Mixing
![Series convergence from Rudin [duplicate]](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhZCTB4-bb-s-32SAm6ytjzFOU06cH9o8q-a_wq_UnH6lcd8hvws23CmBWHM6g256LzTX9yK1EiCCzTC1NSgtxlNk_J9hdr9bA3tD0Nqntbl521j4bLAm_uXPANuWLBhcCKTKzns2gaUodx/s72-c/1.jpg)
Prove for the time derivative of a vector with constant magnitude
Clash Royale CLAN TAG#URR8PPP
up vote
3
down vote
favorite
First of all, sorry for my bad English, I have several doubts about the geometric demonstration made by Kleppner for the derivative of a vector that has a constant magnitude, page 25, which is attached in this photo:
.
Is this proof rigorously validated? I understand that $sin (x) approx x$ for small $x$, by dividing both sides by $Delta t$, and take the limit for the equality. I understand that equality is correct, but is the argument correct? In the book, i understand the intuitive part, but for example what would have happened if I had taken an approximation of higher order, this would not have had any contribution in the limit? How are these terms depreciated rigorously, since I think that depends on how the time angle $θ (t)$ depends, for example, you can say that the vector varies with $Δθ (t)$, very slow or very fast so that the terms of higher order have a contribution in the limit, and therefore the equality raised is not correct, I apologize in case the question is a triviality.
geometry derivatives vectors physics mathematical-physics
asked Jul 17 at 8:18
Jalil Varela Manjarres
184
add a comment |Â
up vote
3
down vote
favorite
First of all, sorry for my bad English, I have several doubts about the geometric demonstration made by Kleppner for the derivative of a vector that has a constant magnitude, page 25, which is attached in this photo:
.
Is this proof rigorously validated? I understand that $sin (x) approx x$ for small $x$, by dividing both sides by $Delta t$, and take the limit for the equality. I understand that equality is correct, but is the argument correct? In the book, i understand the intuitive part, but for example what would have happened if I had taken an approximation of higher order, this would not have had any contribution in the limit? How are these terms depreciated rigorously, since I think that depends on how the time angle $θ (t)$ depends, for example, you can say that the vector varies with $Δθ (t)$, very slow or very fast so that the terms of higher order have a contribution in the limit, and therefore the equality raised is not correct, I apologize in case the question is a triviality.
geometry derivatives vectors physics mathematical-physics
asked Jul 17 at 8:18
Jalil Varela Manjarres
184
3
Your question is perfectly valid. Merely using differentials is a bit hand-wavy here, but the idea is that $Delta theta$ is the difference between the vectors $A(t+Delta t)$ and $A(t)$, and as $Delta t to 0$, $Delta theta to 0$. The rate at which $Delta theta to 0$ is linearly related to the rate at which $Delta t to 0$ by some basic geometry about circles
– Brevan Ellefsen
Jul 17 at 8:31
Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures.
– José Carlos Santos
Jul 17 at 8:38
1
Kleppner's book is a really good introductory mechanics book. Thanks for reminding me of this book.
– Sou
Jul 17 at 18:57
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
First of all, sorry for my bad English, I have several doubts about the geometric demonstration made by Kleppner for the derivative of a vector that has a constant magnitude, page 25, which is attached in this photo:
.
Is this proof rigorously validated? I understand that $sin (x) approx x$ for small $x$, by dividing both sides by $Delta t$, and take the limit for the equality. I understand that equality is correct, but is the argument correct? In the book, i understand the intuitive part, but for example what would have happened if I had taken an approximation of higher order, this would not have had any contribution in the limit? How are these terms depreciated rigorously, since I think that depends on how the time angle $θ (t)$ depends, for example, you can say that the vector varies with $Δθ (t)$, very slow or very fast so that the terms of higher order have a contribution in the limit, and therefore the equality raised is not correct, I apologize in case the question is a triviality.
geometry derivatives vectors physics mathematical-physics
asked Jul 17 at 8:18
Jalil Varela Manjarres
184
First of all, sorry for my bad English, I have several doubts about the geometric demonstration made by Kleppner for the derivative of a vector that has a constant magnitude, page 25, which is attached in this photo:
.
