How to study the critical points of a $2$-variable function?












0












$begingroup$


I am revising some past exam questions and there is one that states:




Study the critical points of the function:
$$f(x,y)=x^2+y^2-x^4-y^4-2x^2y^2.$$




According to my professor, this is what I have to do:




  1. Find the gradient, which is $$nabla f(x,y)=begin{bmatrix} 2x-4x^3-4xy^2 \ 2y-4y^3-4x^2y end{bmatrix}.$$


  2. Find the Hessian, using partial derivatives, which is
    begin{align*}
    f_{xx}(x,y) &= 2-12x^2-4y \
    f_{xy}(x,y) &= -8xy \
    f_{yx}(x,y) &= -8xy \
    f_{yy}(x,y) &= 2-12y^2-4x^2.
    end{align*}



And until here it's all good. However, what confuses me is the next step.



We need to search at first for $Cf$, the set of critical points of $f(x,y)$, so we have to impose $nabla f(x,y) = 0$, namely
$$2x(1-2x^2-2y^2)=0 quad text{and} quad 2y(1-2x^2-2y^2)=0.$$
And here I'm lost. On the answer sheet it states:
$$C*f*={(x,y)inmathbb{R}^2 : x^2+y^2=1/2} cup {(0,0)}.$$
What exactly does this property above mean? Where is the $x^2+y^2=1/2$ coming from and what does $cup {(0,0)}$ mean? Is $cup {(0,0)}$ the critical points, which are $0$?



Please explain in the clearest and simpelst way possible!



Note: this is the first part of the full question that I am interested in. As soon as I understand this, I will proceed with answering the full question.



Thanks, guys!






multivariable-calculus derivatives maxima-minima hessian-matrix







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    I am revising some past exam questions and there is one that states:




    Study the critical points of the function:
    $$f(x,y)=x^2+y^2-x^4-y^4-2x^2y^2.$$




    According to my professor, this is what I have to do:




    1. Find the gradient, which is $$nabla f(x,y)=begin{bmatrix} 2x-4x^3-4xy^2 \ 2y-4y^3-4x^2y end{bmatrix}.$$


    2. Find the Hessian, using partial derivatives, which is
      begin{align*}
      f_{xx}(x,y) &= 2-12x^2-4y \
      f_{xy}(x,y) &= -8xy \
      f_{yx}(x,y) &= -8xy \
      f_{yy}(x,y) &= 2-12y^2-4x^2.
      end{align*}



    And until here it's all good. However, what confuses me is the next step.



    We need to search at first for $Cf$, the set of critical points of $f(x,y)$, so we have to impose $nabla f(x,y) = 0$, namely
    $$2x(1-2x^2-2y^2)=0 quad text{and} quad 2y(1-2x^2-2y^2)=0.$$
    And here I'm lost. On the answer sheet it states:
    $$C*f*={(x,y)inmathbb{R}^2 : x^2+y^2=1/2} cup {(0,0)}.$$
    What exactly does this property above mean? Where is the $x^2+y^2=1/2$ coming from and what does $cup {(0,0)}$ mean? Is $cup {(0,0)}$ the critical points, which are $0$?



    Please explain in the clearest and simpelst way possible!



    Note: this is the first part of the full question that I am interested in. As soon as I understand this, I will proceed with answering the full question.



    Thanks, guys!






    multivariable-calculus derivatives maxima-minima hessian-matrix







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      I am revising some past exam questions and there is one that states:




      Study the critical points of the function:
      $$f(x,y)=x^2+y^2-x^4-y^4-2x^2y^2.$$




      According to my professor, this is what I have to do:




      1. Find the gradient, which is $$nabla f(x,y)=begin{bmatrix} 2x-4x^3-4xy^2 \ 2y-4y^3-4x^2y end{bmatrix}.$$


      2. Find the Hessian, using partial derivatives, which is
        begin{align*}
        f_{xx}(x,y) &= 2-12x^2-4y \
        f_{xy}(x,y) &= -8xy \
        f_{yx}(x,y) &= -8xy \
        f_{yy}(x,y) &= 2-12y^2-4x^2.
        end{align*}



      And until here it's all good. However, what confuses me is the next step.



