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basic potential problem in mechanics
Clash Royale CLAN TAG#URR8PPP
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I am new to this field and I'm having trouble with this question where it asks me to calculate the potential from the basis of the origin. The question is:
$overrightarrowF = (0, -2y + 3y^2,0)$
in $(x, y , z)$
I have no idea what should I do.
physics mathematical-physics
asked Jul 31 at 14:13
Kat
305
add a comment |Â
up vote
0
down vote
favorite
I am new to this field and I'm having trouble with this question where it asks me to calculate the potential from the basis of the origin. The question is:
$overrightarrowF = (0, -2y + 3y^2,0)$
in $(x, y , z)$
I have no idea what should I do.
physics mathematical-physics
asked Jul 31 at 14:13
Kat
305
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Jul 31 at 14:17
Start with $vecF=pmnablaphi$...
– Fakemistake
Jul 31 at 14:23
Are you familiar with line integrals? If so, you can construct a scalar potential function entirely mechanically.
– amd
Jul 31 at 19:57
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am new to this field and I'm having trouble with this question where it asks me to calculate the potential from the basis of the origin. The question is:
$overrightarrowF = (0, -2y + 3y^2,0)$
in $(x, y , z)$
I have no idea what should I do.
physics mathematical-physics
asked Jul 31 at 14:13
Kat
305
I am new to this field and I'm having trouble with this question where it asks me to calculate the potential from the basis of the origin. The question is:
$overrightarrowF = (0, -2y + 3y^2,0)$
in $(x, y , z)$
I have no idea what should I do.
physics mathematical-physics
asked Jul 31 at 14:13
Kat
305
asked Jul 31 at 14:13
Kat
305
asked Jul 31 at 14:13
Kat
305
asked Jul 31 at 14:13
Kat
305
305
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Jul 31 at 14:17
Start with $vecF=pmnablaphi$...
– Fakemistake
Jul 31 at 14:23
Are you familiar with line integrals? If so, you can construct a scalar potential function entirely mechanically.
– amd
Jul 31 at 19:57
add a comment |Â
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Jul 31 at 14:17
Start with $vecF=pmnablaphi$...
– Fakemistake
Jul 31 at 14:23
Are you familiar with line integrals? If so, you can construct a scalar potential function entirely mechanically.
– amd
Jul 31 at 19:57
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Jul 31 at 14:17
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Jul 31 at 14:17
Start with $vecF=pmnablaphi$...
– Fakemistake
Jul 31 at 14:23
Start with $vecF=pmnablaphi$...
– Fakemistake
Jul 31 at 14:23
Are you familiar with line integrals? If so, you can construct a scalar potential function entirely mechanically.
– amd
Jul 31 at 19:57
Are you familiar with line integrals? If so, you can construct a scalar potential function entirely mechanically.
– amd
Jul 31 at 19:57
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
We have that
$$underlineF=-textgradU$$
$$(0, -2y+3x^2, 0)^T=-(partial_xU, partial_yU, partial_zU)^T$$
They are equal if and only if all of their components are equal:
$$0=-partial_xU$$
$$-2y+3y^2=-partial_yU$$
$$0=-partial_zU$$
Where $U$ is the function of the position: $U(x, y, z)$
Now let's pick $1$ out of the $3$ equations, for example the first one:
$$0=partial_xU$$
If we integrate both sides with respect to $x$ we will get that
$$U=f(y, z)$$
Where $f(y, z)$ can be any function of $y$ and $z$.
Now let's substitute this $U$ into another equation, for example into the second one:
$$partial_y U=partial_y f(y, z)=2y-3y^2$$
Integrating both sides with respect to $y$ we will get that
$$f(y, z)=y^2-y^3+g(z)$$
Where $g(z)$ can be any function of $z$. Now let's substitute it into the third equation:
$$partial_zU=partial_z(y^2-y^3+g(z))=g'(z)=0$$
And I think you can solve it for $g$, and you can put everything together to get $U$.
answered Jul 31 at 15:00
Botond
3,8762632
Thanks for your advice. But, what is that T of (0,−2y+3x2,0)T ?
