Parametric function differentiation problem x = a cos ⁡ g ( t ) and y = b sin ⁡ g ( t ) x y 2 d 2 y d x 2 = b 2 d y d x

Question

Parametric function differentiation problem  x  =  a  cos  ⁡  g  (  t  )   and   y  =  b  sin  ⁡  g  (  t  )  x      y    2                      d        2            y              d              x        2              =      b    2              d      y              d      x

Angelo Murray · Accepted Answer

Step 1Evaluate                     d        y                    d        t             and                     d        x                    d        t             . Then,                    d        y                    d        x              =                    d        y                  /                d        t                    d        x                  /                d        t                                          d          2                y                    d                  x          2                      =      d          d      x            (                  d        y                    d        x              )  Now simply substitute x, y into                     d        y                    d        x              , , and                               d          2                y                    d        x             and then all of these into your second equation.

Yesenia Sherman · Accepted Answer

Step 1It is noticed that eliminating g(t) right at the start we need only to deal with two differentiations on a standard ellipse without parameter t:                    (1)                                                        x              2                                      a              2                                      +                                            y              2                                      b              2                                      =        1                    This may not be strictly in line with the aim of the exercise. Also it was on for 6 years.So however to proceed denote &#039;s with resp to x,                    (2)                                          x                          a              2                                      +                                            y                              y                ′                                                    b              2                                      =        0        ;                          y                                                  ′                                      =                                            −              x                              b                2                                                    y                              a                2                                                                            (3)                                          1                          a              2                                      +                                            y                              y                                                                          ″                                                              +              y                              y                                                                          ′                                    2                                                                    b              2                                      =        0        ;                          y                                                  ″                                      =                                            −                              b                2                                            /                                            a                2                            −                              y                                  ′                  2                                                      y                              Plug in 2) and 3) into the given equation                    (4)                    x                  y          2                          y                                                  ″                                      −                  b          2                          y                                                  ′                                      =        0            and simplify LHS:                    (5)                                                        −              x                                                      a                2                            y                                      ⋅        (                                            x              2                                      a              2                                      +                                            y              2                                      b              2                                      −        1        )                 ,            that vanishes.

Parametric function differentiation problem x = a cos ⁡ g (

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