envelope of rays that propagate into the ellipse for $F=u_x^2 + u_y^2 - 1$
$begingroup$
I have the equation $F=u_x^2 + u_y^2 - 1$ and that $u=0$ on the boundary of the ellipse $frac{x^2}{a^2} +frac{y^2}{b^2}=1$
At $u=0$ the ellipse can be parametrised as $x=acos(s), y=bsin(s)$
Charpits eqs give me:
$dx/dt= 2p$
$dy/dt = 2q$
$dp/dt = dq/dt = 0$
$du/dt = 2p^2 + 2q^2$
where $p = frac{partial u}{partial x}$ and $q = frac{partial u}{partial y}$
solving charpits eqs I get:
$p = p_0(s)$
$q = q_0(s)$
$x = 2p_0(s)t + acos(s)$
$y = 2q_0(s)t + bsin(s)$
Since $F(p,q) = p^2 + q^2 - 1$, this implies that $p_0^2 + q_0^2 = 1$
To find the envelope i need to find when the jacobian =0 which is when $2bp_0cos(s)+ 2aq_0sin(s) = 0$
I am confused as to how to draw the envelope of rays that propagate into the ellipse and also how to determine the ridge of discontinuity
ordinary-differential-equations pde
edited Dec 10 '18 at 16:58
Harry49
6,18331132
asked Dec 10 '18 at 12:11
pablo_mathscobarpablo_mathscobar
996
$endgroup$
add a comment |
$begingroup$
I have the equation $F=u_x^2 + u_y^2 - 1$ and that $u=0$ on the boundary of the ellipse $frac{x^2}{a^2} +frac{y^2}{b^2}=1$
At $u=0$ the ellipse can be parametrised as $x=acos(s), y=bsin(s)$
Charpits eqs give me:
$dx/dt= 2p$
$dy/dt = 2q$
$dp/dt = dq/dt = 0$
$du/dt = 2p^2 + 2q^2$
where $p = frac{partial u}{partial x}$ and $q = frac{partial u}{partial y}$
solving charpits eqs I get:
$p = p_0(s)$
$q = q_0(s)$
$x = 2p_0(s)t + acos(s)$
$y = 2q_0(s)t + bsin(s)$
Since $F(p,q) = p^2 + q^2 - 1$, this implies that $p_0^2 + q_0^2 = 1$
To find the envelope i need to find when the jacobian =0 which is when $2bp_0cos(s)+ 2aq_0sin(s) = 0$
I am confused as to how to draw the envelope of rays that propagate into the ellipse and also how to determine the ridge of discontinuity
ordinary-differential-equations pde
edited Dec 10 '18 at 16:58
Harry49
6,18331132
asked Dec 10 '18 at 12:11
pablo_mathscobarpablo_mathscobar
996
$endgroup$
add a comment |
$begingroup$
I have the equation $F=u_x^2 + u_y^2 - 1$ and that $u=0$ on the boundary of the ellipse $frac{x^2}{a^2} +frac{y^2}{b^2}=1$
At $u=0$ the ellipse can be parametrised as $x=acos(s), y=bsin(s)$
Charpits eqs give me:
$dx/dt= 2p$
$dy/dt = 2q$
$dp/dt = dq/dt = 0$
$du/dt = 2p^2 + 2q^2$
where $p = frac{partial u}{partial x}$ and $q = frac{partial u}{partial y}$
solving charpits eqs I get:
$p = p_0(s)$
$q = q_0(s)$
$x = 2p_0(s)t + acos(s)$
$y = 2q_0(s)t + bsin(s)$
Since $F(p,q) = p^2 + q^2 - 1$, this implies that $p_0^2 + q_0^2 = 1$
To find the envelope i need to find when the jacobian =0 which is when $2bp_0cos(s)+ 2aq_0sin(s) = 0$
I am confused as to how to draw the envelope of rays that propagate into the ellipse and also how to determine the ridge of discontinuity
ordinary-differential-equations pde
edited Dec 10 '18 at 16:58
Harry49
6,18331132
asked Dec 10 '18 at 12:11
pablo_mathscobarpablo_mathscobar
996
$endgroup$
I have the equation $F=u_x^2 + u_y^2 - 1$ and that $u=0$ on the boundary of the ellipse $frac{x^2}{a^2} +frac{y^2}{b^2}=1$
At $u=0$ the ellipse can be parametrised as $x=acos(s), y=bsin(s)$
