Calculating the sum of the lengths and plotting all curves on the same map
2
$begingroup$
I have a curve like below;
ParametricPlot[
FromPolarCoordinates[{Exp[(t + 0)*0.5], t}] // Evaluate, {t, 0,
Pi - 0}, PlotRange -> All]
This curve rotates around z axis. Therefore for a rotation of pi/2 gives something like this:
ParametricPlot[
FromPolarCoordinates[{Exp[(t + Pi/2)*0.5], t}] //
Evaluate, {t, -Pi/2, Pi - Pi/2}, PlotRange -> All]
What I want is to plot all the curves in a specified circle when it rotates with 3.6 degrees. And I want the sum of curves in the circle.
I tried this code but it didn't work:
ParametricPlot[Sum[HeavisideTheta[1 -
((Exp[(t + Pi*i/50)*0.5]*Cos[t] -
1)^2 + (Exp[(t + Pi*i/50)*0.5]*Sin[t])^2)]*
FromPolarCoordinates[{Exp[(t + Pi*i/50)*0.5], t}], {i, 1, 100}] //
Evaluate, {t, -Pi*i/50, Pi - Pi*i/50}, PlotRange -> All]
I also tried this code:
ParametricPlot[ Evaluate@Sum[ HeavisideTheta[ 1 - ((Exp[(t + Pi*i/50)*0.5]Cos[t] - 1)^2 + (Exp[(t + Pii/50)*0.5]Sin[t])^2)] FromPolarCoordinates[{Exp[(t + Pi*i/50)*0.5], t}], {i, 1, 100}], {t, -Pi, Pi}, PlotRange -> All]
But it gives something like below:
What I need is something like this:
And the sum of those curves in the circle area.
How should I write the code to achieve what I want?
plotting parametric-functions
asked Feb 17 at 13:18
Alper91Alper91
1355
$endgroup$
add a comment |
2
$begingroup$
I have a curve like below;
ParametricPlot[
FromPolarCoordinates[{Exp[(t + 0)*0.5], t}] // Evaluate, {t, 0,
Pi - 0}, PlotRange -> All]
This curve rotates around z axis. Therefore for a rotation of pi/2 gives something like this:
ParametricPlot[
FromPolarCoordinates[{Exp[(t + Pi/2)*0.5], t}] //
Evaluate, {t, -Pi/2, Pi - Pi/2}, PlotRange -> All]
What I want is to plot all the curves in a specified circle when it rotates with 3.6 degrees. And I want the sum of curves in the circle.
I tried this code but it didn't work:
ParametricPlot[Sum[HeavisideTheta[1 -
((Exp[(t + Pi*i/50)*0.5]*Cos[t] -
1)^2 + (Exp[(t + Pi*i/50)*0.5]*Sin[t])^2)]*
FromPolarCoordinates[{Exp[(t + Pi*i/50)*0.5], t}], {i, 1, 100}] //
Evaluate, {t, -Pi*i/50, Pi - Pi*i/50}, PlotRange -> All]
I also tried this code:
ParametricPlot[ Evaluate@Sum[ HeavisideTheta[ 1 - ((Exp[(t + Pi*i/50)*0.5]Cos[t] - 1)^2 + (Exp[(t + Pii/50)*0.5]Sin[t])^2)] FromPolarCoordinates[{Exp[(t + Pi*i/50)*0.5], t}], {i, 1, 100}], {t, -Pi, Pi}, PlotRange -> All]
But it gives something like below:
What I need is something like this:
And the sum of those curves in the circle area.
How should I write the code to achieve what I want?
plotting parametric-functions
asked Feb 17 at 13:18
Alper91Alper91
1355
$endgroup$
1
$begingroup$
Something like this:ParametricPlot[ Table[FromPolarCoordinates[{Exp[(t + 2 k Pi/100)*0.5], t}], {k, 1, 100, 5}], {t, -Pi, Pi}, PlotRange -> All, PlotStyle -> Red, RegionFunction -> Function[{x, y}, (x - 3)^2 + y^2 < 9]] // Show[#, ContourPlot[(x - 3)^2 + y^2 == 9, {x, 0, 6}, {y, -3, 3}, ContourStyle -> Blue]] &?
$endgroup$
– corey979
Feb 17 at 13:47
$begingroup$
@kglr nope. I editted my answer and showed what your solution is like.
$endgroup$
– Alper91
Feb 17 at 13:47
$begingroup$
@corey979 Yes! Yes! Exactly like that one and the sum of the curves' length within the circle
$endgroup$
– Alper91
Feb 17 at 13:49
$begingroup$
@corey979 However, your solution is not the rotation case. It is the case where radial growth of the spiral increases.
