Plotting contours of the paraboloid $f(x_1, x_2) = 4x_1^2+x_2^2-4x_1x_2-4x_1+2x_2+1$












1












$begingroup$



I have the following function:
$$f(x_1, x_2) = 4x_1^2+x_2^2-4x_1x_2-4x_1+2x_2+1$$
and I need to plot its contours.




My Attempt



I recognized this is a conic section given in its Cartesian form:
$$Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F=0$$
Since $B^2-4AC=16 - 4times 4 = 0$ we know this is a parabola.
Looking at the General Case of a parabola, I rewrite the contours in this way:



$$4x_1^2+x_2^2-4x_1x_2-4x_1+2x_2+1=C quad quad text{for } Cinmathbb{R} $$
$$x_2^2+2x_2(1-2x_1) + 4x_1^2-4x_1+1 - C = 0$$
$$(x_2 + (1-2x_1))^2 - C = 0$$
This would mean that
$$x_2 + 1 - 2x_1 = 0$$
is the directrix of the parabola, i.e. $x_2 = 2x_1 - 1$. Here's a plot of that:
contour



However, I have no idea how to proceed from here. I am not even sure that what I've done makes sense, to be honest.






calculus geometry multivariable-calculus







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Well, I would first ask Wolfram Alpha: wolframalpha.com/input/…
    $endgroup$
    – Matti P.
    Dec 31 '18 at 11:08






  • 1




    $begingroup$
    Wolfram Alpha tells that (replacing $x_1 = y$ and $x_2=y$) the equation is equal to $$ f = (2x-y-1)^2 $$ So therefore, if you set $f=text{constant} Rightarrow (2x-y-1) = sqrt{text{constant}} = text{constant}$ ...
    $endgroup$
    – Matti P.
    Dec 31 '18 at 11:09


















1












$begingroup$



I have the following function:
$$f(x_1, x_2) = 4x_1^2+x_2^2-4x_1x_2-4x_1+2x_2+1$$
and I need to plot its contours.




My Attempt



I recognized this is a conic section given in its Cartesian form:
$$Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F=0$$
Since $B^2-4AC=16 - 4times 4 = 0$ we know this is a parabola.
Looking at the General Case of a parabola, I rewrite the contours in this way:



$$4x_1^2+x_2^2-4x_1x_2-4x_1+2x_2+1=C quad quad text{for } Cinmathbb{R} $$
$$x_2^2+2x_2(1-2x_1) + 4x_1^2-4x_1+1 - C = 0$$
$$(x_2 + (1-2x_1))^2 - C = 0$$
This would mean that
$$x_2 + 1 - 2x_1 = 0$$
is the directrix of the parabola, i.e. $x_2 = 2x_1 - 1$. Here's a plot of that:
contour



However, I have no idea how to proceed from here. I am not even sure that what I've done makes sense, to be honest.






calculus geometry multivariable-calculus







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Well, I would first ask Wolfram Alpha: wolframalpha.com/input/…
    $endgroup$
    – Matti P.
    Dec 31 '18 at 11:08






  • 1




    $begingroup$
    Wolfram Alpha tells that (replacing $x_1 = y$ and $x_2=y$) the equation is equal to $$ f = (2x-y-1)^2 $$ So therefore, if you set $f=text{constant} Rightarrow (2x-y-1) = sqrt{text{constant}} = text{constant}$ ...
    $endgroup$
    – Matti P.
    Dec 31 '18 at 11:09
















1












1








1





$begingroup$



I have the following function:
$$f(x_1, x_2) = 4x_1^2+x_2^2-4x_1x_2-4x_1+2x_2+1$$
and I need to plot its contours.




My Attempt



I recognized this is a conic section given in its Cartesian form:
$$Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F=0$$
Since $B^2-4AC=16 - 4times 4 = 0$ we know this is a parabola.
Looking at the General Case of a parabola, I rewrite the contours in this way:



$$4x_1^2+x_2^2-4x_1x_2-4x_1+2x_2+1=C quad quad text{for } Cinmathbb{R} $$
$$x_2^2+2x_2(1-2x_1) + 4x_1^2-4x_1+1 - C = 0$$
$$(x_2 + (1-2x_1))^2 - C = 0$$
This would mean that
$$x_2 + 1 - 2x_1 = 0$$
is the directrix of the parabola, i.e. $x_2 = 2x_1 - 1$. Here's a plot of that:
contour



However, I have no idea how to proceed from here. I am not even sure that what I've done makes sense, to be honest.






calculus geometry multivariable-calculus







share|cite|improve this question











$endgroup$





I have the following function:
$$f(x_1, x_2) = 4x_1^2+x_2^2-4x_1x_2-4x_1+2x_2+1$$
and I need to plot its contours.