      We need to search at first for $Cf$, the set of critical points of $f(x,y)$, so we have to impose $nabla f(x,y) = 0$, namely
      $$2x(1-2x^2-2y^2)=0 quad text{and} quad 2y(1-2x^2-2y^2)=0.$$
      And here I'm lost. On the answer sheet it states:
      $$C*f*={(x,y)inmathbb{R}^2 : x^2+y^2=1/2} cup {(0,0)}.$$
      What exactly does this property above mean? Where is the $x^2+y^2=1/2$ coming from and what does $cup {(0,0)}$ mean? Is $cup {(0,0)}$ the critical points, which are $0$?



      Please explain in the clearest and simpelst way possible!



      Note: this is the first part of the full question that I am interested in. As soon as I understand this, I will proceed with answering the full question.



      Thanks, guys!






      multivariable-calculus derivatives maxima-minima hessian-matrix







      share|cite|improve this question











      $endgroup$




      I am revising some past exam questions and there is one that states:




      Study the critical points of the function:
      $$f(x,y)=x^2+y^2-x^4-y^4-2x^2y^2.$$




      According to my professor, this is what I have to do:




      1. Find the gradient, which is $$nabla f(x,y)=begin{bmatrix} 2x-4x^3-4xy^2 \ 2y-4y^3-4x^2y end{bmatrix}.$$


      2. Find the Hessian, using partial derivatives, which is
        begin{align*}
        f_{xx}(x,y) &= 2-12x^2-4y \
        f_{xy}(x,y) &= -8xy \
        f_{yx}(x,y) &= -8xy \
        f_{yy}(x,y) &= 2-12y^2-4x^2.
        end{align*}



      And until here it's all good. However, what confuses me is the next step.



      We need to search at first for $Cf$, the set of critical points of $f(x,y)$, so we have to impose $nabla f(x,y) = 0$, namely
      $$2x(1-2x^2-2y^2)=0 quad text{and} quad 2y(1-2x^2-2y^2)=0.$$
      And here I'm lost. On the answer sheet it states:
      $$C*f*={(x,y)inmathbb{R}^2 : x^2+y^2=1/2} cup {(0,0)}.$$
      What exactly does this property above mean? Where is the $x^2+y^2=1/2$ coming from and what does $cup {(0,0)}$ mean? Is $cup {(0,0)}$ the critical points, which are $0$?



      Please explain in the clearest and simpelst way possible!



      Note: this is the first part of the full question that I am interested in. As soon as I understand this, I will proceed with answering the full question.



      Thanks, guys!






      multivariable-calculus derivatives maxima-minima hessian-matrix




      multivariable-calculus derivatives maxima-minima hessian-matrix






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 8 '18 at 14:05









      Rócherz

      2,7762721




      2,7762721










      asked Dec 8 '18 at 13:55









      BM97BM97

      758




      758






















          1 Answer
          1






          active

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          1












          $begingroup$

          $cup$ means union.



          From $$2x(1-2x^2-2y^2)=0 iff 2x=0 text{ or } 1-2x^2-2y^2=0$$
          $$2y(1-2x^2-2y^2)=0 iff 2y=0 text{ or } 1-2x^2-2y^2=0$$



          It is possible that $1-2x^2-2y^2=0 iff x^2+y^2 = frac12$.



          Suppose not, then we must have $x=y=0$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks for the explanation ! So this means that I the U(0,0) is coming from the 2x=0 and the 2y=0, and that the x^2+y^2=1/2 is coming from the 1-2x^2-2y^2, just by re-arranging the equation ? I'm sorry for the too many questions, I have dyscalculia and I need to get even the simplest things straight!
            $endgroup$
            – BM97
            Dec 8 '18 at 14:10










          • $begingroup$
            Yes, they come from those places that you mentioned.
            $endgroup$
            – Siong Thye Goh
            Dec 8 '18 at 14:12










          • $begingroup$
            Thank you !!! You are great!
            $endgroup$
            – BM97
            Dec 8 '18 at 14:13










