Differential equation of Rabbit Fox Problem
$begingroup$
A rabbit is hiding at $(0,1)$ and a fox is standing at the point $(0,-1)$ and a big tree is at $(0,0)$ so that rabbit is safe from the attack of Fox.
Now the Fox moves in horizontal direction at a speed of $1$ unit along the line $y=-1$ towards negative infinity from $(0,-1)$ .
To avoid Detection, rabbit will run towards its right but in a curved path at a constant speed of $2$ units so as to keep Tree between itself and Fox.
Form the Differential equation of the path of Rabbit.
My try:
Let any point on the path of rabbit be $P(r cos theta, r sin theta)$
Also any point on path of Fox is $Q(a,-1)$
according to my thinking the points $P$,$O$ and $Q$ are always collinear.
So
$$begin{vmatrix}
0 & 0 &1 \
a & - 1&1 \
r cos theta&r sin theta & 1
end{vmatrix}=0$$
which gives $$a=-cot theta$$
So the point $Q$ is $Q(-cot theta, -1)$
any clue from here?
algebra-precalculus ordinary-differential-equations analytic-geometry
asked Dec 7 '18 at 12:05
Umesh shankarUmesh shankar
2,61931219
$endgroup$
add a comment |
$begingroup$
A rabbit is hiding at $(0,1)$ and a fox is standing at the point $(0,-1)$ and a big tree is at $(0,0)$ so that rabbit is safe from the attack of Fox.
Now the Fox moves in horizontal direction at a speed of $1$ unit along the line $y=-1$ towards negative infinity from $(0,-1)$ .
To avoid Detection, rabbit will run towards its right but in a curved path at a constant speed of $2$ units so as to keep Tree between itself and Fox.
Form the Differential equation of the path of Rabbit.
My try:
Let any point on the path of rabbit be $P(r cos theta, r sin theta)$
Also any point on path of Fox is $Q(a,-1)$
according to my thinking the points $P$,$O$ and $Q$ are always collinear.
So
$$begin{vmatrix}
0 & 0 &1 \
a & - 1&1 \
r cos theta&r sin theta & 1
end{vmatrix}=0$$
which gives $$a=-cot theta$$
So the point $Q$ is $Q(-cot theta, -1)$
any clue from here?
algebra-precalculus ordinary-differential-equations analytic-geometry
asked Dec 7 '18 at 12:05
Umesh shankarUmesh shankar
2,61931219
$endgroup$
$begingroup$
(0,1) is not on the negative x-axis, so it doesn't make any sense to say "the Fox moves in horizontal direction at a speed of 1 unit along negative x-axis from (0,1)."
$endgroup$
– user10354138
Dec 7 '18 at 12:09
$begingroup$
ok i edited accordingly
$endgroup$
– Umesh shankar
Dec 7 '18 at 12:11
$begingroup$
The points $(1,0)-"Rabit"$, $(0,0)-"Tree"$ and $(0,1)-"Fox"$ are not aligned.
$endgroup$
– Cesareo
Dec 7 '18 at 12:27
$begingroup$
I'm sorry but I really don't get it. The rabbit is not in the line of sight of the fox at the beginning. Then the fox moves away from the rabbit, and so does the rabbit from the fox. The only way to make sense of the problem is to make the tree not a point
$endgroup$
– caverac
Dec 7 '18 at 12:45
$begingroup$
@all i am extremely sorry ,now i have modified the question with proper details
$endgroup$
– Umesh shankar
Dec 7 '18 at 13:03
add a comment |
$begingroup$
A rabbit is hiding at $(0,1)$ and a fox is standing at the point $(0,-1)$ and a big tree is at $(0,0)$ so that rabbit is safe from the attack of Fox.
Now the Fox moves in horizontal direction at a speed of $1$ unit along the line $y=-1$ towards negative infinity from $(0,-1)$ .
To avoid Detection, rabbit will run towards its right but in a curved path at a constant speed of $2$ units so as to keep Tree between itself and Fox.
Form the Differential equation of the path of Rabbit.
