Mixing
Partial Derivative and fraction
In the George F Simmons Calculus with Analytical Geometry, textbook, it is mentioned that the Partial derivatives of a function cannot be treated as fractions as in case of single variable function.
In the example provided it has been proved that,
from
$pV = nRT$
where,
- p - pressure
- T - Temperature
- V - Volume
- n - number of moles
- R - Universal gas constant
$dfrac{partial{p}}{partial{T}} dfrac{partial{T}}{partial{V}} dfrac{partial{V}}{partial{p}} = -1$
which is fine, but in the explanation, it is given as,
as the right hand side is -1 and not 1 we cannot treat the LHS of the above statement as fractions
If I am not wrong the 1st statement would mean, we cannot expand a partial differential as say,
$dfrac{d y}{d t} = dfrac{dy}{dx}dfrac{dx}{dt}$, where, $y=y(x)$ and $x = x(t)$,for a single variable function and we cannot represent the same for partial differentials the same manner if $y = y(x,t)$
What what is the significance of +1 or -1 for a differential to represented as a fraction?
pde partial-derivative
asked Jan 1 at 5:23
RaptorRaptor
3317
add a comment |
In the George F Simmons Calculus with Analytical Geometry, textbook, it is mentioned that the Partial derivatives of a function cannot be treated as fractions as in case of single variable function.
In the example provided it has been proved that,
from
$pV = nRT$
where,
- p - pressure
- T - Temperature
- V - Volume
- n - number of moles
- R - Universal gas constant
$dfrac{partial{p}}{partial{T}} dfrac{partial{T}}{partial{V}} dfrac{partial{V}}{partial{p}} = -1$
which is fine, but in the explanation, it is given as,
as the right hand side is -1 and not 1 we cannot treat the LHS of the above statement as fractions
If I am not wrong the 1st statement would mean, we cannot expand a partial differential as say,
$dfrac{d y}{d t} = dfrac{dy}{dx}dfrac{dx}{dt}$, where, $y=y(x)$ and $x = x(t)$,for a single variable function and we cannot represent the same for partial differentials the same manner if $y = y(x,t)$
What what is the significance of +1 or -1 for a differential to represented as a fraction?
pde partial-derivative
asked Jan 1 at 5:23
RaptorRaptor
3317
add a comment |
In the George F Simmons Calculus with Analytical Geometry, textbook, it is mentioned that the Partial derivatives of a function cannot be treated as fractions as in case of single variable function.
In the example provided it has been proved that,
from
$pV = nRT$
where,
- p - pressure
- T - Temperature
- V - Volume
- n - number of moles
- R - Universal gas constant
$dfrac{partial{p}}{partial{T}} dfrac{partial{T}}{partial{V}} dfrac{partial{V}}{partial{p}} = -1$
which is fine, but in the explanation, it is given as,
as the right hand side is -1 and not 1 we cannot treat the LHS of the above statement as fractions
If I am not wrong the 1st statement would mean, we cannot expand a partial differential as say,
$dfrac{d y}{d t} = dfrac{dy}{dx}dfrac{dx}{dt}$, where, $y=y(x)$ and $x = x(t)$,for a single variable function and we cannot represent the same for partial differentials the same manner if $y = y(x,t)$
What what is the significance of +1 or -1 for a differential to represented as a fraction?
pde partial-derivative
asked Jan 1 at 5:23
RaptorRaptor
3317
In the George F Simmons Calculus with Analytical Geometry, textbook, it is mentioned that the Partial derivatives of a function cannot be treated as fractions as in case of single variable function.
In the example provided it has been proved that,
from
$pV = nRT$
where,
- p - pressure
- T - Temperature
- V - Volume
- n - number of moles
- R - Universal gas constant
$dfrac{partial{p}}{partial{T}} dfrac{partial{T}}{partial{V}} dfrac{partial{V}}{partial{p}} = -1$
which is fine, but in the explanation, it is given as,
as the right hand side is -1 and not 1 we cannot treat the LHS of the above statement as fractions
If I am not wrong the 1st statement would mean, we cannot expand a partial differential as say,
$dfrac{d y}{d t} = dfrac{dy}{dx}dfrac{dx}{dt}$, where, $y=y(x)$ and $x = x(t)$,for a single variable function and we cannot represent the same for partial differentials the same manner if $y = y(x,t)$
What what is the significance of +1 or -1 for a differential to represented as a fraction?
pde partial-derivative
pde partial-derivative
asked Jan 1 at 5:23
RaptorRaptor
3317
asked Jan 1 at 5:23
RaptorRaptor
3317
edited Jan 1 at 5:34
Raptor
asked Jan 1 at 5:23
RaptorRaptor
3317
asked Jan 1 at 5:23
RaptorRaptor
3317
asked Jan 1 at 5:23
RaptorRaptor
3317
3317
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
Yes. $$dfrac{partial{x}}{partial{y}} dfrac{partial{y}}{partial{x}} neq dfrac{partial{x}}{partial{z}}$$
The equation above is an example to this phenomenon.
answered Jan 1 at 5:44
moutheticsmouthetics
50127
Does it just mean that the whole thing wont cancel out to give us 1 thus, cannot be represented as a fraction?