          • $begingroup$
            May i ask you one last thing? What we wrote before implies that that the set of critical points of f(x, y) is given by the origin and the circle or radius √1/2 centered at the origin. When it says to study the hessian at the origin, the matrix Hf(0,0) is a 2x2 matrix , with on top 2,0 and on botthom 0,2.... may i ask you where the 2's are coming from or is it a general formula?
            $endgroup$
            – BM97
            Dec 8 '18 at 14:41










          • $begingroup$
            The top $2$ at the bottom corresponds to computing $f_{xx}$ and evaluate it at $(0,0)$. The bottom $2$ corresponds to computing $f_{yy}$ and evaluate it at $(0,0)$.
            $endgroup$
            – Siong Thye Goh
            Dec 8 '18 at 14:43











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          1 Answer
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          1 Answer
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          active

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          active

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          1












          $begingroup$

          $cup$ means union.



          From $$2x(1-2x^2-2y^2)=0 iff 2x=0 text{ or } 1-2x^2-2y^2=0$$
          $$2y(1-2x^2-2y^2)=0 iff 2y=0 text{ or } 1-2x^2-2y^2=0$$



          It is possible that $1-2x^2-2y^2=0 iff x^2+y^2 = frac12$.



          Suppose not, then we must have $x=y=0$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks for the explanation ! So this means that I the U(0,0) is coming from the 2x=0 and the 2y=0, and that the x^2+y^2=1/2 is coming from the 1-2x^2-2y^2, just by re-arranging the equation ? I'm sorry for the too many questions, I have dyscalculia and I need to get even the simplest things straight!
            $endgroup$
            – BM97
            Dec 8 '18 at 14:10










          • $begingroup$
            Yes, they come from those places that you mentioned.
            $endgroup$
            – Siong Thye Goh
            Dec 8 '18 at 14:12










          • $begingroup$
            Thank you !!! You are great!
            $endgroup$
            – BM97
            Dec 8 '18 at 14:13










          • $begingroup$
            May i ask you one last thing? What we wrote before implies that that the set of critical points of f(x, y) is given by the origin and the circle or radius √1/2 centered at the origin. When it says to study the hessian at the origin, the matrix Hf(0,0) is a 2x2 matrix , with on top 2,0 and on botthom 0,2.... may i ask you where the 2's are coming from or is it a general formula?
            $endgroup$
            – BM97
            Dec 8 '18 at 14:41










          • $begingroup$
            The top $2$ at the bottom corresponds to computing $f_{xx}$ and evaluate it at $(0,0)$. The bottom $2$ corresponds to computing $f_{yy}$ and evaluate it at $(0,0)$.
            $endgroup$
            – Siong Thye Goh
            Dec 8 '18 at 14:43
















          1












          $begingroup$

          $cup$ means union.



          From $$2x(1-2x^2-2y^2)=0 iff 2x=0 text{ or } 1-2x^2-2y^2=0$$
          $$2y(1-2x^2-2y^2)=0 iff 2y=0 text{ or } 1-2x^2-2y^2=0$$



          It is possible that $1-2x^2-2y^2=0 iff x^2+y^2 = frac12$.



          Suppose not, then we must have $x=y=0$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks for the explanation ! So this means that I the U(0,0) is coming from the 2x=0 and the 2y=0, and that the x^2+y^2=1/2 is coming from the 1-2x^2-2y^2, just by re-arranging the equation ? I'm sorry for the too many questions, I have dyscalculia and I need to get even the simplest things straight!
            $endgroup$
            – BM97
            Dec 8 '18 at 14:10










          • $begingroup$
            Yes, they come from those places that you mentioned.
            $endgroup$
            – Siong Thye Goh
            Dec 8 '18 at 14:12










          • $begingroup$
            Thank you !!! You are great!
            $endgroup$
            – BM97
            Dec 8 '18 at 14:13










          • $begingroup$
            May i ask you one last thing? What we wrote before implies that that the set of critical points of f(x, y) is given by the origin and the circle or radius √1/2 centered at the origin. When it says to study the hessian at the origin, the matrix Hf(0,0) is a 2x2 matrix , with on top 2,0 and on botthom 0,2.... may i ask you where the 2's are coming from or is it a general formula?
            $endgroup$
            – BM97
            Dec 8 '18 at 14:41










          • $begingroup$
            The top $2$ at the bottom corresponds to computing $f_{xx}$ and evaluate it at $(0,0)$. The bottom $2$ corresponds to computing $f_{yy}$ and evaluate it at $(0,0)$.
            $endgroup$
            – Siong Thye Goh
            Dec 8 '18 at 14:43














          1












          1








          1





          $begingroup$

          $cup$ means union.



