Mixing
How are the steps to the solution for Arc - Length obtained?
$begingroup$
Can someone please help me follow and understand the steps of the solution marked with $(*)$ and $(@)$? Why is the dot product used and computed with the unit vector. How does this equal the integral? I'm not sure where these steps are coming from.
Example:
Show that in $mathbb{R}^n$, the length of any parametric curve connecting a and b is at least as long as the length of the straight line connecting the points with position vectors a and b.
Solution:
For a parametric curve with $I = [a,b]$ and x(a) = a and x(b) = b and for a unit vector u we have the inequality $$({bf{b}-{bf{a}}})cdot {bf{u}} = int_a^b dot {{bf{x}}} cdot {bf{u}} {it{dt}} ≤int_ b ^a |dot{bf{x}}|dt quad (*)$$
as $dot{bf{x}}·{bf{u} = |dot{x}||u|}cosθ ≤| {bf{dot{x}}}||{bf{u}}| = | {bf{dot{x}}}|$, and by taking
$${bf{u}} =frac{
{bf{b−a} }}{{bf{|b−a|}}} quad @$$
we obtain that
$$|{bf{b−a}}|≤int_{b}^ a |dot{bf{ x}}| dt = L $$so that the length of the curve is at least |b−a|.
vectors proof-explanation calculus-of-variations parametric arc-length
edited Jan 13 at 8:18
Parcly Taxel
41.8k1372101
$endgroup$
add a comment |
$begingroup$
Can someone please help me follow and understand the steps of the solution marked with $(*)$ and $(@)$? Why is the dot product used and computed with the unit vector. How does this equal the integral? I'm not sure where these steps are coming from.
Example:
Show that in $mathbb{R}^n$, the length of any parametric curve connecting a and b is at least as long as the length of the straight line connecting the points with position vectors a and b.
Solution:
For a parametric curve with $I = [a,b]$ and x(a) = a and x(b) = b and for a unit vector u we have the inequality $$({bf{b}-{bf{a}}})cdot {bf{u}} = int_a^b dot {{bf{x}}} cdot {bf{u}} {it{dt}} ≤int_ b ^a |dot{bf{x}}|dt quad (*)$$
as $dot{bf{x}}·{bf{u} = |dot{x}||u|}cosθ ≤| {bf{dot{x}}}||{bf{u}}| = | {bf{dot{x}}}|$, and by taking
$${bf{u}} =frac{
{bf{b−a} }}{{bf{|b−a|}}} quad @$$
we obtain that
$$|{bf{b−a}}|≤int_{b}^ a |dot{bf{ x}}| dt = L $$so that the length of the curve is at least |b−a|.
vectors proof-explanation calculus-of-variations parametric arc-length
edited Jan 13 at 8:18
Parcly Taxel
41.8k1372101
$endgroup$
2
$begingroup$
Gosh... isn't "the shortest path between two points is a straight line," proved 2500 years ago, sufficient?
$endgroup$
– David G. Stork
Jan 13 at 6:30
1
$begingroup$
@DavidG.Stork The problem is, this is the calculus of variations. Intuition needs to be proved here.
$endgroup$
– Parcly Taxel
Jan 13 at 8:18
add a comment |
$begingroup$
Can someone please help me follow and understand the steps of the solution marked with $(*)$ and $(@)$? Why is the dot product used and computed with the unit vector. How does this equal the integral? I'm not sure where these steps are coming from.
Example:
Show that in $mathbb{R}^n$, the length of any parametric curve connecting a and b is at least as long as the length of the straight line connecting the points with position vectors a and b.
Solution:
For a parametric curve with $I = [a,b]$ and x(a) = a and x(b) = b and for a unit vector u we have the inequality $$({bf{b}-{bf{a}}})cdot {bf{u}} = int_a^b dot {{bf{x}}} cdot {bf{u}} {it{dt}} ≤int_ b ^a |dot{bf{x}}|dt quad (*)$$
as $dot{bf{x}}·{bf{u} = |dot{x}||u|}cosθ ≤| {bf{dot{x}}}||{bf{u}}| = | {bf{dot{x}}}|$, and by taking
$${bf{u}} =frac{
{bf{b−a} }}{{bf{|b−a|}}} quad @$$
we obtain that
$$|{bf{b−a}}|≤int_{b}^ a |dot{bf{ x}}| dt = L $$so that the length of the curve is at least |b−a|.
vectors proof-explanation calculus-of-variations parametric arc-length
edited Jan 13 at 8:18
Parcly Taxel
41.8k1372101
$endgroup$
Can someone please help me follow and understand the steps of the solution marked with $(*)$ and $(@)$? Why is the dot product used and computed with the unit vector. How does this equal the integral? I'm not sure where these steps are coming from.
Example:
Show that in $mathbb{R}^n$, the length of any parametric curve connecting a and b is at least as long as the length of the straight line connecting the points with position vectors a and b.
