How to find parallel tangents for a parametric equationFor the function x → ( t ) = ( 2 t + 3 2 − t t 3 − 2 t 2 + t ) t ≥ 0Are there 2 points x → ( t 1 ) , x → ( t 2 ) , such that the function’s tangent vectors at these points are parallel to each other? Find such points, or show that none exist.I know that the derivative of the function is x → ′ ( t ) = ( 2 − 1 3 t 2 − 4 t + 1 ) and that in order for the tangent vectors to be parallel to each other the functions will equal the same value. However, I am not sure how to go about finding the values.

Question

How to find parallel tangents for a parametric equationFor the function            x      →        (  t  )  =      (                            2          t          +          3                                      2          −          t                                                  t            3                    −          2                      t            2                    +          t                      )    t  ≥  0Are there 2 points             x      →        (      t    1    ) ,             x      →        (      t    2    ) , such that the function’s tangent vectors at these points are parallel to each other? Find such points, or show that none exist.I know that the derivative of the function is                    x        →              ′    (  t  )  =      (                            2                                      −          1                                      3                      t            2                    −          4          t          +          1                      )  and that in order for the tangent vectors to be parallel to each other the functions will equal the same value. However, I am not sure how to go about finding the values.

Julius Haley · Accepted Answer

Then we first solve for             u      2        -    12    u    +    11    =    0    →              (      u      -      1      )              (      u      -      12      )        =    0    →    u    =    1                or              u    =    12    →  There are now four solutions:      x    =    ±          1        =    ±    1                or              x    =    ±          12        =    ±    2          3      We&#039;ll have to examine what happens between and outside those points.If we take any value between       -    2          3                    and              -    1   the outcome will be negative, between       -    1                and              +    1   it&#039;s positive, and between       +    1                and              +    2          3       it will be negative again.      &amp;lt;    -    2          3                    and              &amp;gt;    +    2          3       the outcome will be positive.

For the function vec(x)(t)=((2t+3),(2-t),(t^3-2t^2+t)) t >=0 Are there 2 points vec(x) (t_1), vec(x) (t_2), such that the function’s tangent vectors at these points are parallel to each other? Find such points, or show that none exist.

Answered question

Answer & Explanation

New Questions in Precalculus