Mixing
The Euler-Lagrange first integral
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How to calculate $fracdphidz$ from following equation:
$$Afracd^2phidz^2+cos(phi)sin(phi)=0,$$
where $A$ is a constant?
differential-equations derivatives
asked Aug 2 at 9:45
TGram
62
add a comment |Â
up vote
1
down vote
favorite
How to calculate $fracdphidz$ from following equation:
$$Afracd^2phidz^2+cos(phi)sin(phi)=0,$$
where $A$ is a constant?
differential-equations derivatives
asked Aug 2 at 9:45
TGram
62
$int frac1sin x cos xdx$
– Math-fun
Aug 2 at 9:53
@Math-fun How is that relevant?
– Dylan
Aug 2 at 10:11
@Dylan ups, you are right!
– Math-fun
Aug 2 at 10:56
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
How to calculate $fracdphidz$ from following equation:
$$Afracd^2phidz^2+cos(phi)sin(phi)=0,$$
where $A$ is a constant?
differential-equations derivatives
asked Aug 2 at 9:45
TGram
62
How to calculate $fracdphidz$ from following equation:
$$Afracd^2phidz^2+cos(phi)sin(phi)=0,$$
where $A$ is a constant?
differential-equations derivatives
asked Aug 2 at 9:45
TGram
62
asked Aug 2 at 9:45
TGram
62
asked Aug 2 at 9:45
TGram
62
asked Aug 2 at 9:45
TGram
62
62
$int frac1sin x cos xdx$
– Math-fun
Aug 2 at 9:53
@Math-fun How is that relevant?
– Dylan
Aug 2 at 10:11
@Dylan ups, you are right!
– Math-fun
Aug 2 at 10:56
add a comment |Â
$int frac1sin x cos xdx$
– Math-fun
Aug 2 at 9:53
@Math-fun How is that relevant?
– Dylan
Aug 2 at 10:11
@Dylan ups, you are right!
– Math-fun
Aug 2 at 10:56
$int frac1sin x cos xdx$
– Math-fun
Aug 2 at 9:53
$int frac1sin x cos xdx$
– Math-fun
Aug 2 at 9:53
@Math-fun How is that relevant?
– Dylan
Aug 2 at 10:11
@Math-fun How is that relevant?
– Dylan
Aug 2 at 10:11
@Dylan ups, you are right!
– Math-fun
Aug 2 at 10:56
@Dylan ups, you are right!
– Math-fun
Aug 2 at 10:56
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
0
down vote
Multiply both sides by $fracdphidz$ and then integrate (using the chain rule to help...).
answered Aug 2 at 10:02
bobcliffe
435
add a comment |Â
up vote
0
down vote
Since $2Aphi'phi''+phi'sin 2phi=0$, $C:=Aphi'^2-frac12cos 2phi$ is constant. In particular $z=intdfracdphisqrt(C+frac12cos 2phi)/A$, which you can reduce to an elliptic integral of the first kind. Inverting this gives $phi$ in terms of $z$, or if you just want $phi'=(dz/dphi)^-1$ in terms of $phi$ no elliptic integrals are needed.
answered Aug 2 at 10:20
J.G.
12.9k11423
In answer sheet it says that the solution is $left(fracdphidzright)^2=frac1A^2left(frac1C^2-sin^2(phi)right)$ (I assume $C$ comes from integration).
– TGram
Aug 2 at 10:34
@TGram I can't see how a $1/A^2$ ended up in there; either that exponent is a misprint on their part, or in your question you copied out the problem without it. But their answer is basically mine; note $frac12cos 2phi=frac12-sin^2phi$. This is in fact how it reduces to using an elliptic integral.
– J.G.
Aug 2 at 10:46
1
I wrongly copied solution. It should solely by $A$
– TGram
Aug 2 at 10:50
add a comment |Â
up vote
0
down vote
$$Afracd^2phidz^2+cos(phi)sin(phi)=0,$$
Substitute $$p=frac d phidz implies fracd^2phidz^2=pfrac dp d phi $$
$$App'=-cos(phi)sin(phi)$$
$$Aint pdp=-frac 12int sin(2phi) dphi$$
$$Ap^2= frac 12cos (2phi) +K_1$$
$$frac d phidz =pm sqrt frac cos (2phi) +K_1 2A$$
$$.......$$
@tgram
$$Ap^2= frac 12cos (2phi) +K_1$$
note that
$$cos (2phi)=2cos^2 phi -1=2(1-sin^2 phi )-1=1-2sin^2 phi$$
The constant $K_1$ absorb all the constants...
