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 2 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 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

                          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

                              How to label and detect the document text images

                              Image
                              1 $begingroup$ This is what I mean as document text image: I want to label the texts in image as separate blocks and my model should detect these labels as classes. NOTE: This is how the end result should be like: The labels like Block 1, Block 2, Block 3,.. should be Logo, Title, Date,.. Others, etc. Work done: First approach : I tried to implement this method via Object Detection, it didn't work. It didn't even detect any text. Second approach : Then I tried it using PixelLink. As this model is build for scene text detection, it detected each and every text in the image. But this method can detect multiple lines of text if the threshold values are increased. But I have no idea how do I add labels to the text blocks. PIXEL_CLS_WEIGHT_all_ones = 'PIXEL_CLS_WEIGHT_all_ones' PIXEL_C...
                              Read more

                              Vallis Paradisi

                              Image
                              Coordinata: .mw-parser-output .geo-default,.mw-parser-output .geo-dms,.mw-parser-output .geo-dec{display:inline}.mw-parser-output .geo-nondefault,.mw-parser-output .geo-multi-punct{display:none}.mw-parser-output .longitude,.mw-parser-output .latitude{white-space:nowrap} 33°03′S 71°37′W  /  33.05°S 71.617°W  / -33.05; -71.617 Vallis Paradisi Res apud Vicidata repertae: Civitas : Chilia Numerus incolarum : 296 655 Zona horaria : UTC-4 Situs interretialis Nomen officiale : Valparaíso Gubernium Consilium : municipal council of Valparaíso Geographia Superficies : 401 600 000 Metrum quadratum Coniunctiones urbium Urbes gemellae : Ovetum , Badalona , Santa Fe , Ora longa , Sciamhaevum , Quantunia , Callao , Bat Yam , Melaka , Novorossiysk , Busan , Corduba , Odessa , Rosario , Malacca , Barcino , Soteropolis , Franciscopolis (California) , Guanajuato, Silao Tabula aut despectus Despectus in portum sinumque et urbem Vallis Paradisi Vallis Para...
                              Read more

                              Tabula Rosettana

                              Image
                              Tabula multilinguis Rosettana in Museo Britannico ostenditur. Tabula Rosettana, [1] etiam titulo OGIS 90 agnita, est stela decreto de rebus sacris in Aegypto anno 196 a.C.n. lato inscripta. Tabula iuxta Rosettam Aegypti, urbem in delta Nili et ad oram maris Mediterranei iacentem, anno 1799 a milite Francico reperta est. Inventio stelae, linguis duabus scripturisque tribus inscriptae, eruditis Instituti Aegypti statim nuntiata est; ibi enim iussu imperatoris Napoleonis eruditi omnium scientiarum (sub aegide Commissionis Scientiarum et Artium) properaverant cum expeditione Francica. Qua a Britannis mox debellata, tabula Rosettana Londinium missa hodie apud Museum Britannicum iacet. Textus Graecus cito lectus interpretationi textuum Aegyptiorum (in formis hieroglyphica et demotica expressorum) gradatim adiuvit. Denique textum plene interpretatus est Ioannes Franciscus Champollion. Ab opere eruditorum cumulativo coepit hodiernus scripturae hieroglyphicae linguaeque Aegyptiae a...
                              Read more