I want to solve d y / d x for the following: x 2 + y 2 = R 2 where R is a constant.I know to use implicit differentiation, though I have a question. When I derive R 2 , do I obtain 2 R or 0?Additionally, deriving y 2 with respect to x yields 2 y ( d y / d x )? This is different from a partial derivative?Thanks!

Question

I want to solve   d  y      /    d  x for the following:      x    2    +      y    2    =      R    2   where   R is a constant.I know to use implicit differentiation, though I have a question. When I derive       R    2  , do I obtain   2  R or 0?Additionally, deriving       y    2   with respect to x yields   2  y  (  d  y      /    d  x  )? This is different from a partial derivative?Thanks!

bulgarum87 · Accepted Answer

Suppose   R is a function of   x and   y; then                    (1)                              x          2                +                  y          2                =                  R          2                (        x        ,        y        )        ;            if we define                    (2)                    F        (        x        ,        y        )        =                  x          2                +                  y          2                −                  R          2                (        x        ,        y        )        ,            we may also write (1) as                    (3)                    F        (        x        ,        y        )        =                  x          2                +                  y          2                −                  R          2                (        x        ,        y        )        =        0        ;            by the implicit funtion theorem, this equation in fact may be seen as defining   y  (  x  ), a function of   x, provided that                    (4)                                                        ∂              F              (              x              ,              y              )                                      ∂              y                                      ≠        0        ;            we have                    (5)                                                        ∂              F              (              x              ,              y              )                                      ∂              y                                      =        2        y        −        2        R        (        x        ,        y        )                                            ∂              R              (              x              ,              y              )                                      ∂              y                                      ≠        0            provided                    (6)                    y        ≠        R        (        x        ,        y        )                                            ∂              R              (              x              ,              y              )                                      ∂              y                                      ;            under such circumstances, we may affirm   y  (  x  ) is uniquely determined as a differentiable function of   x in some neighborhood of any point   (  x  ,  y  ); then we have                    (7)                    F        (        x        ,        y        )        =                  x          2                +                  y          2                (        x        )        −                  R          2                (        x        ,        y        (        x        )        )        =        0        ;            we may take the total derivative with respect to x to obtain                    (8)                                                        d              F              (              x              ,              y              )                                      d              x                                      =        2        x        +        2        y                                            d              y              (              x              )                                      d              x                                      −        2        R        (        x        ,        y        (        x        )        )                  (                                                    ∂                R                (                x                ,                y                (                x                )                )                                            ∂                x                                              +                                                    ∂                R                (                x                ,                y                (                x                )                )                                            ∂                y                                                                                        d                y                (                x                )                                            d                x                                              )                =        0        ;            a little algebra allows us to isolate the terms containing   d  y  (  x  )      /    d  x:                    (9)                    x        +        y                                            d              y              (              x              )                                      d              x                                      −        R        (        x        ,        y        (        x        )        )                  (                                                    ∂                R                (                x                ,                y                (                x                )                )                                            ∂                x                                              +                                                    ∂                R                (                x                ,                y                (                x                )                )                                            ∂                y                                                                                        d                y                (                x                )                                            d                x                                              )                =        0        ;                                (10)                    y                                            d              y              (              x              )                                      d              x                                      −        R        (        x        ,        y        (        x        )        )                                            ∂              R              (              x              ,              y              (              x              )              )                                      ∂              y                                                                          d              y              (              x              )                                      d              x                                      =        R        (        x        ,        y        (        x        )        )                                            ∂              R              (              x              ,              y              (              x              )              )                                      ∂              x                                      −        x        ;                                (11)                              (          y          −          R          (          x          ,          y          (          x          )          )                                                    ∂                R                (                x                ,                y                (                x                )                                            ∂                y                                              )                                                    d              y              (              x              )                                      d              x                                      =        R        (        x        ,        y        (        x        )        )                                            ∂              R              (              x              ,              y              (              x              )              )                                      ∂              x                                      −        x        ;            for the sake of compactess and brevity, we introduce the subscript notation                    (12)                              R          x                =                                            ∂              R                                      ∂              x                                      ,                etc.        ,            and write (11) in the form                    (13)                              y          ′                (        x        )        =                                            R                              R                x                            −              x                                      y              −              R                              R                y                                                    =        −                                            x              −              R                              R                x                                                    y              −              R                              R                y                                                    ,            which gives a general expression for       y    ′    (  x  ); in the event that   R  (  x  ,  y  ) is constant, we obtain                    (14)                              y          ′                (        x        )        =        −                              x            y                          ,            which the reader may recognize as the slope of the circle                    (15)                              x          2                +                  y          2                =                  R          2                    at any point   (  x  ,  y  ) where   y  ≠  0.

Freddy Friedman · Accepted Answer

By the chaine rule you will get  2  x  +  2  y  ⋅      y    ′    =  0

I want to solve dy/dx for the following: x^2+y^2=R^2 where R is a constant. I know to use implicit differentiation, though I have a question. When I derive R^2, do I obtain 2R or 0? Additionally, deriving y^2 with respect to x yields 2y(dy/dx)? This is different from a partial derivative?

Answered question

Answer & Explanation

New Questions in Calculus 2