Tangent plane to $x^2+y^2+z^2=50$ at $(3,4,5)$












0















Prove that the tangent plane to $x^2+y^2+z^2=50$ at $(3,4,5)$ is $3x+4y+5z=50$




My workings are shown below but get the answer wrong completly, have I made a simple mistake is my method flawed.



So if the surface is said to be $f(x,y,z)$ then
$$frac{partial f}{partial x}=2xqquad frac{partial f}{partial y}=2y$$



Therefore $f_x(3,4)=6$ and $f_y(3,4)=8$



Now using the
$$z=f(a,b)+(x-a)f_x(a,b)+(y-b)f_y(a,b)$$
I get a plane with the equation $$z=5+(x-3)(6)+(y-4)(8)$$
$$z=6x+8y-45$$



Any help on how to get the correct answer would be apprciated.










share|cite|improve this question
























  • I think you want $f(x,y)=sqrt{50-x^2-y^2}$ for your method to work.
    – Lubin
    Jan 3 at 21:39
















0















Prove that the tangent plane to $x^2+y^2+z^2=50$ at $(3,4,5)$ is $3x+4y+5z=50$




My workings are shown below but get the answer wrong completly, have I made a simple mistake is my method flawed.



So if the surface is said to be $f(x,y,z)$ then
$$frac{partial f}{partial x}=2xqquad frac{partial f}{partial y}=2y$$



Therefore $f_x(3,4)=6$ and $f_y(3,4)=8$



Now using the
$$z=f(a,b)+(x-a)f_x(a,b)+(y-b)f_y(a,b)$$
I get a plane with the equation $$z=5+(x-3)(6)+(y-4)(8)$$
$$z=6x+8y-45$$



Any help on how to get the correct answer would be apprciated.










share|cite|improve this question
























  • I think you want $f(x,y)=sqrt{50-x^2-y^2}$ for your method to work.
    – Lubin
    Jan 3 at 21:39














0












0








0








Prove that the tangent plane to $x^2+y^2+z^2=50$ at $(3,4,5)$ is $3x+4y+5z=50$




My workings are shown below but get the answer wrong completly, have I made a simple mistake is my method flawed.



So if the surface is said to be $f(x,y,z)$ then
$$frac{partial f}{partial x}=2xqquad frac{partial f}{partial y}=2y$$



Therefore $f_x(3,4)=6$ and $f_y(3,4)=8$



Now using the
$$z=f(a,b)+(x-a)f_x(a,b)+(y-b)f_y(a,b)$$
I get a plane with the equation $$z=5+(x-3)(6)+(y-4)(8)$$
$$z=6x+8y-45$$



Any help on how to get the correct answer would be apprciated.










share|cite|improve this question
















Prove that the tangent plane to $x^2+y^2+z^2=50$ at $(3,4,5)$ is $3x+4y+5z=50$




My workings are shown below but get the answer wrong completly, have I made a simple mistake is my method flawed.



So if the surface is said to be $f(x,y,z)$ then
$$frac{partial f}{partial x}=2xqquad frac{partial f}{partial y}=2y$$



Therefore $f_x(3,4)=6$ and $f_y(3,4)=8$



Now using the
$$z=f(a,b)+(x-a)f_x(a,b)+(y-b)f_y(a,b)$$
I get a plane with the equation $$z=5+(x-3)(6)+(y-4)(8)$$
$$z=6x+8y-45$$



Any help on how to get the correct answer would be apprciated.







partial-derivative surfaces plane-geometry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 3 at 22:26









pwerth

1,747411




1,747411










asked Jan 3 at 21:32









H.Linkhorn

31912




31912












  • I think you want $f(x,y)=sqrt{50-x^2-y^2}$ for your method to work.
    – Lubin
    Jan 3 at 21:39


















  • I think you want $f(x,y)=sqrt{50-x^2-y^2}$ for your method to work.
    – Lubin
    Jan 3 at 21:39
















I think you want $f(x,y)=sqrt{50-x^2-y^2}$ for your method to work.
– Lubin
Jan 3 at 21:39




I think you want $f(x,y)=sqrt{50-x^2-y^2}$ for your method to work.
– Lubin
Jan 3 at 21:39










4 Answers
4






active

oldest

votes


















2














A few comments:




  • The formula you are referring to is for surfaces which are the graphs of a function $f(x,y)$, not $f(x,y,z)$