– Kat
Jul 31 at 15:15
@Kat T means the transpose. It transforms the row vector to column vector. You can just ignore it.
– Botond
Jul 31 at 15:17
add a comment |Â
up vote
1
down vote
You want to find $phi$ so that
$$(0, -2y+3y^2, 0) = vec F = - nabla phi = - left( fracpartialphipartial x, fracpartialphipartial y, fracpartialphipartial z right) = left( -fracpartialphipartial x, -fracpartialphipartial y, -fracpartialphipartial z right),$$
i.e.
$$beginalign
fracpartialphipartial x &= 0, \
fracpartialphipartial y &= 2y-3y^2, \
fracpartialphipartial z &= 0.
endalign$$
Can you solve this? I recommend starting with $fracpartialphipartial y = 2y-3y^2.$
answered Jul 31 at 14:50
md2perpe
5,6391921
$fracpartialphipartial y = 2y-3y^2$, $fracpartialphipartial y = 2-6y$, $ phi = (0,2-6y,0)$ , right ?
– Kat
Jul 31 at 15:27
@Kat. Can you solve that? What form must $phi$ have?
– md2perpe
Jul 31 at 15:29
I am not sure how to answer it.
– Kat
Jul 31 at 15:35
No, you are taking another derivative of the right hand side without showing that in the left hand side. Also, that is not the correct way. What if you had the following ordinary differential equation instead? Can you solve that? $$f'(y) = 2y - 3y^2$$
– md2perpe
Jul 31 at 15:40
sorry but I can't... how ?
– Kat
Jul 31 at 16:17
 |Â
show 6 more comments
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
We have that
$$underlineF=-textgradU$$
$$(0, -2y+3x^2, 0)^T=-(partial_xU, partial_yU, partial_zU)^T$$
They are equal if and only if all of their components are equal:
$$0=-partial_xU$$
$$-2y+3y^2=-partial_yU$$
$$0=-partial_zU$$
Where $U$ is the function of the position: $U(x, y, z)$
Now let's pick $1$ out of the $3$ equations, for example the first one:
$$0=partial_xU$$
If we integrate both sides with respect to $x$ we will get that
$$U=f(y, z)$$
Where $f(y, z)$ can be any function of $y$ and $z$.
Now let's substitute this $U$ into another equation, for example into the second one:
$$partial_y U=partial_y f(y, z)=2y-3y^2$$
Integrating both sides with respect to $y$ we will get that
$$f(y, z)=y^2-y^3+g(z)$$
Where $g(z)$ can be any function of $z$. Now let's substitute it into the third equation:
$$partial_zU=partial_z(y^2-y^3+g(z))=g'(z)=0$$
And I think you can solve it for $g$, and you can put everything together to get $U$.
answered Jul 31 at 15:00
Botond
3,8762632
Thanks for your advice. But, what is that T of (0,−2y+3x2,0)T ?
– Kat
Jul 31 at 15:15
@Kat T means the transpose. It transforms the row vector to column vector. You can just ignore it.
– Botond
Jul 31 at 15:17
add a comment |Â
up vote
1
down vote
We have that
$$underlineF=-textgradU$$
$$(0, -2y+3x^2, 0)^T=-(partial_xU, partial_yU, partial_zU)^T$$
They are equal if and only if all of their components are equal:
$$0=-partial_xU$$
$$-2y+3y^2=-partial_yU$$
$$0=-partial_zU$$
Where $U$ is the function of the position: $U(x, y, z)$
Now let's pick $1$ out of the $3$ equations, for example the first one:
$$0=partial_xU$$
If we integrate both sides with respect to $x$ we will get that
$$U=f(y, z)$$
Where $f(y, z)$ can be any function of $y$ and $z$.