Charpits eqs give me:
$dx/dt= 2p$
$dy/dt = 2q$
$dp/dt = dq/dt = 0$
$du/dt = 2p^2 + 2q^2$
where $p = frac{partial u}{partial x}$ and $q = frac{partial u}{partial y}$
solving charpits eqs I get:
$p = p_0(s)$
$q = q_0(s)$
$x = 2p_0(s)t + acos(s)$
$y = 2q_0(s)t + bsin(s)$
Since $F(p,q) = p^2 + q^2 - 1$, this implies that $p_0^2 + q_0^2 = 1$
To find the envelope i need to find when the jacobian =0 which is when $2bp_0cos(s)+ 2aq_0sin(s) = 0$
I am confused as to how to draw the envelope of rays that propagate into the ellipse and also how to determine the ridge of discontinuity
ordinary-differential-equations pde
ordinary-differential-equations pde
edited Dec 10 '18 at 16:58
Harry49
6,18331132
asked Dec 10 '18 at 12:11
pablo_mathscobarpablo_mathscobar
996
edited Dec 10 '18 at 16:58
Harry49
6,18331132
asked Dec 10 '18 at 12:11
pablo_mathscobarpablo_mathscobar
996
edited Dec 10 '18 at 16:58
Harry49
6,18331132
edited Dec 10 '18 at 16:58
Harry49
6,18331132
edited Dec 10 '18 at 16:58
Harry49
6,18331132
6,18331132
asked Dec 10 '18 at 12:11
pablo_mathscobarpablo_mathscobar
996
asked Dec 10 '18 at 12:11
pablo_mathscobarpablo_mathscobar
996
asked Dec 10 '18 at 12:11
pablo_mathscobarpablo_mathscobar
996
996
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I think you might have missed a sign in $dot x_0(s)=-asin(s)$. Then the equation for $p_0,q_0$ reads as
$$
0=dot z_0(s)=p_0(s)dot x_0(s)+q_0(s)dot y_0(s)=-ap_0sin(s)+bq_0cos(s).
$$
This tells us that the unit vector $(p_0,q_0)$ has to be orthogonal to $(-asin(s), bcos(s))$, thus $$(p_0,q_0)=pmfrac{(bcos(s),asin(s))}{sqrt{b^2cos^2(s)+a^2sin^2(s)}}.$$
As this should point inwards, chose the negative sign.
To find the characteristic that some point $(x,y)$ lies on, one has to solve
$$
left.begin{aligned}
x&=x_0+2p_0t=acos(s) - r b cos(s)\
y&=y_0+2q_0t=bsin(s) - r a sin(s)
end{aligned}right}
implies frac{x^2}{(a-rb)^2}+frac{y^2}{(b-ar)^2}=1
$$
where $r$ is some positive multiple of $t$.
The last is equivalent to a 4th degree polynomial equation, by the intermediate value theorem there is at least one solution between $r=0$ and $min(frac ba,frac ab)$, there is another solution for $r>max(frac ba,frac ab)$, but that might be inadmissible.
Two close-by rays associated to the angles $s$ and $s+ds$ intersect at
$$
left.begin{aligned}
(a - r b) cos(s) = (a - (r+dr) b) cos(s+ds)\
(b - r a) sin(s) = (b - (r+dr) a) sin(s+ds)
end{aligned}right}
implies
left{begin{aligned}
0=-b,dr,cos(s)-(a-br)sin(s)ds\
0=-a,dr,sin(s)+(b-ar)cos(s)ds
end{aligned}right.
$$
so that necessarily $$a(a-br)sin^2(s)+b(b-ar)cos^2(s)=0implies r = frac{a^2sin^2s+b^2cos^2s}{ab}$$
Plotting that curve against the ellipse gives the picture
answered Dec 10 '18 at 15:55
LutzLLutzL
57.7k42054
$endgroup$
$begingroup$
I thought to find the envelope we need to find where the jacobian = 0 so we can find an expression for t? I realise my expression for the jacobian was wrong but the method of solving for t then putting $p_0,q_0,t$ back into the characteristics to find the envelope is still correct , is it not?
