How to determine that a given map from $mathbb{R}^2 to mathbb{R}^2$ preserves area or not?
I wanted to show whether the following map preserves area or not,
$$f(x,y)=(x+y^2,y+x^2).$$
I tried to draw the graph but as function is not linear, my approach could not work.
Can any one suggest me some approach so that I can work this out?
analysis multivariable-calculus
edited Nov 28 at 9:05
Robert Z
93.2k1061132
asked Nov 27 at 13:13
MathLover
45710
add a comment |
I wanted to show whether the following map preserves area or not,
$$f(x,y)=(x+y^2,y+x^2).$$
I tried to draw the graph but as function is not linear, my approach could not work.
Can any one suggest me some approach so that I can work this out?
analysis multivariable-calculus
edited Nov 28 at 9:05
Robert Z
93.2k1061132
asked Nov 27 at 13:13
MathLover
45710
Do you know the change of variables formula ? Also, are you dealing with Riemann or Lebesgue integration?
– астон вілла олоф мэллбэрг
Nov 27 at 13:21
No Sir .I had just done only real analysis in that reimann steljes intergral only
– MathLover
Nov 27 at 13:22
add a comment |
I wanted to show whether the following map preserves area or not,
$$f(x,y)=(x+y^2,y+x^2).$$
I tried to draw the graph but as function is not linear, my approach could not work.
Can any one suggest me some approach so that I can work this out?
analysis multivariable-calculus
edited Nov 28 at 9:05
Robert Z
93.2k1061132
asked Nov 27 at 13:13
MathLover
45710
I wanted to show whether the following map preserves area or not,
$$f(x,y)=(x+y^2,y+x^2).$$
I tried to draw the graph but as function is not linear, my approach could not work.
Can any one suggest me some approach so that I can work this out?
analysis multivariable-calculus
analysis multivariable-calculus
edited Nov 28 at 9:05
Robert Z
93.2k1061132
asked Nov 27 at 13:13
MathLover
45710
edited Nov 28 at 9:05
Robert Z
93.2k1061132
asked Nov 27 at 13:13
MathLover
45710
edited Nov 28 at 9:05
Robert Z
93.2k1061132
edited Nov 28 at 9:05
Robert Z
93.2k1061132
edited Nov 28 at 9:05
Robert Z
93.2k1061132
93.2k1061132
asked Nov 27 at 13:13
MathLover
45710
asked Nov 27 at 13:13
MathLover
45710
asked Nov 27 at 13:13
MathLover
45710
45710
Do you know the change of variables formula ? Also, are you dealing with Riemann or Lebesgue integration?
– астон вілла олоф мэллбэрг
Nov 27 at 13:21
No Sir .I had just done only real analysis in that reimann steljes intergral only
– MathLover
Nov 27 at 13:22
add a comment |
Do you know the change of variables formula ? Also, are you dealing with Riemann or Lebesgue integration?
– астон вілла олоф мэллбэрг
Nov 27 at 13:21
No Sir .I had just done only real analysis in that reimann steljes intergral only
– MathLover
Nov 27 at 13:22
Do you know the change of variables formula ? Also, are you dealing with Riemann or Lebesgue integration?
– астон вілла олоф мэллбэрг
Nov 27 at 13:21
Do you know the change of variables formula ? Also, are you dealing with Riemann or Lebesgue integration?
– астон вілла олоф мэллбэрг
Nov 27 at 13:21
No Sir .I had just done only real analysis in that reimann steljes intergral only
– MathLover
Nov 27 at 13:22
No Sir .I had just done only real analysis in that reimann steljes intergral only
– MathLover
Nov 27 at 13:22
add a comment |
3 Answers
3
active
oldest
votes
A linear map $A:>{mathbb R}^2to{mathbb R}^2$ multiplies all areas by the factor $bigl|{rm det}(A)bigr|$. It follows that a $C^1$ map
$$f:>{mathbb R}^2to{mathbb R}^2,qquad (x,y)mapsto bigl(u(x,y),v(x,y)bigr)$$
possesses a "local area scaling factor" $bigl|J_f(x,y)bigr|$ that continuously changes from point to point. The so-called Jacobian determinant $J_f(x,y)$ is given by
$$J_f(x,y)={rm det}bigl(df(x,y)bigr)={rm det}left[matrix{u_x&u_ycr v_x&v_ycr}right] .$$
In your example we obtain $bigl|J_f(x,y)bigr|=|1-4xy|$, which is not constant. For an area preserving map we would need $bigl|J_f(x,y)bigr|equiv1$.
