I need to find the potential function of a vector field. Announcing the arrival of Valued...
Right-skewed distribution with mean equals to mode?
Is there a Spanish version of "dot your i's and cross your t's" that includes the letter 'ñ'?
Difference between these two cards?
G-Code for resetting to 100% speed
I need to find the potential function of a vector field.
What is this single-engine low-wing propeller plane?
Gastric acid as a weapon
Did Xerox really develop the first LAN?
How can I fade player when goes inside or outside of the area?
How can I make names more distinctive without making them longer?
What are the motives behind Cersei's orders given to Bronn?
How do I keep my slimes from escaping their pens?
Can a non-EU citizen traveling with me come with me through the EU passport line?
What are the pros and cons of Aerospike nosecones?
Is it true that "carbohydrates are of no use for the basal metabolic need"?
What are 'alternative tunings' of a guitar and why would you use them? Doesn't it make it more difficult to play?
Is a manifold-with-boundary with given interior and non-empty boundary essentially unique?
Is 1 ppb equal to 1 μg/kg?
Why did the IBM 650 use bi-quinary?
ListPlot join points by nearest neighbor rather than order
Why aren't air breathing engines used as small first stages
How do I mention the quality of my school without bragging
Is the address of a local variable a constexpr?
Disable hyphenation for an entire paragraph
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
asked 2 hours ago
UchuukoUchuuko
367
$endgroup$
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
asked 2 hours ago
UchuukoUchuuko
367
$endgroup$
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
asked 2 hours ago
UchuukoUchuuko
367
$endgroup$
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 2 hours ago
UchuukoUchuuko
367
asked 2 hours ago
UchuukoUchuuko
367
asked 2 hours ago
UchuukoUchuuko
367
asked 2 hours ago
UchuukoUchuuko
367
asked 2 hours ago
UchuukoUchuuko
367
367
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You have $frac{partial f}{partial x}= y+ z$ 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, $frac{partial f}{partial y}= x+ g_y(y, z)= x+ z$ 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, $frac{partial f}{partial z}= x+ y+ h'(z)= x+ y$ so that h'(z)= 0. h is a constant, C so that we get f(x, y, z)= xy+ xz+ yz+ C.
answered 2 hours ago
user247327user247327
11.6k1516
$endgroup$
add a comment |
$begingroup$
So far, we have $f(x,y,z) = xy + xz + g(y,z)$. Taking $frac{partial f}{partial x}$ gives us the $x$-component of $textbf{F}$. 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 $mathbf{F}$, is a scalar field $f$ such that $mathbf{F} = - nabla f$. Beware!
answered 2 hours ago
E-muE-mu
1214
$endgroup$
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 $frac{partial f}{partial x}= y+ z$ 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, $frac{partial f}{partial y}= x+ g_y(y, z)= x+ z$ 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, $frac{partial f}{partial z}= x+ y+ h'(z)= x+ y$ so that h'(z)= 0. h is a constant, C so that we get f(x, y, z)= xy+ xz+ yz+ C.
answered 2 hours ago
user247327user247327
11.6k1516
$endgroup$
add a comment |
$begingroup$
You have $frac{partial f}{partial x}= y+ z$ 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, $frac{partial f}{partial y}= x+ g_y(y, z)= x+ z$ 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, $frac{partial f}{partial z}= x+ y+ h'(z)= x+ y$ so that h'(z)= 0. h is a constant, C so that we get f(x, y, z)= xy+ xz+ yz+ C.
answered 2 hours ago
user247327user247327
11.6k1516
$endgroup$
add a comment |
$begingroup$
You have $frac{partial f}{partial x}= y+ z$ 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, $frac{partial f}{partial y}= x+ g_y(y, z)= x+ z$ 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, $frac{partial f}{partial z}= x+ y+ h'(z)= x+ y$ so that h'(z)= 0. h is a constant, C so that we get f(x, y, z)= xy+ xz+ yz+ C.
answered 2 hours ago
user247327user247327
11.6k1516
$endgroup$
You have $frac{partial f}{partial x}= y+ z$ 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, $frac{partial f}{partial y}= x+ g_y(y, z)= x+ z$ 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, $frac{partial f}{partial z}= x+ y+ h'(z)= x+ y$ so that h'(z)= 0. h is a constant, C so that we get f(x, y, z)= xy+ xz+ yz+ C.
answered 2 hours ago
user247327user247327
11.6k1516
answered 2 hours ago
user247327user247327
11.6k1516
answered 2 hours ago
user247327user247327
11.6k1516
answered 2 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 $frac{partial f}{partial x}$ gives us the $x$-component of $textbf{F}$. 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 $mathbf{F}$, is a scalar field $f$ such that $mathbf{F} = - nabla f$. Beware!
answered 2 hours ago
E-muE-mu
1214
$endgroup$
add a comment |
$begingroup$
So far, we have $f(x,y,z) = xy + xz + g(y,z)$. Taking $frac{partial f}{partial x}$ gives us the $x$-component of $textbf{F}$. 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 $mathbf{F}$, is a scalar field $f$ such that $mathbf{F} = - nabla f$. Beware!
answered 2 hours ago
E-muE-mu
1214
$endgroup$
add a comment |
$begingroup$
So far, we have $f(x,y,z) = xy + xz + g(y,z)$. Taking $frac{partial f}{partial x}$ gives us the $x$-component of $textbf{F}$. 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 $mathbf{F}$, is a scalar field $f$ such that $mathbf{F} = - nabla f$. Beware!
answered 2 hours ago
E-muE-mu
1214
$endgroup$
So far, we have $f(x,y,z) = xy + xz + g(y,z)$. Taking $frac{partial f}{partial x}$ gives us the $x$-component of $textbf{F}$. 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 $mathbf{F}$, is a scalar field $f$ such that $mathbf{F} = - nabla f$. Beware!
answered 2 hours ago
E-muE-mu
1214
edited 2 hours ago
answered 2 hours ago
E-muE-mu
1214
answered 2 hours ago
E-muE-mu
1214
answered 2 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