My Attempt



I recognized this is a conic section given in its Cartesian form:
$$Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F=0$$
Since $B^2-4AC=16 - 4times 4 = 0$ we know this is a parabola.
Looking at the General Case of a parabola, I rewrite the contours in this way:



$$4x_1^2+x_2^2-4x_1x_2-4x_1+2x_2+1=C quad quad text{for } Cinmathbb{R} $$
$$x_2^2+2x_2(1-2x_1) + 4x_1^2-4x_1+1 - C = 0$$
$$(x_2 + (1-2x_1))^2 - C = 0$$
This would mean that
$$x_2 + 1 - 2x_1 = 0$$
is the directrix of the parabola, i.e. $x_2 = 2x_1 - 1$. Here's a plot of that:
contour



However, I have no idea how to proceed from here. I am not even sure that what I've done makes sense, to be honest.






calculus geometry multivariable-calculus




calculus geometry multivariable-calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 31 '18 at 16:54









greedoid

45k1157112




45k1157112










asked Dec 31 '18 at 10:57









Euler_SalterEuler_Salter

2,0671336




2,0671336








  • 1




    $begingroup$
    Well, I would first ask Wolfram Alpha: wolframalpha.com/input/…
    $endgroup$
    – Matti P.
    Dec 31 '18 at 11:08






  • 1




    $begingroup$
    Wolfram Alpha tells that (replacing $x_1 = y$ and $x_2=y$) the equation is equal to $$ f = (2x-y-1)^2 $$ So therefore, if you set $f=text{constant} Rightarrow (2x-y-1) = sqrt{text{constant}} = text{constant}$ ...
    $endgroup$
    – Matti P.
    Dec 31 '18 at 11:09
















  • 1




    $begingroup$
    Well, I would first ask Wolfram Alpha: wolframalpha.com/input/…
    $endgroup$
    – Matti P.
    Dec 31 '18 at 11:08






  • 1




    $begingroup$
    Wolfram Alpha tells that (replacing $x_1 = y$ and $x_2=y$) the equation is equal to $$ f = (2x-y-1)^2 $$ So therefore, if you set $f=text{constant} Rightarrow (2x-y-1) = sqrt{text{constant}} = text{constant}$ ...
    $endgroup$
    – Matti P.
    Dec 31 '18 at 11:09










1




1




$begingroup$
Well, I would first ask Wolfram Alpha: wolframalpha.com/input/…
$endgroup$
– Matti P.
Dec 31 '18 at 11:08




$begingroup$
Well, I would first ask Wolfram Alpha: wolframalpha.com/input/…
$endgroup$
– Matti P.
Dec 31 '18 at 11:08




1




1




$begingroup$
Wolfram Alpha tells that (replacing $x_1 = y$ and $x_2=y$) the equation is equal to $$ f = (2x-y-1)^2 $$ So therefore, if you set $f=text{constant} Rightarrow (2x-y-1) = sqrt{text{constant}} = text{constant}$ ...
$endgroup$
– Matti P.
Dec 31 '18 at 11:09






$begingroup$
Wolfram Alpha tells that (replacing $x_1 = y$ and $x_2=y$) the equation is equal to $$ f = (2x-y-1)^2 $$ So therefore, if you set $f=text{constant} Rightarrow (2x-y-1) = sqrt{text{constant}} = text{constant}$ ...
$endgroup$
– Matti P.
Dec 31 '18 at 11:09












2 Answers
2






active

oldest

votes


















2












$begingroup$

$(x_2 + (1-2x_1))^2 - C = 0$ is correct and makes sense. From here, if $C>0$ we get two lines taking along them the same value $C$ for $f$, $r_1:x_2 + (1-2x_1)=sqrt{C}$ and $r_2:x_2 + (1-2x_1)=-sqrt{C}$ When $C=0$ there is only one line and if $C<0$ there are no solutions as $f$ does not reach negative values. Cutting the surface with a plane perpendicular to these lines $2x_2+x_1=K$ ($z$ any) we get the profile of a parabola.



