How do you find the equations of tangent lines at the point where the curve crosses itself x = t 2 - t and y = t 3 - 3 t - 1 ?

Question

How do you find the equations of tangent lines at the point where the curve crosses itself       x    =          t      2        -    t   and       y    =          t      3        -    3    t    -    1   ?

Sage Mcdowell · Accepted Answer

Step 1The gradient of the tangent to a curve at any particular point is given by the derivative of the curve at that point. (If needed, then the normal is perpendicular to the tangent so the product of their gradients is       -    1   ).We have:      x    =          t      2        -    t        y    =          t      3        -    3    t    -    1  Firstly, let us find the coordinates where the curve crosses itself. In which case there will be an ordered pair             t      1        =    α   and             t      2        =    β   with             t      1        ≠          t      2       that simultaneously satisfy the parametric equations, therefore:1)                                                          α      2        -    α    =          β      2        -    β  2)             α      3        -    3    α    -    1    =          β      3        -    3    β    -    1  From (1) we have:            α      2        -          β      2        =    α    -    β        ∴          (      α      +      β      )              (      α      -      β      )        =    α    -    β        ∴    α    +    β    =    1        ∴    β    =    1    -    α  From (2) we have:            α      3        -          β      3        =    3    α    -    3    β        ∴          (      α      -      β      )              (              α        2            +      α      β      +              β        2            )        =    3          (      α      -      β      )            ∴          α      2        +    α    β    +          β      2        =    3        ∴          α      2        +    α          (      1      -      α      )        +                  (        1        -        α        )            2        =    3        ∴          α      2        +    α    -          α      2        +    1    -    2    α    +          α      2        =    3        ∴          α      2        -    α    -    2    =    0        ∴          (      α      +      1      )              (      α      -      2      )        =    0        ∴    α    =    -    1    ,    2  And with these values of twe have:      t    =    -    1    ⇒    x    =    2    ,    y    =    1        t    =                             2    ⇒    x    =    2    ,    y    =    1  Thus the curve touches itself when       t    =    -    1    ,    2   corresponding to the rectangular coordinate             (      2      ,      1      )      Then differentiating implicitly wrt t, and applying the chain rule, gives us:                    d        x                              d          t                      =    2    t   and                     d        y                              d          t                      =    3          t      2        -    3                      d        y                    d        x              =                            d          y                /                  d          t                                      d          x                /                  d          t                      =                  3                  t          2                -        3                    2        t            So at the parametric coordinate       t    =    -    1   , we have;            m      1        =                  d        y                    d        x              =                  3        -        3                    2              =    0  So the tangent passes through             (      2      ,      1      )       and has gradient             m      1        =    0   , so using the point/slope form       y    -          y      1        =    m          (      x      -              x        1            )       the equation we seek is;      y    -    1    =    0        ∴    y    =    1  And, at the parametric coordinate       t    =    2   , we have;            m      2        =                  d        y                    d        x              =                  12        -        3                    4              =          9      4      So the tangent passes through             (      2      ,      1      )       and has gradient             m      2        =          9      4       , so using the point/slope form       y    -          y      1        =    m          (      x      -              x        1            )       the equation we seek is;      y    -    1    =          9      4              (      x      -      2      )            ∴    y    -    1    =          9      4        x    -          9      2            ∴    y    =          9      4        x    -          7      2

How do you find the equations of tangent lines at the point where the curve crosses itself <msty

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