Does the sign matter for proportionality?What does this formula do?what is the constant of proportionality?I...
With a Canadian student visa, can I spend a night at Vancouver before continuing to Toronto?
How can I place the product on a social media post better?
Pass By Reference VS Pass by Value
What happened to Captain America in Endgame?
Combinable filters
How to reduce LED flash rate (frequency)
How can the Zone of Truth spell be defeated without the caster knowing?
Was there a shared-world project before "Thieves World"?
Is there really no use for MD5 anymore?
How to verbalise code in Mathematica?
A Strange Latex Symbol
Is there any limitation with Arduino Nano serial communication distance?
US visa is under administrative processing, I need the passport back ASAP
How do I reattach a shelf to the wall when it ripped out of the wall?
Meaning of Bloch representation
The Defining Moment
Was there a Viking Exchange as well as a Columbian one?
What is the relationship between spectral sequences and obstruction theory?
a sore throat vs a strep throat vs strep throat
Is the 5 MB static resource size limit 5,242,880 bytes or 5,000,000 bytes?
Why other Westeros houses don't use wildfire?
What makes accurate emulation of old systems a difficult task?
What does KSP mean?
Is there a way to get a compiler for the original B programming language?
Does the sign matter for proportionality?
What does this formula do?what is the constant of proportionality?I did not understand the approach needed to solve this problem on inverse proportionalityCalculating Volume for the Perfect CompostWhy does proportionality yield multiplicitiveness and not addition?Why does 1:2 = 1/2?Why does 100% represent the whole of somethingSimple Ratios Shortcut - Why does it work?How do I figure out the proportionality between two aspect ratiosHow is the joint proportionality (joint variation) equation (i.e. $y = kxz$) logically correct?
$begingroup$
This question arose from Physics, where the force on an object attached on a spring is proportional to the displacement to the equilibrium (that is, the rest position). Also, if the displacement to the equilibrium is positive, the force will be negative, as it tries to pull the object back (i.e. if you pull a string, the force is opposite to your direction of pull).
Therefore, it can be said that:
$$F propto -x$$
Where $F$ is the force and $x$ is the displacement from equilibrium
Is this the same as:
$$F propto x$$
According to the relation of proportionality, it should be, but my friend says that putting the second one is not correct.
Both are equivalent right?
ratio
edited 2 hours ago
Brian
1,529516
asked 5 hours ago
Sebi SSebi S
182
New contributor
Sebi S is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
This question arose from Physics, where the force on an object attached on a spring is proportional to the displacement to the equilibrium (that is, the rest position). Also, if the displacement to the equilibrium is positive, the force will be negative, as it tries to pull the object back (i.e. if you pull a string, the force is opposite to your direction of pull).
Therefore, it can be said that:
$$F propto -x$$
Where $F$ is the force and $x$ is the displacement from equilibrium
Is this the same as:
$$F propto x$$
According to the relation of proportionality, it should be, but my friend says that putting the second one is not correct.
Both are equivalent right?
ratio
edited 2 hours ago
Brian
1,529516
asked 5 hours ago
Sebi SSebi S
182
New contributor
Sebi S is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
It depends on definitions/context. Using the definition on wikipedia, yes they are equivalent.
$endgroup$
– David Peterson
5 hours ago
add a comment |
$begingroup$
This question arose from Physics, where the force on an object attached on a spring is proportional to the displacement to the equilibrium (that is, the rest position). Also, if the displacement to the equilibrium is positive, the force will be negative, as it tries to pull the object back (i.e. if you pull a string, the force is opposite to your direction of pull).
Therefore, it can be said that:
$$F propto -x$$
Where $F$ is the force and $x$ is the displacement from equilibrium
Is this the same as:
$$F propto x$$
According to the relation of proportionality, it should be, but my friend says that putting the second one is not correct.
Both are equivalent right?
ratio
edited 2 hours ago
Brian
1,529516
asked 5 hours ago
Sebi SSebi S
182
New contributor
Sebi S is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
This question arose from Physics, where the force on an object attached on a spring is proportional to the displacement to the equilibrium (that is, the rest position). Also, if the displacement to the equilibrium is positive, the force will be negative, as it tries to pull the object back (i.e. if you pull a string, the force is opposite to your direction of pull).
Therefore, it can be said that:
$$F propto -x$$
Where $F$ is the force and $x$ is the displacement from equilibrium
Is this the same as:
$$F propto x$$
According to the relation of proportionality, it should be, but my friend says that putting the second one is not correct.
Both are equivalent right?
ratio
ratio
edited 2 hours ago
Brian
1,529516
asked 5 hours ago
Sebi SSebi S
182
New contributor
Sebi S is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 hours ago
Brian
1,529516
asked 5 hours ago
Sebi SSebi S
182
New contributor
Sebi S is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 hours ago
Brian
1,529516
edited 2 hours ago
Brian
1,529516
edited 2 hours ago
Brian
1,529516
1,529516
asked 5 hours ago
Sebi SSebi S
182
New contributor
Sebi S is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 5 hours ago
Sebi SSebi S
182
asked 5 hours ago
Sebi SSebi S
182
182
New contributor
Sebi S is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Sebi S is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Sebi S is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
It depends on definitions/context. Using the definition on wikipedia, yes they are equivalent.
