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Thread: space curve question

  1. #1
    Senior Member
    Jan 2008

    space curve question

    Two particles travel along the space curves

    r_{1}(t) = (t,t^2,t^3) \ \ r_{2}(t) = (1+2t, 1+6t, 1+ 14t)

    Do the particles collide? Do they intersect?

    so far I have:

    for r_{1}(t) \ \ x=t \ \ y=t^2 \ \ z=t^3
    therefore y=x^2 \ \ z=x^3
    for r_{2}(t) \ \ x= 1 +2t \ \ y = 1+6t \ \ z = 1+14t
    therefore y=3x-2 \ \ z=7x-6

    I'm not sure if I have to take y=y=x^2=3x-2, where x=1 and z=z=x^3=7x-6 where x=1 ?
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  2. #2
    Senior Member TriKri's Avatar
    Nov 2006
    Say they intersect, in that case, particle one intersects in the time t_1 and the second in time t_2. We now have:

    \left(\begin{array}{c}t_1\\ \\t_1^2\\ \\t_1^3\end{array}\right)=\left(\begin{array}{c}1+  2t_2\\ \\1+6t_2\\ \\1+14t_2\end{array}\right)\Longleftrightarrow

    <br />
\left(\begin{array}{c}t_1\\ \\(1+2t_2)^2\\ \\(1+2t_2)^3\end{array}\right)=\left(\begin{array}  {c}1+2t_2\\ \\1+6t_2\\ \\1+14t_2\end{array}\right)\Longleftrightarrow

    \left(\begin{array}{c}t_1\\ \\1+4t_2+4t_2^2\\ \\1+6t_2+12t_2^2+8t_2^3\end{array}\right)=\left(\b  egin{array}{c}1+2t_2\\ \\1+6t_2\\ \\1+14t_2\end{array}\right)\Longleftrightarrow

    <br />
\left(\begin{array}{c}t_1\\ \\t_2\\ \\12t_2^2+8t_2^3\end{array}\right)=\left(\begin{ar  ray}{c}1+2t_2\\ \\0\text{ or }0.5\\ \\8t_2\end{array}\right)\Longleftrightarrow

    <br />
\left(\begin{array}{c}t_1\\ \\t_2\\ \\t_2\end{array}\right)=\left(\begin{array}{c}1+2t  _2\\ \\0\text{ or }0.5\\ \\0\text{ or }0.5\text{ or }2\end{array}\right)\Longleftrightarrow\left(\begi  n{array}{c}t_1\\ \\t_2\end{array}\right)=\left(\begin{array}{c}1+2t  _2\\ \\0\text{ or }0.5\end{array}\right)

    So, they intersect at two places. If they hadn't intersected, the equation(s) wouldn't have been possible to solve. Now if you want to se if they colide, put t_1=t_2. Then we have t_2 = 1+2t_2 \Longleftrightarrow t_2 = -1, which doesn't fit with t_2 = 0\text{ or }0.5.
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