And $$p=frac d phidz$$
answered Aug 2 at 10:33
Isham
10.5k3829
In answer sheet it says that the solution is $left(fracdphidzright)^2=frac1Aleft(frac1C^2-sin^2(phi)right)$ (I assume $C$ comes from integration).
– TGram
Aug 2 at 10:42
It cant be $A^2$ it's A .... you can change the sin function by $1-cos^2 phi$ the constant will absorb the number 1 @TGram
– Isham
Aug 2 at 10:43
@TGram I added some line you should get the same result as that of your book
– Isham
Aug 2 at 10:52
Thanks, however I am still a bit confused regarding the origin of $C^2$.
– TGram
Aug 2 at 10:58
@TGram it's just a constant thats all writing it $1/C$ or C dosent change that much
– Isham
Aug 2 at 11:19
add a comment |Â
up vote
0
down vote
$$
Aphi'phi'' +cosphisinphiphi' = 0Rightarrow frac 12A(phi'^2)'+frac 12(sin^2phi)' = 0
$$
hence
$$
Aphi'^2+sin^2phi = C_1Rightarrow phi' = pmfrac1sqrtAsqrtC_1-sin^2phi
$$
answered Aug 2 at 11:43
Cesareo
5,5012412
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Multiply both sides by $fracdphidz$ and then integrate (using the chain rule to help...).
answered Aug 2 at 10:02
bobcliffe
435
add a comment |Â
up vote
0
down vote
Multiply both sides by $fracdphidz$ and then integrate (using the chain rule to help...).
answered Aug 2 at 10:02
bobcliffe
435
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Multiply both sides by $fracdphidz$ and then integrate (using the chain rule to help...).
answered Aug 2 at 10:02
bobcliffe
435
Multiply both sides by $fracdphidz$ and then integrate (using the chain rule to help...).
answered Aug 2 at 10:02
bobcliffe
435
edited Aug 2 at 10:08
answered Aug 2 at 10:02
bobcliffe
435
answered Aug 2 at 10:02
bobcliffe
435
answered Aug 2 at 10:02
bobcliffe
435
435
add a comment |Â
add a comment |Â
up vote
0
down vote
Since $2Aphi'phi''+phi'sin 2phi=0$, $C:=Aphi'^2-frac12cos 2phi$ is constant. In particular $z=intdfracdphisqrt(C+frac12cos 2phi)/A$, which you can reduce to an elliptic integral of the first kind. Inverting this gives $phi$ in terms of $z$, or if you just want $phi'=(dz/dphi)^-1$ in terms of $phi$ no elliptic integrals are needed.
answered Aug 2 at 10:20
J.G.
12.9k11423
In answer sheet it says that the solution is $left(fracdphidzright)^2=frac1A^2left(frac1C^2-sin^2(phi)right)$ (I assume $C$ comes from integration).
– TGram
Aug 2 at 10:34
@TGram I can't see how a $1/A^2$ ended up in there; either that exponent is a misprint on their part, or in your question you copied out the problem without it. But their answer is basically mine; note $frac12cos 2phi=frac12-sin^2phi$. This is in fact how it reduces to using an elliptic integral.
– J.G.
Aug 2 at 10:46
1
I wrongly copied solution. It should solely by $A$
– TGram
Aug 2 at 10:50
add a comment |Â
up vote
0
down vote
Since $2Aphi'phi''+phi'sin 2phi=0$, $C:=Aphi'^2-frac12cos 2phi$ is constant. In particular $z=intdfracdphisqrt(C+frac12cos 2phi)/A$, which you can reduce to an elliptic integral of the first kind. Inverting this gives $phi$ in terms of $z$, or if you just want $phi'=(dz/dphi)^-1$ in terms of $phi$ no elliptic integrals are needed.
answered Aug 2 at 10:20
J.G.
12.9k11423
In answer sheet it says that the solution is $left(fracdphidzright)^2=frac1A^2left(frac1C^2-sin^2(phi)right)$ (I assume $C$ comes from integration).
– TGram
Aug 2 at 10:34
@TGram I can't see how a $1/A^2$ ended up in there; either that exponent is a misprint on their part, or in your question you copied out the problem without it. But their answer is basically mine; note $frac12cos 2phi=frac12-sin^2phi$. This is in fact how it reduces to using an elliptic integral.
– J.G.
Aug 2 at 10:46
1
I wrongly copied solution. It should solely by $A$
– TGram
Aug 2 at 10:50
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Since $2Aphi'phi''+phi'sin 2phi=0$, $C:=Aphi'^2-frac12cos 2phi$ is constant. In particular $z=intdfracdphisqrt(C+frac12cos 2phi)/A$, which you can reduce to an elliptic integral of the first kind. Inverting this gives $phi$ in terms of $z$, or if you just want $phi'=(dz/dphi)^-1$ in terms of $phi$ no elliptic integrals are needed.
answered Aug 2 at 10:20
J.G.