My try:
Let any point on the path of rabbit be $P(r cos theta, r sin theta)$
Also any point on path of Fox is $Q(a,-1)$
according to my thinking the points $P$,$O$ and $Q$ are always collinear.
So
$$begin{vmatrix}
0 & 0 &1 \
a & - 1&1 \
r cos theta&r sin theta & 1
end{vmatrix}=0$$
which gives $$a=-cot theta$$
So the point $Q$ is $Q(-cot theta, -1)$
any clue from here?
algebra-precalculus ordinary-differential-equations analytic-geometry
asked Dec 7 '18 at 12:05
Umesh shankarUmesh shankar
2,61931219
$endgroup$
A rabbit is hiding at $(0,1)$ and a fox is standing at the point $(0,-1)$ and a big tree is at $(0,0)$ so that rabbit is safe from the attack of Fox.
Now the Fox moves in horizontal direction at a speed of $1$ unit along the line $y=-1$ towards negative infinity from $(0,-1)$ .
To avoid Detection, rabbit will run towards its right but in a curved path at a constant speed of $2$ units so as to keep Tree between itself and Fox.
Form the Differential equation of the path of Rabbit.
My try:
Let any point on the path of rabbit be $P(r cos theta, r sin theta)$
Also any point on path of Fox is $Q(a,-1)$
according to my thinking the points $P$,$O$ and $Q$ are always collinear.
So
$$begin{vmatrix}
0 & 0 &1 \
a & - 1&1 \
r cos theta&r sin theta & 1
end{vmatrix}=0$$
which gives $$a=-cot theta$$
So the point $Q$ is $Q(-cot theta, -1)$
any clue from here?
algebra-precalculus ordinary-differential-equations analytic-geometry
algebra-precalculus ordinary-differential-equations analytic-geometry
asked Dec 7 '18 at 12:05
Umesh shankarUmesh shankar
2,61931219
asked Dec 7 '18 at 12:05
Umesh shankarUmesh shankar
2,61931219
edited Dec 7 '18 at 13:02
Umesh shankar
asked Dec 7 '18 at 12:05
Umesh shankarUmesh shankar
2,61931219
asked Dec 7 '18 at 12:05
Umesh shankarUmesh shankar
2,61931219
asked Dec 7 '18 at 12:05
Umesh shankarUmesh shankar
2,61931219
2,61931219
$begingroup$
(0,1) is not on the negative x-axis, so it doesn't make any sense to say "the Fox moves in horizontal direction at a speed of 1 unit along negative x-axis from (0,1)."
$endgroup$
– user10354138
Dec 7 '18 at 12:09
$begingroup$
ok i edited accordingly
$endgroup$
– Umesh shankar
Dec 7 '18 at 12:11
$begingroup$
The points $(1,0)-"Rabit"$, $(0,0)-"Tree"$ and $(0,1)-"Fox"$ are not aligned.
$endgroup$
– Cesareo
Dec 7 '18 at 12:27
$begingroup$
I'm sorry but I really don't get it. The rabbit is not in the line of sight of the fox at the beginning. Then the fox moves away from the rabbit, and so does the rabbit from the fox. The only way to make sense of the problem is to make the tree not a point
$endgroup$
– caverac
Dec 7 '18 at 12:45
$begingroup$
@all i am extremely sorry ,now i have modified the question with proper details
$endgroup$
– Umesh shankar
Dec 7 '18 at 13:03
add a comment |
$begingroup$
(0,1) is not on the negative x-axis, so it doesn't make any sense to say "the Fox moves in horizontal direction at a speed of 1 unit along negative x-axis from (0,1)."
$endgroup$
– user10354138
Dec 7 '18 at 12:09
$begingroup$
ok i edited accordingly
$endgroup$
– Umesh shankar
Dec 7 '18 at 12:11
$begingroup$
The points $(1,0)-"Rabit"$, $(0,0)-"Tree"$ and $(0,1)-"Fox"$ are not aligned.
