Mixing
Critical values for a projection map
$begingroup$
Let $Xin R^3$ and $f(x,y,z) = x^2+3y^2-z^2-xy+2z$, then $df_{(x,y,z)} = (2x-y, 6y-x, 2-2z)$ and the tangent space $TX = {(a,b,c) in R^3 | (2x-y)a+(6y-x)b+(2-2z)c=0}$. Now let $g:R^3rightarrow R$ s.t. $g((x,y,z))=z$. I am trying to find the critical values of $g$. I know
$dg_{(x,y,z)}:TX_{(x,y,z)}rightarrow R$
So $dg_{(x,y,z)}$ will be surjective provided that $TXneq R^2times {0}$, i.e. when $c=0$, because $dg_{(x,y,z)}$ projects the third coordinate. But I am stumped as to how to get a condition on $x,y,z$ such that $dg_{(x,y,z)}$ is not surjective and thus find critical values. A hint is appreciated. Thanks.
multivariable-calculus differential-geometry differential-topology
asked Jan 31 at 22:49
manifoldedmanifolded
50219
$endgroup$
add a comment |
$begingroup$
Let $Xin R^3$ and $f(x,y,z) = x^2+3y^2-z^2-xy+2z$, then $df_{(x,y,z)} = (2x-y, 6y-x, 2-2z)$ and the tangent space $TX = {(a,b,c) in R^3 | (2x-y)a+(6y-x)b+(2-2z)c=0}$. Now let $g:R^3rightarrow R$ s.t. $g((x,y,z))=z$. I am trying to find the critical values of $g$. I know
$dg_{(x,y,z)}:TX_{(x,y,z)}rightarrow R$
So $dg_{(x,y,z)}$ will be surjective provided that $TXneq R^2times {0}$, i.e. when $c=0$, because $dg_{(x,y,z)}$ projects the third coordinate. But I am stumped as to how to get a condition on $x,y,z$ such that $dg_{(x,y,z)}$ is not surjective and thus find critical values. A hint is appreciated. Thanks.
multivariable-calculus differential-geometry differential-topology
asked Jan 31 at 22:49
manifoldedmanifolded
50219
$endgroup$
$begingroup$
Do you mean $$X = {(x,y,z)in {mathbb R}^3 mid f(x,y,z)= 0}?$$ If so, do you mean to view $g$ as the restriction to $X$, i.e., as $$ gcolon X to {mathbb R}?$$
$endgroup$
– peter a g
Feb 1 at 3:37
$begingroup$
If so, if $f(x,y,z) = 0$, then $$ TX_{x,y,z} = { (a,b,c)in {mathbb R}^3 mid df_{(x,y,z)} (a-x,b-y,c-z) = 0 },$$ which does not quite agree with what you wrote above...
$endgroup$
– peter a g
Feb 1 at 3:47
$begingroup$
Assuming that I have understood what you want: "geometrically," a critical point happen when the corresponding tangent plane of $X$ is PARALLEL to the $x-y$ plane. That means that its normal is parallel to the $z$-axis: the $x$ and $y$ components of the normal are $0$. So you want to solve $f= f_x = f_y = 0$. OK?
$endgroup$
– peter a g
Feb 1 at 18:14
add a comment |
$begingroup$
Let $Xin R^3$ and $f(x,y,z) = x^2+3y^2-z^2-xy+2z$, then $df_{(x,y,z)} = (2x-y, 6y-x, 2-2z)$ and the tangent space $TX = {(a,b,c) in R^3 | (2x-y)a+(6y-x)b+(2-2z)c=0}$. Now let $g:R^3rightarrow R$ s.t. $g((x,y,z))=z$. I am trying to find the critical values of $g$. I know
$dg_{(x,y,z)}:TX_{(x,y,z)}rightarrow R$
So $dg_{(x,y,z)}$ will be surjective provided that $TXneq R^2times {0}$, i.e. when $c=0$, because $dg_{(x,y,z)}$ projects the third coordinate. But I am stumped as to how to get a condition on $x,y,z$ such that $dg_{(x,y,z)}$ is not surjective and thus find critical values. A hint is appreciated. Thanks.
multivariable-calculus differential-geometry differential-topology
asked Jan 31 at 22:49
manifoldedmanifolded
50219
$endgroup$
Let $Xin R^3$ and $f(x,y,z) = x^2+3y^2-z^2-xy+2z$, then $df_{(x,y,z)} = (2x-y, 6y-x, 2-2z)$ and the tangent space $TX = {(a,b,c) in R^3 | (2x-y)a+(6y-x)b+(2-2z)c=0}$. Now let $g:R^3rightarrow R$ s.t. $g((x,y,z))=z$. I am trying to find the critical values of $g$. I know
$dg_{(x,y,z)}:TX_{(x,y,z)}rightarrow R$
So $dg_{(x,y,z)}$ will be surjective provided that $TXneq R^2times {0}$, i.e. when $c=0$, because $dg_{(x,y,z)}$ projects the third coordinate. But I am stumped as to how to get a condition on $x,y,z$ such that $dg_{(x,y,z)}$ is not surjective and thus find critical values. A hint is appreciated. Thanks.
multivariable-calculus differential-geometry differential-topology
multivariable-calculus differential-geometry differential-topology
asked Jan 31 at 22:49
manifoldedmanifolded
50219
asked Jan 31 at 22:49
manifoldedmanifolded
50219
edited Jan 31 at 23:15
manifolded
asked Jan 31 at 22:49
manifoldedmanifolded
50219
asked Jan 31 at 22:49
manifoldedmanifolded
50219
asked Jan 31 at 22:49
manifoldedmanifolded
50219
50219
$begingroup$
Do you mean $$X = {(x,y,z)in {mathbb R}^3 mid f(x,y,z)= 0}?$$ If so, do you mean to view $g$ as the restriction to $X$, i.e., as $$ gcolon X to {mathbb R}?$$
$endgroup$
– peter a g
Feb 1 at 3:37
$begingroup$
If so, if $f(x,y,z) = 0$, then $$ TX_{x,y,z} = { (a,b,c)in {mathbb R}^3 mid df_{(x,y,z)} (a-x,b-y,c-z) = 0 },$$ which does not quite agree with what you wrote above...
