I need to find the potential function of a vector field. Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Calculating the Integral of a non conservative vector fieldFind the potential function of a conservative vector fieldTwo ways of finding a Potential of a Vector FieldFinding potential function for a vector fieldVector Field Conceptual QuestionIs there a specific notation to denote the potential function of a conservative vector field?Every conservative vector field is irrotationalQuestions about the potential of a conservative vector fieldWhy do we need both Divergence and Curl to define a vector field?How to check if a 2 dimensional vector field is irrotational (curl=0)?
Is there a documented rationale why the House Ways and Means chairman can demand tax info?
What happens to sewage if there is no river near by?
How do I mention the quality of my school without bragging
How much radiation do nuclear physics experiments expose researchers to nowadays?
Why did the IBM 650 use bi-quinary?
Right-skewed distribution with mean equals to mode?
Why aren't air breathing engines used as small first stages
Single word antonym of "flightless"
What do you call a phrase that's not an idiom yet?
ListPlot join points by nearest neighbor rather than order
Why is "Captain Marvel" translated as male in Portugal?
The logistics of corpse disposal
Why is black pepper both grey and black?
Proof involving the spectral radius and the Jordan canonical form
What are 'alternative tunings' of a guitar and why would you use them? Doesn't it make it more difficult to play?
How does a Death Domain cleric's Touch of Death feature work with Touch-range spells delivered by familiars?
I am not a queen, who am I?
"Seemed to had" is it correct?
How can players work together to take actions that are otherwise impossible?
Dominant seventh chord in the major scale contains diminished triad of the seventh?
If Jon Snow became King of the Seven Kingdoms what would his regnal number be?
What is the longest distance a 13th-level monk can jump while attacking on the same turn?
Center align columns in table ignoring minus signs?
Bonus calculation: Am I making a mountain out of a molehill?
I need to find the potential function of a vector field.
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Calculating the Integral of a non conservative vector fieldFind the potential function of a conservative vector fieldTwo ways of finding a Potential of a Vector FieldFinding potential function for a vector fieldVector Field Conceptual QuestionIs there a specific notation to denote the potential function of a conservative vector field?Every conservative vector field is irrotationalQuestions about the potential of a conservative vector fieldWhy do we need both Divergence and Curl to define a vector field?How to check if a 2 dimensional vector field is irrotational (curl=0)?
$begingroup$
I was given F = (y+z)i + (x+z)j + (x+y)k. I found said field to be conservative, and I integrated the x partial derivative and got f(x,y,z) = xy + xz + g(y,z). The thing is that I am trying to find g(y,z), and I ended up with something that was expressed in terms of x, y and z (I got x+z-xy-xz). I don't know what to do with this information not that I arrived at something expressed in all three variables.
integration multivariable-calculus vector-fields
$endgroup$
add a comment |
$begingroup$
I was given F = (y+z)i + (x+z)j + (x+y)k. I found said field to be conservative, and I integrated the x partial derivative and got f(x,y,z) = xy + xz + g(y,z). The thing is that I am trying to find g(y,z), and I ended up with something that was expressed in terms of x, y and z (I got x+z-xy-xz). I don't know what to do with this information not that I arrived at something expressed in all three variables.
integration multivariable-calculus vector-fields
$endgroup$
add a comment |
$begingroup$
I was given F = (y+z)i + (x+z)j + (x+y)k. I found said field to be conservative, and I integrated the x partial derivative and got f(x,y,z) = xy + xz + g(y,z). The thing is that I am trying to find g(y,z), and I ended up with something that was expressed in terms of x, y and z (I got x+z-xy-xz). I don't know what to do with this information not that I arrived at something expressed in all three variables.
integration multivariable-calculus vector-fields
$endgroup$
I was given F = (y+z)i + (x+z)j + (x+y)k. I found said field to be conservative, and I integrated the x partial derivative and got f(x,y,z) = xy + xz + g(y,z). The thing is that I am trying to find g(y,z), and I ended up with something that was expressed in terms of x, y and z (I got x+z-xy-xz). I don't know what to do with this information not that I arrived at something expressed in all three variables.
integration multivariable-calculus vector-fields
integration multivariable-calculus vector-fields
asked 4 hours ago
UchuukoUchuuko
367
367
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You have $fracpartial fpartial x= y+ z$ so that $f(x,y,z)= xy+ xz+ g(y,z)$. (Since the differentiation with respect to x treat y and z as constants, the "constant of integration" might in fact be a function of y and z. That is the "g(y, z)".)