– Raptor
Jan 1 at 5:47
Yes true. You can't think of them as fraction, otherwise the product must be $1$ which is not the case.
– mouthetics
Jan 1 at 5:50
add a comment |
Although there's nothing wrong with the answer from mouthetics I would like to expand on something: ultimately what allows you to go from the lhs to rhs is the chain rule. For example, if $y = y(x, t)$ and $x = x(t)$ then
$$
frac{{rm d}y}{{rm d} t} = frac{partial y}{partial x}frac{{rm d}x}{{rm d}t} + frac{partial y}{partial t}
$$
However, consider this other case. Imagine a function $f(x, y, z) = 0$. Now suppose you can invert this relation to obtain $z = z(x, y)$, if you apply the chain rule you will get
$$
{rm d}z = left(frac{partial z}{partial x}right)_{y}{rm d}x + left(frac{partial z}{partial y}right)_x{rm d}y tag{1}
$$
where I have used the script to indicate that variable remains fixed. Suppose now that you move along a path on the surface $f(x,y,z) = 0$, where $z$ is kept constant. In this case you can also invert the function $f$ to obtain $y = y(x)$ and
$$
{rm d}y = left(frac{partial y}{partial x}right)_z{rm d}x tag{2}
$$
Here it is more obvious why this notation is useful, it emphasizes the fact that we keep $z$ constant. Replace that in (1) with ${rm d}z = 0$ (remember, $z$ is constant) and you will get
$$
0 = left(frac{partial z}{partial x}right)_{y}{rm d}x + left(frac{partial z}{partial y}right)_x left(frac{partial y}{partial x}right)_z{rm d}x tag{3}
$$
And from here is trivial to obtain
$$
left(frac{partial x}{partial y}right)_z left(frac{partial y}{partial z}right)_x left(frac{partial z}{partial x}right)_y = -1 tag{4}
$$
Note that this is a result of consistently using the chain rule, and keeping track what function depends on what variables.
In your case just set $x = P, y = V, z = T$, the function $f(x,y,z)$ is called a equation of state
answered Jan 1 at 10:49
caveraccaverac
14.1k21130
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
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%2f3058230%2fpartial-derivative-and-fraction%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Yes. $$dfrac{partial{x}}{partial{y}} dfrac{partial{y}}{partial{x}} neq dfrac{partial{x}}{partial{z}}$$
The equation above is an example to this phenomenon.
answered Jan 1 at 5:44
moutheticsmouthetics
50127
Does it just mean that the whole thing wont cancel out to give us 1 thus, cannot be represented as a fraction?
– Raptor
Jan 1 at 5:47
Yes true. You can't think of them as fraction, otherwise the product must be $1$ which is not the case.
– mouthetics
Jan 1 at 5:50
add a comment |
Yes. $$dfrac{partial{x}}{partial{y}} dfrac{partial{y}}{partial{x}} neq dfrac{partial{x}}{partial{z}}$$
The equation above is an example to this phenomenon.
answered Jan 1 at 5:44
moutheticsmouthetics
50127
Does it just mean that the whole thing wont cancel out to give us 1 thus, cannot be represented as a fraction?
– Raptor
Jan 1 at 5:47
Yes true. You can't think of them as fraction, otherwise the product must be $1$ which is not the case.
– mouthetics
Jan 1 at 5:50
add a comment |
Yes. $$dfrac{partial{x}}{partial{y}} dfrac{partial{y}}{partial{x}} neq dfrac{partial{x}}{partial{z}}$$
The equation above is an example to this phenomenon.
answered Jan 1 at 5:44
moutheticsmouthetics
50127
Yes. $$dfrac{partial{x}}{partial{y}} dfrac{partial{y}}{partial{x}} neq dfrac{partial{x}}{partial{z}}$$
The equation above is an example to this phenomenon.
answered Jan 1 at 5:44
moutheticsmouthetics
50127
answered Jan 1 at 5:44
moutheticsmouthetics
50127
answered Jan 1 at 5:44
moutheticsmouthetics
50127
answered Jan 1 at 5:44
moutheticsmouthetics
50127
50127
Does it just mean that the whole thing wont cancel out to give us 1 thus, cannot be represented as a fraction?