Now let's substitute this $U$ into another equation, for example into the second one:
$$partial_y U=partial_y f(y, z)=2y-3y^2$$
Integrating both sides with respect to $y$ we will get that
$$f(y, z)=y^2-y^3+g(z)$$
Where $g(z)$ can be any function of $z$. Now let's substitute it into the third equation:
$$partial_zU=partial_z(y^2-y^3+g(z))=g'(z)=0$$
And I think you can solve it for $g$, and you can put everything together to get $U$.
answered Jul 31 at 15:00
Botond
3,8762632
Thanks for your advice. But, what is that T of (0,−2y+3x2,0)T ?
– Kat
Jul 31 at 15:15
@Kat T means the transpose. It transforms the row vector to column vector. You can just ignore it.
– Botond
Jul 31 at 15:17
add a comment |Â
up vote
1
down vote
up vote
1
down vote
We have that
$$underlineF=-textgradU$$
$$(0, -2y+3x^2, 0)^T=-(partial_xU, partial_yU, partial_zU)^T$$
They are equal if and only if all of their components are equal:
$$0=-partial_xU$$
$$-2y+3y^2=-partial_yU$$
$$0=-partial_zU$$
Where $U$ is the function of the position: $U(x, y, z)$
Now let's pick $1$ out of the $3$ equations, for example the first one:
$$0=partial_xU$$
If we integrate both sides with respect to $x$ we will get that
$$U=f(y, z)$$
Where $f(y, z)$ can be any function of $y$ and $z$.
Now let's substitute this $U$ into another equation, for example into the second one:
$$partial_y U=partial_y f(y, z)=2y-3y^2$$
Integrating both sides with respect to $y$ we will get that
$$f(y, z)=y^2-y^3+g(z)$$
Where $g(z)$ can be any function of $z$. Now let's substitute it into the third equation:
$$partial_zU=partial_z(y^2-y^3+g(z))=g'(z)=0$$
And I think you can solve it for $g$, and you can put everything together to get $U$.
answered Jul 31 at 15:00
Botond
3,8762632
We have that
$$underlineF=-textgradU$$
$$(0, -2y+3x^2, 0)^T=-(partial_xU, partial_yU, partial_zU)^T$$
They are equal if and only if all of their components are equal:
$$0=-partial_xU$$
$$-2y+3y^2=-partial_yU$$
$$0=-partial_zU$$
Where $U$ is the function of the position: $U(x, y, z)$
Now let's pick $1$ out of the $3$ equations, for example the first one:
$$0=partial_xU$$
If we integrate both sides with respect to $x$ we will get that
$$U=f(y, z)$$
Where $f(y, z)$ can be any function of $y$ and $z$.
Now let's substitute this $U$ into another equation, for example into the second one:
$$partial_y U=partial_y f(y, z)=2y-3y^2$$
Integrating both sides with respect to $y$ we will get that
$$f(y, z)=y^2-y^3+g(z)$$
Where $g(z)$ can be any function of $z$. Now let's substitute it into the third equation:
$$partial_zU=partial_z(y^2-y^3+g(z))=g'(z)=0$$
And I think you can solve it for $g$, and you can put everything together to get $U$.
answered Jul 31 at 15:00
Botond
3,8762632
answered Jul 31 at 15:00
Botond
3,8762632
answered Jul 31 at 15:00
Botond
3,8762632
answered Jul 31 at 15:00
Botond
3,8762632
3,8762632
Thanks for your advice. But, what is that T of (0,−2y+3x2,0)T ?
– Kat
Jul 31 at 15:15
@Kat T means the transpose. It transforms the row vector to column vector. You can just ignore it.
– Botond
Jul 31 at 15:17
add a comment |Â
Thanks for your advice. But, what is that T of (0,−2y+3x2,0)T ?
– Kat
Jul 31 at 15:15
@Kat T means the transpose. It transforms the row vector to column vector. You can just ignore it.
– Botond
Jul 31 at 15:17
Thanks for your advice. But, what is that T of (0,−2y+3x2,0)T ?
– Kat
Jul 31 at 15:15
Thanks for your advice. But, what is that T of (0,−2y+3x2,0)T ?