The plot must consist mostly in two parallel lines.






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    Besides the discriminant, you also have to examine the full $3times3$ determinant of the matrix of this family of conics. It turns out to vanish identically, so the level curves are all degenerate: they are pairs of parallel (or coincident) lines.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I think you can show that the 3 by 3 determinant is just -4 times the discriminant, so that if the discriminant is zero, also the determinant is zero
      $endgroup$
      – Euler_Salter
      Jan 1 at 13:35










    • $begingroup$
      @Euler_Salter That’s not at all true for a general conic section. The discriminant only involves the quadratic part of the equation—it’s a multiple of the principal $2times2$ minor of the matrix. Compare the discriminants and full determinants of $y^2=x$ or $x^2=y^2$, for instance. For the former, the discriminant vanishes, but the $3times3$ determinant doesn’t, and vice-versa for the latter.
      $endgroup$
      – amd
      Jan 1 at 20:22













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

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    2












    $begingroup$

    $(x_2 + (1-2x_1))^2 - C = 0$ is correct and makes sense. From here, if $C>0$ we get two lines taking along them the same value $C$ for $f$, $r_1:x_2 + (1-2x_1)=sqrt{C}$ and $r_2:x_2 + (1-2x_1)=-sqrt{C}$ When $C=0$ there is only one line and if $C<0$ there are no solutions as $f$ does not reach negative values. Cutting the surface with a plane perpendicular to these lines $2x_2+x_1=K$ ($z$ any) we get the profile of a parabola.



    The plot must consist mostly in two parallel lines.






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      $(x_2 + (1-2x_1))^2 - C = 0$ is correct and makes sense. From here, if $C>0$ we get two lines taking along them the same value $C$ for $f$, $r_1:x_2 + (1-2x_1)=sqrt{C}$ and $r_2:x_2 + (1-2x_1)=-sqrt{C}$ When $C=0$ there is only one line and if $C<0$ there are no solutions as $f$ does not reach negative values. Cutting the surface with a plane perpendicular to these lines $2x_2+x_1=K$ ($z$ any) we get the profile of a parabola.



      The plot must consist mostly in two parallel lines.






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        $(x_2 + (1-2x_1))^2 - C = 0$ is correct and makes sense. From here, if $C>0$ we get two lines taking along them the same value $C$ for $f$, $r_1:x_2 + (1-2x_1)=sqrt{C}$ and $r_2:x_2 + (1-2x_1)=-sqrt{C}$ When $C=0$ there is only one line and if $C<0$ there are no solutions as $f$ does not reach negative values. Cutting the surface with a plane perpendicular to these lines $2x_2+x_1=K$ ($z$ any) we get the profile of a parabola.



        The plot must consist mostly in two parallel lines.






        share|cite|improve this answer











        $endgroup$



        $(x_2 + (1-2x_1))^2 - C = 0$ is correct and makes sense. From here, if $C>0$ we get two lines taking along them the same value $C$ for $f$, $r_1:x_2 + (1-2x_1)=sqrt{C}$ and $r_2:x_2 + (1-2x_1)=-sqrt{C}$ When $C=0$ there is only one line and if $C<0$ there are no solutions as $f$ does not reach negative values. Cutting the surface with a plane perpendicular to these lines $2x_2+x_1=K$ ($z$ any) we get the profile of a parabola.



        The plot must consist mostly in two parallel lines.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 31 '18 at 11:42

























        answered Dec 31 '18 at 11:36









        Rafa BudríaRafa Budría

        5,8151825




        5,8151825























            0












            $begingroup$

            Besides the discriminant, you also have to examine the full $3times3$ determinant of the matrix of this family of conics. It turns out to vanish identically, so the level curves are all degenerate: they are pairs of parallel (or coincident) lines.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I think you can show that the 3 by 3 determinant is just -4 times the discriminant, so that if the discriminant is zero, also the determinant is zero
              $endgroup$
              – Euler_Salter
              Jan 1 at 13:35










            • $begingroup$
              @Euler_Salter That’s not at all true for a general conic section. The discriminant only involves the quadratic part of the equation—it’s a multiple of the principal $2times2$ minor of the matrix. Compare the discriminants and full determinants of $y^2=x$ or $x^2=y^2$, for instance. For the former, the discriminant vanishes, but the $3times3$ determinant doesn’t, and vice-versa for the latter.
              $endgroup$
              – amd
              Jan 1 at 20:22


