Is this proof rigorously validated? I understand that $sin (x) approx x$ for small $x$, by dividing both sides by $Delta t$, and take the limit for the equality. I understand that equality is correct, but is the argument correct? In the book, i understand the intuitive part, but for example what would have happened if I had taken an approximation of higher order, this would not have had any contribution in the limit? How are these terms depreciated rigorously, since I think that depends on how the time angle $θ (t)$ depends, for example, you can say that the vector varies with $Δθ (t)$, very slow or very fast so that the terms of higher order have a contribution in the limit, and therefore the equality raised is not correct, I apologize in case the question is a triviality.
geometry derivatives vectors physics mathematical-physics
asked Jul 17 at 8:18
Jalil Varela Manjarres
184
edited Jul 17 at 9:08
asked Jul 17 at 8:18
Jalil Varela Manjarres
184
asked Jul 17 at 8:18
Jalil Varela Manjarres
184
asked Jul 17 at 8:18
Jalil Varela Manjarres
184
184
3
Your question is perfectly valid. Merely using differentials is a bit hand-wavy here, but the idea is that $Delta theta$ is the difference between the vectors $A(t+Delta t)$ and $A(t)$, and as $Delta t to 0$, $Delta theta to 0$. The rate at which $Delta theta to 0$ is linearly related to the rate at which $Delta t to 0$ by some basic geometry about circles
– Brevan Ellefsen
Jul 17 at 8:31
Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures.
– José Carlos Santos
Jul 17 at 8:38
1
Kleppner's book is a really good introductory mechanics book. Thanks for reminding me of this book.
– Sou
Jul 17 at 18:57
add a comment |Â
3
Your question is perfectly valid. Merely using differentials is a bit hand-wavy here, but the idea is that $Delta theta$ is the difference between the vectors $A(t+Delta t)$ and $A(t)$, and as $Delta t to 0$, $Delta theta to 0$. The rate at which $Delta theta to 0$ is linearly related to the rate at which $Delta t to 0$ by some basic geometry about circles
– Brevan Ellefsen
Jul 17 at 8:31
Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures.
– José Carlos Santos
Jul 17 at 8:38
1
Kleppner's book is a really good introductory mechanics book. Thanks for reminding me of this book.
– Sou
Jul 17 at 18:57
3
3
Your question is perfectly valid. Merely using differentials is a bit hand-wavy here, but the idea is that $Delta theta$ is the difference between the vectors $A(t+Delta t)$ and $A(t)$, and as $Delta t to 0$, $Delta theta to 0$. The rate at which $Delta theta to 0$ is linearly related to the rate at which $Delta t to 0$ by some basic geometry about circles
– Brevan Ellefsen
Jul 17 at 8:31
Your question is perfectly valid. Merely using differentials is a bit hand-wavy here, but the idea is that $Delta theta$ is the difference between the vectors $A(t+Delta t)$ and $A(t)$, and as $Delta t to 0$, $Delta theta to 0$. The rate at which $Delta theta to 0$ is linearly related to the rate at which $Delta t to 0$ by some basic geometry about circles
– Brevan Ellefsen
Jul 17 at 8:31
Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures.
– José Carlos Santos
Jul 17 at 8:38
Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures.
– José Carlos Santos
Jul 17 at 8:38
1
1
Kleppner's book is a really good introductory mechanics book. Thanks for reminding me of this book.
– Sou
Jul 17 at 18:57
Kleppner's book is a really good introductory mechanics book. Thanks for reminding me of this book.
– Sou
Jul 17 at 18:57
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
0
down vote
accepted
The argument can be made more rigorous by using the well-known
$displaystylelim_xto0sin xover x=1$:
$$
left|Delta AoverDelta tright|=2Asin(Deltatheta/2)overDelta t
=2Asin(Deltatheta/2)overDeltatheta/2Deltatheta/2overDelta t
=Asin(Deltatheta/2)overDeltatheta/2DeltathetaoverDelta t,
$$
and $displaystylesin(Deltatheta/2)overDeltatheta/2to1$, because $Deltathetato0$ as $Delta tto0$.
answered Jul 17 at 8:39
Aretino
21.8k21342
add a comment |Â
up vote
0
down vote
If you are looking for an intuitive way to justify your book, I repeat my comment above:
Your question is perfectly valid. Merely using differentials is a bit hand-wavy here, but the idea is that $Delta theta$ is the difference between the vectors $A(t+Δt)$ and $A(t)$, and as $Δt→0$, $Δθ→0$. The rate at which $Δθ→0$ is linearly related to the rate at which $Δt→0$ by some basic geometry about circles
I also provide the following if you want an alternative, perhaps more mathematically satisfying approach:
Consider the image at the bottom of this post (from Wikipedia), where we can denote the vector in the diagram with radius $r$ by $bfA$$(t)$ as your textbook does. Then we have $bfA$$(theta)= left[rcostheta,rsinthetaright]$ so $bfv$$(t) = fracdbf A(t)dt = fracdbf A(theta)dtheta fracdthetadt =left[-rsintheta,rcosthetaright]fracdthetadt$ (where we used the chain rule) so that $|bfv(t)| = sqrt(-rsintheta)^2+(rcostheta)^2fracdthetadt = sqrtr^2(sin^2theta+cos^2theta)fracdthetadt = rfracdthetadt$
Which is precisely what your textbook claims.
