Check if the three trajectories are indeed one and the same ellipse

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Consider three time dependent points $P_1(t)$, $P_2(t)$ and $P_3(t)$ in the plane($mathbbR^2$), such that:

$fracdP_1dt(t)=P_2(t)-P_3(t)$

$fracdP_2dt(t)=P_3(t)-P_1(t)$

$fracdP_3dt(t)=P_1(t)-P_2(t)$

This system should look like a triangle rotating but changing its shape.

Check if the images $P_1(mathbbR)=P_2(mathbbR)=P_3(mathbbR)$ and they are an ellipse.

I have simulated this on my computer some time ago, and the trajectories seem to be one and the same ellipse.
I don't know how to solve the differential equations to check it directly.
However, I know for sure that the area of the triangle formed by the points is constant, since the points move along the base line.

differential-equations triangle conic-sections

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asked Jul 18 at 18:35

Cristhian Grundmann

745

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Consider three time dependent points $P_1(t)$, $P_2(t)$ and $P_3(t)$ in the plane($mathbbR^2$), such that:

$fracdP_1dt(t)=P_2(t)-P_3(t)$

$fracdP_2dt(t)=P_3(t)-P_1(t)$

$fracdP_3dt(t)=P_1(t)-P_2(t)$

This system should look like a triangle rotating but changing its shape.

Check if the images $P_1(mathbbR)=P_2(mathbbR)=P_3(mathbbR)$ and they are an ellipse.

I have simulated this on my computer some time ago, and the trajectories seem to be one and the same ellipse.
I don't know how to solve the differential equations to check it directly.
However, I know for sure that the area of the triangle formed by the points is constant, since the points move along the base line.

differential-equations triangle conic-sections

share|cite|improve this question

asked Jul 18 at 18:35

Cristhian Grundmann

745

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

Consider three time dependent points $P_1(t)$, $P_2(t)$ and $P_3(t)$ in the plane($mathbbR^2$), such that:

$fracdP_1dt(t)=P_2(t)-P_3(t)$

$fracdP_2dt(t)=P_3(t)-P_1(t)$

$fracdP_3dt(t)=P_1(t)-P_2(t)$

This system should look like a triangle rotating but changing its shape.

Check if the images $P_1(mathbbR)=P_2(mathbbR)=P_3(mathbbR)$ and they are an ellipse.

I have simulated this on my computer some time ago, and the trajectories seem to be one and the same ellipse.
I don't know how to solve the differential equations to check it directly.
However, I know for sure that the area of the triangle formed by the points is constant, since the points move along the base line.

differential-equations triangle conic-sections

share|cite|improve this question

asked Jul 18 at 18:35

Cristhian Grundmann

745

Consider three time dependent points $P_1(t)$, $P_2(t)$ and $P_3(t)$ in the plane($mathbbR^2$), such that:

$fracdP_1dt(t)=P_2(t)-P_3(t)$

$fracdP_2dt(t)=P_3(t)-P_1(t)$

$fracdP_3dt(t)=P_1(t)-P_2(t)$

This system should look like a triangle rotating but changing its shape.

Check if the images $P_1(mathbbR)=P_2(mathbbR)=P_3(mathbbR)$ and they are an ellipse.

I have simulated this on my computer some time ago, and the trajectories seem to be one and the same ellipse.
I don't know how to solve the differential equations to check it directly.
However, I know for sure that the area of the triangle formed by the points is constant, since the points move along the base line.