– Kat
Jul 31 at 15:15
@Kat T means the transpose. It transforms the row vector to column vector. You can just ignore it.
– Botond
Jul 31 at 15:17
@Kat T means the transpose. It transforms the row vector to column vector. You can just ignore it.
– Botond
Jul 31 at 15:17
add a comment |Â
up vote
1
down vote
You want to find $phi$ so that
$$(0, -2y+3y^2, 0) = vec F = - nabla phi = - left( fracpartialphipartial x, fracpartialphipartial y, fracpartialphipartial z right) = left( -fracpartialphipartial x, -fracpartialphipartial y, -fracpartialphipartial z right),$$
i.e.
$$beginalign
fracpartialphipartial x &= 0, \
fracpartialphipartial y &= 2y-3y^2, \
fracpartialphipartial z &= 0.
endalign$$
Can you solve this? I recommend starting with $fracpartialphipartial y = 2y-3y^2.$
answered Jul 31 at 14:50
md2perpe
5,6391921
$fracpartialphipartial y = 2y-3y^2$, $fracpartialphipartial y = 2-6y$, $ phi = (0,2-6y,0)$ , right ?
– Kat
Jul 31 at 15:27
@Kat. Can you solve that? What form must $phi$ have?
– md2perpe
Jul 31 at 15:29
I am not sure how to answer it.
– Kat
Jul 31 at 15:35
No, you are taking another derivative of the right hand side without showing that in the left hand side. Also, that is not the correct way. What if you had the following ordinary differential equation instead? Can you solve that? $$f'(y) = 2y - 3y^2$$
– md2perpe
Jul 31 at 15:40
sorry but I can't... how ?
– Kat
Jul 31 at 16:17
 |Â
show 6 more comments
up vote
1
down vote
You want to find $phi$ so that
$$(0, -2y+3y^2, 0) = vec F = - nabla phi = - left( fracpartialphipartial x, fracpartialphipartial y, fracpartialphipartial z right) = left( -fracpartialphipartial x, -fracpartialphipartial y, -fracpartialphipartial z right),$$
i.e.
$$beginalign
fracpartialphipartial x &= 0, \
fracpartialphipartial y &= 2y-3y^2, \
fracpartialphipartial z &= 0.
endalign$$
Can you solve this? I recommend starting with $fracpartialphipartial y = 2y-3y^2.$
answered Jul 31 at 14:50
md2perpe
5,6391921
$fracpartialphipartial y = 2y-3y^2$, $fracpartialphipartial y = 2-6y$, $ phi = (0,2-6y,0)$ , right ?
– Kat
Jul 31 at 15:27
@Kat. Can you solve that? What form must $phi$ have?
– md2perpe
Jul 31 at 15:29
I am not sure how to answer it.
– Kat
Jul 31 at 15:35
No, you are taking another derivative of the right hand side without showing that in the left hand side. Also, that is not the correct way. What if you had the following ordinary differential equation instead? Can you solve that? $$f'(y) = 2y - 3y^2$$
– md2perpe
Jul 31 at 15:40
sorry but I can't... how ?
– Kat
Jul 31 at 16:17
 |Â
show 6 more comments
up vote
1
down vote
up vote
1
down vote
You want to find $phi$ so that
$$(0, -2y+3y^2, 0) = vec F = - nabla phi = - left( fracpartialphipartial x, fracpartialphipartial y, fracpartialphipartial z right) = left( -fracpartialphipartial x, -fracpartialphipartial y, -fracpartialphipartial z right),$$
i.e.
$$beginalign
fracpartialphipartial x &= 0, \
fracpartialphipartial y &= 2y-3y^2, \
fracpartialphipartial z &= 0.
endalign$$
Can you solve this? I recommend starting with $fracpartialphipartial y = 2y-3y^2.$
answered Jul 31 at 14:50
md2perpe
5,6391921
You want to find $phi$ so that
$$(0, -2y+3y^2, 0) = vec F = - nabla phi = - left( fracpartialphipartial x, fracpartialphipartial y, fracpartialphipartial z right) = left( -fracpartialphipartial x, -fracpartialphipartial y, -fracpartialphipartial z right),$$
i.e.