            0












            $begingroup$

            Besides the discriminant, you also have to examine the full $3times3$ determinant of the matrix of this family of conics. It turns out to vanish identically, so the level curves are all degenerate: they are pairs of parallel (or coincident) lines.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I think you can show that the 3 by 3 determinant is just -4 times the discriminant, so that if the discriminant is zero, also the determinant is zero
              $endgroup$
              – Euler_Salter
              Jan 1 at 13:35










            • $begingroup$
              @Euler_Salter That’s not at all true for a general conic section. The discriminant only involves the quadratic part of the equation—it’s a multiple of the principal $2times2$ minor of the matrix. Compare the discriminants and full determinants of $y^2=x$ or $x^2=y^2$, for instance. For the former, the discriminant vanishes, but the $3times3$ determinant doesn’t, and vice-versa for the latter.
              $endgroup$
              – amd
              Jan 1 at 20:22
















            0












            0








            0





            $begingroup$

            Besides the discriminant, you also have to examine the full $3times3$ determinant of the matrix of this family of conics. It turns out to vanish identically, so the level curves are all degenerate: they are pairs of parallel (or coincident) lines.






            share|cite|improve this answer









            $endgroup$



            Besides the discriminant, you also have to examine the full $3times3$ determinant of the matrix of this family of conics. It turns out to vanish identically, so the level curves are all degenerate: they are pairs of parallel (or coincident) lines.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 1 at 0:53









            amdamd

            30.7k21050




            30.7k21050












            • $begingroup$
              I think you can show that the 3 by 3 determinant is just -4 times the discriminant, so that if the discriminant is zero, also the determinant is zero
              $endgroup$
              – Euler_Salter
              Jan 1 at 13:35










            • $begingroup$
              @Euler_Salter That’s not at all true for a general conic section. The discriminant only involves the quadratic part of the equation—it’s a multiple of the principal $2times2$ minor of the matrix. Compare the discriminants and full determinants of $y^2=x$ or $x^2=y^2$, for instance. For the former, the discriminant vanishes, but the $3times3$ determinant doesn’t, and vice-versa for the latter.
              $endgroup$
              – amd
              Jan 1 at 20:22




















            • $begingroup$
              I think you can show that the 3 by 3 determinant is just -4 times the discriminant, so that if the discriminant is zero, also the determinant is zero
              $endgroup$
              – Euler_Salter
              Jan 1 at 13:35










            • $begingroup$
              @Euler_Salter That’s not at all true for a general conic section. The discriminant only involves the quadratic part of the equation—it’s a multiple of the principal $2times2$ minor of the matrix. Compare the discriminants and full determinants of $y^2=x$ or $x^2=y^2$, for instance. For the former, the discriminant vanishes, but the $3times3$ determinant doesn’t, and vice-versa for the latter.
              $endgroup$
              – amd
              Jan 1 at 20:22


















            $begingroup$
            I think you can show that the 3 by 3 determinant is just -4 times the discriminant, so that if the discriminant is zero, also the determinant is zero
            $endgroup$
            – Euler_Salter
            Jan 1 at 13:35




            $begingroup$
            I think you can show that the 3 by 3 determinant is just -4 times the discriminant, so that if the discriminant is zero, also the determinant is zero
            $endgroup$
            – Euler_Salter
            Jan 1 at 13:35












            $begingroup$
            @Euler_Salter That’s not at all true for a general conic section. The discriminant only involves the quadratic part of the equation—it’s a multiple of the principal $2times2$ minor of the matrix. Compare the discriminants and full determinants of $y^2=x$ or $x^2=y^2$, for instance. For the former, the discriminant vanishes, but the $3times3$ determinant doesn’t, and vice-versa for the latter.
            $endgroup$
            – amd
            Jan 1 at 20:22






            $begingroup$
            @Euler_Salter That’s not at all true for a general conic section. The discriminant only involves the quadratic part of the equation—it’s a multiple of the principal $2times2$ minor of the matrix. Compare the discriminants and full determinants of $y^2=x$ or $x^2=y^2$, for instance. For the former, the discriminant vanishes, but the $3times3$ determinant doesn’t, and vice-versa for the latter.
            $endgroup$
            – amd
            Jan 1 at 20:22




















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