$endgroup$
– pablo_mathscobar
Dec 12 '18 at 11:15
$begingroup$
You use $dz=p,dx+q,dy$ along the initial condition curve together with all other previously computed relations to determine $(p_0,q_0)$. The zeros of the determinant of the Jacobian of $$(s,t)mapsto(,x_0(s)+2tp_0(s),, y_0(s)+2tq_0(s),)=(,(a+2tf(s),b)cos(s),,(b+2tf(s),a)sin(s),)$$ tell you where to find the envelope. This is the same as me computing the intersection of neighboring rays with $r=2tf(s)$. The points at $s=kfracpi2$ that are inside the ellipse are the borders of the ridge.
$endgroup$
– LutzL
Dec 12 '18 at 11:47
$begingroup$
I see computing the $frac{partial}{partial s} acos(s) + 2tbcos(s)f(s)$ is quite a messy derivative because of the f(s) term so i guess your way is better? also how did you find the intersection between the 2 rays? what did you do remove the cos(ds) terms
$endgroup$
– pablo_mathscobar
Dec 12 '18 at 12:06
$begingroup$
It does not need to be messy if you leave the derivative at $f'(s)$. But yes, by the chain rule you get $$frac{∂(x,y)}{∂(s,t)}=frac{∂(x,y)}{∂(s,r)}frac{∂(s,r)}{∂(s,t)}text{ with }frac{∂(s,r)}{∂(s,t)}=pmatrix{1&0\2tf'(s)&2f(s)},$$ so that the zeros of the determinants give the same points.
$endgroup$
– LutzL
Dec 12 '18 at 12:10
add a comment |
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1 Answer
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1 Answer
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$begingroup$
I think you might have missed a sign in $dot x_0(s)=-asin(s)$. Then the equation for $p_0,q_0$ reads as
$$
0=dot z_0(s)=p_0(s)dot x_0(s)+q_0(s)dot y_0(s)=-ap_0sin(s)+bq_0cos(s).
$$
This tells us that the unit vector $(p_0,q_0)$ has to be orthogonal to $(-asin(s), bcos(s))$, thus $$(p_0,q_0)=pmfrac{(bcos(s),asin(s))}{sqrt{b^2cos^2(s)+a^2sin^2(s)}}.$$
As this should point inwards, chose the negative sign.
To find the characteristic that some point $(x,y)$ lies on, one has to solve
$$
left.begin{aligned}
x&=x_0+2p_0t=acos(s) - r b cos(s)\
y&=y_0+2q_0t=bsin(s) - r a sin(s)
end{aligned}right}
implies frac{x^2}{(a-rb)^2}+frac{y^2}{(b-ar)^2}=1
$$
where $r$ is some positive multiple of $t$.
The last is equivalent to a 4th degree polynomial equation, by the intermediate value theorem there is at least one solution between $r=0$ and $min(frac ba,frac ab)$, there is another solution for $r>max(frac ba,frac ab)$, but that might be inadmissible.
Two close-by rays associated to the angles $s$ and $s+ds$ intersect at
$$
left.begin{aligned}
(a - r b) cos(s) = (a - (r+dr) b) cos(s+ds)\
(b - r a) sin(s) = (b - (r+dr) a) sin(s+ds)
end{aligned}right}
implies
left{begin{aligned}
0=-b,dr,cos(s)-(a-br)sin(s)ds\
0=-a,dr,sin(s)+(b-ar)cos(s)ds
end{aligned}right.
$$
so that necessarily $$a(a-br)sin^2(s)+b(b-ar)cos^2(s)=0implies r = frac{a^2sin^2s+b^2cos^2s}{ab}$$
Plotting that curve against the ellipse gives the picture
answered Dec 10 '18 at 15:55
LutzLLutzL
57.7k42054
$endgroup$
$begingroup$
I thought to find the envelope we need to find where the jacobian = 0 so we can find an expression for t? I realise my expression for the jacobian was wrong but the method of solving for t then putting $p_0,q_0,t$ back into the characteristics to find the envelope is still correct , is it not?
$endgroup$
– pablo_mathscobar
Dec 12 '18 at 11:15
$begingroup$
You use $dz=p,dx+q,dy$ along the initial condition curve together with all other previously computed relations to determine $(p_0,q_0)$. The zeros of the determinant of the Jacobian of $$(s,t)mapsto(,x_0(s)+2tp_0(s),, y_0(s)+2tq_0(s),)=(,(a+2tf(s),b)cos(s),,(b+2tf(s),a)sin(s),)$$ tell you where to find the envelope. This is the same as me computing the intersection of neighboring rays with $r=2tf(s)$. The points at $s=kfracpi2$ that are inside the ellipse are the borders of the ridge.
