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Limiting distance of a chase of two points?
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Gil Kalai's blog post https://gilkalai.wordpress.com/2018/06/29/test-your-intuition-35-what-is-the-limiting-distance/ poses a riddle about a chase of two points.
Point A chases point B at unit velocity. Point B heads right at unit velocity. At $t = 0$, $A(0) = (0,1)$ and $B(0) = (0,0)$.
Question: what is the limiting distance $||B(t) - A(t)||$ as $t$ goes to infinity?
differential-equations analytic-geometry dynamical-systems
edited Jul 15 at 9:14
Dylan
11.4k31026
asked Jul 15 at 2:17
mich
430210
add a comment |Â
up vote
5
down vote
favorite
Gil Kalai's blog post https://gilkalai.wordpress.com/2018/06/29/test-your-intuition-35-what-is-the-limiting-distance/ poses a riddle about a chase of two points.
Point A chases point B at unit velocity. Point B heads right at unit velocity. At $t = 0$, $A(0) = (0,1)$ and $B(0) = (0,0)$.
Question: what is the limiting distance $||B(t) - A(t)||$ as $t$ goes to infinity?
differential-equations analytic-geometry dynamical-systems
edited Jul 15 at 9:14
Dylan
11.4k31026
asked Jul 15 at 2:17
mich
430210
My gut tells me that it's 1...
– Sambo
Jul 15 at 3:43
I tried writing out the differential equations, which failed because they are not linear and I'm not very good at that. Per your guess, it's definitely lower than 1 (since B is running away at a sub-optimal direction and the distance starts at 1). It will also not reach 0.
– mich
Jul 15 at 3:44
The distance between the points is not 1, but $sqrt2$. I would say that the distance is $sqrt2-1$.
– Moti
Jul 15 at 5:59
@Moti: The distance is $1$.
– joriki
Jul 15 at 6:07
You are right... but I still think that the distance is as stated.
– Moti
Jul 15 at 6:23
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
Gil Kalai's blog post https://gilkalai.wordpress.com/2018/06/29/test-your-intuition-35-what-is-the-limiting-distance/ poses a riddle about a chase of two points.
Point A chases point B at unit velocity. Point B heads right at unit velocity. At $t = 0$, $A(0) = (0,1)$ and $B(0) = (0,0)$.
Question: what is the limiting distance $||B(t) - A(t)||$ as $t$ goes to infinity?
differential-equations analytic-geometry dynamical-systems
edited Jul 15 at 9:14
Dylan
11.4k31026
asked Jul 15 at 2:17
mich
430210
Gil Kalai's blog post https://gilkalai.wordpress.com/2018/06/29/test-your-intuition-35-what-is-the-limiting-distance/ poses a riddle about a chase of two points.
Point A chases point B at unit velocity. Point B heads right at unit velocity. At $t = 0$, $A(0) = (0,1)$ and $B(0) = (0,0)$.
Question: what is the limiting distance $||B(t) - A(t)||$ as $t$ goes to infinity?
differential-equations analytic-geometry dynamical-systems
edited Jul 15 at 9:14
Dylan
11.4k31026
asked Jul 15 at 2:17
mich
430210
edited Jul 15 at 9:14
Dylan
11.4k31026
edited Jul 15 at 9:14
Dylan
11.4k31026
edited Jul 15 at 9:14
Dylan
11.4k31026
11.4k31026
asked Jul 15 at 2:17
mich
430210
asked Jul 15 at 2:17
mich
430210
asked Jul 15 at 2:17
mich
430210
430210
My gut tells me that it's 1...
– Sambo
Jul 15 at 3:43
I tried writing out the differential equations, which failed because they are not linear and I'm not very good at that. Per your guess, it's definitely lower than 1 (since B is running away at a sub-optimal direction and the distance starts at 1). It will also not reach 0.
– mich
Jul 15 at 3:44
The distance between the points is not 1, but $sqrt2$. I would say that the distance is $sqrt2-1$.
