Command for vector dot with some power
up vote
4
down vote
favorite
So what I am trying to type is the square of the derivative of vector x. I tried dot{vec{x}}^{,2}
as well as dot{vec{x}^2}
, but the outputs came out to be very offset. Is there a correct way to do this?
The code I used is:
begin{equation}
L=frac{1}{2} m dot{vec{x^2}}
end{equation}
which give me
math-mode symbols accents
New contributor
add a comment |
up vote
4
down vote
favorite
So what I am trying to type is the square of the derivative of vector x. I tried dot{vec{x}}^{,2}
as well as dot{vec{x}^2}
, but the outputs came out to be very offset. Is there a correct way to do this?
The code I used is:
begin{equation}
L=frac{1}{2} m dot{vec{x^2}}
end{equation}
which give me
math-mode symbols accents
New contributor
Welcome to TeX.SE! Can you please, as usual here -- show us a short compilable code and an screenshot of your result?
– Kurt
1 hour ago
add a comment |
up vote
4
down vote
favorite
up vote
4
down vote
favorite
So what I am trying to type is the square of the derivative of vector x. I tried dot{vec{x}}^{,2}
as well as dot{vec{x}^2}
, but the outputs came out to be very offset. Is there a correct way to do this?
The code I used is:
begin{equation}
L=frac{1}{2} m dot{vec{x^2}}
end{equation}
which give me
math-mode symbols accents
New contributor
So what I am trying to type is the square of the derivative of vector x. I tried dot{vec{x}}^{,2}
as well as dot{vec{x}^2}
, but the outputs came out to be very offset. Is there a correct way to do this?
The code I used is:
begin{equation}
L=frac{1}{2} m dot{vec{x^2}}
end{equation}
which give me
math-mode symbols accents
math-mode symbols accents
New contributor
New contributor
edited 53 mins ago
Mico
271k30367755
271k30367755
New contributor
asked 1 hour ago
Kane Billiot
234
234
New contributor
New contributor
Welcome to TeX.SE! Can you please, as usual here -- show us a short compilable code and an screenshot of your result?
– Kurt
1 hour ago
add a comment |
Welcome to TeX.SE! Can you please, as usual here -- show us a short compilable code and an screenshot of your result?
– Kurt
1 hour ago
Welcome to TeX.SE! Can you please, as usual here -- show us a short compilable code and an screenshot of your result?
– Kurt
1 hour ago
Welcome to TeX.SE! Can you please, as usual here -- show us a short compilable code and an screenshot of your result?
– Kurt
1 hour ago
add a comment |
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
I'd probably do
documentclass{article}
usepackage{amsmath}
begin{document}
begin{equation}
L=frac{1}{2} m Dot{vec{x}}^2
end{equation}
end{document}
because the Lagrange function is a function of the square of the time derivative of x (and not the time derivative of the square of x).
That's what I wanted to write, but did not know how to. Thanks.
– Kane Billiot
40 mins ago
Is there a difference between dot and Dot?
– Kane Billiot
38 mins ago
@KaneBilliot Short answer:Dot
works also when you already have something on top of the symbol. So it would not shift.
– marmot
37 mins ago
1
@marmot - For the case at hand, usingdot
andDot
produce the same result.
– Mico
33 mins ago
@Mico Yes, you are right.
– marmot
20 mins ago
add a comment |
up vote
3
down vote
The first or third option below may be close to what you're looking for. Or, switch from Newton-style to Leibniz-style notation for the derivative, as shown by the fourth option (newly fixed to incorporated @marmot's comment). A separate comment: to make the frac{1}{2}
term less visually dominant, consider using tfrac
instead of frac
.
documentclass{article}
usepackage{amsmath} % for tfrac macro and general accent-placement support
begin{document}
[
tfrac{1}{2}m dot{vec{x}} ^2 quad
tfrac{1}{2}m{dot{vec{x}}}^2 quad
tfrac{1}{2}m{dot{vec{x}}}^{,2} quad
tfrac{1}{2}mbigl(tfrac{mathrm{d}vec{x}}{mathrm{d}t}bigr)^{!2}
]
end{document}
@marmot - I know. :-) But sometimes dot-notation becomes so involved that switching to Leibniz-style notation makes things easier...
– Mico
35 mins ago
1
@marmot - Many thanks for this clarification. I've updated the code to make the d's upright.
– Mico
18 mins ago
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
I'd probably do
documentclass{article}
usepackage{amsmath}
begin{document}
begin{equation}
L=frac{1}{2} m Dot{vec{x}}^2
end{equation}
end{document}
because the Lagrange function is a function of the square of the time derivative of x (and not the time derivative of the square of x).
