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gradient of implicit function
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$f : U subset R^3 to R$ is differentiable function.
$f^-(0)$ is regular surface.
Is gradient f(p) $neq$ 0 for at all $p in f^-(0)$ ?
I think the above question is true.
By the way, I do not know where to start the proof.
Give me a hint.
differential-geometry vector-analysis
asked Jul 18 at 6:55
LeeHanWoong
398
add a comment |Â
up vote
0
down vote
favorite
$f : U subset R^3 to R$ is differentiable function.
$f^-(0)$ is regular surface.
Is gradient f(p) $neq$ 0 for at all $p in f^-(0)$ ?
I think the above question is true.
By the way, I do not know where to start the proof.
Give me a hint.
differential-geometry vector-analysis
asked Jul 18 at 6:55
LeeHanWoong
398
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
$f : U subset R^3 to R$ is differentiable function.
$f^-(0)$ is regular surface.
Is gradient f(p) $neq$ 0 for at all $p in f^-(0)$ ?
I think the above question is true.
By the way, I do not know where to start the proof.
Give me a hint.
differential-geometry vector-analysis
asked Jul 18 at 6:55
LeeHanWoong
398
$f : U subset R^3 to R$ is differentiable function.
$f^-(0)$ is regular surface.
Is gradient f(p) $neq$ 0 for at all $p in f^-(0)$ ?
I think the above question is true.
By the way, I do not know where to start the proof.
Give me a hint.
differential-geometry vector-analysis
asked Jul 18 at 6:55
LeeHanWoong
398
asked Jul 18 at 6:55
LeeHanWoong
398
asked Jul 18 at 6:55
LeeHanWoong
398
asked Jul 18 at 6:55
LeeHanWoong
398
398
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Is not true. Take for example the following function
$$
f(x_1, x_2, x_3) := ( |x|^2 - 1 )^2,
$$
where $|x| = sqrtx_1^2 + x_2^2 + x_3^2$.
Clearly $f^-1(0)$ is the unit sphere but you can check that the gradient is zero everywhere on each point of the sphere, because each point is a minimum point.
answered Jul 18 at 7:59
Onil90
1,494521
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Is not true. Take for example the following function
$$
f(x_1, x_2, x_3) := ( |x|^2 - 1 )^2,
$$
where $|x| = sqrtx_1^2 + x_2^2 + x_3^2$.
Clearly $f^-1(0)$ is the unit sphere but you can check that the gradient is zero everywhere on each point of the sphere, because each point is a minimum point.
answered Jul 18 at 7:59
Onil90
1,494521
add a comment |Â
up vote
1
down vote
accepted
Is not true. Take for example the following function
$$
f(x_1, x_2, x_3) := ( |x|^2 - 1 )^2,
$$
where $|x| = sqrtx_1^2 + x_2^2 + x_3^2$.
Clearly $f^-1(0)$ is the unit sphere but you can check that the gradient is zero everywhere on each point of the sphere, because each point is a minimum point.
answered Jul 18 at 7:59
Onil90
1,494521
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Is not true. Take for example the following function
$$
f(x_1, x_2, x_3) := ( |x|^2 - 1 )^2,
$$
where $|x| = sqrtx_1^2 + x_2^2 + x_3^2$.
Clearly $f^-1(0)$ is the unit sphere but you can check that the gradient is zero everywhere on each point of the sphere, because each point is a minimum point.
answered Jul 18 at 7:59
Onil90
1,494521
Is not true. Take for example the following function
$$
f(x_1, x_2, x_3) := ( |x|^2 - 1 )^2,
$$
where $|x| = sqrtx_1^2 + x_2^2 + x_3^2$.
Clearly $f^-1(0)$ is the unit sphere but you can check that the gradient is zero everywhere on each point of the sphere, because each point is a minimum point.
answered Jul 18 at 7:59
Onil90
1,494521
answered Jul 18 at 7:59
Onil90
1,494521
answered Jul 18 at 7:59
Onil90
1,494521
answered Jul 18 at 7:59
Onil90
1,494521
1,494521
add a comment |Â
add a comment |Â
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