Command for vector dot with some power











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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



enter image description here










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Kane Billiot is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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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















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



enter image description here










share|improve this question









New contributor




Kane Billiot is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




















  • 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













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



enter image description here










share|improve this question









New contributor




Kane Billiot is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











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



enter image description here







math-mode symbols accents






share|improve this question









New contributor




Kane Billiot is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|improve this question









New contributor




Kane Billiot is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|improve this question




share|improve this question








edited 53 mins ago









Mico

271k30367755




271k30367755






New contributor




Kane Billiot is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 1 hour ago









Kane Billiot

234




234




New contributor




Kane Billiot is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Kane Billiot is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Kane Billiot is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • 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




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










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}


enter image description here



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).






share|improve this answer





















  • 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, using dot and Dot produce the same result.
    – Mico
    33 mins ago










  • @Mico Yes, you are right.
    – marmot
    21 mins ago


















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.



enter image description here



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}





share|improve this answer























  • @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











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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}


enter image description here



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).






share|improve this answer





















  • 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, using dot and Dot produce the same result.
    – Mico
    33 mins ago










  • @Mico Yes, you are right.
    – marmot
    21 mins ago















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}


enter image description here



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).






share|improve this answer





















  • 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, using dot and Dot produce the same result.
    – Mico
    33 mins ago










  • @Mico Yes, you are right.
    – marmot
    21 mins ago













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}


enter image description here



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).






share|improve this answer












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}


enter image description here



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).







share|improve this answer












share|improve this answer



share|improve this answer










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, using dot and Dot 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










  • 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, using dot and Dot 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










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.



enter image description here



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}





share|improve this answer























  • @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















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.



enter image description here



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}





share|improve this answer























  • @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













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.



enter image description here



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}





share|improve this answer














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.



enter image description here



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}






share|improve this answer














share|improve this answer



share|improve this answer








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


















  • @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










Kane Billiot is a new contributor. Be nice, and check out our Code of Conduct.










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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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