  • The given surface is a sphere of radius $sqrt{50}$ centered at the origin. A sphere cannot be represented as a function $f(x,y)$


However, hope is not lost. The given point $(3,4,5)$ lies on the upper hemisphere, and a hemisphere is the graph of a function $z=f(x,y)$. What function is it? Well in this case we have
$$z^{2}=50-x^{2}-y^{2}$$
so the top hemisphere is given by
$$z=sqrt{50-x^{2}-y^{2}}=(50-x^{2}-y^{2})^{1/2}$$
This is our $f(x,y)$. I'll point out that the lower hemisphere is given by $-f(x,y)$. Now, with this $f$, we have
$$frac{partial f}{partial x}=frac{1}{2}(50-x^{2}-y^{2})^{-1/2}cdot(-2x)$$
and
$$frac{partial f}{partial y}=frac{1}{2}(50-x^{2}-y^{2})^{-1/2}cdot(-2y)$$



Plugging in $x=3,y=4$ we have
$$f_{x}(3,4)=-frac{3}{5}$$
and
$$f_{y}(3,4)=-frac{4}{5}$$
Since $f(3,4)=5$, the tangent plane has equation
$$z=5-frac{3}{5}(x-3)-frac{4}{5}(y-4)$$
which, after some algebra, becomes
$$3x+4y+5z=50$$






share|cite|improve this answer





























    1














    $f(x,y,z)=x^2+y^2+z^2-50=0;$



    Normal vector to surface $f(x,y,z)=0$ is given by



    $vec n= nabla f(x,y,z)=(f_x,f_y,f_z)=(2x,2y,2z)$



    Equation.of tangent plane:



    $vec n cdot (vec r -vec r_0)=0$;



    With $vec n= 2(3,4,5)$ and $vec r_0=(3,4,5)$:



    $2(3,4,5)cdot ((x,y,z)-(3,4,5))=0;$



    $3x+4y+5z -9-16-25=0;$



    $3x+4y+5z =50.$






    share|cite|improve this answer































      1














      In order to calculate the tangent plane at $P = (3,4,5)$, we can consider the parametrization:



      begin{align*}
      varphi(x,y) = left(x,y,sqrt{50 - x^{2} - y^{2}}right)
      end{align*}



      Consequently, the tangent plane $pi$ at the given point can be described by



      begin{align*}
      (x,y,z) = P + alphavarphi_{x}(3,4) + betavarphi_{y}(3,4)
      end{align*}



      where $(alpha,beta)inmathbb{R}^{2}$.






      share|cite|improve this answer































        1














        Why not forget Calculus and look geometrically?



        Having observed that the point $P=(3,4,5)$ is indeed on the sphere, one sees that the segment $overline{Bbb OP}$ is a radius, where $Bbb O$ is the origin, center of the sphere. Now certainly any radius is perpendicular to the tangent plane at the point in question, so that the vector $(3,4,5)$ is orthogonal to the plane, and thus the equation must be $3X+4Y+5Z=c$, for an appropriate $c$, which you find to be $c=50$, in order to guarantee that $P$ be on the plane.






        share|cite|improve this answer





















          Your Answer





          StackExchange.ifUsing("editor", function () {
          return StackExchange.using("mathjaxEditing", function () {
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          });
          });
          }, "mathjax-editing");

          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%2f3061024%2ftangent-plane-to-x2y2z2-50-at-3-4-5%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          4 Answers
          4






          active

          oldest

          votes








          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2














          A few comments:




          • The formula you are referring to is for surfaces which are the graphs of a function $f(x,y)$, not $f(x,y,z)$

          • The given surface is a sphere of radius $sqrt{50}$ centered at the origin. A sphere cannot be represented as a function $f(x,y)$


          However, hope is not lost. The given point $(3,4,5)$ lies on the upper hemisphere, and a hemisphere is the graph of a function $z=f(x,y)$. What function is it? Well in this case we have
          $$z^{2}=50-x^{2}-y^{2}$$
          so the top hemisphere is given by
          $$z=sqrt{50-x^{2}-y^{2}}=(50-x^{2}-y^{2})^{1/2}$$
          This is our $f(x,y)$. I'll point out that the lower hemisphere is given by $-f(x,y)$. Now, with this $f$, we have
          $$frac{partial f}{partial x}=frac{1}{2}(50-x^{2}-y^{2})^{-1/2}cdot(-2x)$$
          and
          $$frac{partial f}{partial y}=frac{1}{2}(50-x^{2}-y^{2})^{-1/2}cdot(-2y)$$



