How do you find the equation of the tangent line to the curve given by parametric equations: x = 1 + ( 1 t 2 ) , y = 1 - ( 3 t ) at the point when t=2?

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How do you find the equation of the tangent line to the curve given by parametric equations:       x    =    1    +          (              1                  t          2                    )      ,       y    =    1    -          (              3        t            )       at the point when t=2?

Travis Fernandez · Accepted Answer

Step 1To find the slope of the tangent line, you need                     d        y                    d        x             evaluated at a specified point.You are given x(t) and y(t). You can use these to compute x&#039;(t) and y&#039;(t) which can be written using the notation                     d        x                    d        t             and                     d        y                    d        t            , respectively.Step 2                    d        x                    d        t              =    -    2          t              -        3                                d        y                    d        t              =    3          t              -        2            From the chain rule, we know that:                    d        y                    d        t              =          (                        d          y                          d          x                    )              (                        d          x                          d          t                    )      Solve for                     d        y                    d        x            :                    d        y                    d        x              =                                        d            y                                d            t                                                            d            x                                d            t                                                  d        y                    d        x              =                  3                  t                      -            2                                      -        2                  t                      -            3                                                  d        y                    d        x              =          (      -              3        2            )        t  To obtain the slope, m, you evaluate the above at t=2:      m    =          (      -              3        2            )              (      2      )        =    -    3  Evaluate x and y at t = 2:      x    =    1    +          1              2        2              =          5      4            y    =    1    -          3      2        =    -          1      2      This is the point             (              5        4            ,      -              1        2            )      Use the point-slope form of the equation of a line:      y    -          y      1        =    m          (      x      -              x        1            )            y    -    -          1      2        =    -    3          (      x      -              5        4            )            y    =    -    3    x    -          17      4

How do you find the equation of the tangent line to the curve given by parametric equations: <ms

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