12.9k11423
Since $2Aphi'phi''+phi'sin 2phi=0$, $C:=Aphi'^2-frac12cos 2phi$ is constant. In particular $z=intdfracdphisqrt(C+frac12cos 2phi)/A$, which you can reduce to an elliptic integral of the first kind. Inverting this gives $phi$ in terms of $z$, or if you just want $phi'=(dz/dphi)^-1$ in terms of $phi$ no elliptic integrals are needed.
answered Aug 2 at 10:20
J.G.
12.9k11423
answered Aug 2 at 10:20
J.G.
12.9k11423
answered Aug 2 at 10:20
J.G.
12.9k11423
answered Aug 2 at 10:20
J.G.
12.9k11423
12.9k11423
In answer sheet it says that the solution is $left(fracdphidzright)^2=frac1A^2left(frac1C^2-sin^2(phi)right)$ (I assume $C$ comes from integration).
– TGram
Aug 2 at 10:34
@TGram I can't see how a $1/A^2$ ended up in there; either that exponent is a misprint on their part, or in your question you copied out the problem without it. But their answer is basically mine; note $frac12cos 2phi=frac12-sin^2phi$. This is in fact how it reduces to using an elliptic integral.
– J.G.
Aug 2 at 10:46
1
I wrongly copied solution. It should solely by $A$
– TGram
Aug 2 at 10:50
add a comment |Â
In answer sheet it says that the solution is $left(fracdphidzright)^2=frac1A^2left(frac1C^2-sin^2(phi)right)$ (I assume $C$ comes from integration).
– TGram
Aug 2 at 10:34
@TGram I can't see how a $1/A^2$ ended up in there; either that exponent is a misprint on their part, or in your question you copied out the problem without it. But their answer is basically mine; note $frac12cos 2phi=frac12-sin^2phi$. This is in fact how it reduces to using an elliptic integral.
– J.G.
Aug 2 at 10:46
1
I wrongly copied solution. It should solely by $A$
– TGram
Aug 2 at 10:50
In answer sheet it says that the solution is $left(fracdphidzright)^2=frac1A^2left(frac1C^2-sin^2(phi)right)$ (I assume $C$ comes from integration).
– TGram
Aug 2 at 10:34
In answer sheet it says that the solution is $left(fracdphidzright)^2=frac1A^2left(frac1C^2-sin^2(phi)right)$ (I assume $C$ comes from integration).
– TGram
Aug 2 at 10:34
@TGram I can't see how a $1/A^2$ ended up in there; either that exponent is a misprint on their part, or in your question you copied out the problem without it. But their answer is basically mine; note $frac12cos 2phi=frac12-sin^2phi$. This is in fact how it reduces to using an elliptic integral.
– J.G.
Aug 2 at 10:46
@TGram I can't see how a $1/A^2$ ended up in there; either that exponent is a misprint on their part, or in your question you copied out the problem without it. But their answer is basically mine; note $frac12cos 2phi=frac12-sin^2phi$. This is in fact how it reduces to using an elliptic integral.
– J.G.
Aug 2 at 10:46
1
1
I wrongly copied solution. It should solely by $A$
– TGram
Aug 2 at 10:50
I wrongly copied solution. It should solely by $A$
– TGram
Aug 2 at 10:50
add a comment |Â
up vote
0
down vote
$$Afracd^2phidz^2+cos(phi)sin(phi)=0,$$
Substitute $$p=frac d phidz implies fracd^2phidz^2=pfrac dp d phi $$
$$App'=-cos(phi)sin(phi)$$
$$Aint pdp=-frac 12int sin(2phi) dphi$$
$$Ap^2= frac 12cos (2phi) +K_1$$
$$frac d phidz =pm sqrt frac cos (2phi) +K_1 2A$$
$$.......$$
@tgram
$$Ap^2= frac 12cos (2phi) +K_1$$
note that
$$cos (2phi)=2cos^2 phi -1=2(1-sin^2 phi )-1=1-2sin^2 phi$$
The constant $K_1$ absorb all the constants...
And $$p=frac d phidz$$
answered Aug 2 at 10:33
Isham
10.5k3829
In answer sheet it says that the solution is $left(fracdphidzright)^2=frac1Aleft(frac1C^2-sin^2(phi)right)$ (I assume $C$ comes from integration).