$endgroup$
– Alper91
Feb 17 at 13:57
$begingroup$
@corey979 I am sorry corey what you provide was right, if I know where my limits are. And this is the code I wanted like you suggested: ParametricPlot[ Table[FromPolarCoordinates[{Exp[(t + 2 k Pi/100)*0.5], t}], {k, 1, 100, 5}], {t, -Pi, Pi}, PlotRange -> All, PlotStyle -> Red, RegionFunction -> Function[{x, y}, (x - (Exp[Pi*0.5] + 1)/2)^2 + y^2 < ((Exp[Pi*0.5] - 1)/2)^2]] // Show[#, ContourPlot[(x - (Exp[Pi*0.5] + 1)/2)^2 + y^2 == ((Exp[Pi*0.5] - 1)/2)^2, {x, 0, 6}, {y, -3, 3}, ContourStyle -> Blue]] & However, I need also sum of the lengths
$endgroup$
– Alper91
Feb 17 at 14:07
add a comment |
2
2
2
$begingroup$
I have a curve like below;
ParametricPlot[
FromPolarCoordinates[{Exp[(t + 0)*0.5], t}] // Evaluate, {t, 0,
Pi - 0}, PlotRange -> All]
This curve rotates around z axis. Therefore for a rotation of pi/2 gives something like this:
ParametricPlot[
FromPolarCoordinates[{Exp[(t + Pi/2)*0.5], t}] //
Evaluate, {t, -Pi/2, Pi - Pi/2}, PlotRange -> All]
What I want is to plot all the curves in a specified circle when it rotates with 3.6 degrees. And I want the sum of curves in the circle.
I tried this code but it didn't work:
ParametricPlot[Sum[HeavisideTheta[1 -
((Exp[(t + Pi*i/50)*0.5]*Cos[t] -
1)^2 + (Exp[(t + Pi*i/50)*0.5]*Sin[t])^2)]*
FromPolarCoordinates[{Exp[(t + Pi*i/50)*0.5], t}], {i, 1, 100}] //
Evaluate, {t, -Pi*i/50, Pi - Pi*i/50}, PlotRange -> All]
I also tried this code:
ParametricPlot[ Evaluate@Sum[ HeavisideTheta[ 1 - ((Exp[(t + Pi*i/50)*0.5]Cos[t] - 1)^2 + (Exp[(t + Pii/50)*0.5]Sin[t])^2)] FromPolarCoordinates[{Exp[(t + Pi*i/50)*0.5], t}], {i, 1, 100}], {t, -Pi, Pi}, PlotRange -> All]
But it gives something like below:
What I need is something like this:
And the sum of those curves in the circle area.
How should I write the code to achieve what I want?
plotting parametric-functions
asked Feb 17 at 13:18
Alper91Alper91
1355
$endgroup$
I have a curve like below;
ParametricPlot[
FromPolarCoordinates[{Exp[(t + 0)*0.5], t}] // Evaluate, {t, 0,
Pi - 0}, PlotRange -> All]
This curve rotates around z axis. Therefore for a rotation of pi/2 gives something like this:
ParametricPlot[
FromPolarCoordinates[{Exp[(t + Pi/2)*0.5], t}] //
Evaluate, {t, -Pi/2, Pi - Pi/2}, PlotRange -> All]
What I want is to plot all the curves in a specified circle when it rotates with 3.6 degrees. And I want the sum of curves in the circle.
I tried this code but it didn't work:
ParametricPlot[Sum[HeavisideTheta[1 -
((Exp[(t + Pi*i/50)*0.5]*Cos[t] -
1)^2 + (Exp[(t + Pi*i/50)*0.5]*Sin[t])^2)]*
FromPolarCoordinates[{Exp[(t + Pi*i/50)*0.5], t}], {i, 1, 100}] //
Evaluate, {t, -Pi*i/50, Pi - Pi*i/50}, PlotRange -> All]
I also tried this code:
ParametricPlot[ Evaluate@Sum[ HeavisideTheta[ 1 - ((Exp[(t + Pi*i/50)*0.5]Cos[t] - 1)^2 + (Exp[(t + Pii/50)*0.5]Sin[t])^2)] FromPolarCoordinates[{Exp[(t + Pi*i/50)*0.5], t}], {i, 1, 100}], {t, -Pi, Pi}, PlotRange -> All]
But it gives something like below:
What I need is something like this:
And the sum of those curves in the circle area.
