Let C be the part of the plane curve defined by y 2 = x 3 − xbetween ( − 1 3 , 4 27 4 )and ( 0 , 0 )oriented from left to right. How would I calculate ∫ C y 2 i → + ( 2 x y + 4 y 3 e y 4 ) j → d sI have already found that the vector field is conservative, I'm just not sure how to proceed from there.

Question

Let C be the part of the plane curve defined by      y    2    =      x    3    −  xbetween  (            −      1              3        ,            4      27        4    )and  (  0  ,  0  )oriented from left to right. How would I calculate      ∫          C            y    2              i      →        +  (  2  x  y  +  4      y    3        e                  y        4              )            j      →        d  sI have already found that the vector field is conservative, I&#039;m just not sure how to proceed from there.

exalantaswo · Accepted Answer

Since the vector field is conservative it has a potential function. Check that the gradient of  f  (  x  ,  y  )  =  x      y    2    +      e                  y        4            is indeed the vector field in the problem. Then use the fundamental theorem of line integrals:      ∫                            a          →                                              b          →                      ∇  f  ⋅  d            r      →        =  f  (            b      →        )  −  f  (            a      →        )to get that the integral equals  f  (  0  ,  0  )  −  f      (    −          1              3              ,                  4        27            4        )    =      11    9    −      e                  4        27

Hugo Stokes · Accepted Answer

F    (  x  ,  y  )  =  (      y    2    ,  2  x  y  +  4      y    3        e                  y        4              )  is the vector field. If F is conservative then       F    =  ∇  G for some scalar field   G  (  x  ,  y  ). And,      ∫    C        F    ⋅  d      r    =      ∫    a    b        F    (      r    (  t  )  )  ⋅            r        ′    (  t  )  d  t  =  G  (      r    (  b  )  )  −  G  (      r    (  a  )  )where       r    (  t  ) is any parametrisation of the curve   C. We have the endpoints       r    (  b  ) and       r    (  a  ) so we only need to find   G      F    =  ∇  G means that                                          ∂            G                                ∂            x                                              =                              y          2                        a        n        d                                      ∂            G                                ∂            y                          =        2        x        y        +        4                  y          3                          e                                    y              4                                                                  ⟹                G                            =                    x                  y          2                +        f        (        y        )        +        C                a        n        d                G        =        x                  y          2                +                  e                                    y              4                                      +        g        (        x        )        +        C            for some functions f(y) and g(x). So, take   f  (  y  )  =      e                  y        4             and   g  (  x  )  =  0 and we have   G  (  x  ,  y  )  =  x      y    2    +      e                  y        4              +  CNow,      ∫    C        F    ⋅  d      r    =  G  (      r    (  b  )  )  −  G  (      r    (  a  )  )  =  G  (  0  ,  0  )  −  G      (    −          1              3              ,                  4        27              )

Let C be the part of the plane curve defined by y^2=x^3−x between ((-1)/(sqrt3), root(4)((4)/(27))) and (0,0) oriented from left to right. How would I calculate int_C y^2 vec(i)+(2xy+4y^3e^(y^4))vec(j) d s

Answered question

Answer & Explanation

New Questions in Precalculus