When $x$ is given, what is $mathrm{d}x$? [closed]
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In the spherical coordinate, we know that $x = rsinthetacosphi$. Why does it imply that
$$mathrm{d}x = rcosthetacosphi mathrm{d}theta - rsin thetasin phi mathrm{d} phi?$$
It is first time I have faced this one. I tried to find some informations on Google, but I do not know what to search. Which rules do you use here?
derivatives infinitesimals
asked Dec 15 '18 at 21:10
UnknownWUnknownW
1,025922
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closed as off-topic by Leucippus, Cesareo, Shailesh, KReiser, Lord Shark the Unknown Dec 16 '18 at 4:38
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Leucippus, Cesareo, Shailesh, KReiser, Lord Shark the Unknown
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
In the spherical coordinate, we know that $x = rsinthetacosphi$. Why does it imply that
$$mathrm{d}x = rcosthetacosphi mathrm{d}theta - rsin thetasin phi mathrm{d} phi?$$
It is first time I have faced this one. I tried to find some informations on Google, but I do not know what to search. Which rules do you use here?
derivatives infinitesimals
asked Dec 15 '18 at 21:10
UnknownWUnknownW
1,025922
$endgroup$
closed as off-topic by Leucippus, Cesareo, Shailesh, KReiser, Lord Shark the Unknown Dec 16 '18 at 4:38
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Leucippus, Cesareo, Shailesh, KReiser, Lord Shark the Unknown
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
What are the derivative of $sin(teta)$ and $cos(phi)$.
$endgroup$
– hamam_Abdallah
Dec 15 '18 at 21:13
$begingroup$
That formula is meaningless end and of itself; it, however, is usually understood to mean what to substitute in a change of variables.
$endgroup$
– Will M.
Dec 15 '18 at 22:32
add a comment |
$begingroup$
In the spherical coordinate, we know that $x = rsinthetacosphi$. Why does it imply that
$$mathrm{d}x = rcosthetacosphi mathrm{d}theta - rsin thetasin phi mathrm{d} phi?$$
It is first time I have faced this one. I tried to find some informations on Google, but I do not know what to search. Which rules do you use here?
derivatives infinitesimals
asked Dec 15 '18 at 21:10
UnknownWUnknownW
1,025922
$endgroup$
In the spherical coordinate, we know that $x = rsinthetacosphi$. Why does it imply that
$$mathrm{d}x = rcosthetacosphi mathrm{d}theta - rsin thetasin phi mathrm{d} phi?$$
It is first time I have faced this one. I tried to find some informations on Google, but I do not know what to search. Which rules do you use here?
derivatives infinitesimals
derivatives infinitesimals
asked Dec 15 '18 at 21:10
UnknownWUnknownW
1,025922
asked Dec 15 '18 at 21:10
UnknownWUnknownW
1,025922
asked Dec 15 '18 at 21:10
UnknownWUnknownW
1,025922
asked Dec 15 '18 at 21:10
UnknownWUnknownW
1,025922
asked Dec 15 '18 at 21:10
UnknownWUnknownW
1,025922
1,025922
closed as off-topic by Leucippus, Cesareo, Shailesh, KReiser, Lord Shark the Unknown Dec 16 '18 at 4:38
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Leucippus, Cesareo, Shailesh, KReiser, Lord Shark the Unknown
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Leucippus, Cesareo, Shailesh, KReiser, Lord Shark the Unknown Dec 16 '18 at 4:38
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Leucippus, Cesareo, Shailesh, KReiser, Lord Shark the Unknown
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
What are the derivative of $sin(teta)$ and $cos(phi)$.
$endgroup$
– hamam_Abdallah
Dec 15 '18 at 21:13
$begingroup$
That formula is meaningless end and of itself; it, however, is usually understood to mean what to substitute in a change of variables.
$endgroup$
– Will M.
Dec 15 '18 at 22:32
add a comment |
$begingroup$
What are the derivative of $sin(teta)$ and $cos(phi)$.
$endgroup$
– hamam_Abdallah
Dec 15 '18 at 21:13
$begingroup$
That formula is meaningless end and of itself; it, however, is usually understood to mean what to substitute in a change of variables.
$endgroup$
– Will M.
Dec 15 '18 at 22:32
$begingroup$
What are the derivative of $sin(teta)$ and $cos(phi)$.
$endgroup$
– hamam_Abdallah
Dec 15 '18 at 21:13
$begingroup$
What are the derivative of $sin(teta)$ and $cos(phi)$.
