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












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









$endgroup$

















    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









    $endgroup$















      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











      share|cite|improve this question




      share|cite|improve this question










      asked 5 hours ago









      UchuukoUchuuko

      367




      367






















          2 Answers
          2






          active

          oldest

          votes


















          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











            $endgroup$














              Your Answer








              StackExchange.ready(function() {
              var channelOptions = {
              tags: "".split(" "),
              id: "69"
              };
              initTagRenderer("".split(" "), "".split(" "), channelOptions);

              StackExchange.using("externalEditor", function() {
              // Have to fire editor after snippets, if snippets enabled
              if (StackExchange.settings.snippets.snippetsEnabled) {
              StackExchange.using("snippets", function() {
              createEditor();
              });
              }
              else {
              createEditor();
              }
              });

              function createEditor() {
              StackExchange.prepareEditor({
              heartbeatType: 'answer',
              autoActivateHeartbeat: false,
              convertImagesToLinks: true,
              noModals: true,
              showLowRepImageUploadWarning: true,
              reputationToPostImages: 10,
              bindNavPrevention: true,
              postfix: "",
              imageUploader: {
              brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
              contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
              allowUrls: true
              },
              noCode: true, onDemand: true,
              discardSelector: ".discard-answer"
              ,immediatelyShowMarkdownHelp:true
              });


              }
              });














              draft saved

              draft discarded


















              StackExchange.ready(
              function () {
              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3189329%2fi-need-to-find-the-potential-function-of-a-vector-field%23new-answer', 'question_page');
              }
              );

              Post as a guest















              Required, but never shown

























              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              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 4 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 4 hours ago

























                          answered 4 hours ago









                          E-muE-mu

                          1214




                          1214






























                              draft saved

                              draft discarded




















































                              Thanks for contributing an answer to Mathematics Stack Exchange!


                              • Please be sure to answer the question. Provide details and share your research!

                              But avoid …



                              • Asking for help, clarification, or responding to other answers.

                              • Making statements based on opinion; back them up with references or personal experience.


                              Use MathJax to format equations. MathJax reference.


                              To learn more, see our tips on writing great answers.




                              draft saved


                              draft discarded














                              StackExchange.ready(
                              function () {
                              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3189329%2fi-need-to-find-the-potential-function-of-a-vector-field%23new-answer', 'question_page');
                              }
                              );

                              Post as a guest















                              Required, but never shown





















































                              Required, but never shown














                              Required, but never shown












                              Required, but never shown







                              Required, but never shown

































                              Required, but never shown














                              Required, but never shown












                              Required, but never shown







                              Required, but never shown







                              -

                              Popular posts from this blog

                              Liste der Baudenkmale in Friedland (Mecklenburg)

                              Image
                              In der Liste der Baudenkmale in Friedland sind alle denkmalgeschützten Bauten der Stadt Friedland (Mecklenburg-Vorpommern) und ihrer Ortsteile aufgelistet. Grundlage ist die Veröffentlichung der Denkmalliste des Landkreises Mecklenburgische Seenplatte (Auszug) vom 28. September 2016. Inhaltsverzeichnis 1 Baudenkmale nach Ortsteilen 1.1 Friedland 1.2 Bresewitz 1.3 Brohm 1.4 Cosa 1.5 Dishley 1.6 Dishley-Ausbau 1.7 Eichhorst 1.8 Glienke 1.9 Heinrichshöh 1.10 Heinrichswalde 1.11 Jatzke 1.12 Jatzke-Meierei 1.13 Liepen 1.14 Ramelow 1.15 Schwanbeck 2 Vorgesehener Denkmalbereich 3 Quelle 4 Weblinks Baudenkmale nach Ortsteilen | Friedland | ID Lage Bezeichnung Beschreibung Bild 281 An der Kleinbahn 2 Windmühle mit Umgang 282 Anklamer Straße alter Friedhof mit Portalwand 282 Anklamer Straße drei Erbbegräbnissen und Gräbern u. a. H.A. Riemann 284
                              Read more

                              Single-Malt-Whisky

                              Image
                              Fünf verschiedene schottische Single-Malt-Whiskys Als Single-Malt-Whisky oder kürzer Single Malt werden Whiskys bezeichnet, die zwei besondere Bedingungen erfüllen: Sie stammen zum ersten aus einer einzigen Brennerei, sind also kein Verschnitt aus mehreren Whiskysorten (daher Single , deutsch: einzeln). Zum zweiten wird als Getreide ausschließlich gemälzte Gerste verwendet (deswegen Malt , deutsch: Malz). Der Zusatz Single Malt wird als Prädikat verstanden, die so ausgezeichneten Whiskys sind in aller Regel besonders hochwertig. Hinzu kommt ihre besonders lange Lagerung, die Reifungszeit währt oft ein Jahrzehnt und länger. Inhaltsverzeichnis 1 Allgemeines 2 Herkunftsländer 2.1 Schottland 2.2 Irland 2.3 Japan 2.4 Kontinentaleuropa 3 Produktionsschritte 3.1 Mälzen 3.2 Brauen 3.3 Brennen 3.4 Reifung und Finish 3.5 Abfüllung 4 Altersangaben 5 Lagerung von Whiskyflaschen 6 Geschmacksklassifikation 7 Sonstig
                              Read more

                              Czorneboh

                              Image
                              Dieser Artikel erläutert den Berg Czorneboh, für die Gottheit siehe Czorneboh (Gottheit). Czorneboh Čornobóh Berggasthof mit Turm Höhe 555,7  m ü. NHN [1] Lage Freistaat Sachsen, Deutschland Gebirge Lausitzer Bergland Dominanz 15,4 km →  Kottmar Koordinaten 51° 7′ 11″  N , 14° 31′ 31″  O 51.119722222222 14.525194444444 555.7 Koordinaten: 51° 7′ 11″  N , 14° 31′ 31″  O Gestein Granodiorit Besonderheiten Aussichtsturm Czorneboh Der Czorneboh , obersorbisch Čornobóh , ist ein Berg zwischen Hochkirch und Cunewalde in der Oberlausitz. Er gehört zu der südöstlich von Bautzen gelegenen Czorneboh-Bergkette und ist mit einer Höhe von 555,7  m ü. NHN [1] der höchste Punkt dieses Ausläufers des Lausitzer Berglandes. Er leitet als Landschaftselement die wellige Granitlandschaft der Lausitzer Vorbergzone ein. Der Gipfel des Czorneboh liegt in der Gemarkung Meschwitz (Gemeinde Hochkirch), die Czornebohbaude jedoch
                              Read more