differential-equations triangle conic-sections

share|cite|improve this question

asked Jul 18 at 18:35

Cristhian Grundmann

745

share|cite|improve this question

share|cite|improve this question

asked Jul 18 at 18:35

Cristhian Grundmann

745

asked Jul 18 at 18:35

Cristhian Grundmann

745

asked Jul 18 at 18:35

Cristhian Grundmann

745

745

add a comment | 

add a comment | 

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accepted

To avoid a proliferation of subscripts, I’m going to rename the points $P$, $Q$ and $R$. The system of equations can be expressed in matrix form as $$beginbmatrixdot P \ dot Q \ dot Rendbmatrix = beginbmatrix0&-1&1\1&0&-1\-1&1&0endbmatrix beginbmatrixP\Q\Rendbmatrix.$$ It’s not terribly difficult to work out the solution to this system using the usual methods: the eigenvalues are $0$ and $pmsqrt3i$, and so the coefficient matrix is similar to $$J=beginbmatrix0&0&0\0&0&-sqrt3\0&sqrt3&0endbmatrix$$ with exponential $$e^tJ = beginbmatrix1&0&0\0&cossqrt3t&-sinsqrt3t\0&sinsqrt3t&cossqrt3tendbmatrix.$$ A corresponding eigenbasis is easily computed (in fact one can find an orthonormal basis—the coefficient matrix is the “cross-product matrix” of $(1,1,1)^T$), and setting $C = frac13(P_0+Q_0+R_0)$, the centroid of the initial configuration of points, gives for the general solution $$P(t)=C+(P_0-C)cossqrt3t+(Q_0-R_0)frac1sqrt3sinsqrt3t \ Q(t) = C+(Q_0-C)cossqrt3t+(R_0-P_0)frac1sqrt3sinsqrt3t \ R(t) = C+(R_0-C)cossqrt3t+(P_0-Q_0)frac1sqrt3sinsqrt3t.$$ Now, set $P_0=(1,0)$, $Q_0=left(-frac12,fracsqrt32right)$ and $R_0=left(-frac12,-fracsqrt32right)$, the vertices of an equilateral triangle centered on the origin. The parametric equations of the three curves simplify to $$P(t) = (1,0)cossqrt3t+(0,1)sinsqrt3t \ Q(t) = left(-frac12,fracsqrt32right)cossqrt3t+left(-fracsqrt32,-frac12right)sinsqrt3t \ R(t) = left(-frac12,-fracsqrt32right)cossqrt3t+left(fracsqrt32,-frac12right)sinsqrt3t, $$ which are all obviously parameterizations of the unit circle. Any other initial point configuration can be obtained from this one by an affine transformation, which also maps the three identical unit circles onto three identical ellipses.

With the above in mind, you could with a bit of tedious trigonometric manipulation verify that the three trajectories in the general solution are phase-shifted versions of each other, i.e., that $Q(t)=P(t+delta)$, $R(t)=Q(t+delta)$ and $P(t)=R(t+delta)$, where $delta = 2piover3sqrt3$.

share|cite|improve this answer

edited Jul 19 at 19:18

answered Jul 18 at 21:24

amd

25.9k2943

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I could’ve gone directly to the unit-circle solution, but I thought seeing the general case would be instructive. The vectors involved have transparent geometric interpretations.
  – amd
  Jul 18 at 23:14
  

  
  
  
  

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

up vote
1
down vote



accepted

To avoid a proliferation of subscripts, I’m going to rename the points $P$, $Q$ and $R$. The system of equations can be expressed in matrix form as $$beginbmatrixdot P \ dot Q \ dot Rendbmatrix = beginbmatrix0&-1&1\1&0&-1\-1&1&0endbmatrix beginbmatrixP\Q\Rendbmatrix.$$ It’s not terribly difficult to work out the solution to this system using the usual methods: the eigenvalues are $0$ and $pmsqrt3i$, and so the coefficient matrix is similar to $$J=beginbmatrix0&0&0\0&0&-sqrt3\0&sqrt3&0endbmatrix$$ with exponential $$e^tJ = beginbmatrix1&0&0\0&cossqrt3t&-sinsqrt3t\0&sinsqrt3t&cossqrt3tendbmatrix.$$ A corresponding eigenbasis is easily computed (in fact one can find an orthonormal basis—the coefficient matrix is the “cross-product matrix” of $(1,1,1)^T$), and setting $C = frac13(P_0+Q_0+R_0)$, the centroid of the initial configuration of points, gives for the general solution $$P(t)=C+(P_0-C)cossqrt3t+(Q_0-R_0)frac1sqrt3sinsqrt3t \ Q(t) = C+(Q_0-C)cossqrt3t+(R_0-P_0)frac1sqrt3sinsqrt3t \ R(t) = C+(R_0-C)cossqrt3t+(P_0-Q_0)frac1sqrt3sinsqrt3t.$$ Now, set $P_0=(1,0)$, $Q_0=left(-frac12,fracsqrt32right)$ and $R_0=left(-frac12,-fracsqrt32right)$, the vertices of an equilateral triangle centered on the origin. The parametric equations of the three curves simplify to $$P(t) = (1,0)cossqrt3t+(0,1)sinsqrt3t \ Q(t) = left(-frac12,fracsqrt32right)cossqrt3t+left(-fracsqrt32,-frac12right)sinsqrt3t \ R(t) = left(-frac12,-fracsqrt32right)cossqrt3t+left(fracsqrt32,-frac12right)sinsqrt3t, $$ which are all obviously parameterizations of the unit circle. Any other initial point configuration can be obtained from this one by an affine transformation, which also maps the three identical unit circles onto three identical ellipses.

