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I need to find the potential function of a vector field.

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I need to find the potential function of a vector field.



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Calculating the Integral of a non conservative vector fieldFind the potential function of a conservative vector fieldTwo ways of finding a Potential of a Vector FieldFinding potential function for a vector fieldVector Field Conceptual QuestionIs there a specific notation to denote the potential function of a conservative vector field?Every conservative vector field is irrotationalQuestions about the potential of a conservative vector fieldWhy do we need both Divergence and Curl to define a vector field?How to check if a 2 dimensional vector field is irrotational (curl=0)?












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I was given F = (y+z)i + (x+z)j + (x+y)k. I found said field to be conservative, and I integrated the x partial derivative and got f(x,y,z) = xy + xz + g(y,z). The thing is that I am trying to find g(y,z), and I ended up with something that was expressed in terms of x, y and z (I got x+z-xy-xz). I don't know what to do with this information not that I arrived at something expressed in all three variables.






integration multivariable-calculus vector-fields







share|cite|improve this question









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    $begingroup$


    I was given F = (y+z)i + (x+z)j + (x+y)k. I found said field to be conservative, and I integrated the x partial derivative and got f(x,y,z) = xy + xz + g(y,z). The thing is that I am trying to find g(y,z), and I ended up with something that was expressed in terms of x, y and z (I got x+z-xy-xz). I don't know what to do with this information not that I arrived at something expressed in all three variables.






    integration multivariable-calculus vector-fields







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      I was given F = (y+z)i + (x+z)j + (x+y)k. I found said field to be conservative, and I integrated the x partial derivative and got f(x,y,z) = xy + xz + g(y,z). The thing is that I am trying to find g(y,z), and I ended up with something that was expressed in terms of x, y and z (I got x+z-xy-xz). I don't know what to do with this information not that I arrived at something expressed in all three variables.






      integration multivariable-calculus vector-fields







      share|cite|improve this question









      $endgroup$




      I was given F = (y+z)i + (x+z)j + (x+y)k. I found said field to be conservative, and I integrated the x partial derivative and got f(x,y,z) = xy + xz + g(y,z). The thing is that I am trying to find g(y,z), and I ended up with something that was expressed in terms of x, y and z (I got x+z-xy-xz). I don't know what to do with this information not that I arrived at something expressed in all three variables.






      integration multivariable-calculus vector-fields




      integration multivariable-calculus vector-fields






      share|cite|improve this question













      share|cite|improve this question











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      asked 3 hours ago









      UchuukoUchuuko

      367




      367






















          2 Answers
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          3












          $begingroup$

          You have $frac{partial f}{partial x}= y+ z$ so that $f(x,y,z)= xy+ xz+ g(y,z)$. (Since the differentiation with respect to x treat y and z as constants, the "constant of integration" might in fact be a function of y and z. That is the "g(y, z)".)



          Differentiating that with respect to y, $frac{partial f}{partial y}= x+ g_y(y, z)= x+ z$ so that $g_y= z$ and $g(y, z)= yz+ h(z)$.



          So f(x,y,z)= xy+ xz+ yz+ h(z). Differentiating that with respect to z, $frac{partial f}{partial z}= x+ y+ h'(z)= x+ y$ so that h'(z)= 0. h is a constant, C so that we get f(x, y, z)= xy+ xz+ yz+ C.






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            So far, we have $f(x,y,z) = xy + xz + g(y,z)$. Taking $frac{partial f}{partial x}$ gives us the $x$-component of $textbf{F}$. To get similar $y$ and $z$-components, we suspect that $g(y,z)$ should be similar to the other terms in $f(x,y,z)$ in some sense. The natural guess is $g(y,z) = yz$, since the other terms in $f(x,y,z)$ are each multiplications of two different independent variables. It can then be verified that the guess for $g$ produces the correct vector field, by computing $nabla f$.



            We now know that we have determined the potential function up to a constant, since if two scalar fields have the same gradient, then they differ by a constant.



            A note of caution: sometimes the convention for what is meant by a potential function for a vector field $mathbf{F}$, is a scalar field $f$ such that $mathbf{F} = - nabla f$. Beware!






            share|cite|improve this answer











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              active

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              3












              $begingroup$

              You have $frac{partial f}{partial x}= y+ z$ so that $f(x,y,z)= xy+ xz+ g(y,z)$. (Since the differentiation with respect to x treat y and z as constants, the "constant of integration" might in fact be a function of y and z. That is the "g(y, z)".)



              Differentiating that with respect to y, $frac{partial f}{partial y}= x+ g_y(y, z)= x+ z$ so that $g_y= z$ and $g(y, z)= yz+ h(z)$.



              So f(x,y,z)= xy+ xz+ yz+ h(z). Differentiating that with respect to z, $frac{partial f}{partial z}= x+ y+ h'(z)= x+ y$ so that h'(z)= 0. h is a constant, C so that we get f(x, y, z)= xy+ xz+ yz+ C.






              share|cite|improve this answer









              $endgroup$


















                3












                $begingroup$

                You have $frac{partial f}{partial x}= y+ z$ so that $f(x,y,z)= xy+ xz+ g(y,z)$. (Since the differentiation with respect to x treat y and z as constants, the "constant of integration" might in fact be a function of y and z. That is the "g(y, z)".)



                Differentiating that with respect to y, $frac{partial f}{partial y}= x+ g_y(y, z)= x+ z$ so that $g_y= z$ and $g(y, z)= yz+ h(z)$.