          Plugging in $x=3,y=4$ we have
          $$f_{x}(3,4)=-frac{3}{5}$$
          and
          $$f_{y}(3,4)=-frac{4}{5}$$
          Since $f(3,4)=5$, the tangent plane has equation
          $$z=5-frac{3}{5}(x-3)-frac{4}{5}(y-4)$$
          which, after some algebra, becomes
          $$3x+4y+5z=50$$






          share|cite|improve this answer


























            2














            A few comments:




            • The formula you are referring to is for surfaces which are the graphs of a function $f(x,y)$, not $f(x,y,z)$

            • The given surface is a sphere of radius $sqrt{50}$ centered at the origin. A sphere cannot be represented as a function $f(x,y)$


            However, hope is not lost. The given point $(3,4,5)$ lies on the upper hemisphere, and a hemisphere is the graph of a function $z=f(x,y)$. What function is it? Well in this case we have
            $$z^{2}=50-x^{2}-y^{2}$$
            so the top hemisphere is given by
            $$z=sqrt{50-x^{2}-y^{2}}=(50-x^{2}-y^{2})^{1/2}$$
            This is our $f(x,y)$. I'll point out that the lower hemisphere is given by $-f(x,y)$. Now, with this $f$, we have
            $$frac{partial f}{partial x}=frac{1}{2}(50-x^{2}-y^{2})^{-1/2}cdot(-2x)$$
            and
            $$frac{partial f}{partial y}=frac{1}{2}(50-x^{2}-y^{2})^{-1/2}cdot(-2y)$$



            Plugging in $x=3,y=4$ we have
            $$f_{x}(3,4)=-frac{3}{5}$$
            and
            $$f_{y}(3,4)=-frac{4}{5}$$
            Since $f(3,4)=5$, the tangent plane has equation
            $$z=5-frac{3}{5}(x-3)-frac{4}{5}(y-4)$$
            which, after some algebra, becomes
            $$3x+4y+5z=50$$






            share|cite|improve this answer
























              2












              2








              2






              A few comments:




              • The formula you are referring to is for surfaces which are the graphs of a function $f(x,y)$, not $f(x,y,z)$

              • The given surface is a sphere of radius $sqrt{50}$ centered at the origin. A sphere cannot be represented as a function $f(x,y)$


              However, hope is not lost. The given point $(3,4,5)$ lies on the upper hemisphere, and a hemisphere is the graph of a function $z=f(x,y)$. What function is it? Well in this case we have
              $$z^{2}=50-x^{2}-y^{2}$$
              so the top hemisphere is given by
              $$z=sqrt{50-x^{2}-y^{2}}=(50-x^{2}-y^{2})^{1/2}$$
              This is our $f(x,y)$. I'll point out that the lower hemisphere is given by $-f(x,y)$. Now, with this $f$, we have
              $$frac{partial f}{partial x}=frac{1}{2}(50-x^{2}-y^{2})^{-1/2}cdot(-2x)$$
              and
              $$frac{partial f}{partial y}=frac{1}{2}(50-x^{2}-y^{2})^{-1/2}cdot(-2y)$$



              Plugging in $x=3,y=4$ we have
              $$f_{x}(3,4)=-frac{3}{5}$$
              and
              $$f_{y}(3,4)=-frac{4}{5}$$
              Since $f(3,4)=5$, the tangent plane has equation
              $$z=5-frac{3}{5}(x-3)-frac{4}{5}(y-4)$$
              which, after some algebra, becomes
              $$3x+4y+5z=50$$






              share|cite|improve this answer












              A few comments:




              • The formula you are referring to is for surfaces which are the graphs of a function $f(x,y)$, not $f(x,y,z)$

              • The given surface is a sphere of radius $sqrt{50}$ centered at the origin. A sphere cannot be represented as a function $f(x,y)$


              However, hope is not lost. The given point $(3,4,5)$ lies on the upper hemisphere, and a hemisphere is the graph of a function $z=f(x,y)$. What function is it? Well in this case we have
              $$z^{2}=50-x^{2}-y^{2}$$
              so the top hemisphere is given by
              $$z=sqrt{50-x^{2}-y^{2}}=(50-x^{2}-y^{2})^{1/2}$$
              This is our $f(x,y)$. I'll point out that the lower hemisphere is given by $-f(x,y)$. Now, with this $f$, we have
              $$frac{partial f}{partial x}=frac{1}{2}(50-x^{2}-y^{2})^{-1/2}cdot(-2x)$$
              and
              $$frac{partial f}{partial y}=frac{1}{2}(50-x^{2}-y^{2})^{-1/2}cdot(-2y)$$