answered Jul 17 at 8:48
Brevan Ellefsen
11.4k31449
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
The argument can be made more rigorous by using the well-known
$displaystylelim_xto0sin xover x=1$:
$$
left|Delta AoverDelta tright|=2Asin(Deltatheta/2)overDelta t
=2Asin(Deltatheta/2)overDeltatheta/2Deltatheta/2overDelta t
=Asin(Deltatheta/2)overDeltatheta/2DeltathetaoverDelta t,
$$
and $displaystylesin(Deltatheta/2)overDeltatheta/2to1$, because $Deltathetato0$ as $Delta tto0$.
answered Jul 17 at 8:39
Aretino
21.8k21342
add a comment |Â
up vote
0
down vote
accepted
The argument can be made more rigorous by using the well-known
$displaystylelim_xto0sin xover x=1$:
$$
left|Delta AoverDelta tright|=2Asin(Deltatheta/2)overDelta t
=2Asin(Deltatheta/2)overDeltatheta/2Deltatheta/2overDelta t
=Asin(Deltatheta/2)overDeltatheta/2DeltathetaoverDelta t,
$$
and $displaystylesin(Deltatheta/2)overDeltatheta/2to1$, because $Deltathetato0$ as $Delta tto0$.
answered Jul 17 at 8:39
Aretino
21.8k21342
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
The argument can be made more rigorous by using the well-known
$displaystylelim_xto0sin xover x=1$:
$$
left|Delta AoverDelta tright|=2Asin(Deltatheta/2)overDelta t
=2Asin(Deltatheta/2)overDeltatheta/2Deltatheta/2overDelta t
=Asin(Deltatheta/2)overDeltatheta/2DeltathetaoverDelta t,
$$
and $displaystylesin(Deltatheta/2)overDeltatheta/2to1$, because $Deltathetato0$ as $Delta tto0$.
answered Jul 17 at 8:39
Aretino
21.8k21342
The argument can be made more rigorous by using the well-known
$displaystylelim_xto0sin xover x=1$:
$$
left|Delta AoverDelta tright|=2Asin(Deltatheta/2)overDelta t
=2Asin(Deltatheta/2)overDeltatheta/2Deltatheta/2overDelta t
=Asin(Deltatheta/2)overDeltatheta/2DeltathetaoverDelta t,
$$
and $displaystylesin(Deltatheta/2)overDeltatheta/2to1$, because $Deltathetato0$ as $Delta tto0$.
answered Jul 17 at 8:39
Aretino
21.8k21342
answered Jul 17 at 8:39
Aretino
21.8k21342
answered Jul 17 at 8:39
Aretino
21.8k21342
answered Jul 17 at 8:39
Aretino
21.8k21342
21.8k21342
add a comment |Â
add a comment |Â
up vote
0
down vote
If you are looking for an intuitive way to justify your book, I repeat my comment above:
Your question is perfectly valid. Merely using differentials is a bit hand-wavy here, but the idea is that $Delta theta$ is the difference between the vectors $A(t+Δt)$ and $A(t)$, and as $Δt→0$, $Δθ→0$. The rate at which $Δθ→0$ is linearly related to the rate at which $Δt→0$ by some basic geometry about circles
I also provide the following if you want an alternative, perhaps more mathematically satisfying approach:
Consider the image at the bottom of this post (from Wikipedia), where we can denote the vector in the diagram with radius $r$ by $bfA$$(t)$ as your textbook does. Then we have $bfA$$(theta)= left[rcostheta,rsinthetaright]$ so $bfv$$(t) = fracdbf A(t)dt = fracdbf A(theta)dtheta fracdthetadt =left[-rsintheta,rcosthetaright]fracdthetadt$ (where we used the chain rule) so that $|bfv(t)| = sqrt(-rsintheta)^2+(rcostheta)^2fracdthetadt = sqrtr^2(sin^2theta+cos^2theta)fracdthetadt = rfracdthetadt$
Which is precisely what your textbook claims.