– Moti
Jul 15 at 5:59
@Moti: The distance is $1$.
– joriki
Jul 15 at 6:07
You are right... but I still think that the distance is as stated.
– Moti
Jul 15 at 6:23
add a comment |Â
My gut tells me that it's 1...
– Sambo
Jul 15 at 3:43
I tried writing out the differential equations, which failed because they are not linear and I'm not very good at that. Per your guess, it's definitely lower than 1 (since B is running away at a sub-optimal direction and the distance starts at 1). It will also not reach 0.
– mich
Jul 15 at 3:44
The distance between the points is not 1, but $sqrt2$. I would say that the distance is $sqrt2-1$.
– Moti
Jul 15 at 5:59
@Moti: The distance is $1$.
– joriki
Jul 15 at 6:07
You are right... but I still think that the distance is as stated.
– Moti
Jul 15 at 6:23
My gut tells me that it's 1...
– Sambo
Jul 15 at 3:43
My gut tells me that it's 1...
– Sambo
Jul 15 at 3:43
I tried writing out the differential equations, which failed because they are not linear and I'm not very good at that. Per your guess, it's definitely lower than 1 (since B is running away at a sub-optimal direction and the distance starts at 1). It will also not reach 0.
– mich
Jul 15 at 3:44
I tried writing out the differential equations, which failed because they are not linear and I'm not very good at that. Per your guess, it's definitely lower than 1 (since B is running away at a sub-optimal direction and the distance starts at 1). It will also not reach 0.
– mich
Jul 15 at 3:44
The distance between the points is not 1, but $sqrt2$. I would say that the distance is $sqrt2-1$.
– Moti
Jul 15 at 5:59
The distance between the points is not 1, but $sqrt2$. I would say that the distance is $sqrt2-1$.
– Moti
Jul 15 at 5:59
@Moti: The distance is $1$.
– joriki
Jul 15 at 6:07
@Moti: The distance is $1$.
– joriki
Jul 15 at 6:07
You are right... but I still think that the distance is as stated.
– Moti
Jul 15 at 6:23
You are right... but I still think that the distance is as stated.
– Moti
Jul 15 at 6:23
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
4
down vote
accepted
The problem becomes more managable if you transform to a coordinate system where $mathbf B$ is at rest. Then $mathbf A$ has velocity
$$
pmatrixdot x\dot y=-pmatrixfrac xr+1\frac yr;.
$$
Now transform to polar coordinates $r=sqrtx^2+y^2$ and $phi=arctanleft(frac yxright)$, with
$$
dot r=fracxdot xr+fracydot yr=-1-frac xr=-1-cosphi
$$
and
$$
dotphi=fracfracdot yx-fracdot xyy^21+left(frac yxright)^2=fracxdot y-ydot xx^2+y^2=frac yr^2=fracsinphir;.
$$
So we have
$$
fracmathrm drmathrm dphi=fracdot rdotphi=-rcdotfrac1+cosphisinphi=-rfraccosfracphi2sinfracphi2;.
$$
Dividing by $r$ and integrating both sides yields
$$
log r=C-2logsinfracphi2;.
$$
The initial value is $rleft(fracpi2right)=1$, which yields $C=-log 2$, so
$$
r=frac12sin^2fracphi2;.
$$
In the limit $ttoinfty$ we have $phitopi$, and thus $rtofrac12$.
answered Jul 15 at 6:43
joriki
165k10180328
You mixed up $phi$ with $x$ towards the end
– Dylan
Jul 15 at 8:25
@Dylan: Indeed I did, thanks -- fixed.
– joriki
Jul 15 at 8:37
add a comment |Â
up vote
2
down vote
There is a mismatch of notations between your wording and the wording in https://gilkalai.wordpress.com/2018/06/29/test-your-intuition-35-what-is-the-limiting-distance/ . I will refer to the notations from the link instead of the notations in the above question because the wording in the link is more clear.
The figure below shows the result of numerical simulation.
The result isn't $1$ but $frac12$. This is proved thanks to analytical solving.