$endgroup$
– Cesareo
Dec 7 '18 at 12:27
$begingroup$
I'm sorry but I really don't get it. The rabbit is not in the line of sight of the fox at the beginning. Then the fox moves away from the rabbit, and so does the rabbit from the fox. The only way to make sense of the problem is to make the tree not a point
$endgroup$
– caverac
Dec 7 '18 at 12:45
$begingroup$
@all i am extremely sorry ,now i have modified the question with proper details
$endgroup$
– Umesh shankar
Dec 7 '18 at 13:03
$begingroup$
(0,1) is not on the negative x-axis, so it doesn't make any sense to say "the Fox moves in horizontal direction at a speed of 1 unit along negative x-axis from (0,1)."
$endgroup$
– user10354138
Dec 7 '18 at 12:09
$begingroup$
(0,1) is not on the negative x-axis, so it doesn't make any sense to say "the Fox moves in horizontal direction at a speed of 1 unit along negative x-axis from (0,1)."
$endgroup$
– user10354138
Dec 7 '18 at 12:09
$begingroup$
ok i edited accordingly
$endgroup$
– Umesh shankar
Dec 7 '18 at 12:11
$begingroup$
ok i edited accordingly
$endgroup$
– Umesh shankar
Dec 7 '18 at 12:11
$begingroup$
The points $(1,0)-"Rabit"$, $(0,0)-"Tree"$ and $(0,1)-"Fox"$ are not aligned.
$endgroup$
– Cesareo
Dec 7 '18 at 12:27
$begingroup$
The points $(1,0)-"Rabit"$, $(0,0)-"Tree"$ and $(0,1)-"Fox"$ are not aligned.
$endgroup$
– Cesareo
Dec 7 '18 at 12:27
$begingroup$
I'm sorry but I really don't get it. The rabbit is not in the line of sight of the fox at the beginning. Then the fox moves away from the rabbit, and so does the rabbit from the fox. The only way to make sense of the problem is to make the tree not a point
$endgroup$
– caverac
Dec 7 '18 at 12:45
$begingroup$
I'm sorry but I really don't get it. The rabbit is not in the line of sight of the fox at the beginning. Then the fox moves away from the rabbit, and so does the rabbit from the fox. The only way to make sense of the problem is to make the tree not a point
$endgroup$
– caverac
Dec 7 '18 at 12:45
$begingroup$
@all i am extremely sorry ,now i have modified the question with proper details
$endgroup$
– Umesh shankar
Dec 7 '18 at 13:03
$begingroup$
@all i am extremely sorry ,now i have modified the question with proper details
$endgroup$
– Umesh shankar
Dec 7 '18 at 13:03
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The movement is described by
$$
dot x^2+dot y^2 = v_r^2\
frac{x}{y} = frac{-v_f t}{-1}
$$
here $v_r = 2, v_f = 1$
Attached the rabbit evasion curve in green
In dashed red the opposition at $t = 0.5$
The MATHEMATICA script
parms = {vf -> 1, vr -> 2};
equs = {x'[t]^2 + y'[t]^2 == vr^2, x'[t] == D[y[t] vf t, t]} /. parms
tmax = 1;
soly = NDSolve[{equs, x[0] == 0, y[0] == 1}, {x, y}, {t, 0, tmax}][[2]];
gr1 = ParametricPlot[Evaluate[{x[t], y[t]} /. soly], {t, 0, tmax}, PlotStyle -> {Green, Thick}];
