Mixing
Planar Curve $(acos(theta), asin(theta), f(theta))$
$begingroup$
Find $f$ such that following represents a planar Curve $(acos(theta), asin(theta), f(theta))$ for parameter $theta$.
I have a gut feeling that $f(theta)= constant$ as otherwise it would become similar to helix for infinitesimal change in $theta$, thus rendering the curve non planar. But I am not able to formally derive any result. Any hint is much appreciated. Thanks
vector-analysis 3d plane-curves
edited Jun 15 '18 at 8:42
José Carlos Santos
174k23133242
asked Jun 15 '18 at 8:35
markovchainmarkovchain
24818
$endgroup$
add a comment |
$begingroup$
Find $f$ such that following represents a planar Curve $(acos(theta), asin(theta), f(theta))$ for parameter $theta$.
I have a gut feeling that $f(theta)= constant$ as otherwise it would become similar to helix for infinitesimal change in $theta$, thus rendering the curve non planar. But I am not able to formally derive any result. Any hint is much appreciated. Thanks
vector-analysis 3d plane-curves
edited Jun 15 '18 at 8:42
José Carlos Santos
174k23133242
asked Jun 15 '18 at 8:35
markovchainmarkovchain
24818
$endgroup$
$begingroup$
How does the second order approximation lock like?
$endgroup$
– Stefan
Jun 15 '18 at 8:42
add a comment |
$begingroup$
Find $f$ such that following represents a planar Curve $(acos(theta), asin(theta), f(theta))$ for parameter $theta$.
I have a gut feeling that $f(theta)= constant$ as otherwise it would become similar to helix for infinitesimal change in $theta$, thus rendering the curve non planar. But I am not able to formally derive any result. Any hint is much appreciated. Thanks
vector-analysis 3d plane-curves
edited Jun 15 '18 at 8:42
José Carlos Santos
174k23133242
asked Jun 15 '18 at 8:35
markovchainmarkovchain
24818
$endgroup$
Find $f$ such that following represents a planar Curve $(acos(theta), asin(theta), f(theta))$ for parameter $theta$.
I have a gut feeling that $f(theta)= constant$ as otherwise it would become similar to helix for infinitesimal change in $theta$, thus rendering the curve non planar. But I am not able to formally derive any result. Any hint is much appreciated. Thanks
vector-analysis 3d plane-curves
vector-analysis 3d plane-curves
edited Jun 15 '18 at 8:42
José Carlos Santos
174k23133242
asked Jun 15 '18 at 8:35
markovchainmarkovchain
24818
edited Jun 15 '18 at 8:42
José Carlos Santos
174k23133242
asked Jun 15 '18 at 8:35
markovchainmarkovchain
24818
edited Jun 15 '18 at 8:42
José Carlos Santos
174k23133242
edited Jun 15 '18 at 8:42
José Carlos Santos
174k23133242
edited Jun 15 '18 at 8:42
José Carlos Santos
174k23133242
174k23133242
asked Jun 15 '18 at 8:35
markovchainmarkovchain
24818
asked Jun 15 '18 at 8:35
markovchainmarkovchain
24818
asked Jun 15 '18 at 8:35
markovchainmarkovchain
24818
24818
$begingroup$
How does the second order approximation lock like?
$endgroup$
– Stefan
Jun 15 '18 at 8:42
add a comment |
$begingroup$
How does the second order approximation lock like?
$endgroup$
– Stefan
Jun 15 '18 at 8:42
$begingroup$
How does the second order approximation lock like?
$endgroup$
– Stefan
Jun 15 '18 at 8:42
$begingroup$
How does the second order approximation lock like?
$endgroup$
– Stefan
Jun 15 '18 at 8:42
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
That is not true. For instance, it is clear that$$thetamapstobigl(acos(theta),asin(theta),acos(theta)bigr)$$is also a plane curve. In fact, you can take $f(theta)=alpha+betacos(theta)+gammasin(theta)$ (and these are the only solutions).
answered Jun 15 '18 at 8:41
José Carlos SantosJosé Carlos Santos
174k23133242
$endgroup$
$begingroup$
Thanks. I can see why above $f(theta)$ represents a planar curve. I am using the fact that 4x4 determinant of any four points on this curve has to be zero identically for it to represent a planar curve. For above parameterization this is true as two columns of determinant becomes same or linearly dependent. But how can we say these are the only possible solutions?