                So f(x,y,z)= xy+ xz+ yz+ h(z). Differentiating that with respect to z, $frac{partial f}{partial z}= x+ y+ h'(z)= x+ y$ so that h'(z)= 0. h is a constant, C so that we get f(x, y, z)= xy+ xz+ yz+ C.






                share|cite|improve this answer









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  You have $frac{partial f}{partial x}= y+ z$ so that $f(x,y,z)= xy+ xz+ g(y,z)$. (Since the differentiation with respect to x treat y and z as constants, the "constant of integration" might in fact be a function of y and z. That is the "g(y, z)".)



                  Differentiating that with respect to y, $frac{partial f}{partial y}= x+ g_y(y, z)= x+ z$ so that $g_y= z$ and $g(y, z)= yz+ h(z)$.



                  So f(x,y,z)= xy+ xz+ yz+ h(z). Differentiating that with respect to z, $frac{partial f}{partial z}= x+ y+ h'(z)= x+ y$ so that h'(z)= 0. h is a constant, C so that we get f(x, y, z)= xy+ xz+ yz+ C.






                  share|cite|improve this answer









                  $endgroup$



                  You have $frac{partial f}{partial x}= y+ z$ so that $f(x,y,z)= xy+ xz+ g(y,z)$. (Since the differentiation with respect to x treat y and z as constants, the "constant of integration" might in fact be a function of y and z. That is the "g(y, z)".)



                  Differentiating that with respect to y, $frac{partial f}{partial y}= x+ g_y(y, z)= x+ z$ so that $g_y= z$ and $g(y, z)= yz+ h(z)$.



                  So f(x,y,z)= xy+ xz+ yz+ h(z). Differentiating that with respect to z, $frac{partial f}{partial z}= x+ y+ h'(z)= x+ y$ so that h'(z)= 0. h is a constant, C so that we get f(x, y, z)= xy+ xz+ yz+ C.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 2 hours ago









                  user247327user247327

                  11.6k1516




                  11.6k1516























                      1












                      $begingroup$

                      So far, we have $f(x,y,z) = xy + xz + g(y,z)$. Taking $frac{partial f}{partial x}$ gives us the $x$-component of $textbf{F}$. To get similar $y$ and $z$-components, we suspect that $g(y,z)$ should be similar to the other terms in $f(x,y,z)$ in some sense. The natural guess is $g(y,z) = yz$, since the other terms in $f(x,y,z)$ are each multiplications of two different independent variables. It can then be verified that the guess for $g$ produces the correct vector field, by computing $nabla f$.



                      We now know that we have determined the potential function up to a constant, since if two scalar fields have the same gradient, then they differ by a constant.



                      A note of caution: sometimes the convention for what is meant by a potential function for a vector field $mathbf{F}$, is a scalar field $f$ such that $mathbf{F} = - nabla f$. Beware!






                      share|cite|improve this answer











                      $endgroup$


















                        1












                        $begingroup$

                        So far, we have $f(x,y,z) = xy + xz + g(y,z)$. Taking $frac{partial f}{partial x}$ gives us the $x$-component of $textbf{F}$. To get similar $y$ and $z$-components, we suspect that $g(y,z)$ should be similar to the other terms in $f(x,y,z)$ in some sense. The natural guess is $g(y,z) = yz$, since the other terms in $f(x,y,z)$ are each multiplications of two different independent variables. It can then be verified that the guess for $g$ produces the correct vector field, by computing $nabla f$.



                        We now know that we have determined the potential function up to a constant, since if two scalar fields have the same gradient, then they differ by a constant.



                        A note of caution: sometimes the convention for what is meant by a potential function for a vector field $mathbf{F}$, is a scalar field $f$ such that $mathbf{F} = - nabla f$. Beware!






                        share|cite|improve this answer











                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          So far, we have $f(x,y,z) = xy + xz + g(y,z)$. Taking $frac{partial f}{partial x}$ gives us the $x$-component of $textbf{F}$. To get similar $y$ and $z$-components, we suspect that $g(y,z)$ should be similar to the other terms in $f(x,y,z)$ in some sense. The natural guess is $g(y,z) = yz$, since the other terms in $f(x,y,z)$ are each multiplications of two different independent variables. It can then be verified that the guess for $g$ produces the correct vector field, by computing $nabla f$.



                          We now know that we have determined the potential function up to a constant, since if two scalar fields have the same gradient, then they differ by a constant.



                          A note of caution: sometimes the convention for what is meant by a potential function for a vector field $mathbf{F}$, is a scalar field $f$ such that $mathbf{F} = - nabla f$. Beware!






                          share|cite|improve this answer











                          $endgroup$



                          So far, we have $f(x,y,z) = xy + xz + g(y,z)$. Taking $frac{partial f}{partial x}$ gives us the $x$-component of $textbf{F}$. To get similar $y$ and $z$-components, we suspect that $g(y,z)$ should be similar to the other terms in $f(x,y,z)$ in some sense. The natural guess is $g(y,z) = yz$, since the other terms in $f(x,y,z)$ are each multiplications of two different independent variables. It can then be verified that the guess for $g$ produces the correct vector field, by computing $nabla f$.



                          We now know that we have determined the potential function up to a constant, since if two scalar fields have the same gradient, then they differ by a constant.



                          A note of caution: sometimes the convention for what is meant by a potential function for a vector field $mathbf{F}$, is a scalar field $f$ such that $mathbf{F} = - nabla f$. Beware!







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited 2 hours ago

























                          answered 2 hours ago









                          E-muE-mu

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