With the above in mind, you could with a bit of tedious trigonometric manipulation verify that the three trajectories in the general solution are phase-shifted versions of each other, i.e., that $Q(t)=P(t+delta)$, $R(t)=Q(t+delta)$ and $P(t)=R(t+delta)$, where $delta = 2piover3sqrt3$.

share|cite|improve this answer

edited Jul 19 at 19:18

answered Jul 18 at 21:24

amd

25.9k2943

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I could’ve gone directly to the unit-circle solution, but I thought seeing the general case would be instructive. The vectors involved have transparent geometric interpretations.
  – amd
  Jul 18 at 23:14
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

To avoid a proliferation of subscripts, I’m going to rename the points $P$, $Q$ and $R$. The system of equations can be expressed in matrix form as $$beginbmatrixdot P \ dot Q \ dot Rendbmatrix = beginbmatrix0&-1&1\1&0&-1\-1&1&0endbmatrix beginbmatrixP\Q\Rendbmatrix.$$ It’s not terribly difficult to work out the solution to this system using the usual methods: the eigenvalues are $0$ and $pmsqrt3i$, and so the coefficient matrix is similar to $$J=beginbmatrix0&0&0\0&0&-sqrt3\0&sqrt3&0endbmatrix$$ with exponential $$e^tJ = beginbmatrix1&0&0\0&cossqrt3t&-sinsqrt3t\0&sinsqrt3t&cossqrt3tendbmatrix.$$ A corresponding eigenbasis is easily computed (in fact one can find an orthonormal basis—the coefficient matrix is the “cross-product matrix” of $(1,1,1)^T$), and setting $C = frac13(P_0+Q_0+R_0)$, the centroid of the initial configuration of points, gives for the general solution $$P(t)=C+(P_0-C)cossqrt3t+(Q_0-R_0)frac1sqrt3sinsqrt3t \ Q(t) = C+(Q_0-C)cossqrt3t+(R_0-P_0)frac1sqrt3sinsqrt3t \ R(t) = C+(R_0-C)cossqrt3t+(P_0-Q_0)frac1sqrt3sinsqrt3t.$$ Now, set $P_0=(1,0)$, $Q_0=left(-frac12,fracsqrt32right)$ and $R_0=left(-frac12,-fracsqrt32right)$, the vertices of an equilateral triangle centered on the origin. The parametric equations of the three curves simplify to $$P(t) = (1,0)cossqrt3t+(0,1)sinsqrt3t \ Q(t) = left(-frac12,fracsqrt32right)cossqrt3t+left(-fracsqrt32,-frac12right)sinsqrt3t \ R(t) = left(-frac12,-fracsqrt32right)cossqrt3t+left(fracsqrt32,-frac12right)sinsqrt3t, $$ which are all obviously parameterizations of the unit circle. Any other initial point configuration can be obtained from this one by an affine transformation, which also maps the three identical unit circles onto three identical ellipses.

With the above in mind, you could with a bit of tedious trigonometric manipulation verify that the three trajectories in the general solution are phase-shifted versions of each other, i.e., that $Q(t)=P(t+delta)$, $R(t)=Q(t+delta)$ and $P(t)=R(t+delta)$, where $delta = 2piover3sqrt3$.

share|cite|improve this answer

edited Jul 19 at 19:18

answered Jul 18 at 21:24

amd

25.9k2943

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I could’ve gone directly to the unit-circle solution, but I thought seeing the general case would be instructive. The vectors involved have transparent geometric interpretations.
  – amd
  Jul 18 at 23:14
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

To avoid a proliferation of subscripts, I’m going to rename the points $P$, $Q$ and $R$. The system of equations can be expressed in matrix form as $$beginbmatrixdot P \ dot Q \ dot Rendbmatrix = beginbmatrix0&-1&1\1&0&-1\-1&1&0endbmatrix beginbmatrixP\Q\Rendbmatrix.$$ It’s not terribly difficult to work out the solution to this system using the usual methods: the eigenvalues are $0$ and $pmsqrt3i$, and so the coefficient matrix is similar to $$J=beginbmatrix0&0&0\0&0&-sqrt3\0&sqrt3&0endbmatrix$$ with exponential $$e^tJ = beginbmatrix1&0&0\0&cossqrt3t&-sinsqrt3t\0&sinsqrt3t&cossqrt3tendbmatrix.$$ A corresponding eigenbasis is easily computed (in fact one can find an orthonormal basis—the coefficient matrix is the “cross-product matrix” of $(1,1,1)^T$), and setting $C = frac13(P_0+Q_0+R_0)$, the centroid of the initial configuration of points, gives for the general solution $$P(t)=C+(P_0-C)cossqrt3t+(Q_0-R_0)frac1sqrt3sinsqrt3t \ Q(t) = C+(Q_0-C)cossqrt3t+(R_0-P_0)frac1sqrt3sinsqrt3t \ R(t) = C+(R_0-C)cossqrt3t+(P_0-Q_0)frac1sqrt3sinsqrt3t.$$ Now, set $P_0=(1,0)$, $Q_0=left(-frac12,fracsqrt32right)$ and $R_0=left(-frac12,-fracsqrt32right)$, the vertices of an equilateral triangle centered on the origin. The parametric equations of the three curves simplify to $$P(t) = (1,0)cossqrt3t+(0,1)sinsqrt3t \ Q(t) = left(-frac12,fracsqrt32right)cossqrt3t+left(-fracsqrt32,-frac12right)sinsqrt3t \ R(t) = left(-frac12,-fracsqrt32right)cossqrt3t+left(fracsqrt32,-frac12right)sinsqrt3t, $$ which are all obviously parameterizations of the unit circle. Any other initial point configuration can be obtained from this one by an affine transformation, which also maps the three identical unit circles onto three identical ellipses.