$endgroup$
– markovchain
Jun 15 '18 at 8:51
1
$begingroup$
Because the torsion of your curve is $a^2bigl(f'(theta)+f'''(theta)bigr)$, which is $0$ if and only if $f'$ is a linear combination of $cos(theta)$ and $sin(theta)$.
$endgroup$
– José Carlos Santos
Jun 15 '18 at 9:03
add a comment |
$begingroup$
Let
$$r(t)=left(,acos t,,,,asin t,,,,f(t),right)implies r'(t)=left(,-asin t,,acos t,,f'(t),right),$$
$$r''(t)=left(,-acos t,,-asin t,,f''(t),right),,,,r'''(t)=left(,asin t,,-acos t,,f'''(t),right)implies$$
$$ r'times r''=begin{vmatrix}i&j&k\
-asin t&acos t&f'(t)\
-acos t&-asin t&f''(t)end{vmatrix}=left(,a(f''cos t+f'sin t),,,,a(f''sin t-f'cos t),,,,a^2,right)$$
so
$$(r'times r'')cdot r'''=require{cancel}cancel{a^2f''sin tcos t}+a^2f'sin^2t-cancel{a^2f''cos tsin t}+a^2f'cos^2t+a^2f'''=$$
$$=a^2(f'+f''')$$
Since $;r;$ is planar iff $;tau=0;$ (its torsion), we get that it is planar iff
$$f'+f'''=0iff f''+f=C,,,C=text{constant}$$
and now you solve this easy differential equation.
answered Jun 15 '18 at 9:05
DonAntonioDonAntonio
180k1494233
$endgroup$
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%2f2820472%2fplanar-curve-a-cos-theta-a-sin-theta-f-theta%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
$begingroup$
That is not true. For instance, it is clear that$$thetamapstobigl(acos(theta),asin(theta),acos(theta)bigr)$$is also a plane curve. In fact, you can take $f(theta)=alpha+betacos(theta)+gammasin(theta)$ (and these are the only solutions).
answered Jun 15 '18 at 8:41
José Carlos SantosJosé Carlos Santos
174k23133242
$endgroup$
$begingroup$
Thanks. I can see why above $f(theta)$ represents a planar curve. I am using the fact that 4x4 determinant of any four points on this curve has to be zero identically for it to represent a planar curve. For above parameterization this is true as two columns of determinant becomes same or linearly dependent. But how can we say these are the only possible solutions?
$endgroup$
– markovchain
Jun 15 '18 at 8:51
1
$begingroup$
Because the torsion of your curve is $a^2bigl(f'(theta)+f'''(theta)bigr)$, which is $0$ if and only if $f'$ is a linear combination of $cos(theta)$ and $sin(theta)$.
$endgroup$
– José Carlos Santos
Jun 15 '18 at 9:03
add a comment |
$begingroup$
That is not true. For instance, it is clear that$$thetamapstobigl(acos(theta),asin(theta),acos(theta)bigr)$$is also a plane curve. In fact, you can take $f(theta)=alpha+betacos(theta)+gammasin(theta)$ (and these are the only solutions).
answered Jun 15 '18 at 8:41
José Carlos SantosJosé Carlos Santos
174k23133242
$endgroup$
$begingroup$
Thanks. I can see why above $f(theta)$ represents a planar curve. I am using the fact that 4x4 determinant of any four points on this curve has to be zero identically for it to represent a planar curve. For above parameterization this is true as two columns of determinant becomes same or linearly dependent. But how can we say these are the only possible solutions?
$endgroup$
– markovchain
Jun 15 '18 at 8:51
1
$begingroup$
Because the torsion of your curve is $a^2bigl(f'(theta)+f'''(theta)bigr)$, which is $0$ if and only if $f'$ is a linear combination of $cos(theta)$ and $sin(theta)$.