$$beginalign
fracpartialphipartial x &= 0, \
fracpartialphipartial y &= 2y-3y^2, \
fracpartialphipartial z &= 0.
endalign$$
Can you solve this? I recommend starting with $fracpartialphipartial y = 2y-3y^2.$
answered Jul 31 at 14:50
md2perpe
5,6391921
edited Jul 31 at 15:04
answered Jul 31 at 14:50
md2perpe
5,6391921
answered Jul 31 at 14:50
md2perpe
5,6391921
answered Jul 31 at 14:50
md2perpe
5,6391921
5,6391921
$fracpartialphipartial y = 2y-3y^2$, $fracpartialphipartial y = 2-6y$, $ phi = (0,2-6y,0)$ , right ?
– Kat
Jul 31 at 15:27
@Kat. Can you solve that? What form must $phi$ have?
– md2perpe
Jul 31 at 15:29
I am not sure how to answer it.
– Kat
Jul 31 at 15:35
No, you are taking another derivative of the right hand side without showing that in the left hand side. Also, that is not the correct way. What if you had the following ordinary differential equation instead? Can you solve that? $$f'(y) = 2y - 3y^2$$
– md2perpe
Jul 31 at 15:40
sorry but I can't... how ?
– Kat
Jul 31 at 16:17
 |Â
show 6 more comments
$fracpartialphipartial y = 2y-3y^2$, $fracpartialphipartial y = 2-6y$, $ phi = (0,2-6y,0)$ , right ?
– Kat
Jul 31 at 15:27
@Kat. Can you solve that? What form must $phi$ have?
– md2perpe
Jul 31 at 15:29
I am not sure how to answer it.
– Kat
Jul 31 at 15:35
No, you are taking another derivative of the right hand side without showing that in the left hand side. Also, that is not the correct way. What if you had the following ordinary differential equation instead? Can you solve that? $$f'(y) = 2y - 3y^2$$
– md2perpe
Jul 31 at 15:40
sorry but I can't... how ?
– Kat
Jul 31 at 16:17
$fracpartialphipartial y = 2y-3y^2$, $fracpartialphipartial y = 2-6y$, $ phi = (0,2-6y,0)$ , right ?
– Kat
Jul 31 at 15:27
$fracpartialphipartial y = 2y-3y^2$, $fracpartialphipartial y = 2-6y$, $ phi = (0,2-6y,0)$ , right ?
– Kat
Jul 31 at 15:27
@Kat. Can you solve that? What form must $phi$ have?
– md2perpe
Jul 31 at 15:29
@Kat. Can you solve that? What form must $phi$ have?
– md2perpe
Jul 31 at 15:29
I am not sure how to answer it.
– Kat
Jul 31 at 15:35
I am not sure how to answer it.
– Kat
Jul 31 at 15:35
No, you are taking another derivative of the right hand side without showing that in the left hand side. Also, that is not the correct way. What if you had the following ordinary differential equation instead? Can you solve that? $$f'(y) = 2y - 3y^2$$
– md2perpe
Jul 31 at 15:40
No, you are taking another derivative of the right hand side without showing that in the left hand side. Also, that is not the correct way. What if you had the following ordinary differential equation instead? Can you solve that? $$f'(y) = 2y - 3y^2$$
– md2perpe
Jul 31 at 15:40
sorry but I can't... how ?
– Kat
Jul 31 at 16:17
sorry but I can't... how ?
– Kat
Jul 31 at 16:17
 |Â
show 6 more comments
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Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Jul 31 at 14:17
Start with $vecF=pmnablaphi$...
– Fakemistake
Jul 31 at 14:23
Are you familiar with line integrals? If so, you can construct a scalar potential function entirely mechanically.
– amd
Jul 31 at 19:57