The coordinates of the moving points are B$(x,y)$ and A$(t,0)$.
Since the absolute velocity of B is $=1$ in the direction of $overrightarrowBA$ :
$$begincases
fracdxdt=fract-xsqrt(t-x)^2+y^2 \
fracdydt=frac-ysqrt(t-x)^2+y^2
endcases$$
$fracdydx=frac-yt-x$
Change of variable : $x=u+t$
$fracdxdt=fracdudt+1=frac-usqrtu^2+y^2$
$fracdudt=-fracusqrtu^2+y^2-1$
$fracdydt=fracdydufracdudt=fracdyduleft(-fracusqrtu^2+y^2-1right)=frac-ysqrtu^2+y^2$
$$yfracdudy=sqrtu^2+y^2+u$$
This is a fist order homogeneous ODE easy to solve. The general solution is :
$$u=ysinhleft(c+ln(y)right)$$
The initial conditions $t=0$ , $x=0$ , $y=1$ , $u=x-t=0$ which implies $c=0$ thus :
$$u=ysinhleft(ln(y)right)=yfracy-frac1y2=fracy^2-12$$
$$t-x=-u=frac1-y^22$$
When $xtoinfty quad;quad yto 0quad;quad (t-x)tofrac12$.
$$|AB[=sqrt(x-t)^2+y^2tofrac12$$
answered Jul 15 at 10:54
JJacquelin
40.1k21649
I think in the displayed equation right before "This is a separable ODE" you're missing a factor $sqrtu^2+y^2$? The one in the denominator of the right-hand side of the previous equation. (By the way, my comment "The distance is $1$" under the question referred to Moti's comment that the (original) distance between the points is $sqrt2$ and not $1$; I wasn't talking about the limit distance there.)
– joriki
Jul 15 at 11:04
This was fixed meanwhile. Referring of the wording of the problem from the link, the initial distance is 1. I don't consider the wording of the mich's question.
– JJacquelin
Jul 15 at 11:20
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
The problem becomes more managable if you transform to a coordinate system where $mathbf B$ is at rest. Then $mathbf A$ has velocity
$$
pmatrixdot x\dot y=-pmatrixfrac xr+1\frac yr;.
$$
Now transform to polar coordinates $r=sqrtx^2+y^2$ and $phi=arctanleft(frac yxright)$, with
$$
dot r=fracxdot xr+fracydot yr=-1-frac xr=-1-cosphi
$$
and
$$
dotphi=fracfracdot yx-fracdot xyy^21+left(frac yxright)^2=fracxdot y-ydot xx^2+y^2=frac yr^2=fracsinphir;.
$$
So we have
$$
fracmathrm drmathrm dphi=fracdot rdotphi=-rcdotfrac1+cosphisinphi=-rfraccosfracphi2sinfracphi2;.
$$
Dividing by $r$ and integrating both sides yields
$$
log r=C-2logsinfracphi2;.
$$
The initial value is $rleft(fracpi2right)=1$, which yields $C=-log 2$, so
$$
r=frac12sin^2fracphi2;.
$$
In the limit $ttoinfty$ we have $phitopi$, and thus $rtofrac12$.
answered Jul 15 at 6:43
joriki
165k10180328
You mixed up $phi$ with $x$ towards the end
– Dylan
Jul 15 at 8:25
@Dylan: Indeed I did, thanks -- fixed.
– joriki
Jul 15 at 8:37
add a comment |Â
up vote
4
down vote
accepted
The problem becomes more managable if you transform to a coordinate system where $mathbf B$ is at rest. Then $mathbf A$ has velocity
$$
pmatrixdot x\dot y=-pmatrixfrac xr+1\frac yr;.
$$
Now transform to polar coordinates $r=sqrtx^2+y^2$ and $phi=arctanleft(frac yxright)$, with
$$
dot r=fracxdot xr+fracydot yr=-1-frac xr=-1-cosphi
$$
and
$$
dotphi=fracfracdot yx-fracdot xyy^21+left(frac yxright)^2=fracxdot y-ydot xx^2+y^2=frac yr^2=fracsinphir;.