$endgroup$
– José Carlos Santos
Jun 15 '18 at 9:03
add a comment |
$begingroup$
That is not true. For instance, it is clear that$$thetamapstobigl(acos(theta),asin(theta),acos(theta)bigr)$$is also a plane curve. In fact, you can take $f(theta)=alpha+betacos(theta)+gammasin(theta)$ (and these are the only solutions).
answered Jun 15 '18 at 8:41
José Carlos SantosJosé Carlos Santos
174k23133242
$endgroup$
That is not true. For instance, it is clear that$$thetamapstobigl(acos(theta),asin(theta),acos(theta)bigr)$$is also a plane curve. In fact, you can take $f(theta)=alpha+betacos(theta)+gammasin(theta)$ (and these are the only solutions).
answered Jun 15 '18 at 8:41
José Carlos SantosJosé Carlos Santos
174k23133242
edited Feb 1 at 13:13
answered Jun 15 '18 at 8:41
José Carlos SantosJosé Carlos Santos
174k23133242
answered Jun 15 '18 at 8:41
José Carlos SantosJosé Carlos Santos
174k23133242
answered Jun 15 '18 at 8:41
José Carlos SantosJosé Carlos Santos
174k23133242
174k23133242
$begingroup$
Thanks. I can see why above $f(theta)$ represents a planar curve. I am using the fact that 4x4 determinant of any four points on this curve has to be zero identically for it to represent a planar curve. For above parameterization this is true as two columns of determinant becomes same or linearly dependent. But how can we say these are the only possible solutions?
$endgroup$
– markovchain
Jun 15 '18 at 8:51
1
$begingroup$
Because the torsion of your curve is $a^2bigl(f'(theta)+f'''(theta)bigr)$, which is $0$ if and only if $f'$ is a linear combination of $cos(theta)$ and $sin(theta)$.
$endgroup$
– José Carlos Santos
Jun 15 '18 at 9:03
add a comment |
$begingroup$
Thanks. I can see why above $f(theta)$ represents a planar curve. I am using the fact that 4x4 determinant of any four points on this curve has to be zero identically for it to represent a planar curve. For above parameterization this is true as two columns of determinant becomes same or linearly dependent. But how can we say these are the only possible solutions?
$endgroup$
– markovchain
Jun 15 '18 at 8:51
1
$begingroup$
Because the torsion of your curve is $a^2bigl(f'(theta)+f'''(theta)bigr)$, which is $0$ if and only if $f'$ is a linear combination of $cos(theta)$ and $sin(theta)$.
$endgroup$
– José Carlos Santos
Jun 15 '18 at 9:03
$begingroup$
Thanks. I can see why above $f(theta)$ represents a planar curve. I am using the fact that 4x4 determinant of any four points on this curve has to be zero identically for it to represent a planar curve. For above parameterization this is true as two columns of determinant becomes same or linearly dependent. But how can we say these are the only possible solutions?
$endgroup$
– markovchain
Jun 15 '18 at 8:51
$begingroup$
Thanks. I can see why above $f(theta)$ represents a planar curve. I am using the fact that 4x4 determinant of any four points on this curve has to be zero identically for it to represent a planar curve. For above parameterization this is true as two columns of determinant becomes same or linearly dependent. But how can we say these are the only possible solutions?
$endgroup$
– markovchain
Jun 15 '18 at 8:51
1
1
$begingroup$
Because the torsion of your curve is $a^2bigl(f'(theta)+f'''(theta)bigr)$, which is $0$ if and only if $f'$ is a linear combination of $cos(theta)$ and $sin(theta)$.
$endgroup$
– José Carlos Santos
Jun 15 '18 at 9:03
$begingroup$
Because the torsion of your curve is $a^2bigl(f'(theta)+f'''(theta)bigr)$, which is $0$ if and only if $f'$ is a linear combination of $cos(theta)$ and $sin(theta)$.
$endgroup$
– José Carlos Santos
Jun 15 '18 at 9:03
add a comment |
$begingroup$
Let
$$r(t)=left(,acos t,,,,asin t,,,,f(t),right)implies r'(t)=left(,-asin t,,acos t,,f'(t),right),$$
$$r''(t)=left(,-acos t,,-asin t,,f''(t),right),,,,r'''(t)=left(,asin t,,-acos t,,f'''(t),right)implies$$
$$ r'times r''=begin{vmatrix}i&j&k\
-asin t&acos t&f'(t)\
-acos t&-asin t&f''(t)end{vmatrix}=left(,a(f''cos t+f'sin t),,,,a(f''sin t-f'cos t),,,,a^2,right)$$
so
$$(r'times r'')cdot r'''=require{cancel}cancel{a^2f''sin tcos t}+a^2f'sin^2t-cancel{a^2f''cos tsin t}+a^2f'cos^2t+a^2f'''=$$
$$=a^2(f'+f''')$$
Since $;r;$ is planar iff $;tau=0;$ (its torsion), we get that it is planar iff
$$f'+f'''=0iff f''+f=C,,,C=text{constant}$$
and now you solve this easy differential equation.