Solution:
For a parametric curve with $I = [a,b]$ and x(a) = a and x(b) = b and for a unit vector u we have the inequality $$({bf{b}-{bf{a}}})cdot {bf{u}} = int_a^b dot {{bf{x}}} cdot {bf{u}} {it{dt}} ≤int_ b ^a |dot{bf{x}}|dt quad (*)$$
as $dot{bf{x}}·{bf{u} = |dot{x}||u|}cosθ ≤| {bf{dot{x}}}||{bf{u}}| = | {bf{dot{x}}}|$, and by taking
$${bf{u}} =frac{
{bf{b−a} }}{{bf{|b−a|}}} quad @$$
we obtain that
$$|{bf{b−a}}|≤int_{b}^ a |dot{bf{ x}}| dt = L $$so that the length of the curve is at least |b−a|.
vectors proof-explanation calculus-of-variations parametric arc-length
vectors proof-explanation calculus-of-variations parametric arc-length
edited Jan 13 at 8:18
Parcly Taxel
41.8k1372101
edited Jan 13 at 8:18
Parcly Taxel
41.8k1372101
edited Jan 13 at 8:18
Parcly Taxel
41.8k1372101
edited Jan 13 at 8:18
Parcly Taxel
41.8k1372101
edited Jan 13 at 8:18
Parcly Taxel
41.8k1372101
41.8k1372101
asked Jan 13 at 6:22
user503154
2
$begingroup$
Gosh... isn't "the shortest path between two points is a straight line," proved 2500 years ago, sufficient?
$endgroup$
– David G. Stork
Jan 13 at 6:30
1
$begingroup$
@DavidG.Stork The problem is, this is the calculus of variations. Intuition needs to be proved here.
$endgroup$
– Parcly Taxel
Jan 13 at 8:18
add a comment |
2
$begingroup$
Gosh... isn't "the shortest path between two points is a straight line," proved 2500 years ago, sufficient?
$endgroup$
– David G. Stork
Jan 13 at 6:30
1
$begingroup$
@DavidG.Stork The problem is, this is the calculus of variations. Intuition needs to be proved here.
$endgroup$
– Parcly Taxel
Jan 13 at 8:18
2
2
$begingroup$
Gosh... isn't "the shortest path between two points is a straight line," proved 2500 years ago, sufficient?
$endgroup$
– David G. Stork
Jan 13 at 6:30
$begingroup$
Gosh... isn't "the shortest path between two points is a straight line," proved 2500 years ago, sufficient?
$endgroup$
– David G. Stork
Jan 13 at 6:30
1
1
$begingroup$
@DavidG.Stork The problem is, this is the calculus of variations. Intuition needs to be proved here.
$endgroup$
– Parcly Taxel
Jan 13 at 8:18
$begingroup$
@DavidG.Stork The problem is, this is the calculus of variations. Intuition needs to be proved here.
$endgroup$
– Parcly Taxel
Jan 13 at 8:18
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The dot product is taken because you want to estimate the length $|{bf b} - {bf a}|$.
Now:
$$|{bf b} - {bf a}|^2 = ({bf b}-{bf a}) cdot ({bf b} - {bf a})$$
Thus:
$$|{bf b} - {bf a}| = frac{({bf b}-{bf a}) cdot ({bf b} - {bf a}) }{ |{bf b} - {bf a}|} = ({bf b}-{bf a}) cdot {bf u}$$
Note that the unit vector $bf{u}$ is constant. If you spell out the inner product, you'll get
$$
int_a^b {dot{bf{x}}} {cdot} {bf{u}} , dt =
int_a^b sum_{i=1}^n frac {dx_i(t)} {dt} u_i , dt =
sum_{i=1}^n u_i int_a^b frac {dx_i(t)}{dt} dt
$$
The integral on the right amounts to the total difference $x_i(b) - x_i(a)$. Thus,
$$
sum_{i=1}^n u_i int_a^b frac {dx_i(t)}{dt} dt = sum_{i=1}^n u_i , (b_i - a_i) = {bf u} cdot ({bf b} - {bf a})
$$
answered Jan 13 at 15:56
jgbjgb
14813
$endgroup$
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%2f3071748%2fhow-are-the-steps-to-the-solution-for-arc-length-obtained%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 dot product is taken because you want to estimate the length $|{bf b} - {bf a}|$.
