Plotting contours of the paraboloid $f(x_1, x_2) = 4x_1^2+x_2^2-4x_1x_2-4x_1+2x_2+1$
$begingroup$
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:
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
edited Dec 31 '18 at 16:54
greedoid
45k1157112
asked Dec 31 '18 at 10:57
Euler_SalterEuler_Salter
2,0671336
$endgroup$
add a comment |
$begingroup$
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:
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
edited Dec 31 '18 at 16:54
greedoid
45k1157112
asked Dec 31 '18 at 10:57
Euler_SalterEuler_Salter
2,0671336
$endgroup$
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
add a comment |
$begingroup$
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:
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
edited Dec 31 '18 at 16:54
greedoid
45k1157112
asked Dec 31 '18 at 10:57
Euler_SalterEuler_Salter
2,0671336
$endgroup$
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:
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
edited Dec 31 '18 at 16:54
greedoid
45k1157112
asked Dec 31 '18 at 10:57
Euler_SalterEuler_Salter
2,0671336
edited Dec 31 '18 at 16:54
greedoid
45k1157112
asked Dec 31 '18 at 10:57
Euler_SalterEuler_Salter
2,0671336
edited Dec 31 '18 at 16:54
greedoid
45k1157112
edited Dec 31 '18 at 16:54
greedoid
45k1157112
edited Dec 31 '18 at 16:54
greedoid
45k1157112
45k1157112
asked Dec 31 '18 at 10:57
Euler_SalterEuler_Salter
2,0671336
asked Dec 31 '18 at 10:57
Euler_SalterEuler_Salter
2,0671336
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
add a comment |
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
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$(x_2 + (1-2x_1))^2 - C = 0$ 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.
answered Dec 31 '18 at 11:36
Rafa BudríaRafa Budría
5,8151825
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 1 at 0:53
amdamd
30.7k21050
$endgroup$
$begingroup$
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
add a comment |
Your Answer
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2 Answers
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active
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2 Answers
2
active
oldest
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votes
$begingroup$
$(x_2 + (1-2x_1))^2 - C = 0$ 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.
answered Dec 31 '18 at 11:36
Rafa BudríaRafa Budría
5,8151825
$endgroup$
add a comment |
$begingroup$
$(x_2 + (1-2x_1))^2 - C = 0$ 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.
answered Dec 31 '18 at 11:36
Rafa BudríaRafa Budría
5,8151825
$endgroup$
add a comment |
$begingroup$
$(x_2 + (1-2x_1))^2 - C = 0$ 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.
answered Dec 31 '18 at 11:36
Rafa BudríaRafa Budría
5,8151825
$endgroup$
$(x_2 + (1-2x_1))^2 - C = 0$ 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.
answered Dec 31 '18 at 11:36
Rafa BudríaRafa Budría
5,8151825
edited Dec 31 '18 at 11:42
answered Dec 31 '18 at 11:36
Rafa BudríaRafa Budría
5,8151825
answered Dec 31 '18 at 11:36
Rafa BudríaRafa Budría
5,8151825
answered Dec 31 '18 at 11:36
Rafa BudríaRafa Budría
5,8151825
5,8151825
add a comment |
add a comment |
$begingroup$
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.
answered Jan 1 at 0:53
amdamd
30.7k21050
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Jan 1 at 0:53
amdamd
30.7k21050
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Jan 1 at 0:53
amdamd
30.7k21050
$endgroup$
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.
answered Jan 1 at 0:53
amdamd
30.7k21050
answered Jan 1 at 0:53
amdamd
30.7k21050
answered Jan 1 at 0:53
amdamd
30.7k21050
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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