The vector field is perpendicular to the family of level surfaces
$begingroup$
The vector field given by $F(x,y,z)=leftlangle-x,-y,-z rightrangle $ is perpendicular to the family of level surfaces is composed of
$(1)$ spheres
$(2)$ ellipsoid,
$(3)$ planes,
$(4)$ None of the above.
Answer:
The option $(3)$ is true.
Am I right?
Help me
calculus
asked Dec 14 '18 at 0:43
arifamatharifamath
1176
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add a comment |
$begingroup$
The vector field given by $F(x,y,z)=leftlangle-x,-y,-z rightrangle $ is perpendicular to the family of level surfaces is composed of
$(1)$ spheres
$(2)$ ellipsoid,
$(3)$ planes,
$(4)$ None of the above.
Answer:
The option $(3)$ is true.
Am I right?
Help me
calculus
asked Dec 14 '18 at 0:43
arifamatharifamath
1176
$endgroup$
add a comment |
$begingroup$
The vector field given by $F(x,y,z)=leftlangle-x,-y,-z rightrangle $ is perpendicular to the family of level surfaces is composed of
$(1)$ spheres
$(2)$ ellipsoid,
$(3)$ planes,
$(4)$ None of the above.
Answer:
The option $(3)$ is true.
Am I right?
Help me
calculus
asked Dec 14 '18 at 0:43
arifamatharifamath
1176
$endgroup$
The vector field given by $F(x,y,z)=leftlangle-x,-y,-z rightrangle $ is perpendicular to the family of level surfaces is composed of
$(1)$ spheres
$(2)$ ellipsoid,
$(3)$ planes,
$(4)$ None of the above.
Answer:
The option $(3)$ is true.
Am I right?
Help me
calculus
calculus
asked Dec 14 '18 at 0:43
arifamatharifamath
1176
asked Dec 14 '18 at 0:43
arifamatharifamath
1176
edited Dec 14 '18 at 7:57
arifamath
asked Dec 14 '18 at 0:43
arifamatharifamath
1176
asked Dec 14 '18 at 0:43
arifamatharifamath
1176
asked Dec 14 '18 at 0:43
arifamatharifamath
1176
1176
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
The result is:
The gradient of $f = f(x, y, z)$ is perpendicular to the level curves of $f$.
In your case you know the gradient ${bf F}(x, y, z) = -(x, y, z) = -{bf r}$, where ${bf r}$ is the radial vector. So the question becomes: what is the surface to which the radial vector is perpendicular everywhere? And the solution is a sphere. In other words, the function level curves of $f$ are spheres centered at the origin.
Note that in you post you calculate the divergence of ${bf}$: ${bf nabla} cdot {bf F} = -3$, but that doesn't have any particular meaning in this problem.
answered Dec 14 '18 at 1:57
caveraccaverac
14.2k21130
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1 Answer
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1 Answer
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$begingroup$
The result is:
The gradient of $f = f(x, y, z)$ is perpendicular to the level curves of $f$.
In your case you know the gradient ${bf F}(x, y, z) = -(x, y, z) = -{bf r}$, where ${bf r}$ is the radial vector. So the question becomes: what is the surface to which the radial vector is perpendicular everywhere? And the solution is a sphere. In other words, the function level curves of $f$ are spheres centered at the origin.
Note that in you post you calculate the divergence of ${bf}$: ${bf nabla} cdot {bf F} = -3$, but that doesn't have any particular meaning in this problem.
answered Dec 14 '18 at 1:57
caveraccaverac
14.2k21130
$endgroup$
add a comment |
$begingroup$
The result is:
The gradient of $f = f(x, y, z)$ is perpendicular to the level curves of $f$.
In your case you know the gradient ${bf F}(x, y, z) = -(x, y, z) = -{bf r}$, where ${bf r}$ is the radial vector. So the question becomes: what is the surface to which the radial vector is perpendicular everywhere? And the solution is a sphere. In other words, the function level curves of $f$ are spheres centered at the origin.
Note that in you post you calculate the divergence of ${bf}$: ${bf nabla} cdot {bf F} = -3$, but that doesn't have any particular meaning in this problem.
answered Dec 14 '18 at 1:57
caveraccaverac
14.2k21130
$endgroup$
add a comment |
$begingroup$
The result is:
The gradient of $f = f(x, y, z)$ is perpendicular to the level curves of $f$.
In your case you know the gradient ${bf F}(x, y, z) = -(x, y, z) = -{bf r}$, where ${bf r}$ is the radial vector. So the question becomes: what is the surface to which the radial vector is perpendicular everywhere? And the solution is a sphere. In other words, the function level curves of $f$ are spheres centered at the origin.
Note that in you post you calculate the divergence of ${bf}$: ${bf nabla} cdot {bf F} = -3$, but that doesn't have any particular meaning in this problem.
answered Dec 14 '18 at 1:57
caveraccaverac
14.2k21130
$endgroup$
The result is:
The gradient of $f = f(x, y, z)$ is perpendicular to the level curves of $f$.
In your case you know the gradient ${bf F}(x, y, z) = -(x, y, z) = -{bf r}$, where ${bf r}$ is the radial vector. So the question becomes: what is the surface to which the radial vector is perpendicular everywhere? And the solution is a sphere. In other words, the function level curves of $f$ are spheres centered at the origin.
Note that in you post you calculate the divergence of ${bf}$: ${bf nabla} cdot {bf F} = -3$, but that doesn't have any particular meaning in this problem.
answered Dec 14 '18 at 1:57
caveraccaverac
14.2k21130
answered Dec 14 '18 at 1:57
caveraccaverac
14.2k21130
answered Dec 14 '18 at 1:57
caveraccaverac
14.2k21130
answered Dec 14 '18 at 1:57
caveraccaverac
14.2k21130
14.2k21130
add a comment |
add a comment |
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