answered Jul 17 at 8:48
Brevan Ellefsen
11.4k31449
add a comment |Â
up vote
0
down vote
If you are looking for an intuitive way to justify your book, I repeat my comment above:
Your question is perfectly valid. Merely using differentials is a bit hand-wavy here, but the idea is that $Delta theta$ is the difference between the vectors $A(t+Δt)$ and $A(t)$, and as $Δt→0$, $Δθ→0$. The rate at which $Δθ→0$ is linearly related to the rate at which $Δt→0$ by some basic geometry about circles
I also provide the following if you want an alternative, perhaps more mathematically satisfying approach:
Consider the image at the bottom of this post (from Wikipedia), where we can denote the vector in the diagram with radius $r$ by $bfA$$(t)$ as your textbook does. Then we have $bfA$$(theta)= left[rcostheta,rsinthetaright]$ so $bfv$$(t) = fracdbf A(t)dt = fracdbf A(theta)dtheta fracdthetadt =left[-rsintheta,rcosthetaright]fracdthetadt$ (where we used the chain rule) so that $|bfv(t)| = sqrt(-rsintheta)^2+(rcostheta)^2fracdthetadt = sqrtr^2(sin^2theta+cos^2theta)fracdthetadt = rfracdthetadt$
Which is precisely what your textbook claims.
answered Jul 17 at 8:48
Brevan Ellefsen
11.4k31449
add a comment |Â
up vote
0
down vote
up vote
0
down vote
If you are looking for an intuitive way to justify your book, I repeat my comment above:
Your question is perfectly valid. Merely using differentials is a bit hand-wavy here, but the idea is that $Delta theta$ is the difference between the vectors $A(t+Δt)$ and $A(t)$, and as $Δt→0$, $Δθ→0$. The rate at which $Δθ→0$ is linearly related to the rate at which $Δt→0$ by some basic geometry about circles
I also provide the following if you want an alternative, perhaps more mathematically satisfying approach:
Consider the image at the bottom of this post (from Wikipedia), where we can denote the vector in the diagram with radius $r$ by $bfA$$(t)$ as your textbook does. Then we have $bfA$$(theta)= left[rcostheta,rsinthetaright]$ so $bfv$$(t) = fracdbf A(t)dt = fracdbf A(theta)dtheta fracdthetadt =left[-rsintheta,rcosthetaright]fracdthetadt$ (where we used the chain rule) so that $|bfv(t)| = sqrt(-rsintheta)^2+(rcostheta)^2fracdthetadt = sqrtr^2(sin^2theta+cos^2theta)fracdthetadt = rfracdthetadt$
Which is precisely what your textbook claims.
answered Jul 17 at 8:48
Brevan Ellefsen
11.4k31449
If you are looking for an intuitive way to justify your book, I repeat my comment above:
Your question is perfectly valid. Merely using differentials is a bit hand-wavy here, but the idea is that $Delta theta$ is the difference between the vectors $A(t+Δt)$ and $A(t)$, and as $Δt→0$, $Δθ→0$. The rate at which $Δθ→0$ is linearly related to the rate at which $Δt→0$ by some basic geometry about circles
I also provide the following if you want an alternative, perhaps more mathematically satisfying approach:
Consider the image at the bottom of this post (from Wikipedia), where we can denote the vector in the diagram with radius $r$ by $bfA$$(t)$ as your textbook does. Then we have $bfA$$(theta)= left[rcostheta,rsinthetaright]$ so $bfv$$(t) = fracdbf A(t)dt = fracdbf A(theta)dtheta fracdthetadt =left[-rsintheta,rcosthetaright]fracdthetadt$ (where we used the chain rule) so that $|bfv(t)| = sqrt(-rsintheta)^2+(rcostheta)^2fracdthetadt = sqrtr^2(sin^2theta+cos^2theta)fracdthetadt = rfracdthetadt$
Which is precisely what your textbook claims.
answered Jul 17 at 8:48
Brevan Ellefsen
11.4k31449
answered Jul 17 at 8:48
Brevan Ellefsen
11.4k31449
answered Jul 17 at 8:48
Brevan Ellefsen
11.4k31449
answered Jul 17 at 8:48
Brevan Ellefsen
11.4k31449
11.4k31449
add a comment |Â
add a comment |Â
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
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2854291%2fprove-for-the-time-derivative-of-a-vector-with-constant-magnitude%23new-answer', 'question_page');
);
Post as a guest
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
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
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
3
Your question is perfectly valid. Merely using differentials is a bit hand-wavy here, but the idea is that $Delta theta$ is the difference between the vectors $A(t+Delta t)$ and $A(t)$, and as $Delta t to 0$, $Delta theta to 0$. The rate at which $Delta theta to 0$ is linearly related to the rate at which $Delta t to 0$ by some basic geometry about circles
– Brevan Ellefsen
Jul 17 at 8:31
Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures.
– José Carlos Santos
Jul 17 at 8:38
1
Kleppner's book is a really good introductory mechanics book. Thanks for reminding me of this book.
– Sou
Jul 17 at 18:57