gr2 = ParametricPlot[{-t, -1}, {t, 0, tmax}, PlotStyle -> {Blue, Thick}];
p1 = Evaluate[{x[t], y[t]} /. soly] /. {t -> 0.5};
p2 = {-1 t, -1} /. {t -> 0.5};
p0 = {0, 0};
grp1 = Graphics[{Red, Disk[p1, 0.02]}];
grp2 = Graphics[{Red, Disk[p2, 0.02]}];
grp0 = Graphics[{Red, Disk[p0, 0.02]}];
grp3 = ListLinePlot[{p1, p2}, PlotStyle -> {Red, Dashed}];
Show[gr1, gr2, grp1, grp2, grp3, grp0, PlotRange -> All, AxesOrigin -> {0, 0}]
answered Dec 7 '18 at 18:54
CesareoCesareo
8,6243516
$endgroup$
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
$begingroup$
The movement is described by
$$
dot x^2+dot y^2 = v_r^2\
frac{x}{y} = frac{-v_f t}{-1}
$$
here $v_r = 2, v_f = 1$
Attached the rabbit evasion curve in green
In dashed red the opposition at $t = 0.5$
The MATHEMATICA script
parms = {vf -> 1, vr -> 2};
equs = {x'[t]^2 + y'[t]^2 == vr^2, x'[t] == D[y[t] vf t, t]} /. parms
tmax = 1;
soly = NDSolve[{equs, x[0] == 0, y[0] == 1}, {x, y}, {t, 0, tmax}][[2]];
gr1 = ParametricPlot[Evaluate[{x[t], y[t]} /. soly], {t, 0, tmax}, PlotStyle -> {Green, Thick}];
gr2 = ParametricPlot[{-t, -1}, {t, 0, tmax}, PlotStyle -> {Blue, Thick}];
p1 = Evaluate[{x[t], y[t]} /. soly] /. {t -> 0.5};
p2 = {-1 t, -1} /. {t -> 0.5};
p0 = {0, 0};
grp1 = Graphics[{Red, Disk[p1, 0.02]}];
grp2 = Graphics[{Red, Disk[p2, 0.02]}];
grp0 = Graphics[{Red, Disk[p0, 0.02]}];
grp3 = ListLinePlot[{p1, p2}, PlotStyle -> {Red, Dashed}];
Show[gr1, gr2, grp1, grp2, grp3, grp0, PlotRange -> All, AxesOrigin -> {0, 0}]
answered Dec 7 '18 at 18:54
CesareoCesareo
8,6243516
$endgroup$
add a comment |
$begingroup$
The movement is described by
$$
dot x^2+dot y^2 = v_r^2\
frac{x}{y} = frac{-v_f t}{-1}
$$
here $v_r = 2, v_f = 1$
Attached the rabbit evasion curve in green
In dashed red the opposition at $t = 0.5$
The MATHEMATICA script
parms = {vf -> 1, vr -> 2};
equs = {x'[t]^2 + y'[t]^2 == vr^2, x'[t] == D[y[t] vf t, t]} /. parms
tmax = 1;
soly = NDSolve[{equs, x[0] == 0, y[0] == 1}, {x, y}, {t, 0, tmax}][[2]];
gr1 = ParametricPlot[Evaluate[{x[t], y[t]} /. soly], {t, 0, tmax}, PlotStyle -> {Green, Thick}];
gr2 = ParametricPlot[{-t, -1}, {t, 0, tmax}, PlotStyle -> {Blue, Thick}];
p1 = Evaluate[{x[t], y[t]} /. soly] /. {t -> 0.5};
p2 = {-1 t, -1} /. {t -> 0.5};
p0 = {0, 0};
grp1 = Graphics[{Red, Disk[p1, 0.02]}];
grp2 = Graphics[{Red, Disk[p2, 0.02]}];
grp0 = Graphics[{Red, Disk[p0, 0.02]}];
grp3 = ListLinePlot[{p1, p2}, PlotStyle -> {Red, Dashed}];
Show[gr1, gr2, grp1, grp2, grp3, grp0, PlotRange -> All, AxesOrigin -> {0, 0}]
answered Dec 7 '18 at 18:54
CesareoCesareo
8,6243516
$endgroup$
add a comment |
$begingroup$
The movement is described by
$$
dot x^2+dot y^2 = v_r^2\
frac{x}{y} = frac{-v_f t}{-1}
$$
here $v_r = 2, v_f = 1$
Attached the rabbit evasion curve in green
In dashed red the opposition at $t = 0.5$
The MATHEMATICA script
parms = {vf -> 1, vr -> 2};
equs = {x'[t]^2 + y'[t]^2 == vr^2, x'[t] == D[y[t] vf t, t]} /. parms