              Plugging in $x=3,y=4$ we have
              $$f_{x}(3,4)=-frac{3}{5}$$
              and
              $$f_{y}(3,4)=-frac{4}{5}$$
              Since $f(3,4)=5$, the tangent plane has equation
              $$z=5-frac{3}{5}(x-3)-frac{4}{5}(y-4)$$
              which, after some algebra, becomes
              $$3x+4y+5z=50$$







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Jan 3 at 21:46









              pwerth

              1,747411




              1,747411























                  1














                  $f(x,y,z)=x^2+y^2+z^2-50=0;$



                  Normal vector to surface $f(x,y,z)=0$ is given by



                  $vec n= nabla f(x,y,z)=(f_x,f_y,f_z)=(2x,2y,2z)$



                  Equation.of tangent plane:



                  $vec n cdot (vec r -vec r_0)=0$;



                  With $vec n= 2(3,4,5)$ and $vec r_0=(3,4,5)$:



                  $2(3,4,5)cdot ((x,y,z)-(3,4,5))=0;$



                  $3x+4y+5z -9-16-25=0;$



                  $3x+4y+5z =50.$






                  share|cite|improve this answer




























                    1














                    $f(x,y,z)=x^2+y^2+z^2-50=0;$



                    Normal vector to surface $f(x,y,z)=0$ is given by



                    $vec n= nabla f(x,y,z)=(f_x,f_y,f_z)=(2x,2y,2z)$



                    Equation.of tangent plane:



                    $vec n cdot (vec r -vec r_0)=0$;



                    With $vec n= 2(3,4,5)$ and $vec r_0=(3,4,5)$:



                    $2(3,4,5)cdot ((x,y,z)-(3,4,5))=0;$



                    $3x+4y+5z -9-16-25=0;$



                    $3x+4y+5z =50.$






                    share|cite|improve this answer


























                      1












                      1








                      1






                      $f(x,y,z)=x^2+y^2+z^2-50=0;$



                      Normal vector to surface $f(x,y,z)=0$ is given by



                      $vec n= nabla f(x,y,z)=(f_x,f_y,f_z)=(2x,2y,2z)$



                      Equation.of tangent plane:



                      $vec n cdot (vec r -vec r_0)=0$;



                      With $vec n= 2(3,4,5)$ and $vec r_0=(3,4,5)$:



                      $2(3,4,5)cdot ((x,y,z)-(3,4,5))=0;$



                      $3x+4y+5z -9-16-25=0;$



                      $3x+4y+5z =50.$






                      share|cite|improve this answer














                      $f(x,y,z)=x^2+y^2+z^2-50=0;$



                      Normal vector to surface $f(x,y,z)=0$ is given by



                      $vec n= nabla f(x,y,z)=(f_x,f_y,f_z)=(2x,2y,2z)$



                      Equation.of tangent plane:



                      $vec n cdot (vec r -vec r_0)=0$;



                      With $vec n= 2(3,4,5)$ and $vec r_0=(3,4,5)$:



                      $2(3,4,5)cdot ((x,y,z)-(3,4,5))=0;$



                      $3x+4y+5z -9-16-25=0;$



                      $3x+4y+5z =50.$







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Jan 3 at 22:13

























                      answered Jan 3 at 22:07









                      Peter Szilas

                      10.7k2720




                      10.7k2720























                          1














                          In order to calculate the tangent plane at $P = (3,4,5)$, we can consider the parametrization:



                          begin{align*}
                          varphi(x,y) = left(x,y,sqrt{50 - x^{2} - y^{2}}right)
                          end{align*}



                          Consequently, the tangent plane $pi$ at the given point can be described by



                          begin{align*}
                          (x,y,z) = P + alphavarphi_{x}(3,4) + betavarphi_{y}(3,4)
                          end{align*}



                          where $(alpha,beta)inmathbb{R}^{2}$.