With the above in mind, you could with a bit of tedious trigonometric manipulation verify that the three trajectories in the general solution are phase-shifted versions of each other, i.e., that $Q(t)=P(t+delta)$, $R(t)=Q(t+delta)$ and $P(t)=R(t+delta)$, where $delta = 2piover3sqrt3$.

share|cite|improve this answer

edited Jul 19 at 19:18

answered Jul 18 at 21:24

amd

25.9k2943

To avoid a proliferation of subscripts, I’m going to rename the points $P$, $Q$ and $R$. The system of equations can be expressed in matrix form as $$beginbmatrixdot P \ dot Q \ dot Rendbmatrix = beginbmatrix0&-1&1\1&0&-1\-1&1&0endbmatrix beginbmatrixP\Q\Rendbmatrix.$$ It’s not terribly difficult to work out the solution to this system using the usual methods: the eigenvalues are $0$ and $pmsqrt3i$, and so the coefficient matrix is similar to $$J=beginbmatrix0&0&0\0&0&-sqrt3\0&sqrt3&0endbmatrix$$ with exponential $$e^tJ = beginbmatrix1&0&0\0&cossqrt3t&-sinsqrt3t\0&sinsqrt3t&cossqrt3tendbmatrix.$$ A corresponding eigenbasis is easily computed (in fact one can find an orthonormal basis—the coefficient matrix is the “cross-product matrix” of $(1,1,1)^T$), and setting $C = frac13(P_0+Q_0+R_0)$, the centroid of the initial configuration of points, gives for the general solution $$P(t)=C+(P_0-C)cossqrt3t+(Q_0-R_0)frac1sqrt3sinsqrt3t \ Q(t) = C+(Q_0-C)cossqrt3t+(R_0-P_0)frac1sqrt3sinsqrt3t \ R(t) = C+(R_0-C)cossqrt3t+(P_0-Q_0)frac1sqrt3sinsqrt3t.$$ Now, set $P_0=(1,0)$, $Q_0=left(-frac12,fracsqrt32right)$ and $R_0=left(-frac12,-fracsqrt32right)$, the vertices of an equilateral triangle centered on the origin. The parametric equations of the three curves simplify to $$P(t) = (1,0)cossqrt3t+(0,1)sinsqrt3t \ Q(t) = left(-frac12,fracsqrt32right)cossqrt3t+left(-fracsqrt32,-frac12right)sinsqrt3t \ R(t) = left(-frac12,-fracsqrt32right)cossqrt3t+left(fracsqrt32,-frac12right)sinsqrt3t, $$ which are all obviously parameterizations of the unit circle. Any other initial point configuration can be obtained from this one by an affine transformation, which also maps the three identical unit circles onto three identical ellipses.

With the above in mind, you could with a bit of tedious trigonometric manipulation verify that the three trajectories in the general solution are phase-shifted versions of each other, i.e., that $Q(t)=P(t+delta)$, $R(t)=Q(t+delta)$ and $P(t)=R(t+delta)$, where $delta = 2piover3sqrt3$.

share|cite|improve this answer

edited Jul 19 at 19:18

answered Jul 18 at 21:24

amd

25.9k2943

share|cite|improve this answer

share|cite|improve this answer

edited Jul 19 at 19:18

edited Jul 19 at 19:18

edited Jul 19 at 19:18

answered Jul 18 at 21:24

amd

25.9k2943

answered Jul 18 at 21:24

amd

25.9k2943

answered Jul 18 at 21:24

amd

25.9k2943

25.9k2943

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I could’ve gone directly to the unit-circle solution, but I thought seeing the general case would be instructive. The vectors involved have transparent geometric interpretations.
  – amd
  Jul 18 at 23:14
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I could’ve gone directly to the unit-circle solution, but I thought seeing the general case would be instructive. The vectors involved have transparent geometric interpretations.
  – amd
  Jul 18 at 23:14
  

  
  
  
  

I could’ve gone directly to the unit-circle solution, but I thought seeing the general case would be instructive. The vectors involved have transparent geometric interpretations.
– amd
Jul 18 at 23:14

I could’ve gone directly to the unit-circle solution, but I thought seeing the general case would be instructive. The vectors involved have transparent geometric interpretations.
– amd
Jul 18 at 23:14

add a comment | 

 

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