Multivariable integration $int_{y_0}^{y} alpha (x,y) dy + int_{x_0}^{x} beta (y,z) dx = 0$
$begingroup$
We got two functions given as:
$alpha(x,y)=a_1 + a_2Delta y + a_3Delta x$
$beta (y,z)= b_1 + b_2Delta y +b_3Delta z$
and I need to figure out if the coefficients $a_i$ and $b_i$ are
somehow dependent.
We also know that the $alpha$ and $beta$ functions satistify two following relationships:
$alpha (x,y) = frac {1}{z} (frac{partial z}{partial y})_x$ $;$(meaning $x$ is held constant)
$beta (y,z) = - frac {1}{z} (frac{partial z}{partial x})_y$ $;$(meaning $y$ is held constant)
And we also have that $Delta x = (x - x_0)$,$;$ $Delta y = (y - y_0)$$;$ and$;$ $Delta z = (z - z_0)$ and we examine the behaviour around the point $(x_0, y_0, z_0)$.
Attempt of the solution:
I wanted to express $z$ in both functions by the integration and then sum them.
$frac {dz}{z} = alpha (x,y) dy $ $;$ and $;$ $-frac {dz}{z} = beta (y,z) dx$
$alpha (x,y) dy + beta (y,z) dx = 0$
(So we got 2-form equal to a zero function, if it is possible to view it like this)
$int_{y_0}^{y} alpha (x,y) dy + int_{x_0}^{x} beta (y,z) dx = 0$
$ alpha (x,y)(y-y_0) + A(x) + beta (y,z) (x-x_0) + B(y,z) = 0$
where $A$ and $B$ are integration constants.
(My first question is, if I can integrate it like the $alpha$ and $beta$ functions first and after that put the expression of them in it. And how can I get the $A$, $B$)
Then I would study the dependence of the coefficients $a_i$, $b_i$:
I was thinking about putting $(x_0,y_0,z_0) = (0,0,0)$ for simplicity, if I can. Then It would look like:
$ alpha (x,y)(y) + A(x) + beta (y,z) (x) + B(y,z) = T(a_1 + a_2y + a_3x) + x(b_1 + b_2 y +b_3z)=0$
OR
$ alpha (x,y)(y-y_0) + A(x) + beta (y,z) (x-x_0) + B(y,z) = (y-y_0)(a_1 + a_2(y-y_0) + a_3(x-x_0)) + (x-x_0)(b_1 + b_2 (y-y_0) +b_3(z-z_0))=0$
and try what would happen If $x=x_0$ and so on for example.
This is a part of the thermodynamics problem actually, where the math is maybe a little bit advanced for me. I have an idea what to do, but I struggle how to make it correctly mathematically. So I would appreciate any help! $(x,y,z)$ actually corresponds to $(p,V,T)$.
real-analysis calculus integration multivariable-calculus differential-forms
asked Jan 13 at 11:56
LeifLeif
628314
$endgroup$
add a comment |
$begingroup$
We got two functions given as:
$alpha(x,y)=a_1 + a_2Delta y + a_3Delta x$
$beta (y,z)= b_1 + b_2Delta y +b_3Delta z$
and I need to figure out if the coefficients $a_i$ and $b_i$ are
somehow dependent.
We also know that the $alpha$ and $beta$ functions satistify two following relationships:
$alpha (x,y) = frac {1}{z} (frac{partial z}{partial y})_x$ $;$(meaning $x$ is held constant)
$beta (y,z) = - frac {1}{z} (frac{partial z}{partial x})_y$ $;$(meaning $y$ is held constant)
And we also have that $Delta x = (x - x_0)$,$;$ $Delta y = (y - y_0)$$;$ and$;$ $Delta z = (z - z_0)$ and we examine the behaviour around the point $(x_0, y_0, z_0)$.
Attempt of the solution:
I wanted to express $z$ in both functions by the integration and then sum them.
$frac {dz}{z} = alpha (x,y) dy $ $;$ and $;$ $-frac {dz}{z} = beta (y,z) dx$
$alpha (x,y) dy + beta (y,z) dx = 0$
(So we got 2-form equal to a zero function, if it is possible to view it like this)
$int_{y_0}^{y} alpha (x,y) dy + int_{x_0}^{x} beta (y,z) dx = 0$
$ alpha (x,y)(y-y_0) + A(x) + beta (y,z) (x-x_0) + B(y,z) = 0$
where $A$ and $B$ are integration constants.
