Mixing
Line integral method calculate work done by a particle
$begingroup$
I'm having trouble knowing how to go about solving this question:
Q: The force on a particle at a point with position vector $r = xi + yj + zk$ exerted by a charge at the origin is $F(r)=left(frac{P(r)}{|r|^2}right)$
in which P is constant. Calculate the work done as the particle moves in a straight line from (1, 0, 0) to (1, 2, 3).
What I think I need to do:
$r_1=i$, $quad$ $r_2=i+2j+3k$ $quad$ so let $r(t)=i+2tj+3tk$,$quad$ $0lt t lt 1$
Then $frac{dr}{dt}=2+3=5$ and $|r|^2=1+13t^2$
Therefore, $$F(r)=frac{P(i+2tj+3tk)}{1+13t^2}$$
As such, $int_C F(r) cdot dr=int_0^1 frac{P(i+2tj+3tk)}{1+13t^2} cdot5 dt=int_0^1 frac{5P(1+5t)}{1+13t^2}dt=5Pint_0^1 frac{1+5t}{1+13t^2}dt$
Is this the right way to go about answering this question, or am I doing something completely wrong? Any advice would be greatly appreciated.
vectors line-integrals
asked Feb 1 at 13:01
AnonAnon
172
$endgroup$
add a comment |
$begingroup$
I'm having trouble knowing how to go about solving this question:
Q: The force on a particle at a point with position vector $r = xi + yj + zk$ exerted by a charge at the origin is $F(r)=left(frac{P(r)}{|r|^2}right)$
in which P is constant. Calculate the work done as the particle moves in a straight line from (1, 0, 0) to (1, 2, 3).
What I think I need to do:
$r_1=i$, $quad$ $r_2=i+2j+3k$ $quad$ so let $r(t)=i+2tj+3tk$,$quad$ $0lt t lt 1$
Then $frac{dr}{dt}=2+3=5$ and $|r|^2=1+13t^2$
Therefore, $$F(r)=frac{P(i+2tj+3tk)}{1+13t^2}$$
As such, $int_C F(r) cdot dr=int_0^1 frac{P(i+2tj+3tk)}{1+13t^2} cdot5 dt=int_0^1 frac{5P(1+5t)}{1+13t^2}dt=5Pint_0^1 frac{1+5t}{1+13t^2}dt$
Is this the right way to go about answering this question, or am I doing something completely wrong? Any advice would be greatly appreciated.
vectors line-integrals
asked Feb 1 at 13:01
AnonAnon
172
$endgroup$
$begingroup$
Note that $r(t)=(1,2t,3t)$ and $r'(t)$ is also a vector.
$endgroup$
– PierreCarre
Feb 1 at 13:26
add a comment |
$begingroup$
I'm having trouble knowing how to go about solving this question:
Q: The force on a particle at a point with position vector $r = xi + yj + zk$ exerted by a charge at the origin is $F(r)=left(frac{P(r)}{|r|^2}right)$
in which P is constant. Calculate the work done as the particle moves in a straight line from (1, 0, 0) to (1, 2, 3).
What I think I need to do:
$r_1=i$, $quad$ $r_2=i+2j+3k$ $quad$ so let $r(t)=i+2tj+3tk$,$quad$ $0lt t lt 1$
Then $frac{dr}{dt}=2+3=5$ and $|r|^2=1+13t^2$
Therefore, $$F(r)=frac{P(i+2tj+3tk)}{1+13t^2}$$
As such, $int_C F(r) cdot dr=int_0^1 frac{P(i+2tj+3tk)}{1+13t^2} cdot5 dt=int_0^1 frac{5P(1+5t)}{1+13t^2}dt=5Pint_0^1 frac{1+5t}{1+13t^2}dt$
Is this the right way to go about answering this question, or am I doing something completely wrong? Any advice would be greatly appreciated.
vectors line-integrals
asked Feb 1 at 13:01
AnonAnon
172
$endgroup$
I'm having trouble knowing how to go about solving this question:
Q: The force on a particle at a point with position vector $r = xi + yj + zk$ exerted by a charge at the origin is $F(r)=left(frac{P(r)}{|r|^2}right)$
in which P is constant. Calculate the work done as the particle moves in a straight line from (1, 0, 0) to (1, 2, 3).
What I think I need to do:
$r_1=i$, $quad$ $r_2=i+2j+3k$ $quad$ so let $r(t)=i+2tj+3tk$,$quad$ $0lt t lt 1$
Then $frac{dr}{dt}=2+3=5$ and $|r|^2=1+13t^2$
Therefore, $$F(r)=frac{P(i+2tj+3tk)}{1+13t^2}$$
As such, $int_C F(r) cdot dr=int_0^1 frac{P(i+2tj+3tk)}{1+13t^2} cdot5 dt=int_0^1 frac{5P(1+5t)}{1+13t^2}dt=5Pint_0^1 frac{1+5t}{1+13t^2}dt$
Is this the right way to go about answering this question, or am I doing something completely wrong? Any advice would be greatly appreciated.
vectors line-integrals
vectors line-integrals
asked Feb 1 at 13:01
AnonAnon
172
asked Feb 1 at 13:01
AnonAnon
172
asked Feb 1 at 13:01
AnonAnon
172
asked Feb 1 at 13:01
AnonAnon
172
asked Feb 1 at 13:01
AnonAnon
172
172
$begingroup$
Note that $r(t)=(1,2t,3t)$ and $r'(t)$ is also a vector.
$endgroup$
– PierreCarre
Feb 1 at 13:26
add a comment |
$begingroup$
Note that $r(t)=(1,2t,3t)$ and $r'(t)$ is also a vector.
