Does the FTC apply on a function of two variables?
I am doing a project on traffic flow, and am deriving the conservation equation between density and flux. I have got to this line in my calculations, and am unsure how to justify the step.
$$q(x_1,t)-q(x_2,t)=-int_{x_1}^{x_2}frac{partial q(x,t)}{partial x} dx $$
Does the FTC apply? I wasn't sure since we have a function of two variables in this case. In all of the sources I have used, they have taken this step as purely trivial, but I would like to know for certain why we are allowed to do this. Out of curiosity, is this always possible?
Many thanks in advance!
calculus real-analysis pde
asked Nov 28 '18 at 20:39
kinga120
112
add a comment |
I am doing a project on traffic flow, and am deriving the conservation equation between density and flux. I have got to this line in my calculations, and am unsure how to justify the step.
$$q(x_1,t)-q(x_2,t)=-int_{x_1}^{x_2}frac{partial q(x,t)}{partial x} dx $$
Does the FTC apply? I wasn't sure since we have a function of two variables in this case. In all of the sources I have used, they have taken this step as purely trivial, but I would like to know for certain why we are allowed to do this. Out of curiosity, is this always possible?
Many thanks in advance!
calculus real-analysis pde
asked Nov 28 '18 at 20:39
kinga120
112
1
It would be $q(x_2,t) - q(x_1,t)$, not $q(x_1,t)-q(x_2,t)$. Yes fundamental theorem of calculus applies (under mild assumptions that the derivative is continuous that are likely automatically satisfied for your problem) since if we fix time $t$ then we can write $f(x)=q(x,t)$ and $f'(x)$ is the partial derivative you use.
– Michael
Nov 28 '18 at 20:41
Yes sorry, my error, the integral was supposed to have a minus sign in front! Many thanks for your help!
– kinga120
Nov 28 '18 at 20:52
add a comment |
I am doing a project on traffic flow, and am deriving the conservation equation between density and flux. I have got to this line in my calculations, and am unsure how to justify the step.
$$q(x_1,t)-q(x_2,t)=-int_{x_1}^{x_2}frac{partial q(x,t)}{partial x} dx $$
Does the FTC apply? I wasn't sure since we have a function of two variables in this case. In all of the sources I have used, they have taken this step as purely trivial, but I would like to know for certain why we are allowed to do this. Out of curiosity, is this always possible?
Many thanks in advance!
calculus real-analysis pde
asked Nov 28 '18 at 20:39
kinga120
112
I am doing a project on traffic flow, and am deriving the conservation equation between density and flux. I have got to this line in my calculations, and am unsure how to justify the step.
$$q(x_1,t)-q(x_2,t)=-int_{x_1}^{x_2}frac{partial q(x,t)}{partial x} dx $$
Does the FTC apply? I wasn't sure since we have a function of two variables in this case. In all of the sources I have used, they have taken this step as purely trivial, but I would like to know for certain why we are allowed to do this. Out of curiosity, is this always possible?
Many thanks in advance!
calculus real-analysis pde
calculus real-analysis pde
asked Nov 28 '18 at 20:39
kinga120
112
asked Nov 28 '18 at 20:39
kinga120
112
edited Nov 28 '18 at 20:53
asked Nov 28 '18 at 20:39
kinga120
112
asked Nov 28 '18 at 20:39
kinga120
112
asked Nov 28 '18 at 20:39
kinga120
112
112
1
It would be $q(x_2,t) - q(x_1,t)$, not $q(x_1,t)-q(x_2,t)$. Yes fundamental theorem of calculus applies (under mild assumptions that the derivative is continuous that are likely automatically satisfied for your problem) since if we fix time $t$ then we can write $f(x)=q(x,t)$ and $f'(x)$ is the partial derivative you use.
– Michael
Nov 28 '18 at 20:41
Yes sorry, my error, the integral was supposed to have a minus sign in front! Many thanks for your help!
– kinga120
Nov 28 '18 at 20:52
add a comment |
1
It would be $q(x_2,t) - q(x_1,t)$, not $q(x_1,t)-q(x_2,t)$. Yes fundamental theorem of calculus applies (under mild assumptions that the derivative is continuous that are likely automatically satisfied for your problem) since if we fix time $t$ then we can write $f(x)=q(x,t)$ and $f'(x)$ is the partial derivative you use.
– Michael
Nov 28 '18 at 20:41
Yes sorry, my error, the integral was supposed to have a minus sign in front! Many thanks for your help!
– kinga120
Nov 28 '18 at 20:52
1
1
It would be $q(x_2,t) - q(x_1,t)$, not $q(x_1,t)-q(x_2,t)$. Yes fundamental theorem of calculus applies (under mild assumptions that the derivative is continuous that are likely automatically satisfied for your problem) since if we fix time $t$ then we can write $f(x)=q(x,t)$ and $f'(x)$ is the partial derivative you use.
– Michael
Nov 28 '18 at 20:41
It would be $q(x_2,t) - q(x_1,t)$, not $q(x_1,t)-q(x_2,t)$. Yes fundamental theorem of calculus applies (under mild assumptions that the derivative is continuous that are likely automatically satisfied for your problem) since if we fix time $t$ then we can write $f(x)=q(x,t)$ and $f'(x)$ is the partial derivative you use.
– Michael
Nov 28 '18 at 20:41
Yes sorry, my error, the integral was supposed to have a minus sign in front! Many thanks for your help!
– kinga120
Nov 28 '18 at 20:52
Yes sorry, my error, the integral was supposed to have a minus sign in front! Many thanks for your help!
– kinga120
Nov 28 '18 at 20:52
add a comment |
0
active
oldest
votes
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%2f3017682%2fdoes-the-ftc-apply-on-a-function-of-two-variables%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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3017682%2fdoes-the-ftc-apply-on-a-function-of-two-variables%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
1
It would be $q(x_2,t) - q(x_1,t)$, not $q(x_1,t)-q(x_2,t)$. Yes fundamental theorem of calculus applies (under mild assumptions that the derivative is continuous that are likely automatically satisfied for your problem) since if we fix time $t$ then we can write $f(x)=q(x,t)$ and $f'(x)$ is the partial derivative you use.
– Michael
Nov 28 '18 at 20:41
Yes sorry, my error, the integral was supposed to have a minus sign in front! Many thanks for your help!
– kinga120
Nov 28 '18 at 20:52