                          share|cite|improve this answer




























                            1














                            In order to calculate the tangent plane at $P = (3,4,5)$, we can consider the parametrization:



                            begin{align*}
                            varphi(x,y) = left(x,y,sqrt{50 - x^{2} - y^{2}}right)
                            end{align*}



                            Consequently, the tangent plane $pi$ at the given point can be described by



                            begin{align*}
                            (x,y,z) = P + alphavarphi_{x}(3,4) + betavarphi_{y}(3,4)
                            end{align*}



                            where $(alpha,beta)inmathbb{R}^{2}$.






                            share|cite|improve this answer


























                              1












                              1








                              1






                              In order to calculate the tangent plane at $P = (3,4,5)$, we can consider the parametrization:



                              begin{align*}
                              varphi(x,y) = left(x,y,sqrt{50 - x^{2} - y^{2}}right)
                              end{align*}



                              Consequently, the tangent plane $pi$ at the given point can be described by



                              begin{align*}
                              (x,y,z) = P + alphavarphi_{x}(3,4) + betavarphi_{y}(3,4)
                              end{align*}



                              where $(alpha,beta)inmathbb{R}^{2}$.






                              share|cite|improve this answer














                              In order to calculate the tangent plane at $P = (3,4,5)$, we can consider the parametrization:



                              begin{align*}
                              varphi(x,y) = left(x,y,sqrt{50 - x^{2} - y^{2}}right)
                              end{align*}



                              Consequently, the tangent plane $pi$ at the given point can be described by



                              begin{align*}
                              (x,y,z) = P + alphavarphi_{x}(3,4) + betavarphi_{y}(3,4)
                              end{align*}



                              where $(alpha,beta)inmathbb{R}^{2}$.







                              share|cite|improve this answer














                              share|cite|improve this answer



                              share|cite|improve this answer








                              edited Jan 4 at 0:26

























                              answered Jan 3 at 21:42









                              APC89

                              1,950418




                              1,950418























                                  1














                                  Why not forget Calculus and look geometrically?



                                  Having observed that the point $P=(3,4,5)$ is indeed on the sphere, one sees that the segment $overline{Bbb OP}$ is a radius, where $Bbb O$ is the origin, center of the sphere. Now certainly any radius is perpendicular to the tangent plane at the point in question, so that the vector $(3,4,5)$ is orthogonal to the plane, and thus the equation must be $3X+4Y+5Z=c$, for an appropriate $c$, which you find to be $c=50$, in order to guarantee that $P$ be on the plane.






                                  share|cite|improve this answer


























                                    1














                                    Why not forget Calculus and look geometrically?



                                    Having observed that the point $P=(3,4,5)$ is indeed on the sphere, one sees that the segment $overline{Bbb OP}$ is a radius, where $Bbb O$ is the origin, center of the sphere. Now certainly any radius is perpendicular to the tangent plane at the point in question, so that the vector $(3,4,5)$ is orthogonal to the plane, and thus the equation must be $3X+4Y+5Z=c$, for an appropriate $c$, which you find to be $c=50$, in order to guarantee that $P$ be on the plane.






                                    share|cite|improve this answer
























                                      1












                                      1








                                      1






                                      Why not forget Calculus and look geometrically?



                                      Having observed that the point $P=(3,4,5)$ is indeed on the sphere, one sees that the segment $overline{Bbb OP}$ is a radius, where $Bbb O$ is the origin, center of the sphere. Now certainly any radius is perpendicular to the tangent plane at the point in question, so that the vector $(3,4,5)$ is orthogonal to the plane, and thus the equation must be $3X+4Y+5Z=c$, for an appropriate $c$, which you find to be $c=50$, in order to guarantee that $P$ be on the plane.






                                      share|cite|improve this answer












                                      Why not forget Calculus and look geometrically?



                                      Having observed that the point $P=(3,4,5)$ is indeed on the sphere, one sees that the segment $overline{Bbb OP}$ is a radius, where $Bbb O$ is the origin, center of the sphere. Now certainly any radius is perpendicular to the tangent plane at the point in question, so that the vector $(3,4,5)$ is orthogonal to the plane, and thus the equation must be $3X+4Y+5Z=c$, for an appropriate $c$, which you find to be $c=50$, in order to guarantee that $P$ be on the plane.







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Jan 4 at 1:00









                                      Lubin

                                      43.8k44585




                                      43.8k44585






























                                          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.





                                          Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                                          Please pay close attention to the following guidance:


                                          • 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.


                                          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%2f3061024%2ftangent-plane-to-x2y2z2-50-at-3-4-5%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

                                          1300-talet

                                          1300-talet

                                          Display a custom attribute below product name in the front-end Magento 1.9.3.8