$endgroup$
– LutzL
Dec 12 '18 at 11:47
$begingroup$
I see computing the $frac{partial}{partial s} acos(s) + 2tbcos(s)f(s)$ is quite a messy derivative because of the f(s) term so i guess your way is better? also how did you find the intersection between the 2 rays? what did you do remove the cos(ds) terms
$endgroup$
– pablo_mathscobar
Dec 12 '18 at 12:06
$begingroup$
It does not need to be messy if you leave the derivative at $f'(s)$. But yes, by the chain rule you get $$frac{∂(x,y)}{∂(s,t)}=frac{∂(x,y)}{∂(s,r)}frac{∂(s,r)}{∂(s,t)}text{ with }frac{∂(s,r)}{∂(s,t)}=pmatrix{1&0\2tf'(s)&2f(s)},$$ so that the zeros of the determinants give the same points.
$endgroup$
– LutzL
Dec 12 '18 at 12:10
add a comment |
$begingroup$
I think you might have missed a sign in $dot x_0(s)=-asin(s)$. Then the equation for $p_0,q_0$ reads as
$$
0=dot z_0(s)=p_0(s)dot x_0(s)+q_0(s)dot y_0(s)=-ap_0sin(s)+bq_0cos(s).
$$
This tells us that the unit vector $(p_0,q_0)$ has to be orthogonal to $(-asin(s), bcos(s))$, thus $$(p_0,q_0)=pmfrac{(bcos(s),asin(s))}{sqrt{b^2cos^2(s)+a^2sin^2(s)}}.$$
As this should point inwards, chose the negative sign.
To find the characteristic that some point $(x,y)$ lies on, one has to solve
$$
left.begin{aligned}
x&=x_0+2p_0t=acos(s) - r b cos(s)\
y&=y_0+2q_0t=bsin(s) - r a sin(s)
end{aligned}right}
implies frac{x^2}{(a-rb)^2}+frac{y^2}{(b-ar)^2}=1
$$
where $r$ is some positive multiple of $t$.
The last is equivalent to a 4th degree polynomial equation, by the intermediate value theorem there is at least one solution between $r=0$ and $min(frac ba,frac ab)$, there is another solution for $r>max(frac ba,frac ab)$, but that might be inadmissible.
Two close-by rays associated to the angles $s$ and $s+ds$ intersect at
$$
left.begin{aligned}
(a - r b) cos(s) = (a - (r+dr) b) cos(s+ds)\
(b - r a) sin(s) = (b - (r+dr) a) sin(s+ds)
end{aligned}right}
implies
left{begin{aligned}
0=-b,dr,cos(s)-(a-br)sin(s)ds\
0=-a,dr,sin(s)+(b-ar)cos(s)ds
end{aligned}right.
$$
so that necessarily $$a(a-br)sin^2(s)+b(b-ar)cos^2(s)=0implies r = frac{a^2sin^2s+b^2cos^2s}{ab}$$
Plotting that curve against the ellipse gives the picture
answered Dec 10 '18 at 15:55
LutzLLutzL
57.7k42054
$endgroup$
$begingroup$
I thought to find the envelope we need to find where the jacobian = 0 so we can find an expression for t? I realise my expression for the jacobian was wrong but the method of solving for t then putting $p_0,q_0,t$ back into the characteristics to find the envelope is still correct , is it not?
$endgroup$
– pablo_mathscobar
Dec 12 '18 at 11:15
$begingroup$
You use $dz=p,dx+q,dy$ along the initial condition curve together with all other previously computed relations to determine $(p_0,q_0)$. The zeros of the determinant of the Jacobian of $$(s,t)mapsto(,x_0(s)+2tp_0(s),, y_0(s)+2tq_0(s),)=(,(a+2tf(s),b)cos(s),,(b+2tf(s),a)sin(s),)$$ tell you where to find the envelope. This is the same as me computing the intersection of neighboring rays with $r=2tf(s)$. The points at $s=kfracpi2$ that are inside the ellipse are the borders of the ridge.
