=1 + 6s\), \(t^{3}=1+14s\). From \(t = 1 + 2s\), we can substitute \(t\) into the second - equation. So \((1 + 2s)^{2}=1 + 6s\).
Expanding \((1 + 2s)^{2}\), we get \(1 + 4s+4s^{2}=1 + 6s\).
Rearranging gives \(4s^{2}-2s = 0\), factoring out \(2s\) gives \(2s(2s - 1)=0\). So \(s = 0\) or \(s=\frac{1}{2}\).

Step2: Check values of \(s\)

When \(s = 0\), \(t=1 + 2s=1\).
For \(t = 1\), \(t^{3}=1\) and \(1+14s=1\) (since \(s = 0\)).
When \(s=\frac{1}{2}\), \(t=1 + 2s=2\).
For \(t = 2\), \(t^{3}=8\) and \(1+14s=1 + 7=8\).

Step3: Find intersection points

When \(t = 1\) (corresponding to \(s = 0\)), the point on \(r_1(t)\) is \((t,t^{2},t^{3})=(1,1,1)\).
When \(t = 2\) (corresponding to \(s=\frac{1}{2}\)), the point on \(r_1(t)\) is \((t,t^{2},t^{3})=(2,4,8)\).

Step4: Check for collision

For collision, the same parameter value \(t\) should satisfy all three equations.
We check if the values of \(t\) that give intersection also give the same position for both curves at the same time.
From \(t = 1+2s\), \(t^{2}=1 + 6s\), \(t^{3}=1+14s\). Solving the system, we find that when \(t = 1\) and \(s = 0\) and when \(t = 2\) and \(s=\frac{1}{2}\).
The particles collide when \(t = 1\) because when \(t = 1\), \(r_1(1)=\langle1,1,1
angle\) and from \(r_2(t)\) with \(t = 1\) (since \(t = 1+2s\) and \(s = 0\)), \(r_2(0)=\langle1,1,1
angle\).

Answer:

\((x,y,z)=(1,1,1)\) (smaller \(x\) - value)
\((x,y,z)=(2,4,8)\) (larger \(x\) - value)
\(t = 1\)