$$
So we have
$$
fracmathrm drmathrm dphi=fracdot rdotphi=-rcdotfrac1+cosphisinphi=-rfraccosfracphi2sinfracphi2;.
$$
Dividing by $r$ and integrating both sides yields
$$
log r=C-2logsinfracphi2;.
$$
The initial value is $rleft(fracpi2right)=1$, which yields $C=-log 2$, so
$$
r=frac12sin^2fracphi2;.
$$
In the limit $ttoinfty$ we have $phitopi$, and thus $rtofrac12$.
answered Jul 15 at 6:43
joriki
165k10180328
You mixed up $phi$ with $x$ towards the end
– Dylan
Jul 15 at 8:25
@Dylan: Indeed I did, thanks -- fixed.
– joriki
Jul 15 at 8:37
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
The problem becomes more managable if you transform to a coordinate system where $mathbf B$ is at rest. Then $mathbf A$ has velocity
$$
pmatrixdot x\dot y=-pmatrixfrac xr+1\frac yr;.
$$
Now transform to polar coordinates $r=sqrtx^2+y^2$ and $phi=arctanleft(frac yxright)$, with
$$
dot r=fracxdot xr+fracydot yr=-1-frac xr=-1-cosphi
$$
and
$$
dotphi=fracfracdot yx-fracdot xyy^21+left(frac yxright)^2=fracxdot y-ydot xx^2+y^2=frac yr^2=fracsinphir;.
$$
So we have
$$
fracmathrm drmathrm dphi=fracdot rdotphi=-rcdotfrac1+cosphisinphi=-rfraccosfracphi2sinfracphi2;.
$$
Dividing by $r$ and integrating both sides yields
$$
log r=C-2logsinfracphi2;.
$$
The initial value is $rleft(fracpi2right)=1$, which yields $C=-log 2$, so
$$
r=frac12sin^2fracphi2;.
$$
In the limit $ttoinfty$ we have $phitopi$, and thus $rtofrac12$.
answered Jul 15 at 6:43
joriki
165k10180328
The problem becomes more managable if you transform to a coordinate system where $mathbf B$ is at rest. Then $mathbf A$ has velocity
$$
pmatrixdot x\dot y=-pmatrixfrac xr+1\frac yr;.
$$
Now transform to polar coordinates $r=sqrtx^2+y^2$ and $phi=arctanleft(frac yxright)$, with
$$
dot r=fracxdot xr+fracydot yr=-1-frac xr=-1-cosphi
$$
and
$$
dotphi=fracfracdot yx-fracdot xyy^21+left(frac yxright)^2=fracxdot y-ydot xx^2+y^2=frac yr^2=fracsinphir;.
$$
So we have
$$
fracmathrm drmathrm dphi=fracdot rdotphi=-rcdotfrac1+cosphisinphi=-rfraccosfracphi2sinfracphi2;.
$$
Dividing by $r$ and integrating both sides yields
$$
log r=C-2logsinfracphi2;.
$$
The initial value is $rleft(fracpi2right)=1$, which yields $C=-log 2$, so
$$
r=frac12sin^2fracphi2;.
$$
In the limit $ttoinfty$ we have $phitopi$, and thus $rtofrac12$.
answered Jul 15 at 6:43
joriki
165k10180328
edited Jul 15 at 8:37
answered Jul 15 at 6:43
joriki
165k10180328
answered Jul 15 at 6:43
joriki
165k10180328
answered Jul 15 at 6:43
joriki
165k10180328
165k10180328
You mixed up $phi$ with $x$ towards the end
– Dylan
Jul 15 at 8:25
@Dylan: Indeed I did, thanks -- fixed.
– joriki
Jul 15 at 8:37
add a comment |Â
You mixed up $phi$ with $x$ towards the end
– Dylan
Jul 15 at 8:25
@Dylan: Indeed I did, thanks -- fixed.