– Raptor
Jan 1 at 5:47
Yes true. You can't think of them as fraction, otherwise the product must be $1$ which is not the case.
– mouthetics
Jan 1 at 5:50
add a comment |
Does it just mean that the whole thing wont cancel out to give us 1 thus, cannot be represented as a fraction?
– Raptor
Jan 1 at 5:47
Yes true. You can't think of them as fraction, otherwise the product must be $1$ which is not the case.
– mouthetics
Jan 1 at 5:50
Does it just mean that the whole thing wont cancel out to give us 1 thus, cannot be represented as a fraction?
– Raptor
Jan 1 at 5:47
Does it just mean that the whole thing wont cancel out to give us 1 thus, cannot be represented as a fraction?
– Raptor
Jan 1 at 5:47
Yes true. You can't think of them as fraction, otherwise the product must be $1$ which is not the case.
– mouthetics
Jan 1 at 5:50
Yes true. You can't think of them as fraction, otherwise the product must be $1$ which is not the case.
– mouthetics
Jan 1 at 5:50
add a comment |
Although there's nothing wrong with the answer from mouthetics I would like to expand on something: ultimately what allows you to go from the lhs to rhs is the chain rule. For example, if $y = y(x, t)$ and $x = x(t)$ then
$$
frac{{rm d}y}{{rm d} t} = frac{partial y}{partial x}frac{{rm d}x}{{rm d}t} + frac{partial y}{partial t}
$$
However, consider this other case. Imagine a function $f(x, y, z) = 0$. Now suppose you can invert this relation to obtain $z = z(x, y)$, if you apply the chain rule you will get
$$
{rm d}z = left(frac{partial z}{partial x}right)_{y}{rm d}x + left(frac{partial z}{partial y}right)_x{rm d}y tag{1}
$$
where I have used the script to indicate that variable remains fixed. Suppose now that you move along a path on the surface $f(x,y,z) = 0$, where $z$ is kept constant. In this case you can also invert the function $f$ to obtain $y = y(x)$ and
$$
{rm d}y = left(frac{partial y}{partial x}right)_z{rm d}x tag{2}
$$
Here it is more obvious why this notation is useful, it emphasizes the fact that we keep $z$ constant. Replace that in (1) with ${rm d}z = 0$ (remember, $z$ is constant) and you will get
$$
0 = left(frac{partial z}{partial x}right)_{y}{rm d}x + left(frac{partial z}{partial y}right)_x left(frac{partial y}{partial x}right)_z{rm d}x tag{3}
$$
And from here is trivial to obtain
$$
left(frac{partial x}{partial y}right)_z left(frac{partial y}{partial z}right)_x left(frac{partial z}{partial x}right)_y = -1 tag{4}
$$
Note that this is a result of consistently using the chain rule, and keeping track what function depends on what variables.
In your case just set $x = P, y = V, z = T$, the function $f(x,y,z)$ is called a equation of state
answered Jan 1 at 10:49
caveraccaverac
14.1k21130
add a comment |
Although there's nothing wrong with the answer from mouthetics I would like to expand on something: ultimately what allows you to go from the lhs to rhs is the chain rule. For example, if $y = y(x, t)$ and $x = x(t)$ then
$$
frac{{rm d}y}{{rm d} t} = frac{partial y}{partial x}frac{{rm d}x}{{rm d}t} + frac{partial y}{partial t}
$$
However, consider this other case. Imagine a function $f(x, y, z) = 0$. Now suppose you can invert this relation to obtain $z = z(x, y)$, if you apply the chain rule you will get
$$
{rm d}z = left(frac{partial z}{partial x}right)_{y}{rm d}x + left(frac{partial z}{partial y}right)_x{rm d}y tag{1}
$$
where I have used the script to indicate that variable remains fixed. Suppose now that you move along a path on the surface $f(x,y,z) = 0$, where $z$ is kept constant. In this case you can also invert the function $f$ to obtain $y = y(x)$ and
$$
{rm d}y = left(frac{partial y}{partial x}right)_z{rm d}x tag{2}
$$
Here it is more obvious why this notation is useful, it emphasizes the fact that we keep $z$ constant. Replace that in (1) with ${rm d}z = 0$ (remember, $z$ is constant) and you will get
$$
0 = left(frac{partial z}{partial x}right)_{y}{rm d}x + left(frac{partial z}{partial y}right)_x left(frac{partial y}{partial x}right)_z{rm d}x tag{3}
$$
And from here is trivial to obtain
$$
left(frac{partial x}{partial y}right)_z left(frac{partial y}{partial z}right)_x left(frac{partial z}{partial x}right)_y = -1 tag{4}
$$
Note that this is a result of consistently using the chain rule, and keeping track what function depends on what variables.