$endgroup$
– PierreCarre
Feb 1 at 13:26
$begingroup$
Note that $r(t)=(1,2t,3t)$ and $r'(t)$ is also a vector.
$endgroup$
– PierreCarre
Feb 1 at 13:26
$begingroup$
Note that $r(t)=(1,2t,3t)$ and $r'(t)$ is also a vector.
$endgroup$
– PierreCarre
Feb 1 at 13:26
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The work is computed as the line integral:
$$
int_C F(r)cdot dr = int_0^1 F(r(t))cdot r'(t) dt = int_0^1 frac{P}{1+4t^2+9t^2} (1,2t,3t)cdot(0,2,3) dt = int_0^1 frac{13 P ,t}{1+13t^2}dt
$$
answered Feb 1 at 13:24
PierreCarrePierreCarre
2,103214
$endgroup$
$begingroup$
Ah ok, that makes more sense thanks. One thing I don't get though is why there is no P in the last integral, shouldn't it be 13Pt as the numerator of the integral?
$endgroup$
– Anon
Feb 1 at 13:58
$begingroup$
Gone with the wind... I'll edit the post!
$endgroup$
– PierreCarre
Feb 1 at 16:06
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%2f3096203%2fline-integral-method-calculate-work-done-by-a-particle%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The work is computed as the line integral:
$$
int_C F(r)cdot dr = int_0^1 F(r(t))cdot r'(t) dt = int_0^1 frac{P}{1+4t^2+9t^2} (1,2t,3t)cdot(0,2,3) dt = int_0^1 frac{13 P ,t}{1+13t^2}dt
$$
answered Feb 1 at 13:24
PierreCarrePierreCarre
2,103214
$endgroup$
$begingroup$
Ah ok, that makes more sense thanks. One thing I don't get though is why there is no P in the last integral, shouldn't it be 13Pt as the numerator of the integral?
$endgroup$
– Anon
Feb 1 at 13:58
$begingroup$
Gone with the wind... I'll edit the post!
$endgroup$
– PierreCarre
Feb 1 at 16:06
add a comment |
$begingroup$
The work is computed as the line integral:
$$
int_C F(r)cdot dr = int_0^1 F(r(t))cdot r'(t) dt = int_0^1 frac{P}{1+4t^2+9t^2} (1,2t,3t)cdot(0,2,3) dt = int_0^1 frac{13 P ,t}{1+13t^2}dt
$$
answered Feb 1 at 13:24
PierreCarrePierreCarre
2,103214
$endgroup$
$begingroup$
Ah ok, that makes more sense thanks. One thing I don't get though is why there is no P in the last integral, shouldn't it be 13Pt as the numerator of the integral?
$endgroup$
– Anon
Feb 1 at 13:58
$begingroup$
Gone with the wind... I'll edit the post!
$endgroup$
– PierreCarre
Feb 1 at 16:06
add a comment |
$begingroup$
The work is computed as the line integral:
$$
int_C F(r)cdot dr = int_0^1 F(r(t))cdot r'(t) dt = int_0^1 frac{P}{1+4t^2+9t^2} (1,2t,3t)cdot(0,2,3) dt = int_0^1 frac{13 P ,t}{1+13t^2}dt
$$
answered Feb 1 at 13:24
PierreCarrePierreCarre
2,103214
$endgroup$
The work is computed as the line integral:
$$
int_C F(r)cdot dr = int_0^1 F(r(t))cdot r'(t) dt = int_0^1 frac{P}{1+4t^2+9t^2} (1,2t,3t)cdot(0,2,3) dt = int_0^1 frac{13 P ,t}{1+13t^2}dt
$$
answered Feb 1 at 13:24
PierreCarrePierreCarre
2,103214
edited Feb 1 at 16:07
answered Feb 1 at 13:24
PierreCarrePierreCarre
2,103214
answered Feb 1 at 13:24
PierreCarrePierreCarre
2,103214
answered Feb 1 at 13:24
PierreCarrePierreCarre
2,103214
2,103214
$begingroup$
Ah ok, that makes more sense thanks. One thing I don't get though is why there is no P in the last integral, shouldn't it be 13Pt as the numerator of the integral?
$endgroup$
– Anon
Feb 1 at 13:58
$begingroup$
Gone with the wind... I'll edit the post!
$endgroup$
– PierreCarre
Feb 1 at 16:06
add a comment |
$begingroup$
Ah ok, that makes more sense thanks. One thing I don't get though is why there is no P in the last integral, shouldn't it be 13Pt as the numerator of the integral?
$endgroup$
– Anon
Feb 1 at 13:58
$begingroup$
Gone with the wind... I'll edit the post!
$endgroup$
– PierreCarre
Feb 1 at 16:06
$begingroup$
Ah ok, that makes more sense thanks. One thing I don't get though is why there is no P in the last integral, shouldn't it be 13Pt as the numerator of the integral?
$endgroup$
– Anon
Feb 1 at 13:58
$begingroup$
Ah ok, that makes more sense thanks. One thing I don't get though is why there is no P in the last integral, shouldn't it be 13Pt as the numerator of the integral?
$endgroup$
– Anon
Feb 1 at 13:58
$begingroup$
Gone with the wind... I'll edit the post!
$endgroup$
– PierreCarre
Feb 1 at 16:06
$begingroup$
Gone with the wind... I'll edit the post!
$endgroup$
– PierreCarre
Feb 1 at 16:06
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%2f3096203%2fline-integral-method-calculate-work-done-by-a-particle%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
$begingroup$
Note that $r(t)=(1,2t,3t)$ and $r'(t)$ is also a vector.
$endgroup$
– PierreCarre
Feb 1 at 13:26