(My first question is, if I can integrate it like the $alpha$ and $beta$ functions first and after that put the expression of them in it. And how can I get the $A$, $B$)
Then I would study the dependence of the coefficients $a_i$, $b_i$:
I was thinking about putting $(x_0,y_0,z_0) = (0,0,0)$ for simplicity, if I can. Then It would look like:
$ alpha (x,y)(y) + A(x) + beta (y,z) (x) + B(y,z) = T(a_1 + a_2y + a_3x) + x(b_1 + b_2 y +b_3z)=0$
OR
$ alpha (x,y)(y-y_0) + A(x) + beta (y,z) (x-x_0) + B(y,z) = (y-y_0)(a_1 + a_2(y-y_0) + a_3(x-x_0)) + (x-x_0)(b_1 + b_2 (y-y_0) +b_3(z-z_0))=0$
and try what would happen If $x=x_0$ and so on for example.
This is a part of the thermodynamics problem actually, where the math is maybe a little bit advanced for me. I have an idea what to do, but I struggle how to make it correctly mathematically. So I would appreciate any help! $(x,y,z)$ actually corresponds to $(p,V,T)$.
real-analysis calculus integration multivariable-calculus differential-forms
asked Jan 13 at 11:56
LeifLeif
628314
$endgroup$
$begingroup$
Something seems to be wrong here: When writing $int_{y_0}^{y}...dy$ you are using the variable $y$ twice! The same is true for the variable $x$ in the second integral.
$endgroup$
– Martin Rosenau
Jan 13 at 12:45
$begingroup$
Yes, that is actually another problem. Because the task is already given with known $(y-y_0)$. It confuses me, because it looks like it was already integrated.
$endgroup$
– Leif
Jan 13 at 12:54
add a comment |
$begingroup$
We got two functions given as:
$alpha(x,y)=a_1 + a_2Delta y + a_3Delta x$
$beta (y,z)= b_1 + b_2Delta y +b_3Delta z$
and I need to figure out if the coefficients $a_i$ and $b_i$ are
somehow dependent.
We also know that the $alpha$ and $beta$ functions satistify two following relationships:
$alpha (x,y) = frac {1}{z} (frac{partial z}{partial y})_x$ $;$(meaning $x$ is held constant)
$beta (y,z) = - frac {1}{z} (frac{partial z}{partial x})_y$ $;$(meaning $y$ is held constant)
And we also have that $Delta x = (x - x_0)$,$;$ $Delta y = (y - y_0)$$;$ and$;$ $Delta z = (z - z_0)$ and we examine the behaviour around the point $(x_0, y_0, z_0)$.
Attempt of the solution:
I wanted to express $z$ in both functions by the integration and then sum them.
$frac {dz}{z} = alpha (x,y) dy $ $;$ and $;$ $-frac {dz}{z} = beta (y,z) dx$
$alpha (x,y) dy + beta (y,z) dx = 0$
(So we got 2-form equal to a zero function, if it is possible to view it like this)
$int_{y_0}^{y} alpha (x,y) dy + int_{x_0}^{x} beta (y,z) dx = 0$
$ alpha (x,y)(y-y_0) + A(x) + beta (y,z) (x-x_0) + B(y,z) = 0$
where $A$ and $B$ are integration constants.
(My first question is, if I can integrate it like the $alpha$ and $beta$ functions first and after that put the expression of them in it. And how can I get the $A$, $B$)
Then I would study the dependence of the coefficients $a_i$, $b_i$:
I was thinking about putting $(x_0,y_0,z_0) = (0,0,0)$ for simplicity, if I can. Then It would look like:
$ alpha (x,y)(y) + A(x) + beta (y,z) (x) + B(y,z) = T(a_1 + a_2y + a_3x) + x(b_1 + b_2 y +b_3z)=0$
OR
$ alpha (x,y)(y-y_0) + A(x) + beta (y,z) (x-x_0) + B(y,z) = (y-y_0)(a_1 + a_2(y-y_0) + a_3(x-x_0)) + (x-x_0)(b_1 + b_2 (y-y_0) +b_3(z-z_0))=0$
and try what would happen If $x=x_0$ and so on for example.
This is a part of the thermodynamics problem actually, where the math is maybe a little bit advanced for me. I have an idea what to do, but I struggle how to make it correctly mathematically. So I would appreciate any help! $(x,y,z)$ actually corresponds to $(p,V,T)$.
real-analysis calculus integration multivariable-calculus differential-forms
asked Jan 13 at 11:56
LeifLeif
628314
$endgroup$
We got two functions given as:
$alpha(x,y)=a_1 + a_2Delta y + a_3Delta x$
$beta (y,z)= b_1 + b_2Delta y +b_3Delta z$
and I need to figure out if the coefficients $a_i$ and $b_i$ are
somehow dependent.
We also know that the $alpha$ and $beta$ functions satistify two following relationships:
$alpha (x,y) = frac {1}{z} (frac{partial z}{partial y})_x$ $;$(meaning $x$ is held constant)
$beta (y,z) = - frac {1}{z} (frac{partial z}{partial x})_y$ $;$(meaning $y$ is held constant)
And we also have that $Delta x = (x - x_0)$,$;$ $Delta y = (y - y_0)$$;$ and$;$ $Delta z = (z - z_0)$ and we examine the behaviour around the point $(x_0, y_0, z_0)$.