$endgroup$
– hamam_Abdallah
Dec 15 '18 at 21:13
$begingroup$
That formula is meaningless end and of itself; it, however, is usually understood to mean what to substitute in a change of variables.
$endgroup$
– Will M.
Dec 15 '18 at 22:32
$begingroup$
That formula is meaningless end and of itself; it, however, is usually understood to mean what to substitute in a change of variables.
$endgroup$
– Will M.
Dec 15 '18 at 22:32
add a comment |
3 Answers
3
active
oldest
votes
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This is called the total derivative. If $f=f(x_1,dots,x_n)$ then
$$df=sum_{i=1}^nfrac{partial f}{partial x_i}dx_i.$$
You can take a look at the Wikipedia entry: Total dervative for more information.
answered Dec 15 '18 at 21:17
Will FisherWill Fisher
4,04811032
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add a comment |
$begingroup$
hint
If $A=f(u,v,w)$
then under some conditions
$$dA=frac{partial f}{partial u}du+...$$
answered Dec 15 '18 at 21:17
hamam_Abdallahhamam_Abdallah
38.2k21634
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add a comment |
$begingroup$
To understand this, you must know that $x,r,theta,phi$ are (almost) smooth functions defined on the usual Euclidean space, and $dx,dr,dtheta,dphi$ are the derivatives of these functions.
Derivatives abide by a few rules:
Linearity: if $u,v$ are smooth and if $t$ is a scalar, then $d(u+tv)=du+tcdot dv$.
Leibniz: if $u,v$ are smooth, then $d(uv)=u dv+vdu$.
Chain rule: if $f$ is smooth and $u$ is a smooth function from $mathbb{R}$ to itself, then $d(u circ f)=u’circ f df
As a consequence, constant functions’ derivatives vanish. So to get your formula, you just need to apply Leibniz’s rule and assume that in your context $r$ is a constant function.
answered Dec 15 '18 at 21:18
MindlackMindlack
4,885211
$endgroup$
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This is called the total derivative. If $f=f(x_1,dots,x_n)$ then
$$df=sum_{i=1}^nfrac{partial f}{partial x_i}dx_i.$$
You can take a look at the Wikipedia entry: Total dervative for more information.
answered Dec 15 '18 at 21:17
Will FisherWill Fisher
4,04811032
$endgroup$
add a comment |
$begingroup$
This is called the total derivative. If $f=f(x_1,dots,x_n)$ then
$$df=sum_{i=1}^nfrac{partial f}{partial x_i}dx_i.$$
You can take a look at the Wikipedia entry: Total dervative for more information.
answered Dec 15 '18 at 21:17
Will FisherWill Fisher
4,04811032
$endgroup$
add a comment |
$begingroup$
This is called the total derivative. If $f=f(x_1,dots,x_n)$ then
$$df=sum_{i=1}^nfrac{partial f}{partial x_i}dx_i.$$
You can take a look at the Wikipedia entry: Total dervative for more information.
answered Dec 15 '18 at 21:17
Will FisherWill Fisher
4,04811032
$endgroup$
This is called the total derivative. If $f=f(x_1,dots,x_n)$ then
$$df=sum_{i=1}^nfrac{partial f}{partial x_i}dx_i.$$
You can take a look at the Wikipedia entry: Total dervative for more information.
answered Dec 15 '18 at 21:17
Will FisherWill Fisher
4,04811032
edited Dec 15 '18 at 22:22
answered Dec 15 '18 at 21:17
Will FisherWill Fisher
4,04811032
answered Dec 15 '18 at 21:17
Will FisherWill Fisher
4,04811032
answered Dec 15 '18 at 21:17
Will FisherWill Fisher
4,04811032
4,04811032
add a comment |
add a comment |
$begingroup$
hint
If $A=f(u,v,w)$
then under some conditions
$$dA=frac{partial f}{partial u}du+...$$
answered Dec 15 '18 at 21:17
hamam_Abdallahhamam_Abdallah
38.2k21634
$endgroup$
add a comment |
$begingroup$
hint
If $A=f(u,v,w)$
then under some conditions
$$dA=frac{partial f}{partial u}du+...$$
answered Dec 15 '18 at 21:17
hamam_Abdallahhamam_Abdallah
38.2k21634
$endgroup$
add a comment |
$begingroup$
hint
If $A=f(u,v,w)$
then under some conditions
$$dA=frac{partial f}{partial u}du+...$$
answered Dec 15 '18 at 21:17
hamam_Abdallahhamam_Abdallah
38.2k21634
$endgroup$
hint
If $A=f(u,v,w)$
then under some conditions
$$dA=frac{partial f}{partial u}du+...$$
answered Dec 15 '18 at 21:17
hamam_Abdallahhamam_Abdallah
38.2k21634
answered Dec 15 '18 at 21:17
hamam_Abdallahhamam_Abdallah
38.2k21634
answered Dec 15 '18 at 21:17
hamam_Abdallahhamam_Abdallah
38.2k21634
answered Dec 15 '18 at 21:17
hamam_Abdallahhamam_Abdallah
38.2k21634
38.2k21634
add a comment |
add a comment |
$begingroup$
To understand this, you must know that $x,r,theta,phi$ are (almost) smooth functions defined on the usual Euclidean space, and $dx,dr,dtheta,dphi$ are the derivatives of these functions.