$endgroup$
– peter a g
Feb 1 at 3:47
$begingroup$
Assuming that I have understood what you want: "geometrically," a critical point happen when the corresponding tangent plane of $X$ is PARALLEL to the $x-y$ plane. That means that its normal is parallel to the $z$-axis: the $x$ and $y$ components of the normal are $0$. So you want to solve $f= f_x = f_y = 0$. OK?
$endgroup$
– peter a g
Feb 1 at 18:14
add a comment |
$begingroup$
Do you mean $$X = {(x,y,z)in {mathbb R}^3 mid f(x,y,z)= 0}?$$ If so, do you mean to view $g$ as the restriction to $X$, i.e., as $$ gcolon X to {mathbb R}?$$
$endgroup$
– peter a g
Feb 1 at 3:37
$begingroup$
If so, if $f(x,y,z) = 0$, then $$ TX_{x,y,z} = { (a,b,c)in {mathbb R}^3 mid df_{(x,y,z)} (a-x,b-y,c-z) = 0 },$$ which does not quite agree with what you wrote above...
$endgroup$
– peter a g
Feb 1 at 3:47
$begingroup$
Assuming that I have understood what you want: "geometrically," a critical point happen when the corresponding tangent plane of $X$ is PARALLEL to the $x-y$ plane. That means that its normal is parallel to the $z$-axis: the $x$ and $y$ components of the normal are $0$. So you want to solve $f= f_x = f_y = 0$. OK?
$endgroup$
– peter a g
Feb 1 at 18:14
$begingroup$
Do you mean $$X = {(x,y,z)in {mathbb R}^3 mid f(x,y,z)= 0}?$$ If so, do you mean to view $g$ as the restriction to $X$, i.e., as $$ gcolon X to {mathbb R}?$$
$endgroup$
– peter a g
Feb 1 at 3:37
$begingroup$
Do you mean $$X = {(x,y,z)in {mathbb R}^3 mid f(x,y,z)= 0}?$$ If so, do you mean to view $g$ as the restriction to $X$, i.e., as $$ gcolon X to {mathbb R}?$$
$endgroup$
– peter a g
Feb 1 at 3:37
$begingroup$
If so, if $f(x,y,z) = 0$, then $$ TX_{x,y,z} = { (a,b,c)in {mathbb R}^3 mid df_{(x,y,z)} (a-x,b-y,c-z) = 0 },$$ which does not quite agree with what you wrote above...
$endgroup$
– peter a g
Feb 1 at 3:47
$begingroup$
If so, if $f(x,y,z) = 0$, then $$ TX_{x,y,z} = { (a,b,c)in {mathbb R}^3 mid df_{(x,y,z)} (a-x,b-y,c-z) = 0 },$$ which does not quite agree with what you wrote above...
$endgroup$
– peter a g
Feb 1 at 3:47
$begingroup$
Assuming that I have understood what you want: "geometrically," a critical point happen when the corresponding tangent plane of $X$ is PARALLEL to the $x-y$ plane. That means that its normal is parallel to the $z$-axis: the $x$ and $y$ components of the normal are $0$. So you want to solve $f= f_x = f_y = 0$. OK?
$endgroup$
– peter a g
Feb 1 at 18:14
$begingroup$
Assuming that I have understood what you want: "geometrically," a critical point happen when the corresponding tangent plane of $X$ is PARALLEL to the $x-y$ plane. That means that its normal is parallel to the $z$-axis: the $x$ and $y$ components of the normal are $0$. So you want to solve $f= f_x = f_y = 0$. OK?
$endgroup$
– peter a g
Feb 1 at 18:14
add a comment |
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$begingroup$
Do you mean $$X = {(x,y,z)in {mathbb R}^3 mid f(x,y,z)= 0}?$$ If so, do you mean to view $g$ as the restriction to $X$, i.e., as $$ gcolon X to {mathbb R}?$$
$endgroup$
– peter a g
Feb 1 at 3:37
$begingroup$
If so, if $f(x,y,z) = 0$, then $$ TX_{x,y,z} = { (a,b,c)in {mathbb R}^3 mid df_{(x,y,z)} (a-x,b-y,c-z) = 0 },$$ which does not quite agree with what you wrote above...
$endgroup$
– peter a g
Feb 1 at 3:47
$begingroup$
Assuming that I have understood what you want: "geometrically," a critical point happen when the corresponding tangent plane of $X$ is PARALLEL to the $x-y$ plane. That means that its normal is parallel to the $z$-axis: the $x$ and $y$ components of the normal are $0$. So you want to solve $f= f_x = f_y = 0$. OK?
$endgroup$
– peter a g
Feb 1 at 18:14