Attempt of the solution:
I wanted to express $z$ in both functions by the integration and then sum them.
$frac {dz}{z} = alpha (x,y) dy $ $;$ and $;$ $-frac {dz}{z} = beta (y,z) dx$
$alpha (x,y) dy + beta (y,z) dx = 0$
(So we got 2-form equal to a zero function, if it is possible to view it like this)
$int_{y_0}^{y} alpha (x,y) dy + int_{x_0}^{x} beta (y,z) dx = 0$
$ alpha (x,y)(y-y_0) + A(x) + beta (y,z) (x-x_0) + B(y,z) = 0$
where $A$ and $B$ are integration constants.
(My first question is, if I can integrate it like the $alpha$ and $beta$ functions first and after that put the expression of them in it. And how can I get the $A$, $B$)
Then I would study the dependence of the coefficients $a_i$, $b_i$:
I was thinking about putting $(x_0,y_0,z_0) = (0,0,0)$ for simplicity, if I can. Then It would look like:
$ alpha (x,y)(y) + A(x) + beta (y,z) (x) + B(y,z) = T(a_1 + a_2y + a_3x) + x(b_1 + b_2 y +b_3z)=0$
OR
$ alpha (x,y)(y-y_0) + A(x) + beta (y,z) (x-x_0) + B(y,z) = (y-y_0)(a_1 + a_2(y-y_0) + a_3(x-x_0)) + (x-x_0)(b_1 + b_2 (y-y_0) +b_3(z-z_0))=0$
and try what would happen If $x=x_0$ and so on for example.
This is a part of the thermodynamics problem actually, where the math is maybe a little bit advanced for me. I have an idea what to do, but I struggle how to make it correctly mathematically. So I would appreciate any help! $(x,y,z)$ actually corresponds to $(p,V,T)$.
real-analysis calculus integration multivariable-calculus differential-forms
real-analysis calculus integration multivariable-calculus differential-forms
asked Jan 13 at 11:56
LeifLeif
628314
asked Jan 13 at 11:56
LeifLeif
628314
edited Jan 13 at 12:35
Leif
asked Jan 13 at 11:56
LeifLeif
628314
asked Jan 13 at 11:56
LeifLeif
628314
asked Jan 13 at 11:56
LeifLeif
628314
628314
$begingroup$
Something seems to be wrong here: When writing $int_{y_0}^{y}...dy$ you are using the variable $y$ twice! The same is true for the variable $x$ in the second integral.
$endgroup$
– Martin Rosenau
Jan 13 at 12:45
$begingroup$
Yes, that is actually another problem. Because the task is already given with known $(y-y_0)$. It confuses me, because it looks like it was already integrated.
$endgroup$
– Leif
Jan 13 at 12:54
add a comment |
$begingroup$
Something seems to be wrong here: When writing $int_{y_0}^{y}...dy$ you are using the variable $y$ twice! The same is true for the variable $x$ in the second integral.
$endgroup$
– Martin Rosenau
Jan 13 at 12:45
$begingroup$
Yes, that is actually another problem. Because the task is already given with known $(y-y_0)$. It confuses me, because it looks like it was already integrated.
$endgroup$
– Leif
Jan 13 at 12:54
$begingroup$
Something seems to be wrong here: When writing $int_{y_0}^{y}...dy$ you are using the variable $y$ twice! The same is true for the variable $x$ in the second integral.
$endgroup$
– Martin Rosenau
Jan 13 at 12:45
$begingroup$
Something seems to be wrong here: When writing $int_{y_0}^{y}...dy$ you are using the variable $y$ twice! The same is true for the variable $x$ in the second integral.
$endgroup$
– Martin Rosenau
Jan 13 at 12:45
$begingroup$
Yes, that is actually another problem. Because the task is already given with known $(y-y_0)$. It confuses me, because it looks like it was already integrated.
$endgroup$
– Leif
Jan 13 at 12:54
$begingroup$
Yes, that is actually another problem. Because the task is already given with known $(y-y_0)$. It confuses me, because it looks like it was already integrated.
$endgroup$
– Leif
Jan 13 at 12:54
add a comment |
0
active
oldest
votes
Your Answer
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3071932%2fmultivariable-integration-int-y-0y-alpha-x-y-dy-int-x-0x-beta%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3071932%2fmultivariable-integration-int-y-0y-alpha-x-y-dy-int-x-0x-beta%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
Something seems to be wrong here: When writing $int_{y_0}^{y}...dy$ you are using the variable $y$ twice! The same is true for the variable $x$ in the second integral.
$endgroup$
– Martin Rosenau
Jan 13 at 12:45
$begingroup$
Yes, that is actually another problem. Because the task is already given with known $(y-y_0)$. It confuses me, because it looks like it was already integrated.
$endgroup$
– Leif
Jan 13 at 12:54