– TGram
Aug 2 at 10:42
It cant be $A^2$ it's A .... you can change the sin function by $1-cos^2 phi$ the constant will absorb the number 1 @TGram
– Isham
Aug 2 at 10:43
@TGram I added some line you should get the same result as that of your book
– Isham
Aug 2 at 10:52
Thanks, however I am still a bit confused regarding the origin of $C^2$.
– TGram
Aug 2 at 10:58
@TGram it's just a constant thats all writing it $1/C$ or C dosent change that much
– Isham
Aug 2 at 11:19
add a comment |Â
up vote
0
down vote
$$Afracd^2phidz^2+cos(phi)sin(phi)=0,$$
Substitute $$p=frac d phidz implies fracd^2phidz^2=pfrac dp d phi $$
$$App'=-cos(phi)sin(phi)$$
$$Aint pdp=-frac 12int sin(2phi) dphi$$
$$Ap^2= frac 12cos (2phi) +K_1$$
$$frac d phidz =pm sqrt frac cos (2phi) +K_1 2A$$
$$.......$$
@tgram
$$Ap^2= frac 12cos (2phi) +K_1$$
note that
$$cos (2phi)=2cos^2 phi -1=2(1-sin^2 phi )-1=1-2sin^2 phi$$
The constant $K_1$ absorb all the constants...
And $$p=frac d phidz$$
answered Aug 2 at 10:33
Isham
10.5k3829
In answer sheet it says that the solution is $left(fracdphidzright)^2=frac1Aleft(frac1C^2-sin^2(phi)right)$ (I assume $C$ comes from integration).
– TGram
Aug 2 at 10:42
It cant be $A^2$ it's A .... you can change the sin function by $1-cos^2 phi$ the constant will absorb the number 1 @TGram
– Isham
Aug 2 at 10:43
@TGram I added some line you should get the same result as that of your book
– Isham
Aug 2 at 10:52
Thanks, however I am still a bit confused regarding the origin of $C^2$.
– TGram
Aug 2 at 10:58
@TGram it's just a constant thats all writing it $1/C$ or C dosent change that much
– Isham
Aug 2 at 11:19
add a comment |Â
up vote
0
down vote
up vote
0
down vote
$$Afracd^2phidz^2+cos(phi)sin(phi)=0,$$
Substitute $$p=frac d phidz implies fracd^2phidz^2=pfrac dp d phi $$
$$App'=-cos(phi)sin(phi)$$
$$Aint pdp=-frac 12int sin(2phi) dphi$$
$$Ap^2= frac 12cos (2phi) +K_1$$
$$frac d phidz =pm sqrt frac cos (2phi) +K_1 2A$$
$$.......$$
@tgram
$$Ap^2= frac 12cos (2phi) +K_1$$
note that
$$cos (2phi)=2cos^2 phi -1=2(1-sin^2 phi )-1=1-2sin^2 phi$$
The constant $K_1$ absorb all the constants...
And $$p=frac d phidz$$
answered Aug 2 at 10:33
Isham
10.5k3829
$$Afracd^2phidz^2+cos(phi)sin(phi)=0,$$
Substitute $$p=frac d phidz implies fracd^2phidz^2=pfrac dp d phi $$
$$App'=-cos(phi)sin(phi)$$
$$Aint pdp=-frac 12int sin(2phi) dphi$$
$$Ap^2= frac 12cos (2phi) +K_1$$
$$frac d phidz =pm sqrt frac cos (2phi) +K_1 2A$$
$$.......$$
@tgram
$$Ap^2= frac 12cos (2phi) +K_1$$
note that
$$cos (2phi)=2cos^2 phi -1=2(1-sin^2 phi )-1=1-2sin^2 phi$$
The constant $K_1$ absorb all the constants...
And $$p=frac d phidz$$
answered Aug 2 at 10:33
Isham
10.5k3829
edited Aug 2 at 10:51
answered Aug 2 at 10:33
Isham
10.5k3829
answered Aug 2 at 10:33
Isham
10.5k3829
answered Aug 2 at 10:33
Isham
10.5k3829
10.5k3829
In answer sheet it says that the solution is $left(fracdphidzright)^2=frac1Aleft(frac1C^2-sin^2(phi)right)$ (I assume $C$ comes from integration).
– TGram
Aug 2 at 10:42
It cant be $A^2$ it's A .... you can change the sin function by $1-cos^2 phi$ the constant will absorb the number 1 @TGram
– Isham
Aug 2 at 10:43
@TGram I added some line you should get the same result as that of your book
– Isham
Aug 2 at 10:52
Thanks, however I am still a bit confused regarding the origin of $C^2$.