          From $$2x(1-2x^2-2y^2)=0 iff 2x=0 text{ or } 1-2x^2-2y^2=0$$
          $$2y(1-2x^2-2y^2)=0 iff 2y=0 text{ or } 1-2x^2-2y^2=0$$



          It is possible that $1-2x^2-2y^2=0 iff x^2+y^2 = frac12$.



          Suppose not, then we must have $x=y=0$.






          share|cite|improve this answer









          $endgroup$



          $cup$ means union.



          From $$2x(1-2x^2-2y^2)=0 iff 2x=0 text{ or } 1-2x^2-2y^2=0$$
          $$2y(1-2x^2-2y^2)=0 iff 2y=0 text{ or } 1-2x^2-2y^2=0$$



          It is possible that $1-2x^2-2y^2=0 iff x^2+y^2 = frac12$.



          Suppose not, then we must have $x=y=0$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 8 '18 at 13:59









          Siong Thye GohSiong Thye Goh

          100k1466117




          100k1466117












          • $begingroup$
            Thanks for the explanation ! So this means that I the U(0,0) is coming from the 2x=0 and the 2y=0, and that the x^2+y^2=1/2 is coming from the 1-2x^2-2y^2, just by re-arranging the equation ? I'm sorry for the too many questions, I have dyscalculia and I need to get even the simplest things straight!
            $endgroup$
            – BM97
            Dec 8 '18 at 14:10










          • $begingroup$
            Yes, they come from those places that you mentioned.
            $endgroup$
            – Siong Thye Goh
            Dec 8 '18 at 14:12










          • $begingroup$
            Thank you !!! You are great!
            $endgroup$
            – BM97
            Dec 8 '18 at 14:13










          • $begingroup$
            May i ask you one last thing? What we wrote before implies that that the set of critical points of f(x, y) is given by the origin and the circle or radius √1/2 centered at the origin. When it says to study the hessian at the origin, the matrix Hf(0,0) is a 2x2 matrix , with on top 2,0 and on botthom 0,2.... may i ask you where the 2's are coming from or is it a general formula?
            $endgroup$
            – BM97
            Dec 8 '18 at 14:41










          • $begingroup$
            The top $2$ at the bottom corresponds to computing $f_{xx}$ and evaluate it at $(0,0)$. The bottom $2$ corresponds to computing $f_{yy}$ and evaluate it at $(0,0)$.
            $endgroup$
            – Siong Thye Goh
            Dec 8 '18 at 14:43


















          • $begingroup$
            Thanks for the explanation ! So this means that I the U(0,0) is coming from the 2x=0 and the 2y=0, and that the x^2+y^2=1/2 is coming from the 1-2x^2-2y^2, just by re-arranging the equation ? I'm sorry for the too many questions, I have dyscalculia and I need to get even the simplest things straight!
            $endgroup$
            – BM97
            Dec 8 '18 at 14:10










          • $begingroup$
            Yes, they come from those places that you mentioned.
            $endgroup$
            – Siong Thye Goh
            Dec 8 '18 at 14:12










          • $begingroup$
            Thank you !!! You are great!
            $endgroup$
            – BM97
            Dec 8 '18 at 14:13










          • $begingroup$
            May i ask you one last thing? What we wrote before implies that that the set of critical points of f(x, y) is given by the origin and the circle or radius √1/2 centered at the origin. When it says to study the hessian at the origin, the matrix Hf(0,0) is a 2x2 matrix , with on top 2,0 and on botthom 0,2.... may i ask you where the 2's are coming from or is it a general formula?
            $endgroup$
            – BM97
            Dec 8 '18 at 14:41










          • $begingroup$
            The top $2$ at the bottom corresponds to computing $f_{xx}$ and evaluate it at $(0,0)$. The bottom $2$ corresponds to computing $f_{yy}$ and evaluate it at $(0,0)$.
            $endgroup$
            – Siong Thye Goh
            Dec 8 '18 at 14:43
