answered Nov 27 at 15:33
Christian Blatter
172k7112325
Nice explanation (+1).
– Robert Z
Nov 28 at 9:03
add a comment |
If it preserved areas, then $f'(a,b)$ would also preserve areas for each $(a,b)inmathbb{R}^2$. But $det f'left(frac12,frac12right)=0$.
answered Nov 27 at 13:22
José Carlos Santos
150k22120221
Why Sir? if some function preserve area then that means its derivative also preserve same?
– MathLover
Nov 27 at 13:24
Indeed. Besides, since $(a,b)inmathbb{R}^2$ and since $f(x,y)$ behaves as $fleft(frac12,frac12right)+f'left(frac12,frac12right)left(x-frac12,y-frac12right)$ near $left(frac12,frac12right)$ and $det f'left(frac12,frac12right)=0$, it is clear that $f$ cannot be area-preserving, at least near $left(frac12,frac12right)$.
– José Carlos Santos
Nov 27 at 13:28
add a comment |
Take a look at the image under $f$ of the unit square $(0,0), (1,0), (1,1), (0,1)$.
answered Nov 28 at 10:11
gandalf61
7,693623
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
A linear map $A:>{mathbb R}^2to{mathbb R}^2$ multiplies all areas by the factor $bigl|{rm det}(A)bigr|$. It follows that a $C^1$ map
$$f:>{mathbb R}^2to{mathbb R}^2,qquad (x,y)mapsto bigl(u(x,y),v(x,y)bigr)$$
possesses a "local area scaling factor" $bigl|J_f(x,y)bigr|$ that continuously changes from point to point. The so-called Jacobian determinant $J_f(x,y)$ is given by
$$J_f(x,y)={rm det}bigl(df(x,y)bigr)={rm det}left[matrix{u_x&u_ycr v_x&v_ycr}right] .$$
In your example we obtain $bigl|J_f(x,y)bigr|=|1-4xy|$, which is not constant. For an area preserving map we would need $bigl|J_f(x,y)bigr|equiv1$.
answered Nov 27 at 15:33
Christian Blatter
172k7112325
Nice explanation (+1).
– Robert Z
Nov 28 at 9:03
add a comment |
A linear map $A:>{mathbb R}^2to{mathbb R}^2$ multiplies all areas by the factor $bigl|{rm det}(A)bigr|$. It follows that a $C^1$ map
$$f:>{mathbb R}^2to{mathbb R}^2,qquad (x,y)mapsto bigl(u(x,y),v(x,y)bigr)$$
possesses a "local area scaling factor" $bigl|J_f(x,y)bigr|$ that continuously changes from point to point. The so-called Jacobian determinant $J_f(x,y)$ is given by
$$J_f(x,y)={rm det}bigl(df(x,y)bigr)={rm det}left[matrix{u_x&u_ycr v_x&v_ycr}right] .$$
In your example we obtain $bigl|J_f(x,y)bigr|=|1-4xy|$, which is not constant. For an area preserving map we would need $bigl|J_f(x,y)bigr|equiv1$.
answered Nov 27 at 15:33
Christian Blatter
172k7112325
Nice explanation (+1).