answered Jun 15 '18 at 9:05
DonAntonioDonAntonio
180k1494233
$endgroup$
add a comment |
$begingroup$
Let
$$r(t)=left(,acos t,,,,asin t,,,,f(t),right)implies r'(t)=left(,-asin t,,acos t,,f'(t),right),$$
$$r''(t)=left(,-acos t,,-asin t,,f''(t),right),,,,r'''(t)=left(,asin t,,-acos t,,f'''(t),right)implies$$
$$ r'times r''=begin{vmatrix}i&j&k\
-asin t&acos t&f'(t)\
-acos t&-asin t&f''(t)end{vmatrix}=left(,a(f''cos t+f'sin t),,,,a(f''sin t-f'cos t),,,,a^2,right)$$
so
$$(r'times r'')cdot r'''=require{cancel}cancel{a^2f''sin tcos t}+a^2f'sin^2t-cancel{a^2f''cos tsin t}+a^2f'cos^2t+a^2f'''=$$
$$=a^2(f'+f''')$$
Since $;r;$ is planar iff $;tau=0;$ (its torsion), we get that it is planar iff
$$f'+f'''=0iff f''+f=C,,,C=text{constant}$$
and now you solve this easy differential equation.
answered Jun 15 '18 at 9:05
DonAntonioDonAntonio
180k1494233
$endgroup$
add a comment |
$begingroup$
Let
$$r(t)=left(,acos t,,,,asin t,,,,f(t),right)implies r'(t)=left(,-asin t,,acos t,,f'(t),right),$$
$$r''(t)=left(,-acos t,,-asin t,,f''(t),right),,,,r'''(t)=left(,asin t,,-acos t,,f'''(t),right)implies$$
$$ r'times r''=begin{vmatrix}i&j&k\
-asin t&acos t&f'(t)\
-acos t&-asin t&f''(t)end{vmatrix}=left(,a(f''cos t+f'sin t),,,,a(f''sin t-f'cos t),,,,a^2,right)$$
so
$$(r'times r'')cdot r'''=require{cancel}cancel{a^2f''sin tcos t}+a^2f'sin^2t-cancel{a^2f''cos tsin t}+a^2f'cos^2t+a^2f'''=$$
$$=a^2(f'+f''')$$
Since $;r;$ is planar iff $;tau=0;$ (its torsion), we get that it is planar iff
$$f'+f'''=0iff f''+f=C,,,C=text{constant}$$
and now you solve this easy differential equation.
answered Jun 15 '18 at 9:05
DonAntonioDonAntonio
180k1494233
$endgroup$
Let
$$r(t)=left(,acos t,,,,asin t,,,,f(t),right)implies r'(t)=left(,-asin t,,acos t,,f'(t),right),$$
$$r''(t)=left(,-acos t,,-asin t,,f''(t),right),,,,r'''(t)=left(,asin t,,-acos t,,f'''(t),right)implies$$
$$ r'times r''=begin{vmatrix}i&j&k\
-asin t&acos t&f'(t)\
-acos t&-asin t&f''(t)end{vmatrix}=left(,a(f''cos t+f'sin t),,,,a(f''sin t-f'cos t),,,,a^2,right)$$
so
$$(r'times r'')cdot r'''=require{cancel}cancel{a^2f''sin tcos t}+a^2f'sin^2t-cancel{a^2f''cos tsin t}+a^2f'cos^2t+a^2f'''=$$
$$=a^2(f'+f''')$$
Since $;r;$ is planar iff $;tau=0;$ (its torsion), we get that it is planar iff
$$f'+f'''=0iff f''+f=C,,,C=text{constant}$$
and now you solve this easy differential equation.
answered Jun 15 '18 at 9:05
DonAntonioDonAntonio
180k1494233
answered Jun 15 '18 at 9:05
DonAntonioDonAntonio
180k1494233
answered Jun 15 '18 at 9:05
DonAntonioDonAntonio
180k1494233
answered Jun 15 '18 at 9:05
DonAntonioDonAntonio
180k1494233
180k1494233
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%2f2820472%2fplanar-curve-a-cos-theta-a-sin-theta-f-theta%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
$begingroup$
How does the second order approximation lock like?
$endgroup$
– Stefan
Jun 15 '18 at 8:42