Differentiating that with respect to y, $fracpartial fpartial y= x+ g_y(y, z)= x+ z$ so that $g_y= z$ and $g(y, z)= yz+ h(z)$.
So f(x,y,z)= xy+ xz+ yz+ h(z). Differentiating that with respect to z, $fracpartial fpartial z= x+ y+ h'(z)= x+ y$ so that h'(z)= 0. h is a constant, C so that we get f(x, y, z)= xy+ xz+ yz+ C.
$endgroup$
add a comment |
$begingroup$
So far, we have $f(x,y,z) = xy + xz + g(y,z)$. Taking $fracpartial fpartial x$ gives us the $x$-component of $textbfF$. To get similar $y$ and $z$-components, we suspect that $g(y,z)$ should be similar to the other terms in $f(x,y,z)$ in some sense. The natural guess is $g(y,z) = yz$, since the other terms in $f(x,y,z)$ are each multiplications of two different independent variables. It can then be verified that the guess for $g$ produces the correct vector field, by computing $nabla f$.
We now know that we have determined the potential function up to a constant, since if two scalar fields have the same gradient, then they differ by a constant.
A note of caution: sometimes the convention for what is meant by a potential function for a vector field $mathbfF$, is a scalar field $f$ such that $mathbfF = - nabla f$. Beware!
$endgroup$
add a comment |
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%2f3189329%2fi-need-to-find-the-potential-function-of-a-vector-field%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
$begingroup$
You have $fracpartial fpartial x= y+ z$ so that $f(x,y,z)= xy+ xz+ g(y,z)$. (Since the differentiation with respect to x treat y and z as constants, the "constant of integration" might in fact be a function of y and z. That is the "g(y, z)".)
Differentiating that with respect to y, $fracpartial fpartial y= x+ g_y(y, z)= x+ z$ so that $g_y= z$ and $g(y, z)= yz+ h(z)$.
So f(x,y,z)= xy+ xz+ yz+ h(z). Differentiating that with respect to z, $fracpartial fpartial z= x+ y+ h'(z)= x+ y$ so that h'(z)= 0. h is a constant, C so that we get f(x, y, z)= xy+ xz+ yz+ C.
$endgroup$
add a comment |
$begingroup$
You have $fracpartial fpartial x= y+ z$ so that $f(x,y,z)= xy+ xz+ g(y,z)$. (Since the differentiation with respect to x treat y and z as constants, the "constant of integration" might in fact be a function of y and z. That is the "g(y, z)".)
Differentiating that with respect to y, $fracpartial fpartial y= x+ g_y(y, z)= x+ z$ so that $g_y= z$ and $g(y, z)= yz+ h(z)$.
So f(x,y,z)= xy+ xz+ yz+ h(z). Differentiating that with respect to z, $fracpartial fpartial z= x+ y+ h'(z)= x+ y$ so that h'(z)= 0. h is a constant, C so that we get f(x, y, z)= xy+ xz+ yz+ C.
$endgroup$
add a comment |
$begingroup$
You have $fracpartial fpartial x= y+ z$ so that $f(x,y,z)= xy+ xz+ g(y,z)$. (Since the differentiation with respect to x treat y and z as constants, the "constant of integration" might in fact be a function of y and z. That is the "g(y, z)".)
Differentiating that with respect to y, $fracpartial fpartial y= x+ g_y(y, z)= x+ z$ so that $g_y= z$ and $g(y, z)= yz+ h(z)$.
So f(x,y,z)= xy+ xz+ yz+ h(z). Differentiating that with respect to z, $fracpartial fpartial z= x+ y+ h'(z)= x+ y$ so that h'(z)= 0. h is a constant, C so that we get f(x, y, z)= xy+ xz+ yz+ C.