$endgroup$
– David Peterson
5 hours ago
add a comment |
$begingroup$
It depends on definitions/context. Using the definition on wikipedia, yes they are equivalent.
$endgroup$
– David Peterson
5 hours ago
$begingroup$
It depends on definitions/context. Using the definition on wikipedia, yes they are equivalent.
$endgroup$
– David Peterson
5 hours ago
$begingroup$
It depends on definitions/context. Using the definition on wikipedia, yes they are equivalent.
$endgroup$
– David Peterson
5 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Yes, proportionality does not tell anything about the sign of the proportionality constant. This is probably done in physics so that the proportionality constant $k$ can be considered to be always positive. This has a physical interpretation , so it is convenient for it to be positive.
answered 5 hours ago
John DoeJohn Doe
12.3k11341
$endgroup$
add a comment |
$begingroup$
Mathematically they are equivalent, but physycist may want to differentiate between a force in the same direction as the displacement, and a force opposite to the displacement, as they lead to physically different behavior of the system. Thus the minus sign matters.
answered 5 hours ago
Adam LatosińskiAdam Latosiński
9809
$endgroup$
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
});
}
});
Sebi S is a new contributor. Be nice, and check out our Code of Conduct.
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%2f3204950%2fdoes-the-sign-matter-for-proportionality%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
$begingroup$
Yes, proportionality does not tell anything about the sign of the proportionality constant. This is probably done in physics so that the proportionality constant $k$ can be considered to be always positive. This has a physical interpretation , so it is convenient for it to be positive.
answered 5 hours ago
John DoeJohn Doe
12.3k11341
$endgroup$
add a comment |
$begingroup$
Yes, proportionality does not tell anything about the sign of the proportionality constant. This is probably done in physics so that the proportionality constant $k$ can be considered to be always positive. This has a physical interpretation , so it is convenient for it to be positive.
answered 5 hours ago
John DoeJohn Doe
12.3k11341
$endgroup$
add a comment |
$begingroup$
Yes, proportionality does not tell anything about the sign of the proportionality constant. This is probably done in physics so that the proportionality constant $k$ can be considered to be always positive. This has a physical interpretation , so it is convenient for it to be positive.
answered 5 hours ago
John DoeJohn Doe
12.3k11341
$endgroup$
Yes, proportionality does not tell anything about the sign of the proportionality constant. This is probably done in physics so that the proportionality constant $k$ can be considered to be always positive. This has a physical interpretation , so it is convenient for it to be positive.
answered 5 hours ago
John DoeJohn Doe
12.3k11341
answered 5 hours ago
John DoeJohn Doe
12.3k11341
answered 5 hours ago
John DoeJohn Doe
12.3k11341
answered 5 hours ago
John DoeJohn Doe
12.3k11341
12.3k11341
add a comment |
add a comment |
$begingroup$
Mathematically they are equivalent, but physycist may want to differentiate between a force in the same direction as the displacement, and a force opposite to the displacement, as they lead to physically different behavior of the system. Thus the minus sign matters.
answered 5 hours ago
Adam LatosińskiAdam Latosiński
9809
$endgroup$
add a comment |
$begingroup$
Mathematically they are equivalent, but physycist may want to differentiate between a force in the same direction as the displacement, and a force opposite to the displacement, as they lead to physically different behavior of the system. Thus the minus sign matters.
answered 5 hours ago
Adam LatosińskiAdam Latosiński
9809
$endgroup$
add a comment |
$begingroup$
Mathematically they are equivalent, but physycist may want to differentiate between a force in the same direction as the displacement, and a force opposite to the displacement, as they lead to physically different behavior of the system. Thus the minus sign matters.
answered 5 hours ago
Adam LatosińskiAdam Latosiński
9809
$endgroup$
Mathematically they are equivalent, but physycist may want to differentiate between a force in the same direction as the displacement, and a force opposite to the displacement, as they lead to physically different behavior of the system. Thus the minus sign matters.
answered 5 hours ago
Adam LatosińskiAdam Latosiński
9809
answered 5 hours ago
Adam LatosińskiAdam Latosiński
9809
answered 5 hours ago
Adam LatosińskiAdam Latosiński
9809
answered 5 hours ago
Adam LatosińskiAdam Latosiński
9809
9809
add a comment |
add a comment |
Sebi S is a new contributor. Be nice, and check out our Code of Conduct.
Sebi S is a new contributor. Be nice, and check out our Code of Conduct.
Sebi S is a new contributor. Be nice, and check out our Code of Conduct.
Sebi S is a new contributor. Be nice, and check out our Code of Conduct.
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%2f3204950%2fdoes-the-sign-matter-for-proportionality%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
$begingroup$
It depends on definitions/context. Using the definition on wikipedia, yes they are equivalent.
$endgroup$
– David Peterson
5 hours ago