– joriki
Jul 15 at 8:37
You mixed up $phi$ with $x$ towards the end
– Dylan
Jul 15 at 8:25
You mixed up $phi$ with $x$ towards the end
– Dylan
Jul 15 at 8:25
@Dylan: Indeed I did, thanks -- fixed.
– joriki
Jul 15 at 8:37
@Dylan: Indeed I did, thanks -- fixed.
– joriki
Jul 15 at 8:37
add a comment |Â
up vote
2
down vote
There is a mismatch of notations between your wording and the wording in https://gilkalai.wordpress.com/2018/06/29/test-your-intuition-35-what-is-the-limiting-distance/ . I will refer to the notations from the link instead of the notations in the above question because the wording in the link is more clear.
The figure below shows the result of numerical simulation.
The result isn't $1$ but $frac12$. This is proved thanks to analytical solving.
The coordinates of the moving points are B$(x,y)$ and A$(t,0)$.
Since the absolute velocity of B is $=1$ in the direction of $overrightarrowBA$ :
$$begincases
fracdxdt=fract-xsqrt(t-x)^2+y^2 \
fracdydt=frac-ysqrt(t-x)^2+y^2
endcases$$
$fracdydx=frac-yt-x$
Change of variable : $x=u+t$
$fracdxdt=fracdudt+1=frac-usqrtu^2+y^2$
$fracdudt=-fracusqrtu^2+y^2-1$
$fracdydt=fracdydufracdudt=fracdyduleft(-fracusqrtu^2+y^2-1right)=frac-ysqrtu^2+y^2$
$$yfracdudy=sqrtu^2+y^2+u$$
This is a fist order homogeneous ODE easy to solve. The general solution is :
$$u=ysinhleft(c+ln(y)right)$$
The initial conditions $t=0$ , $x=0$ , $y=1$ , $u=x-t=0$ which implies $c=0$ thus :
$$u=ysinhleft(ln(y)right)=yfracy-frac1y2=fracy^2-12$$
$$t-x=-u=frac1-y^22$$
When $xtoinfty quad;quad yto 0quad;quad (t-x)tofrac12$.
$$|AB[=sqrt(x-t)^2+y^2tofrac12$$
answered Jul 15 at 10:54
JJacquelin
40.1k21649
I think in the displayed equation right before "This is a separable ODE" you're missing a factor $sqrtu^2+y^2$? The one in the denominator of the right-hand side of the previous equation. (By the way, my comment "The distance is $1$" under the question referred to Moti's comment that the (original) distance between the points is $sqrt2$ and not $1$; I wasn't talking about the limit distance there.)
– joriki
Jul 15 at 11:04
This was fixed meanwhile. Referring of the wording of the problem from the link, the initial distance is 1. I don't consider the wording of the mich's question.
– JJacquelin
Jul 15 at 11:20
add a comment |Â
up vote
2
down vote
There is a mismatch of notations between your wording and the wording in https://gilkalai.wordpress.com/2018/06/29/test-your-intuition-35-what-is-the-limiting-distance/ . I will refer to the notations from the link instead of the notations in the above question because the wording in the link is more clear.
The figure below shows the result of numerical simulation.
The result isn't $1$ but $frac12$. This is proved thanks to analytical solving.
The coordinates of the moving points are B$(x,y)$ and A$(t,0)$.
Since the absolute velocity of B is $=1$ in the direction of $overrightarrowBA$ :
$$begincases
fracdxdt=fract-xsqrt(t-x)^2+y^2 \
fracdydt=frac-ysqrt(t-x)^2+y^2
endcases$$
$fracdydx=frac-yt-x$
Change of variable : $x=u+t$
$fracdxdt=fracdudt+1=frac-usqrtu^2+y^2$
$fracdudt=-fracusqrtu^2+y^2-1$
$fracdydt=fracdydufracdudt=fracdyduleft(-fracusqrtu^2+y^2-1right)=frac-ysqrtu^2+y^2$
$$yfracdudy=sqrtu^2+y^2+u$$
This is a fist order homogeneous ODE easy to solve. The general solution is :
$$u=ysinhleft(c+ln(y)right)$$
The initial conditions $t=0$ , $x=0$ , $y=1$ , $u=x-t=0$ which implies $c=0$ thus :
$$u=ysinhleft(ln(y)right)=yfracy-frac1y2=fracy^2-12$$
$$t-x=-u=frac1-y^22$$
When $xtoinfty quad;quad yto 0quad;quad (t-x)tofrac12$.