That's what I wanted to write, but did not know how to. Thanks.
– Kane Billiot
40 mins ago
Is there a difference between dot and Dot?
– Kane Billiot
38 mins ago
@KaneBilliot Short answer:Dot
works also when you already have something on top of the symbol. So it would not shift.
– marmot
37 mins ago
1
@marmot - For the case at hand, usingdot
andDot
produce the same result.
– Mico
33 mins ago
@Mico Yes, you are right.
– marmot
20 mins ago
add a comment |
up vote
3
down vote
accepted
I'd probably do
documentclass{article}
usepackage{amsmath}
begin{document}
begin{equation}
L=frac{1}{2} m Dot{vec{x}}^2
end{equation}
end{document}
because the Lagrange function is a function of the square of the time derivative of x (and not the time derivative of the square of x).
That's what I wanted to write, but did not know how to. Thanks.
– Kane Billiot
40 mins ago
Is there a difference between dot and Dot?
– Kane Billiot
38 mins ago
@KaneBilliot Short answer:Dot
works also when you already have something on top of the symbol. So it would not shift.
– marmot
37 mins ago
1
@marmot - For the case at hand, usingdot
andDot
produce the same result.
– Mico
33 mins ago
@Mico Yes, you are right.
– marmot
20 mins ago
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
I'd probably do
documentclass{article}
usepackage{amsmath}
begin{document}
begin{equation}
L=frac{1}{2} m Dot{vec{x}}^2
end{equation}
end{document}
because the Lagrange function is a function of the square of the time derivative of x (and not the time derivative of the square of x).
I'd probably do
documentclass{article}
usepackage{amsmath}
begin{document}
begin{equation}
L=frac{1}{2} m Dot{vec{x}}^2
end{equation}
end{document}
because the Lagrange function is a function of the square of the time derivative of x (and not the time derivative of the square of x).
answered 41 mins ago
marmot
81.3k491173
81.3k491173
That's what I wanted to write, but did not know how to. Thanks.
– Kane Billiot
40 mins ago
Is there a difference between dot and Dot?
– Kane Billiot
38 mins ago
@KaneBilliot Short answer:Dot
works also when you already have something on top of the symbol. So it would not shift.
– marmot
37 mins ago
1
@marmot - For the case at hand, usingdot
andDot
produce the same result.
– Mico
33 mins ago
@Mico Yes, you are right.
– marmot
20 mins ago
add a comment |
That's what I wanted to write, but did not know how to. Thanks.
– Kane Billiot
40 mins ago
Is there a difference between dot and Dot?
– Kane Billiot
38 mins ago
@KaneBilliot Short answer:Dot
works also when you already have something on top of the symbol. So it would not shift.
– marmot
37 mins ago
1
@marmot - For the case at hand, usingdot
andDot
produce the same result.
– Mico
33 mins ago
@Mico Yes, you are right.
– marmot
20 mins ago
That's what I wanted to write, but did not know how to. Thanks.
– Kane Billiot
40 mins ago
That's what I wanted to write, but did not know how to. Thanks.
– Kane Billiot
40 mins ago
Is there a difference between dot and Dot?
– Kane Billiot
38 mins ago
Is there a difference between dot and Dot?
– Kane Billiot
38 mins ago
@KaneBilliot Short answer:
Dot
works also when you already have something on top of the symbol. So it would not shift.– marmot
37 mins ago
@KaneBilliot Short answer:
Dot
works also when you already have something on top of the symbol. So it would not shift.– marmot
37 mins ago
1
1
@marmot - For the case at hand, using
dot
and Dot
produce the same result.– Mico
33 mins ago
@marmot - For the case at hand, using
dot
and Dot
produce the same result.– Mico
33 mins ago
@Mico Yes, you are right.
– marmot
20 mins ago
@Mico Yes, you are right.
– marmot
20 mins ago
add a comment |
up vote
3
down vote
The first or third option below may be close to what you're looking for. Or, switch from Newton-style to Leibniz-style notation for the derivative, as shown by the fourth option (newly fixed to incorporated @marmot's comment). A separate comment: to make the frac{1}{2}
term less visually dominant, consider using tfrac
instead of frac
.
documentclass{article}
usepackage{amsmath} % for tfrac macro and general accent-placement support
begin{document}
[
tfrac{1}{2}m dot{vec{x}} ^2 quad
tfrac{1}{2}m{dot{vec{x}}}^2 quad
tfrac{1}{2}m{dot{vec{x}}}^{,2} quad
tfrac{1}{2}mbigl(tfrac{mathrm{d}vec{x}}{mathrm{d}t}bigr)^{!2}
]
end{document}
@marmot - I know. :-) But sometimes dot-notation becomes so involved that switching to Leibniz-style notation makes things easier...