tmax = 1;
soly = NDSolve[{equs, x[0] == 0, y[0] == 1}, {x, y}, {t, 0, tmax}][[2]];
gr1 = ParametricPlot[Evaluate[{x[t], y[t]} /. soly], {t, 0, tmax}, PlotStyle -> {Green, Thick}];
gr2 = ParametricPlot[{-t, -1}, {t, 0, tmax}, PlotStyle -> {Blue, Thick}];
p1 = Evaluate[{x[t], y[t]} /. soly] /. {t -> 0.5};
p2 = {-1 t, -1} /. {t -> 0.5};
p0 = {0, 0};
grp1 = Graphics[{Red, Disk[p1, 0.02]}];
grp2 = Graphics[{Red, Disk[p2, 0.02]}];
grp0 = Graphics[{Red, Disk[p0, 0.02]}];
grp3 = ListLinePlot[{p1, p2}, PlotStyle -> {Red, Dashed}];
Show[gr1, gr2, grp1, grp2, grp3, grp0, PlotRange -> All, AxesOrigin -> {0, 0}]
answered Dec 7 '18 at 18:54
CesareoCesareo
8,6243516
$endgroup$
The movement is described by
$$
dot x^2+dot y^2 = v_r^2\
frac{x}{y} = frac{-v_f t}{-1}
$$
here $v_r = 2, v_f = 1$
Attached the rabbit evasion curve in green
In dashed red the opposition at $t = 0.5$
The MATHEMATICA script
parms = {vf -> 1, vr -> 2};
equs = {x'[t]^2 + y'[t]^2 == vr^2, x'[t] == D[y[t] vf t, t]} /. parms
tmax = 1;
soly = NDSolve[{equs, x[0] == 0, y[0] == 1}, {x, y}, {t, 0, tmax}][[2]];
gr1 = ParametricPlot[Evaluate[{x[t], y[t]} /. soly], {t, 0, tmax}, PlotStyle -> {Green, Thick}];
gr2 = ParametricPlot[{-t, -1}, {t, 0, tmax}, PlotStyle -> {Blue, Thick}];
p1 = Evaluate[{x[t], y[t]} /. soly] /. {t -> 0.5};
p2 = {-1 t, -1} /. {t -> 0.5};
p0 = {0, 0};
grp1 = Graphics[{Red, Disk[p1, 0.02]}];
grp2 = Graphics[{Red, Disk[p2, 0.02]}];
grp0 = Graphics[{Red, Disk[p0, 0.02]}];
grp3 = ListLinePlot[{p1, p2}, PlotStyle -> {Red, Dashed}];
Show[gr1, gr2, grp1, grp2, grp3, grp0, PlotRange -> All, AxesOrigin -> {0, 0}]
answered Dec 7 '18 at 18:54
CesareoCesareo
8,6243516
edited Dec 8 '18 at 15:52
answered Dec 7 '18 at 18:54
CesareoCesareo
8,6243516
answered Dec 7 '18 at 18:54
CesareoCesareo
8,6243516
answered Dec 7 '18 at 18:54
CesareoCesareo
8,6243516
8,6243516
add a comment |
add a comment |
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$begingroup$
(0,1) is not on the negative x-axis, so it doesn't make any sense to say "the Fox moves in horizontal direction at a speed of 1 unit along negative x-axis from (0,1)."
$endgroup$
– user10354138
Dec 7 '18 at 12:09
$begingroup$
ok i edited accordingly
$endgroup$
– Umesh shankar
Dec 7 '18 at 12:11
$begingroup$
The points $(1,0)-"Rabit"$, $(0,0)-"Tree"$ and $(0,1)-"Fox"$ are not aligned.
$endgroup$
– Cesareo
Dec 7 '18 at 12:27
$begingroup$
I'm sorry but I really don't get it. The rabbit is not in the line of sight of the fox at the beginning. Then the fox moves away from the rabbit, and so does the rabbit from the fox. The only way to make sense of the problem is to make the tree not a point
$endgroup$
– caverac
Dec 7 '18 at 12:45
$begingroup$
@all i am extremely sorry ,now i have modified the question with proper details
$endgroup$
– Umesh shankar
Dec 7 '18 at 13:03