– Robert Z
Nov 28 at 9:03
add a comment |
A linear map $A:>{mathbb R}^2to{mathbb R}^2$ multiplies all areas by the factor $bigl|{rm det}(A)bigr|$. It follows that a $C^1$ map
$$f:>{mathbb R}^2to{mathbb R}^2,qquad (x,y)mapsto bigl(u(x,y),v(x,y)bigr)$$
possesses a "local area scaling factor" $bigl|J_f(x,y)bigr|$ that continuously changes from point to point. The so-called Jacobian determinant $J_f(x,y)$ is given by
$$J_f(x,y)={rm det}bigl(df(x,y)bigr)={rm det}left[matrix{u_x&u_ycr v_x&v_ycr}right] .$$
In your example we obtain $bigl|J_f(x,y)bigr|=|1-4xy|$, which is not constant. For an area preserving map we would need $bigl|J_f(x,y)bigr|equiv1$.
answered Nov 27 at 15:33
Christian Blatter
172k7112325
A linear map $A:>{mathbb R}^2to{mathbb R}^2$ multiplies all areas by the factor $bigl|{rm det}(A)bigr|$. It follows that a $C^1$ map
$$f:>{mathbb R}^2to{mathbb R}^2,qquad (x,y)mapsto bigl(u(x,y),v(x,y)bigr)$$
possesses a "local area scaling factor" $bigl|J_f(x,y)bigr|$ that continuously changes from point to point. The so-called Jacobian determinant $J_f(x,y)$ is given by
$$J_f(x,y)={rm det}bigl(df(x,y)bigr)={rm det}left[matrix{u_x&u_ycr v_x&v_ycr}right] .$$
In your example we obtain $bigl|J_f(x,y)bigr|=|1-4xy|$, which is not constant. For an area preserving map we would need $bigl|J_f(x,y)bigr|equiv1$.
answered Nov 27 at 15:33
Christian Blatter
172k7112325
answered Nov 27 at 15:33
Christian Blatter
172k7112325
answered Nov 27 at 15:33
Christian Blatter
172k7112325
answered Nov 27 at 15:33
Christian Blatter
172k7112325
172k7112325
Nice explanation (+1).
– Robert Z
Nov 28 at 9:03
add a comment |
Nice explanation (+1).
– Robert Z
Nov 28 at 9:03
Nice explanation (+1).
– Robert Z
Nov 28 at 9:03
Nice explanation (+1).
– Robert Z
Nov 28 at 9:03
add a comment |
If it preserved areas, then $f'(a,b)$ would also preserve areas for each $(a,b)inmathbb{R}^2$. But $det f'left(frac12,frac12right)=0$.
answered Nov 27 at 13:22
José Carlos Santos
150k22120221
Why Sir? if some function preserve area then that means its derivative also preserve same?
– MathLover
Nov 27 at 13:24
Indeed. Besides, since $(a,b)inmathbb{R}^2$ and since $f(x,y)$ behaves as $fleft(frac12,frac12right)+f'left(frac12,frac12right)left(x-frac12,y-frac12right)$ near $left(frac12,frac12right)$ and $det f'left(frac12,frac12right)=0$, it is clear that $f$ cannot be area-preserving, at least near $left(frac12,frac12right)$.
– José Carlos Santos
Nov 27 at 13:28
add a comment |
If it preserved areas, then $f'(a,b)$ would also preserve areas for each $(a,b)inmathbb{R}^2$. But $det f'left(frac12,frac12right)=0$.
answered Nov 27 at 13:22
José Carlos Santos
150k22120221
Why Sir? if some function preserve area then that means its derivative also preserve same?
– MathLover
Nov 27 at 13:24
Indeed. Besides, since $(a,b)inmathbb{R}^2$ and since $f(x,y)$ behaves as $fleft(frac12,frac12right)+f'left(frac12,frac12right)left(x-frac12,y-frac12right)$ near $left(frac12,frac12right)$ and $det f'left(frac12,frac12right)=0$, it is clear that $f$ cannot be area-preserving, at least near $left(frac12,frac12right)$.