– Mico
35 mins ago
1
@marmot - Many thanks for this clarification. I've updated the code to make the d's upright.
– Mico
18 mins ago
add a comment |
up vote
3
down vote
The first or third option below may be close to what you're looking for. Or, switch from Newton-style to Leibniz-style notation for the derivative, as shown by the fourth option (newly fixed to incorporated @marmot's comment). A separate comment: to make the frac{1}{2}
term less visually dominant, consider using tfrac
instead of frac
.
documentclass{article}
usepackage{amsmath} % for tfrac macro and general accent-placement support
begin{document}
[
tfrac{1}{2}m dot{vec{x}} ^2 quad
tfrac{1}{2}m{dot{vec{x}}}^2 quad
tfrac{1}{2}m{dot{vec{x}}}^{,2} quad
tfrac{1}{2}mbigl(tfrac{mathrm{d}vec{x}}{mathrm{d}t}bigr)^{!2}
]
end{document}
@marmot - I know. :-) But sometimes dot-notation becomes so involved that switching to Leibniz-style notation makes things easier...
– Mico
35 mins ago
1
@marmot - Many thanks for this clarification. I've updated the code to make the d's upright.
– Mico
18 mins ago
add a comment |
up vote
3
down vote
up vote
3
down vote
The first or third option below may be close to what you're looking for. Or, switch from Newton-style to Leibniz-style notation for the derivative, as shown by the fourth option (newly fixed to incorporated @marmot's comment). A separate comment: to make the frac{1}{2}
term less visually dominant, consider using tfrac
instead of frac
.
documentclass{article}
usepackage{amsmath} % for tfrac macro and general accent-placement support
begin{document}
[
tfrac{1}{2}m dot{vec{x}} ^2 quad
tfrac{1}{2}m{dot{vec{x}}}^2 quad
tfrac{1}{2}m{dot{vec{x}}}^{,2} quad
tfrac{1}{2}mbigl(tfrac{mathrm{d}vec{x}}{mathrm{d}t}bigr)^{!2}
]
end{document}
The first or third option below may be close to what you're looking for. Or, switch from Newton-style to Leibniz-style notation for the derivative, as shown by the fourth option (newly fixed to incorporated @marmot's comment). A separate comment: to make the frac{1}{2}
term less visually dominant, consider using tfrac
instead of frac
.
documentclass{article}
usepackage{amsmath} % for tfrac macro and general accent-placement support
begin{document}
[
tfrac{1}{2}m dot{vec{x}} ^2 quad
tfrac{1}{2}m{dot{vec{x}}}^2 quad
tfrac{1}{2}m{dot{vec{x}}}^{,2} quad
tfrac{1}{2}mbigl(tfrac{mathrm{d}vec{x}}{mathrm{d}t}bigr)^{!2}
]
end{document}
edited 19 mins ago
answered 40 mins ago
Mico
271k30367755
271k30367755
@marmot - I know. :-) But sometimes dot-notation becomes so involved that switching to Leibniz-style notation makes things easier...
– Mico
35 mins ago
1
@marmot - Many thanks for this clarification. I've updated the code to make the d's upright.
– Mico
18 mins ago
add a comment |
@marmot - I know. :-) But sometimes dot-notation becomes so involved that switching to Leibniz-style notation makes things easier...
– Mico
35 mins ago
1
@marmot - Many thanks for this clarification. I've updated the code to make the d's upright.
– Mico
18 mins ago
@marmot - I know. :-) But sometimes dot-notation becomes so involved that switching to Leibniz-style notation makes things easier...
– Mico
35 mins ago
@marmot - I know. :-) But sometimes dot-notation becomes so involved that switching to Leibniz-style notation makes things easier...
– Mico
35 mins ago
1
1
@marmot - Many thanks for this clarification. I've updated the code to make the d's upright.
– Mico
18 mins ago
@marmot - Many thanks for this clarification. I've updated the code to make the d's upright.
– Mico
18 mins ago
add a comment |
Kane Billiot is a new contributor. Be nice, and check out our Code of Conduct.
Kane Billiot is a new contributor. Be nice, and check out our Code of Conduct.
Kane Billiot is a new contributor. Be nice, and check out our Code of Conduct.
Kane Billiot is a new contributor. Be nice, and check out our Code of Conduct.
Thanks for contributing an answer to TeX - LaTeX 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.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2ftex.stackexchange.com%2fquestions%2f463627%2fcommand-for-vector-dot-with-some-power%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
Welcome to TeX.SE! Can you please, as usual here -- show us a short compilable code and an screenshot of your result?
– Kurt
1 hour ago