$endgroup$
– LutzL
Dec 12 '18 at 11:47
$begingroup$
I see computing the $frac{partial}{partial s} acos(s) + 2tbcos(s)f(s)$ is quite a messy derivative because of the f(s) term so i guess your way is better? also how did you find the intersection between the 2 rays? what did you do remove the cos(ds) terms
$endgroup$
– pablo_mathscobar
Dec 12 '18 at 12:06
$begingroup$
It does not need to be messy if you leave the derivative at $f'(s)$. But yes, by the chain rule you get $$frac{∂(x,y)}{∂(s,t)}=frac{∂(x,y)}{∂(s,r)}frac{∂(s,r)}{∂(s,t)}text{ with }frac{∂(s,r)}{∂(s,t)}=pmatrix{1&0\2tf'(s)&2f(s)},$$ so that the zeros of the determinants give the same points.
$endgroup$
– LutzL
Dec 12 '18 at 12:10
add a comment |
$begingroup$
I think you might have missed a sign in $dot x_0(s)=-asin(s)$. Then the equation for $p_0,q_0$ reads as
$$
0=dot z_0(s)=p_0(s)dot x_0(s)+q_0(s)dot y_0(s)=-ap_0sin(s)+bq_0cos(s).
$$
This tells us that the unit vector $(p_0,q_0)$ has to be orthogonal to $(-asin(s), bcos(s))$, thus $$(p_0,q_0)=pmfrac{(bcos(s),asin(s))}{sqrt{b^2cos^2(s)+a^2sin^2(s)}}.$$
As this should point inwards, chose the negative sign.
To find the characteristic that some point $(x,y)$ lies on, one has to solve
$$
left.begin{aligned}
x&=x_0+2p_0t=acos(s) - r b cos(s)\
y&=y_0+2q_0t=bsin(s) - r a sin(s)
end{aligned}right}
implies frac{x^2}{(a-rb)^2}+frac{y^2}{(b-ar)^2}=1
$$
where $r$ is some positive multiple of $t$.
The last is equivalent to a 4th degree polynomial equation, by the intermediate value theorem there is at least one solution between $r=0$ and $min(frac ba,frac ab)$, there is another solution for $r>max(frac ba,frac ab)$, but that might be inadmissible.
Two close-by rays associated to the angles $s$ and $s+ds$ intersect at
$$
left.begin{aligned}
(a - r b) cos(s) = (a - (r+dr) b) cos(s+ds)\
(b - r a) sin(s) = (b - (r+dr) a) sin(s+ds)
end{aligned}right}
implies
left{begin{aligned}
0=-b,dr,cos(s)-(a-br)sin(s)ds\
0=-a,dr,sin(s)+(b-ar)cos(s)ds
end{aligned}right.
$$
so that necessarily $$a(a-br)sin^2(s)+b(b-ar)cos^2(s)=0implies r = frac{a^2sin^2s+b^2cos^2s}{ab}$$
Plotting that curve against the ellipse gives the picture
answered Dec 10 '18 at 15:55
LutzLLutzL
57.7k42054
$endgroup$
I think you might have missed a sign in $dot x_0(s)=-asin(s)$. Then the equation for $p_0,q_0$ reads as
$$
0=dot z_0(s)=p_0(s)dot x_0(s)+q_0(s)dot y_0(s)=-ap_0sin(s)+bq_0cos(s).
$$
This tells us that the unit vector $(p_0,q_0)$ has to be orthogonal to $(-asin(s), bcos(s))$, thus $$(p_0,q_0)=pmfrac{(bcos(s),asin(s))}{sqrt{b^2cos^2(s)+a^2sin^2(s)}}.$$
As this should point inwards, chose the negative sign.
To find the characteristic that some point $(x,y)$ lies on, one has to solve
$$
left.begin{aligned}
x&=x_0+2p_0t=acos(s) - r b cos(s)\
y&=y_0+2q_0t=bsin(s) - r a sin(s)
end{aligned}right}
implies frac{x^2}{(a-rb)^2}+frac{y^2}{(b-ar)^2}=1
$$
where $r$ is some positive multiple of $t$.
The last is equivalent to a 4th degree polynomial equation, by the intermediate value theorem there is at least one solution between $r=0$ and $min(frac ba,frac ab)$, there is another solution for $r>max(frac ba,frac ab)$, but that might be inadmissible.