$$|AB[=sqrt(x-t)^2+y^2tofrac12$$
answered Jul 15 at 10:54
JJacquelin
40.1k21649
I think in the displayed equation right before "This is a separable ODE" you're missing a factor $sqrtu^2+y^2$? The one in the denominator of the right-hand side of the previous equation. (By the way, my comment "The distance is $1$" under the question referred to Moti's comment that the (original) distance between the points is $sqrt2$ and not $1$; I wasn't talking about the limit distance there.)
– joriki
Jul 15 at 11:04
This was fixed meanwhile. Referring of the wording of the problem from the link, the initial distance is 1. I don't consider the wording of the mich's question.
– JJacquelin
Jul 15 at 11:20
add a comment |Â
up vote
2
down vote
up vote
2
down vote
There is a mismatch of notations between your wording and the wording in https://gilkalai.wordpress.com/2018/06/29/test-your-intuition-35-what-is-the-limiting-distance/ . I will refer to the notations from the link instead of the notations in the above question because the wording in the link is more clear.
The figure below shows the result of numerical simulation.
The result isn't $1$ but $frac12$. This is proved thanks to analytical solving.
The coordinates of the moving points are B$(x,y)$ and A$(t,0)$.
Since the absolute velocity of B is $=1$ in the direction of $overrightarrowBA$ :
$$begincases
fracdxdt=fract-xsqrt(t-x)^2+y^2 \
fracdydt=frac-ysqrt(t-x)^2+y^2
endcases$$
$fracdydx=frac-yt-x$
Change of variable : $x=u+t$
$fracdxdt=fracdudt+1=frac-usqrtu^2+y^2$
$fracdudt=-fracusqrtu^2+y^2-1$
$fracdydt=fracdydufracdudt=fracdyduleft(-fracusqrtu^2+y^2-1right)=frac-ysqrtu^2+y^2$
$$yfracdudy=sqrtu^2+y^2+u$$
This is a fist order homogeneous ODE easy to solve. The general solution is :
$$u=ysinhleft(c+ln(y)right)$$
The initial conditions $t=0$ , $x=0$ , $y=1$ , $u=x-t=0$ which implies $c=0$ thus :
$$u=ysinhleft(ln(y)right)=yfracy-frac1y2=fracy^2-12$$
$$t-x=-u=frac1-y^22$$
When $xtoinfty quad;quad yto 0quad;quad (t-x)tofrac12$.
$$|AB[=sqrt(x-t)^2+y^2tofrac12$$
answered Jul 15 at 10:54
JJacquelin
40.1k21649
There is a mismatch of notations between your wording and the wording in https://gilkalai.wordpress.com/2018/06/29/test-your-intuition-35-what-is-the-limiting-distance/ . I will refer to the notations from the link instead of the notations in the above question because the wording in the link is more clear.
The figure below shows the result of numerical simulation.
The result isn't $1$ but $frac12$. This is proved thanks to analytical solving.
The coordinates of the moving points are B$(x,y)$ and A$(t,0)$.