Derivatives abide by a few rules:
Linearity: if $u,v$ are smooth and if $t$ is a scalar, then $d(u+tv)=du+tcdot dv$.
Leibniz: if $u,v$ are smooth, then $d(uv)=u dv+vdu$.
Chain rule: if $f$ is smooth and $u$ is a smooth function from $mathbb{R}$ to itself, then $d(u circ f)=u’circ f df
As a consequence, constant functions’ derivatives vanish. So to get your formula, you just need to apply Leibniz’s rule and assume that in your context $r$ is a constant function.
answered Dec 15 '18 at 21:18
MindlackMindlack
4,885211
$endgroup$
add a comment |
$begingroup$
To understand this, you must know that $x,r,theta,phi$ are (almost) smooth functions defined on the usual Euclidean space, and $dx,dr,dtheta,dphi$ are the derivatives of these functions.
Derivatives abide by a few rules:
Linearity: if $u,v$ are smooth and if $t$ is a scalar, then $d(u+tv)=du+tcdot dv$.
Leibniz: if $u,v$ are smooth, then $d(uv)=u dv+vdu$.
Chain rule: if $f$ is smooth and $u$ is a smooth function from $mathbb{R}$ to itself, then $d(u circ f)=u’circ f df
As a consequence, constant functions’ derivatives vanish. So to get your formula, you just need to apply Leibniz’s rule and assume that in your context $r$ is a constant function.
answered Dec 15 '18 at 21:18
MindlackMindlack
4,885211
$endgroup$
add a comment |
$begingroup$
To understand this, you must know that $x,r,theta,phi$ are (almost) smooth functions defined on the usual Euclidean space, and $dx,dr,dtheta,dphi$ are the derivatives of these functions.
Derivatives abide by a few rules:
Linearity: if $u,v$ are smooth and if $t$ is a scalar, then $d(u+tv)=du+tcdot dv$.
Leibniz: if $u,v$ are smooth, then $d(uv)=u dv+vdu$.
Chain rule: if $f$ is smooth and $u$ is a smooth function from $mathbb{R}$ to itself, then $d(u circ f)=u’circ f df
As a consequence, constant functions’ derivatives vanish. So to get your formula, you just need to apply Leibniz’s rule and assume that in your context $r$ is a constant function.
answered Dec 15 '18 at 21:18
MindlackMindlack
4,885211
$endgroup$
To understand this, you must know that $x,r,theta,phi$ are (almost) smooth functions defined on the usual Euclidean space, and $dx,dr,dtheta,dphi$ are the derivatives of these functions.
Derivatives abide by a few rules:
Linearity: if $u,v$ are smooth and if $t$ is a scalar, then $d(u+tv)=du+tcdot dv$.
Leibniz: if $u,v$ are smooth, then $d(uv)=u dv+vdu$.
Chain rule: if $f$ is smooth and $u$ is a smooth function from $mathbb{R}$ to itself, then $d(u circ f)=u’circ f df
As a consequence, constant functions’ derivatives vanish. So to get your formula, you just need to apply Leibniz’s rule and assume that in your context $r$ is a constant function.
answered Dec 15 '18 at 21:18
MindlackMindlack
4,885211
answered Dec 15 '18 at 21:18
MindlackMindlack
4,885211
answered Dec 15 '18 at 21:18
MindlackMindlack
4,885211
answered Dec 15 '18 at 21:18
MindlackMindlack
4,885211
4,885211
add a comment |
add a comment |
$begingroup$
What are the derivative of $sin(teta)$ and $cos(phi)$.
$endgroup$
– hamam_Abdallah
Dec 15 '18 at 21:13
$begingroup$
That formula is meaningless end and of itself; it, however, is usually understood to mean what to substitute in a change of variables.
$endgroup$
– Will M.
Dec 15 '18 at 22:32