Now:
$$|{bf b} - {bf a}|^2 = ({bf b}-{bf a}) cdot ({bf b} - {bf a})$$
Thus:
$$|{bf b} - {bf a}| = frac{({bf b}-{bf a}) cdot ({bf b} - {bf a}) }{ |{bf b} - {bf a}|} = ({bf b}-{bf a}) cdot {bf u}$$
Note that the unit vector $bf{u}$ is constant. If you spell out the inner product, you'll get
$$
int_a^b {dot{bf{x}}} {cdot} {bf{u}} , dt =
int_a^b sum_{i=1}^n frac {dx_i(t)} {dt} u_i , dt =
sum_{i=1}^n u_i int_a^b frac {dx_i(t)}{dt} dt
$$
The integral on the right amounts to the total difference $x_i(b) - x_i(a)$. Thus,
$$
sum_{i=1}^n u_i int_a^b frac {dx_i(t)}{dt} dt = sum_{i=1}^n u_i , (b_i - a_i) = {bf u} cdot ({bf b} - {bf a})
$$
answered Jan 13 at 15:56
jgbjgb
14813
$endgroup$
add a comment |
$begingroup$
The dot product is taken because you want to estimate the length $|{bf b} - {bf a}|$.
Now:
$$|{bf b} - {bf a}|^2 = ({bf b}-{bf a}) cdot ({bf b} - {bf a})$$
Thus:
$$|{bf b} - {bf a}| = frac{({bf b}-{bf a}) cdot ({bf b} - {bf a}) }{ |{bf b} - {bf a}|} = ({bf b}-{bf a}) cdot {bf u}$$
Note that the unit vector $bf{u}$ is constant. If you spell out the inner product, you'll get
$$
int_a^b {dot{bf{x}}} {cdot} {bf{u}} , dt =
int_a^b sum_{i=1}^n frac {dx_i(t)} {dt} u_i , dt =
sum_{i=1}^n u_i int_a^b frac {dx_i(t)}{dt} dt
$$
The integral on the right amounts to the total difference $x_i(b) - x_i(a)$. Thus,
$$
sum_{i=1}^n u_i int_a^b frac {dx_i(t)}{dt} dt = sum_{i=1}^n u_i , (b_i - a_i) = {bf u} cdot ({bf b} - {bf a})
$$
answered Jan 13 at 15:56
jgbjgb
14813
$endgroup$
add a comment |
$begingroup$
The dot product is taken because you want to estimate the length $|{bf b} - {bf a}|$.
Now:
$$|{bf b} - {bf a}|^2 = ({bf b}-{bf a}) cdot ({bf b} - {bf a})$$
Thus:
$$|{bf b} - {bf a}| = frac{({bf b}-{bf a}) cdot ({bf b} - {bf a}) }{ |{bf b} - {bf a}|} = ({bf b}-{bf a}) cdot {bf u}$$
Note that the unit vector $bf{u}$ is constant. If you spell out the inner product, you'll get
$$
int_a^b {dot{bf{x}}} {cdot} {bf{u}} , dt =
int_a^b sum_{i=1}^n frac {dx_i(t)} {dt} u_i , dt =
sum_{i=1}^n u_i int_a^b frac {dx_i(t)}{dt} dt
$$
The integral on the right amounts to the total difference $x_i(b) - x_i(a)$. Thus,
$$
sum_{i=1}^n u_i int_a^b frac {dx_i(t)}{dt} dt = sum_{i=1}^n u_i , (b_i - a_i) = {bf u} cdot ({bf b} - {bf a})
$$
answered Jan 13 at 15:56
jgbjgb
14813
$endgroup$
The dot product is taken because you want to estimate the length $|{bf b} - {bf a}|$.
Now:
$$|{bf b} - {bf a}|^2 = ({bf b}-{bf a}) cdot ({bf b} - {bf a})$$
Thus:
$$|{bf b} - {bf a}| = frac{({bf b}-{bf a}) cdot ({bf b} - {bf a}) }{ |{bf b} - {bf a}|} = ({bf b}-{bf a}) cdot {bf u}$$
Note that the unit vector $bf{u}$ is constant. If you spell out the inner product, you'll get
$$
int_a^b {dot{bf{x}}} {cdot} {bf{u}} , dt =
int_a^b sum_{i=1}^n frac {dx_i(t)} {dt} u_i , dt =
sum_{i=1}^n u_i int_a^b frac {dx_i(t)}{dt} dt
$$
The integral on the right amounts to the total difference $x_i(b) - x_i(a)$. Thus,
$$
sum_{i=1}^n u_i int_a^b frac {dx_i(t)}{dt} dt = sum_{i=1}^n u_i , (b_i - a_i) = {bf u} cdot ({bf b} - {bf a})
$$
answered Jan 13 at 15:56
jgbjgb
14813
edited Jan 14 at 3:29
answered Jan 13 at 15:56
jgbjgb
14813
answered Jan 13 at 15:56
jgbjgb
14813
answered Jan 13 at 15:56
jgbjgb
14813
14813
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%2f3071748%2fhow-are-the-steps-to-the-solution-for-arc-length-obtained%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
2
$begingroup$
Gosh... isn't "the shortest path between two points is a straight line," proved 2500 years ago, sufficient?
$endgroup$
– David G. Stork
Jan 13 at 6:30
1
$begingroup$
@DavidG.Stork The problem is, this is the calculus of variations. Intuition needs to be proved here.
$endgroup$
– Parcly Taxel
Jan 13 at 8:18