Two close-by rays associated to the angles $s$ and $s+ds$ intersect at
$$
left.begin{aligned}
(a - r b) cos(s) = (a - (r+dr) b) cos(s+ds)\
(b - r a) sin(s) = (b - (r+dr) a) sin(s+ds)
end{aligned}right}
implies
left{begin{aligned}
0=-b,dr,cos(s)-(a-br)sin(s)ds\
0=-a,dr,sin(s)+(b-ar)cos(s)ds
end{aligned}right.
$$
so that necessarily $$a(a-br)sin^2(s)+b(b-ar)cos^2(s)=0implies r = frac{a^2sin^2s+b^2cos^2s}{ab}$$
Plotting that curve against the ellipse gives the picture
answered Dec 10 '18 at 15:55
LutzLLutzL
57.7k42054
answered Dec 10 '18 at 15:55
LutzLLutzL
57.7k42054
answered Dec 10 '18 at 15:55
LutzLLutzL
57.7k42054
answered Dec 10 '18 at 15:55
LutzLLutzL
57.7k42054
57.7k42054
$begingroup$
I thought to find the envelope we need to find where the jacobian = 0 so we can find an expression for t? I realise my expression for the jacobian was wrong but the method of solving for t then putting $p_0,q_0,t$ back into the characteristics to find the envelope is still correct , is it not?
$endgroup$
– pablo_mathscobar
Dec 12 '18 at 11:15
$begingroup$
You use $dz=p,dx+q,dy$ along the initial condition curve together with all other previously computed relations to determine $(p_0,q_0)$. The zeros of the determinant of the Jacobian of $$(s,t)mapsto(,x_0(s)+2tp_0(s),, y_0(s)+2tq_0(s),)=(,(a+2tf(s),b)cos(s),,(b+2tf(s),a)sin(s),)$$ tell you where to find the envelope. This is the same as me computing the intersection of neighboring rays with $r=2tf(s)$. The points at $s=kfracpi2$ that are inside the ellipse are the borders of the ridge.
$endgroup$
– LutzL
Dec 12 '18 at 11:47
$begingroup$
I see computing the $frac{partial}{partial s} acos(s) + 2tbcos(s)f(s)$ is quite a messy derivative because of the f(s) term so i guess your way is better? also how did you find the intersection between the 2 rays? what did you do remove the cos(ds) terms
$endgroup$
– pablo_mathscobar
Dec 12 '18 at 12:06
$begingroup$
It does not need to be messy if you leave the derivative at $f'(s)$. But yes, by the chain rule you get $$frac{∂(x,y)}{∂(s,t)}=frac{∂(x,y)}{∂(s,r)}frac{∂(s,r)}{∂(s,t)}text{ with }frac{∂(s,r)}{∂(s,t)}=pmatrix{1&0\2tf'(s)&2f(s)},$$ so that the zeros of the determinants give the same points.
$endgroup$
– LutzL
Dec 12 '18 at 12:10
add a comment |
$begingroup$
I thought to find the envelope we need to find where the jacobian = 0 so we can find an expression for t? I realise my expression for the jacobian was wrong but the method of solving for t then putting $p_0,q_0,t$ back into the characteristics to find the envelope is still correct , is it not?
$endgroup$
– pablo_mathscobar
Dec 12 '18 at 11:15
$begingroup$
You use $dz=p,dx+q,dy$ along the initial condition curve together with all other previously computed relations to determine $(p_0,q_0)$. The zeros of the determinant of the Jacobian of $$(s,t)mapsto(,x_0(s)+2tp_0(s),, y_0(s)+2tq_0(s),)=(,(a+2tf(s),b)cos(s),,(b+2tf(s),a)sin(s),)$$ tell you where to find the envelope. This is the same as me computing the intersection of neighboring rays with $r=2tf(s)$. The points at $s=kfracpi2$ that are inside the ellipse are the borders of the ridge.