In your case just set $x = P, y = V, z = T$, the function $f(x,y,z)$ is called a equation of state
answered Jan 1 at 10:49
caveraccaverac
14.1k21130
add a comment |
Although there's nothing wrong with the answer from mouthetics I would like to expand on something: ultimately what allows you to go from the lhs to rhs is the chain rule. For example, if $y = y(x, t)$ and $x = x(t)$ then
$$
frac{{rm d}y}{{rm d} t} = frac{partial y}{partial x}frac{{rm d}x}{{rm d}t} + frac{partial y}{partial t}
$$
However, consider this other case. Imagine a function $f(x, y, z) = 0$. Now suppose you can invert this relation to obtain $z = z(x, y)$, if you apply the chain rule you will get
$$
{rm d}z = left(frac{partial z}{partial x}right)_{y}{rm d}x + left(frac{partial z}{partial y}right)_x{rm d}y tag{1}
$$
where I have used the script to indicate that variable remains fixed. Suppose now that you move along a path on the surface $f(x,y,z) = 0$, where $z$ is kept constant. In this case you can also invert the function $f$ to obtain $y = y(x)$ and
$$
{rm d}y = left(frac{partial y}{partial x}right)_z{rm d}x tag{2}
$$
Here it is more obvious why this notation is useful, it emphasizes the fact that we keep $z$ constant. Replace that in (1) with ${rm d}z = 0$ (remember, $z$ is constant) and you will get
$$
0 = left(frac{partial z}{partial x}right)_{y}{rm d}x + left(frac{partial z}{partial y}right)_x left(frac{partial y}{partial x}right)_z{rm d}x tag{3}
$$
And from here is trivial to obtain
$$
left(frac{partial x}{partial y}right)_z left(frac{partial y}{partial z}right)_x left(frac{partial z}{partial x}right)_y = -1 tag{4}
$$
Note that this is a result of consistently using the chain rule, and keeping track what function depends on what variables.
In your case just set $x = P, y = V, z = T$, the function $f(x,y,z)$ is called a equation of state
answered Jan 1 at 10:49
caveraccaverac
14.1k21130
Although there's nothing wrong with the answer from mouthetics I would like to expand on something: ultimately what allows you to go from the lhs to rhs is the chain rule. For example, if $y = y(x, t)$ and $x = x(t)$ then
$$
frac{{rm d}y}{{rm d} t} = frac{partial y}{partial x}frac{{rm d}x}{{rm d}t} + frac{partial y}{partial t}
$$
However, consider this other case. Imagine a function $f(x, y, z) = 0$. Now suppose you can invert this relation to obtain $z = z(x, y)$, if you apply the chain rule you will get
$$
{rm d}z = left(frac{partial z}{partial x}right)_{y}{rm d}x + left(frac{partial z}{partial y}right)_x{rm d}y tag{1}
$$
where I have used the script to indicate that variable remains fixed. Suppose now that you move along a path on the surface $f(x,y,z) = 0$, where $z$ is kept constant. In this case you can also invert the function $f$ to obtain $y = y(x)$ and
$$
{rm d}y = left(frac{partial y}{partial x}right)_z{rm d}x tag{2}
$$
Here it is more obvious why this notation is useful, it emphasizes the fact that we keep $z$ constant. Replace that in (1) with ${rm d}z = 0$ (remember, $z$ is constant) and you will get
$$
0 = left(frac{partial z}{partial x}right)_{y}{rm d}x + left(frac{partial z}{partial y}right)_x left(frac{partial y}{partial x}right)_z{rm d}x tag{3}
$$
And from here is trivial to obtain
$$
left(frac{partial x}{partial y}right)_z left(frac{partial y}{partial z}right)_x left(frac{partial z}{partial x}right)_y = -1 tag{4}
$$
Note that this is a result of consistently using the chain rule, and keeping track what function depends on what variables.
In your case just set $x = P, y = V, z = T$, the function $f(x,y,z)$ is called a equation of state
answered Jan 1 at 10:49
caveraccaverac
14.1k21130
answered Jan 1 at 10:49
caveraccaverac
14.1k21130
answered Jan 1 at 10:49
caveraccaverac
14.1k21130
answered Jan 1 at 10:49
caveraccaverac
14.1k21130
14.1k21130
add a comment |
add a comment |
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%2f3058230%2fpartial-derivative-and-fraction%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
