A specific example of $F$-related vector fields
I need to prove the following:
Let $F:Bbb{R}toBbb{R}^2$ be the smooth map $F(t)=(cos t,sin t)$. Then $d/dtinmathcal{T}(Bbb{R})$ is $F$-related to the vector field $Zinmathcal{T}(Bbb{R}^2)$ defined by $$Z=xfrac{∂}{∂y}-yfrac{∂}{∂x}.$$
So for any smooth $f$ defined on an open subset of $Bbb{R}^2$ we want $$frac{d}{dt}(fcirc F)=(xfrac{∂}{∂y}f-yfrac{∂}{∂x}f)circ F.$$ That is,$$frac{∂}{∂x}ffrac{d}{dt}F+frac{∂}{∂y}ffrac{d}{dt}F=(xfrac{∂}{∂y}f-yfrac{∂}{∂x}f)circ F.$$ What should I do next?
differential-geometry differential-topology
asked Dec 8 '13 at 14:48
figurafigura
9515
add a comment |
I need to prove the following:
Let $F:Bbb{R}toBbb{R}^2$ be the smooth map $F(t)=(cos t,sin t)$. Then $d/dtinmathcal{T}(Bbb{R})$ is $F$-related to the vector field $Zinmathcal{T}(Bbb{R}^2)$ defined by $$Z=xfrac{∂}{∂y}-yfrac{∂}{∂x}.$$
So for any smooth $f$ defined on an open subset of $Bbb{R}^2$ we want $$frac{d}{dt}(fcirc F)=(xfrac{∂}{∂y}f-yfrac{∂}{∂x}f)circ F.$$ That is,$$frac{∂}{∂x}ffrac{d}{dt}F+frac{∂}{∂y}ffrac{d}{dt}F=(xfrac{∂}{∂y}f-yfrac{∂}{∂x}f)circ F.$$ What should I do next?
differential-geometry differential-topology
asked Dec 8 '13 at 14:48
figurafigura
9515
add a comment |
I need to prove the following:
Let $F:Bbb{R}toBbb{R}^2$ be the smooth map $F(t)=(cos t,sin t)$. Then $d/dtinmathcal{T}(Bbb{R})$ is $F$-related to the vector field $Zinmathcal{T}(Bbb{R}^2)$ defined by $$Z=xfrac{∂}{∂y}-yfrac{∂}{∂x}.$$
So for any smooth $f$ defined on an open subset of $Bbb{R}^2$ we want $$frac{d}{dt}(fcirc F)=(xfrac{∂}{∂y}f-yfrac{∂}{∂x}f)circ F.$$ That is,$$frac{∂}{∂x}ffrac{d}{dt}F+frac{∂}{∂y}ffrac{d}{dt}F=(xfrac{∂}{∂y}f-yfrac{∂}{∂x}f)circ F.$$ What should I do next?
differential-geometry differential-topology
asked Dec 8 '13 at 14:48
figurafigura
9515
I need to prove the following:
Let $F:Bbb{R}toBbb{R}^2$ be the smooth map $F(t)=(cos t,sin t)$. Then $d/dtinmathcal{T}(Bbb{R})$ is $F$-related to the vector field $Zinmathcal{T}(Bbb{R}^2)$ defined by $$Z=xfrac{∂}{∂y}-yfrac{∂}{∂x}.$$
So for any smooth $f$ defined on an open subset of $Bbb{R}^2$ we want $$frac{d}{dt}(fcirc F)=(xfrac{∂}{∂y}f-yfrac{∂}{∂x}f)circ F.$$ That is,$$frac{∂}{∂x}ffrac{d}{dt}F+frac{∂}{∂y}ffrac{d}{dt}F=(xfrac{∂}{∂y}f-yfrac{∂}{∂x}f)circ F.$$ What should I do next?
differential-geometry differential-topology
differential-geometry differential-topology
asked Dec 8 '13 at 14:48
figurafigura
9515
asked Dec 8 '13 at 14:48
figurafigura
9515
asked Dec 8 '13 at 14:48
figurafigura
9515
asked Dec 8 '13 at 14:48
figurafigura
9515
asked Dec 8 '13 at 14:48
figurafigura
9515
9515
add a comment |
add a comment |
1 Answer
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LHS:
$$frac{d}{dt}(fcirc F)= frac{d}{dt}(f(cos t, sin t))=f_x cdot (-sin t)+f_y cdot cos t$$
RHS:
$$(xfrac{∂}{∂y}f-yfrac{∂}{∂x}f)circ F= (xcirc F) f_x-(ycirc F) f_y= cos t cdot f_y - sin t cdot f_x$$
So they are equal.
edited Nov 29 '18 at 14:24
amWhy
192k28225439
answered Apr 27 '14 at 19:42
WWKWWK
1,003722
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
LHS:
$$frac{d}{dt}(fcirc F)= frac{d}{dt}(f(cos t, sin t))=f_x cdot (-sin t)+f_y cdot cos t$$
RHS:
$$(xfrac{∂}{∂y}f-yfrac{∂}{∂x}f)circ F= (xcirc F) f_x-(ycirc F) f_y= cos t cdot f_y - sin t cdot f_x$$
So they are equal.
edited Nov 29 '18 at 14:24
amWhy
192k28225439
answered Apr 27 '14 at 19:42
WWKWWK
1,003722
add a comment |
LHS:
$$frac{d}{dt}(fcirc F)= frac{d}{dt}(f(cos t, sin t))=f_x cdot (-sin t)+f_y cdot cos t$$
RHS:
$$(xfrac{∂}{∂y}f-yfrac{∂}{∂x}f)circ F= (xcirc F) f_x-(ycirc F) f_y= cos t cdot f_y - sin t cdot f_x$$
So they are equal.
edited Nov 29 '18 at 14:24
amWhy
192k28225439
answered Apr 27 '14 at 19:42
WWKWWK
1,003722
add a comment |
LHS:
$$frac{d}{dt}(fcirc F)= frac{d}{dt}(f(cos t, sin t))=f_x cdot (-sin t)+f_y cdot cos t$$
RHS:
$$(xfrac{∂}{∂y}f-yfrac{∂}{∂x}f)circ F= (xcirc F) f_x-(ycirc F) f_y= cos t cdot f_y - sin t cdot f_x$$
So they are equal.
edited Nov 29 '18 at 14:24
amWhy
192k28225439
answered Apr 27 '14 at 19:42
WWKWWK
1,003722
LHS:
$$frac{d}{dt}(fcirc F)= frac{d}{dt}(f(cos t, sin t))=f_x cdot (-sin t)+f_y cdot cos t$$
RHS:
$$(xfrac{∂}{∂y}f-yfrac{∂}{∂x}f)circ F= (xcirc F) f_x-(ycirc F) f_y= cos t cdot f_y - sin t cdot f_x$$
So they are equal.
edited Nov 29 '18 at 14:24
amWhy
192k28225439
answered Apr 27 '14 at 19:42
WWKWWK
1,003722
edited Nov 29 '18 at 14:24
amWhy
192k28225439
edited Nov 29 '18 at 14:24
amWhy
192k28225439
edited Nov 29 '18 at 14:24
amWhy
192k28225439
192k28225439
answered Apr 27 '14 at 19:42
WWKWWK
1,003722
answered Apr 27 '14 at 19:42
WWKWWK
1,003722
answered Apr 27 '14 at 19:42
WWKWWK
1,003722
1,003722
add a comment |
add a comment |
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