$endgroup$
You have $fracpartial fpartial x= y+ z$ so that $f(x,y,z)= xy+ xz+ g(y,z)$. (Since the differentiation with respect to x treat y and z as constants, the "constant of integration" might in fact be a function of y and z. That is the "g(y, z)".)
Differentiating that with respect to y, $fracpartial fpartial y= x+ g_y(y, z)= x+ z$ so that $g_y= z$ and $g(y, z)= yz+ h(z)$.
So f(x,y,z)= xy+ xz+ yz+ h(z). Differentiating that with respect to z, $fracpartial fpartial z= x+ y+ h'(z)= x+ y$ so that h'(z)= 0. h is a constant, C so that we get f(x, y, z)= xy+ xz+ yz+ C.
answered 3 hours ago
user247327user247327
11.6k1516
11.6k1516
add a comment |
add a comment |
$begingroup$
So far, we have $f(x,y,z) = xy + xz + g(y,z)$. Taking $fracpartial fpartial x$ gives us the $x$-component of $textbfF$. To get similar $y$ and $z$-components, we suspect that $g(y,z)$ should be similar to the other terms in $f(x,y,z)$ in some sense. The natural guess is $g(y,z) = yz$, since the other terms in $f(x,y,z)$ are each multiplications of two different independent variables. It can then be verified that the guess for $g$ produces the correct vector field, by computing $nabla f$.
We now know that we have determined the potential function up to a constant, since if two scalar fields have the same gradient, then they differ by a constant.
A note of caution: sometimes the convention for what is meant by a potential function for a vector field $mathbfF$, is a scalar field $f$ such that $mathbfF = - nabla f$. Beware!
$endgroup$
add a comment |
$begingroup$
So far, we have $f(x,y,z) = xy + xz + g(y,z)$. Taking $fracpartial fpartial x$ gives us the $x$-component of $textbfF$. To get similar $y$ and $z$-components, we suspect that $g(y,z)$ should be similar to the other terms in $f(x,y,z)$ in some sense. The natural guess is $g(y,z) = yz$, since the other terms in $f(x,y,z)$ are each multiplications of two different independent variables. It can then be verified that the guess for $g$ produces the correct vector field, by computing $nabla f$.
We now know that we have determined the potential function up to a constant, since if two scalar fields have the same gradient, then they differ by a constant.
A note of caution: sometimes the convention for what is meant by a potential function for a vector field $mathbfF$, is a scalar field $f$ such that $mathbfF = - nabla f$. Beware!
$endgroup$
add a comment |
$begingroup$
So far, we have $f(x,y,z) = xy + xz + g(y,z)$. Taking $fracpartial fpartial x$ gives us the $x$-component of $textbfF$. To get similar $y$ and $z$-components, we suspect that $g(y,z)$ should be similar to the other terms in $f(x,y,z)$ in some sense. The natural guess is $g(y,z) = yz$, since the other terms in $f(x,y,z)$ are each multiplications of two different independent variables. It can then be verified that the guess for $g$ produces the correct vector field, by computing $nabla f$.
We now know that we have determined the potential function up to a constant, since if two scalar fields have the same gradient, then they differ by a constant.
A note of caution: sometimes the convention for what is meant by a potential function for a vector field $mathbfF$, is a scalar field $f$ such that $mathbfF = - nabla f$. Beware!
$endgroup$
So far, we have $f(x,y,z) = xy + xz + g(y,z)$. Taking $fracpartial fpartial x$ gives us the $x$-component of $textbfF$. To get similar $y$ and $z$-components, we suspect that $g(y,z)$ should be similar to the other terms in $f(x,y,z)$ in some sense. The natural guess is $g(y,z) = yz$, since the other terms in $f(x,y,z)$ are each multiplications of two different independent variables. It can then be verified that the guess for $g$ produces the correct vector field, by computing $nabla f$.
We now know that we have determined the potential function up to a constant, since if two scalar fields have the same gradient, then they differ by a constant.
A note of caution: sometimes the convention for what is meant by a potential function for a vector field $mathbfF$, is a scalar field $f$ such that $mathbfF = - nabla f$. Beware!
edited 3 hours ago
answered 3 hours ago
E-muE-mu
1214
1214
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%2f3189329%2fi-need-to-find-the-potential-function-of-a-vector-field%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