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
41 mins ago
Is there a difference between dot and Dot?
– Kane Billiot
39 mins ago
@KaneBilliot Short answer:Dot
works also when you already have something on top of the symbol. So it would not shift.
– marmot
38 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
21 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
36 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
41 mins ago
Is there a difference between dot and Dot?
– Kane Billiot
39 mins ago
@KaneBilliot Short answer:Dot
works also when you already have something on top of the symbol. So it would not shift.
– marmot
38 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
21 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
41 mins ago
Is there a difference between dot and Dot?
– Kane Billiot
39 mins ago
@KaneBilliot Short answer:Dot
works also when you already have something on top of the symbol. So it would not shift.
– marmot
38 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
21 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 42 mins ago
marmot
81.3k491173
81.3k491173
That's what I wanted to write, but did not know how to. Thanks.
– Kane Billiot
41 mins ago
Is there a difference between dot and Dot?
– Kane Billiot
39 mins ago
@KaneBilliot Short answer:Dot
works also when you already have something on top of the symbol. So it would not shift.
– marmot
38 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
21 mins ago
add a comment |
That's what I wanted to write, but did not know how to. Thanks.
– Kane Billiot
41 mins ago
Is there a difference between dot and Dot?
– Kane Billiot
39 mins ago
@KaneBilliot Short answer:Dot
works also when you already have something on top of the symbol. So it would not shift.
– marmot
38 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
21 mins ago
That's what I wanted to write, but did not know how to. Thanks.
– Kane Billiot
41 mins ago
That's what I wanted to write, but did not know how to. Thanks.
– Kane Billiot
41 mins ago
Is there a difference between dot and Dot?
– Kane Billiot
39 mins ago
Is there a difference between dot and Dot?
– Kane Billiot
39 mins ago
@KaneBilliot Short answer:
Dot
works also when you already have something on top of the symbol. So it would not shift.– marmot
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
38 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
21 mins ago
@Mico Yes, you are right.
– marmot
21 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
36 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
36 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
36 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
36 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
36 mins ago
@marmot - I know. :-) But sometimes dot-notation becomes so involved that switching to Leibniz-style notation makes things easier...
– Mico
36 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.
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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