$endgroup$
– LutzL
Dec 12 '18 at 11:47
$begingroup$
I see computing the $frac{partial}{partial s} acos(s) + 2tbcos(s)f(s)$ is quite a messy derivative because of the f(s) term so i guess your way is better? also how did you find the intersection between the 2 rays? what did you do remove the cos(ds) terms
$endgroup$
– pablo_mathscobar
Dec 12 '18 at 12:06
$begingroup$
It does not need to be messy if you leave the derivative at $f'(s)$. But yes, by the chain rule you get $$frac{∂(x,y)}{∂(s,t)}=frac{∂(x,y)}{∂(s,r)}frac{∂(s,r)}{∂(s,t)}text{ with }frac{∂(s,r)}{∂(s,t)}=pmatrix{1&0\2tf'(s)&2f(s)},$$ so that the zeros of the determinants give the same points.
$endgroup$
– LutzL
Dec 12 '18 at 12:10
$begingroup$
I thought to find the envelope we need to find where the jacobian = 0 so we can find an expression for t? I realise my expression for the jacobian was wrong but the method of solving for t then putting $p_0,q_0,t$ back into the characteristics to find the envelope is still correct , is it not?
$endgroup$
– pablo_mathscobar
Dec 12 '18 at 11:15
$begingroup$
I thought to find the envelope we need to find where the jacobian = 0 so we can find an expression for t? I realise my expression for the jacobian was wrong but the method of solving for t then putting $p_0,q_0,t$ back into the characteristics to find the envelope is still correct , is it not?
$endgroup$
– pablo_mathscobar
Dec 12 '18 at 11:15
$begingroup$
You use $dz=p,dx+q,dy$ along the initial condition curve together with all other previously computed relations to determine $(p_0,q_0)$. The zeros of the determinant of the Jacobian of $$(s,t)mapsto(,x_0(s)+2tp_0(s),, y_0(s)+2tq_0(s),)=(,(a+2tf(s),b)cos(s),,(b+2tf(s),a)sin(s),)$$ tell you where to find the envelope. This is the same as me computing the intersection of neighboring rays with $r=2tf(s)$. The points at $s=kfracpi2$ that are inside the ellipse are the borders of the ridge.
$endgroup$
– LutzL
Dec 12 '18 at 11:47
$begingroup$
You use $dz=p,dx+q,dy$ along the initial condition curve together with all other previously computed relations to determine $(p_0,q_0)$. The zeros of the determinant of the Jacobian of $$(s,t)mapsto(,x_0(s)+2tp_0(s),, y_0(s)+2tq_0(s),)=(,(a+2tf(s),b)cos(s),,(b+2tf(s),a)sin(s),)$$ tell you where to find the envelope. This is the same as me computing the intersection of neighboring rays with $r=2tf(s)$. The points at $s=kfracpi2$ that are inside the ellipse are the borders of the ridge.
$endgroup$
– LutzL
Dec 12 '18 at 11:47
$begingroup$
I see computing the $frac{partial}{partial s} acos(s) + 2tbcos(s)f(s)$ is quite a messy derivative because of the f(s) term so i guess your way is better? also how did you find the intersection between the 2 rays? what did you do remove the cos(ds) terms
$endgroup$
– pablo_mathscobar
Dec 12 '18 at 12:06
$begingroup$
I see computing the $frac{partial}{partial s} acos(s) + 2tbcos(s)f(s)$ is quite a messy derivative because of the f(s) term so i guess your way is better? also how did you find the intersection between the 2 rays? what did you do remove the cos(ds) terms
$endgroup$
– pablo_mathscobar
Dec 12 '18 at 12:06
$begingroup$
It does not need to be messy if you leave the derivative at $f'(s)$. But yes, by the chain rule you get $$frac{∂(x,y)}{∂(s,t)}=frac{∂(x,y)}{∂(s,r)}frac{∂(s,r)}{∂(s,t)}text{ with }frac{∂(s,r)}{∂(s,t)}=pmatrix{1&0\2tf'(s)&2f(s)},$$ so that the zeros of the determinants give the same points.
$endgroup$
– LutzL
Dec 12 '18 at 12:10
$begingroup$
It does not need to be messy if you leave the derivative at $f'(s)$. But yes, by the chain rule you get $$frac{∂(x,y)}{∂(s,t)}=frac{∂(x,y)}{∂(s,r)}frac{∂(s,r)}{∂(s,t)}text{ with }frac{∂(s,r)}{∂(s,t)}=pmatrix{1&0\2tf'(s)&2f(s)},$$ so that the zeros of the determinants give the same points.
$endgroup$
– LutzL
Dec 12 '18 at 12:10
add a comment |
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