– TGram
Aug 2 at 10:58
@TGram it's just a constant thats all writing it $1/C$ or C dosent change that much
– Isham
Aug 2 at 11:19
add a comment |Â
In answer sheet it says that the solution is $left(fracdphidzright)^2=frac1Aleft(frac1C^2-sin^2(phi)right)$ (I assume $C$ comes from integration).
– TGram
Aug 2 at 10:42
It cant be $A^2$ it's A .... you can change the sin function by $1-cos^2 phi$ the constant will absorb the number 1 @TGram
– Isham
Aug 2 at 10:43
@TGram I added some line you should get the same result as that of your book
– Isham
Aug 2 at 10:52
Thanks, however I am still a bit confused regarding the origin of $C^2$.
– TGram
Aug 2 at 10:58
@TGram it's just a constant thats all writing it $1/C$ or C dosent change that much
– Isham
Aug 2 at 11:19
In answer sheet it says that the solution is $left(fracdphidzright)^2=frac1Aleft(frac1C^2-sin^2(phi)right)$ (I assume $C$ comes from integration).
– TGram
Aug 2 at 10:42
In answer sheet it says that the solution is $left(fracdphidzright)^2=frac1Aleft(frac1C^2-sin^2(phi)right)$ (I assume $C$ comes from integration).
– TGram
Aug 2 at 10:42
It cant be $A^2$ it's A .... you can change the sin function by $1-cos^2 phi$ the constant will absorb the number 1 @TGram
– Isham
Aug 2 at 10:43
It cant be $A^2$ it's A .... you can change the sin function by $1-cos^2 phi$ the constant will absorb the number 1 @TGram
– Isham
Aug 2 at 10:43
@TGram I added some line you should get the same result as that of your book
– Isham
Aug 2 at 10:52
@TGram I added some line you should get the same result as that of your book
– Isham
Aug 2 at 10:52
Thanks, however I am still a bit confused regarding the origin of $C^2$.
– TGram
Aug 2 at 10:58
Thanks, however I am still a bit confused regarding the origin of $C^2$.
– TGram
Aug 2 at 10:58
@TGram it's just a constant thats all writing it $1/C$ or C dosent change that much
– Isham
Aug 2 at 11:19
@TGram it's just a constant thats all writing it $1/C$ or C dosent change that much
– Isham
Aug 2 at 11:19
add a comment |Â
up vote
0
down vote
$$
Aphi'phi'' +cosphisinphiphi' = 0Rightarrow frac 12A(phi'^2)'+frac 12(sin^2phi)' = 0
$$
hence
$$
Aphi'^2+sin^2phi = C_1Rightarrow phi' = pmfrac1sqrtAsqrtC_1-sin^2phi
$$
answered Aug 2 at 11:43
Cesareo
5,5012412
add a comment |Â
up vote
0
down vote
$$
Aphi'phi'' +cosphisinphiphi' = 0Rightarrow frac 12A(phi'^2)'+frac 12(sin^2phi)' = 0
$$
hence
$$
Aphi'^2+sin^2phi = C_1Rightarrow phi' = pmfrac1sqrtAsqrtC_1-sin^2phi
$$
answered Aug 2 at 11:43
Cesareo
5,5012412
add a comment |Â
up vote
0
down vote
up vote
0
down vote
$$
Aphi'phi'' +cosphisinphiphi' = 0Rightarrow frac 12A(phi'^2)'+frac 12(sin^2phi)' = 0
$$
hence
$$
Aphi'^2+sin^2phi = C_1Rightarrow phi' = pmfrac1sqrtAsqrtC_1-sin^2phi
$$
answered Aug 2 at 11:43
Cesareo
5,5012412
$$
Aphi'phi'' +cosphisinphiphi' = 0Rightarrow frac 12A(phi'^2)'+frac 12(sin^2phi)' = 0
$$
hence
$$
Aphi'^2+sin^2phi = C_1Rightarrow phi' = pmfrac1sqrtAsqrtC_1-sin^2phi
$$
answered Aug 2 at 11:43
Cesareo
5,5012412
answered Aug 2 at 11:43
Cesareo
5,5012412
answered Aug 2 at 11:43
Cesareo
5,5012412
answered Aug 2 at 11:43
Cesareo
5,5012412
5,5012412
add a comment |Â
add a comment |Â
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$int frac1sin x cos xdx$
– Math-fun
Aug 2 at 9:53
@Math-fun How is that relevant?
– Dylan
Aug 2 at 10:11
@Dylan ups, you are right!
– Math-fun
Aug 2 at 10:56