How do you find the parametric equations for the tangent line to the curve x = t 4 − 1 , y = t 2 + 1 , z = t 3 at the point (15, 5, 8)?

Question

How do you find the parametric equations for the tangent line to the curve       x    =          t      4        −    1   ,       y    =          t      2        +    1   ,       z    =          t      3       at the point (15, 5, 8)?

stellak012s7aoc · Accepted Answer

Step 1For             r      →        =          (                                                  t              4                        -            1                                                              t              2                        +            1                                                              t              3                                          )        =          (                                    15                                                5                                                8                              )       , we can see that       t    =    4  the tangent vector (or velocity vector) is the first derivative:            r      →              ′              t        =        4              =                  (                                            4                              t                3                                                                        2              t                                                          3                              t                2                                                    )                    t        =        4              =          (                                    4                          (                              4                3                            )                                                            2                          (              4              )                                                            3                          (                              4                2                            )                                          )        =          (                                    32                                                1                                                6                              )      so for generalised line             r      →        =                  r        →            o        +    λ          r      →        ′   we have here            r      →        =          (                                    15                                                5                                                8                              )        +    λ          (                                    32                                                1                                                6                              )

How do you find the parametric equations for the tangent line to the curve <mstyle displaystyle=

Answered question

Answer & Explanation

New Questions in Integral Calculus