          $begingroup$
          Thanks for the explanation ! So this means that I the U(0,0) is coming from the 2x=0 and the 2y=0, and that the x^2+y^2=1/2 is coming from the 1-2x^2-2y^2, just by re-arranging the equation ? I'm sorry for the too many questions, I have dyscalculia and I need to get even the simplest things straight!
          $endgroup$
          – BM97
          Dec 8 '18 at 14:10




          $begingroup$
          Thanks for the explanation ! So this means that I the U(0,0) is coming from the 2x=0 and the 2y=0, and that the x^2+y^2=1/2 is coming from the 1-2x^2-2y^2, just by re-arranging the equation ? I'm sorry for the too many questions, I have dyscalculia and I need to get even the simplest things straight!
          $endgroup$
          – BM97
          Dec 8 '18 at 14:10












          $begingroup$
          Yes, they come from those places that you mentioned.
          $endgroup$
          – Siong Thye Goh
          Dec 8 '18 at 14:12




          $begingroup$
          Yes, they come from those places that you mentioned.
          $endgroup$
          – Siong Thye Goh
          Dec 8 '18 at 14:12












          $begingroup$
          Thank you !!! You are great!
          $endgroup$
          – BM97
          Dec 8 '18 at 14:13




          $begingroup$
          Thank you !!! You are great!
          $endgroup$
          – BM97
          Dec 8 '18 at 14:13












          $begingroup$
          May i ask you one last thing? What we wrote before implies that that the set of critical points of f(x, y) is given by the origin and the circle or radius √1/2 centered at the origin. When it says to study the hessian at the origin, the matrix Hf(0,0) is a 2x2 matrix , with on top 2,0 and on botthom 0,2.... may i ask you where the 2's are coming from or is it a general formula?
          $endgroup$
          – BM97
          Dec 8 '18 at 14:41




          $begingroup$
          May i ask you one last thing? What we wrote before implies that that the set of critical points of f(x, y) is given by the origin and the circle or radius √1/2 centered at the origin. When it says to study the hessian at the origin, the matrix Hf(0,0) is a 2x2 matrix , with on top 2,0 and on botthom 0,2.... may i ask you where the 2's are coming from or is it a general formula?
          $endgroup$
          – BM97
          Dec 8 '18 at 14:41












          $begingroup$
          The top $2$ at the bottom corresponds to computing $f_{xx}$ and evaluate it at $(0,0)$. The bottom $2$ corresponds to computing $f_{yy}$ and evaluate it at $(0,0)$.
          $endgroup$
          – Siong Thye Goh
          Dec 8 '18 at 14:43




          $begingroup$
          The top $2$ at the bottom corresponds to computing $f_{xx}$ and evaluate it at $(0,0)$. The bottom $2$ corresponds to computing $f_{yy}$ and evaluate it at $(0,0)$.
          $endgroup$
          – Siong Thye Goh
          Dec 8 '18 at 14:43


















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          Marschland

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          Kilometerweite Marsch in den Vier- und Marschlanden am Elbdeich Die Wedeler Marschlandschaft in Schleswig-Holstein Der Hadelner Kanal zur Entwässerung der Marsch Die tief gelegene Marsch und ein Entwässerungskanal Das Schöpfwerk Otterndorf im Land Hadeln zur Entwässerung der Medem verfügte damals bereits über die größte Kreiselpumpe Europas. Das Schöpfwerk in Neuhaus für die Aue und des Neuhaus-Bülkauer Kanals Als Marsch(land) (v. niederdt., altsächs. mersc ) – auch Masch , Mersch oder Schwemmland genannt – bezeichnet man eine nacheiszeitlich entstandene geomorphologische Landform im Gebiet der nordwestdeutschen Küsten und Flüsse sowie vergleichbare Landformen weltweit. Inhaltsverzeichnis 1 Landschaft 2 Entstehung 3 Marschböden 4 Nutzung 5 Marschgebiete in Deutschland 6 Siehe auch 7 Literatur 8 Weblinks 9 Einzelnachweise Landschaft | Marschen sind generell flache Landstriche ohne natürliche Erhebungen. Si
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          Dieringhausen

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          Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
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