– José Carlos Santos
Nov 27 at 13:28
add a comment |
If it preserved areas, then $f'(a,b)$ would also preserve areas for each $(a,b)inmathbb{R}^2$. But $det f'left(frac12,frac12right)=0$.
answered Nov 27 at 13:22
José Carlos Santos
150k22120221
If it preserved areas, then $f'(a,b)$ would also preserve areas for each $(a,b)inmathbb{R}^2$. But $det f'left(frac12,frac12right)=0$.
answered Nov 27 at 13:22
José Carlos Santos
150k22120221
answered Nov 27 at 13:22
José Carlos Santos
150k22120221
answered Nov 27 at 13:22
José Carlos Santos
150k22120221
answered Nov 27 at 13:22
José Carlos Santos
150k22120221
150k22120221
Why Sir? if some function preserve area then that means its derivative also preserve same?
– MathLover
Nov 27 at 13:24
Indeed. Besides, since $(a,b)inmathbb{R}^2$ and since $f(x,y)$ behaves as $fleft(frac12,frac12right)+f'left(frac12,frac12right)left(x-frac12,y-frac12right)$ near $left(frac12,frac12right)$ and $det f'left(frac12,frac12right)=0$, it is clear that $f$ cannot be area-preserving, at least near $left(frac12,frac12right)$.
– José Carlos Santos
Nov 27 at 13:28
add a comment |
Why Sir? if some function preserve area then that means its derivative also preserve same?
– MathLover
Nov 27 at 13:24
Indeed. Besides, since $(a,b)inmathbb{R}^2$ and since $f(x,y)$ behaves as $fleft(frac12,frac12right)+f'left(frac12,frac12right)left(x-frac12,y-frac12right)$ near $left(frac12,frac12right)$ and $det f'left(frac12,frac12right)=0$, it is clear that $f$ cannot be area-preserving, at least near $left(frac12,frac12right)$.
– José Carlos Santos
Nov 27 at 13:28
Why Sir? if some function preserve area then that means its derivative also preserve same?
– MathLover
Nov 27 at 13:24
Why Sir? if some function preserve area then that means its derivative also preserve same?
– MathLover
Nov 27 at 13:24
Indeed. Besides, since $(a,b)inmathbb{R}^2$ and since $f(x,y)$ behaves as $fleft(frac12,frac12right)+f'left(frac12,frac12right)left(x-frac12,y-frac12right)$ near $left(frac12,frac12right)$ and $det f'left(frac12,frac12right)=0$, it is clear that $f$ cannot be area-preserving, at least near $left(frac12,frac12right)$.
– José Carlos Santos
Nov 27 at 13:28
Indeed. Besides, since $(a,b)inmathbb{R}^2$ and since $f(x,y)$ behaves as $fleft(frac12,frac12right)+f'left(frac12,frac12right)left(x-frac12,y-frac12right)$ near $left(frac12,frac12right)$ and $det f'left(frac12,frac12right)=0$, it is clear that $f$ cannot be area-preserving, at least near $left(frac12,frac12right)$.
– José Carlos Santos
Nov 27 at 13:28
add a comment |
Take a look at the image under $f$ of the unit square $(0,0), (1,0), (1,1), (0,1)$.
answered Nov 28 at 10:11
gandalf61
7,693623
add a comment |
Take a look at the image under $f$ of the unit square $(0,0), (1,0), (1,1), (0,1)$.
answered Nov 28 at 10:11
gandalf61
7,693623
add a comment |
Take a look at the image under $f$ of the unit square $(0,0), (1,0), (1,1), (0,1)$.
answered Nov 28 at 10:11
gandalf61
7,693623
Take a look at the image under $f$ of the unit square $(0,0), (1,0), (1,1), (0,1)$.
answered Nov 28 at 10:11
gandalf61
7,693623
answered Nov 28 at 10:11
gandalf61
7,693623
answered Nov 28 at 10:11
gandalf61
7,693623
answered Nov 28 at 10:11
gandalf61
7,693623
7,693623
add a comment |
add a comment |
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Do you know the change of variables formula ? Also, are you dealing with Riemann or Lebesgue integration?
– астон вілла олоф мэллбэрг
Nov 27 at 13:21
No Sir .I had just done only real analysis in that reimann steljes intergral only
– MathLover
Nov 27 at 13:22