Since the absolute velocity of B is $=1$ in the direction of $overrightarrowBA$ :
$$begincases
fracdxdt=fract-xsqrt(t-x)^2+y^2 \
fracdydt=frac-ysqrt(t-x)^2+y^2
endcases$$
$fracdydx=frac-yt-x$
Change of variable : $x=u+t$
$fracdxdt=fracdudt+1=frac-usqrtu^2+y^2$
$fracdudt=-fracusqrtu^2+y^2-1$
$fracdydt=fracdydufracdudt=fracdyduleft(-fracusqrtu^2+y^2-1right)=frac-ysqrtu^2+y^2$
$$yfracdudy=sqrtu^2+y^2+u$$
This is a fist order homogeneous ODE easy to solve. The general solution is :
$$u=ysinhleft(c+ln(y)right)$$
The initial conditions $t=0$ , $x=0$ , $y=1$ , $u=x-t=0$ which implies $c=0$ thus :
$$u=ysinhleft(ln(y)right)=yfracy-frac1y2=fracy^2-12$$
$$t-x=-u=frac1-y^22$$
When $xtoinfty quad;quad yto 0quad;quad (t-x)tofrac12$.
$$|AB[=sqrt(x-t)^2+y^2tofrac12$$
answered Jul 15 at 10:54
JJacquelin
40.1k21649
edited Jul 15 at 17:52
answered Jul 15 at 10:54
JJacquelin
40.1k21649
answered Jul 15 at 10:54
JJacquelin
40.1k21649
answered Jul 15 at 10:54
JJacquelin
40.1k21649
40.1k21649
I think in the displayed equation right before "This is a separable ODE" you're missing a factor $sqrtu^2+y^2$? The one in the denominator of the right-hand side of the previous equation. (By the way, my comment "The distance is $1$" under the question referred to Moti's comment that the (original) distance between the points is $sqrt2$ and not $1$; I wasn't talking about the limit distance there.)
– joriki
Jul 15 at 11:04
This was fixed meanwhile. Referring of the wording of the problem from the link, the initial distance is 1. I don't consider the wording of the mich's question.
– JJacquelin
Jul 15 at 11:20
add a comment |Â
I think in the displayed equation right before "This is a separable ODE" you're missing a factor $sqrtu^2+y^2$? The one in the denominator of the right-hand side of the previous equation. (By the way, my comment "The distance is $1$" under the question referred to Moti's comment that the (original) distance between the points is $sqrt2$ and not $1$; I wasn't talking about the limit distance there.)
– joriki
Jul 15 at 11:04
This was fixed meanwhile. Referring of the wording of the problem from the link, the initial distance is 1. I don't consider the wording of the mich's question.
– JJacquelin
Jul 15 at 11:20
I think in the displayed equation right before "This is a separable ODE" you're missing a factor $sqrtu^2+y^2$? The one in the denominator of the right-hand side of the previous equation. (By the way, my comment "The distance is $1$" under the question referred to Moti's comment that the (original) distance between the points is $sqrt2$ and not $1$; I wasn't talking about the limit distance there.)
– joriki
Jul 15 at 11:04
I think in the displayed equation right before "This is a separable ODE" you're missing a factor $sqrtu^2+y^2$? The one in the denominator of the right-hand side of the previous equation. (By the way, my comment "The distance is $1$" under the question referred to Moti's comment that the (original) distance between the points is $sqrt2$ and not $1$; I wasn't talking about the limit distance there.)
– joriki
Jul 15 at 11:04
This was fixed meanwhile. Referring of the wording of the problem from the link, the initial distance is 1. I don't consider the wording of the mich's question.
– JJacquelin
Jul 15 at 11:20
This was fixed meanwhile. Referring of the wording of the problem from the link, the initial distance is 1. I don't consider the wording of the mich's question.
– JJacquelin
Jul 15 at 11:20
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My gut tells me that it's 1...
– Sambo
Jul 15 at 3:43
I tried writing out the differential equations, which failed because they are not linear and I'm not very good at that. Per your guess, it's definitely lower than 1 (since B is running away at a sub-optimal direction and the distance starts at 1). It will also not reach 0.
– mich
Jul 15 at 3:44
The distance between the points is not 1, but $sqrt2$. I would say that the distance is $sqrt2-1$.
– Moti
Jul 15 at 5:59
@Moti: The distance is $1$.
– joriki
Jul 15 at 6:07
You are right... but I still think that the distance is as stated.
– Moti
Jul 15 at 6:23