suppose two particles are traveling along the curves given by the vector functions (r_1(t)=langle…

suppose two particles are traveling along the curves given by the vector functions (r_1(t)=langle t,t^{2},t^{3}\rangle) (r_2(t)=langle1 + 3t,1 + 15t,1 + 63t\rangle) do the particles collide? in other words, are they ever in the same position at the same time (t)? if the particles collide, enter the least value of (t) when they collide. if they never collide, enter none. do the paths of the particles intersect? if the paths intersect, enter the intersection points as ordered triples ((a,b,c)) and separate multiple answers with commas. if there are no such points, enter none.

suppose two particles are traveling along the curves given by the vector functions (r_1(t)=langle t,t^{2},t^{3}\rangle) (r_2(t)=langle1 + 3t,1 + 15t,1 + 63t\rangle) do the particles collide? in other words, are they ever in the same position at the same time (t)? if the particles collide, enter the least value of (t) when they collide. if they never collide, enter none. do the paths of the particles intersect? if the paths intersect, enter the intersection points as ordered triples ((a,b,c)) and separate multiple answers with commas. if there are no such points, enter none.

Answer

Explanation:

Step1: Set up equations for collision

For the particles to collide, (r_1(t)=r_2(t)). So we have the system of equations: (t = 1 + 3t), (t^{2}=1 + 15t), (t^{3}=1+63t). From (t = 1 + 3t), we get (2t=-1), so (t =-\frac{1}{2}). Substitute (t =-\frac{1}{2}) into (t^{2}=1 + 15t), (\left(-\frac{1}{2}\right)^{2}=\frac{1}{4}) and (1+15\times\left(-\frac{1}{2}\right)=1-\frac{15}{2}=-\frac{13}{2}). Since (\frac{1}{4}\neq-\frac{13}{2}), the particles never collide.

Answer:

none

Explanation for intersection:

Let (r_1(s)=r_2(t)). Then we have the system of equations (\left{\begin{array}{l}s = 1+3t\s^{2}=1 + 15t\s^{3}=1+63t\end{array}\right.). From (s = 1+3t), we have (t=\frac{s - 1}{3}). Substitute (t=\frac{s - 1}{3}) into (s^{2}=1 + 15t), we get (s^{2}=1+15\times\frac{s - 1}{3}=1 + 5(s - 1)=5s-4). So (s^{2}-5s + 4=0). Factoring gives ((s - 1)(s - 4)=0). If (s = 1), then (t = 0). When (s = 1) and (t = 0), (r_1(1)=(1,1,1)) and (r_2(0)=(1,1,1)). If (s = 4), then (t = 1). When (s = 4) and (t = 1), (r_1(4)=(4,16,64)) and (r_2(1)=(4,16,64)).

Answer:

(1,1,1),(4,16,64)