How should I write the code to achieve what I want?
plotting parametric-functions
plotting parametric-functions
asked Feb 17 at 13:18
Alper91Alper91
1355
asked Feb 17 at 13:18
Alper91Alper91
1355
edited Feb 17 at 13:47
Alper91
asked Feb 17 at 13:18
Alper91Alper91
1355
asked Feb 17 at 13:18
Alper91Alper91
1355
asked Feb 17 at 13:18
Alper91Alper91
1355
1355
1
$begingroup$
Something like this:ParametricPlot[ Table[FromPolarCoordinates[{Exp[(t + 2 k Pi/100)*0.5], t}], {k, 1, 100, 5}], {t, -Pi, Pi}, PlotRange -> All, PlotStyle -> Red, RegionFunction -> Function[{x, y}, (x - 3)^2 + y^2 < 9]] // Show[#, ContourPlot[(x - 3)^2 + y^2 == 9, {x, 0, 6}, {y, -3, 3}, ContourStyle -> Blue]] &?
$endgroup$
– corey979
Feb 17 at 13:47
$begingroup$
@kglr nope. I editted my answer and showed what your solution is like.
$endgroup$
– Alper91
Feb 17 at 13:47
$begingroup$
@corey979 Yes! Yes! Exactly like that one and the sum of the curves' length within the circle
$endgroup$
– Alper91
Feb 17 at 13:49
$begingroup$
@corey979 However, your solution is not the rotation case. It is the case where radial growth of the spiral increases.
$endgroup$
– Alper91
Feb 17 at 13:57
$begingroup$
@corey979 I am sorry corey what you provide was right, if I know where my limits are. And this is the code I wanted like you suggested: ParametricPlot[ Table[FromPolarCoordinates[{Exp[(t + 2 k Pi/100)*0.5], t}], {k, 1, 100, 5}], {t, -Pi, Pi}, PlotRange -> All, PlotStyle -> Red, RegionFunction -> Function[{x, y}, (x - (Exp[Pi*0.5] + 1)/2)^2 + y^2 < ((Exp[Pi*0.5] - 1)/2)^2]] // Show[#, ContourPlot[(x - (Exp[Pi*0.5] + 1)/2)^2 + y^2 == ((Exp[Pi*0.5] - 1)/2)^2, {x, 0, 6}, {y, -3, 3}, ContourStyle -> Blue]] & However, I need also sum of the lengths
$endgroup$
– Alper91
Feb 17 at 14:07
add a comment |
1
$begingroup$
Something like this:ParametricPlot[ Table[FromPolarCoordinates[{Exp[(t + 2 k Pi/100)*0.5], t}], {k, 1, 100, 5}], {t, -Pi, Pi}, PlotRange -> All, PlotStyle -> Red, RegionFunction -> Function[{x, y}, (x - 3)^2 + y^2 < 9]] // Show[#, ContourPlot[(x - 3)^2 + y^2 == 9, {x, 0, 6}, {y, -3, 3}, ContourStyle -> Blue]] &?
$endgroup$
– corey979
Feb 17 at 13:47
$begingroup$
@kglr nope. I editted my answer and showed what your solution is like.
$endgroup$
– Alper91
Feb 17 at 13:47
$begingroup$
@corey979 Yes! Yes! Exactly like that one and the sum of the curves' length within the circle
$endgroup$
– Alper91
Feb 17 at 13:49
$begingroup$
@corey979 However, your solution is not the rotation case. It is the case where radial growth of the spiral increases.
$endgroup$
– Alper91
Feb 17 at 13:57
$begingroup$
@corey979 I am sorry corey what you provide was right, if I know where my limits are. And this is the code I wanted like you suggested: ParametricPlot[ Table[FromPolarCoordinates[{Exp[(t + 2 k Pi/100)*0.5], t}], {k, 1, 100, 5}], {t, -Pi, Pi}, PlotRange -> All, PlotStyle -> Red, RegionFunction -> Function[{x, y}, (x - (Exp[Pi*0.5] + 1)/2)^2 + y^2 < ((Exp[Pi*0.5] - 1)/2)^2]] // Show[#, ContourPlot[(x - (Exp[Pi*0.5] + 1)/2)^2 + y^2 == ((Exp[Pi*0.5] - 1)/2)^2, {x, 0, 6}, {y, -3, 3}, ContourStyle -> Blue]] & However, I need also sum of the lengths
$endgroup$
– Alper91
Feb 17 at 14:07
1
1
$begingroup$
Something like this:
ParametricPlot[ Table[FromPolarCoordinates[{Exp[(t + 2 k Pi/100)*0.5], t}], {k, 1, 100, 5}], {t, -Pi, Pi}, PlotRange -> All, PlotStyle -> Red, RegionFunction -> Function[{x, y}, (x - 3)^2 + y^2 < 9]] // Show[#, ContourPlot[(x - 3)^2 + y^2 == 9, {x, 0, 6}, {y, -3, 3}, ContourStyle -> Blue]] &?$endgroup$
– corey979
Feb 17 at 13:47
$begingroup$
Something like this:
ParametricPlot[ Table[FromPolarCoordinates[{Exp[(t + 2 k Pi/100)*0.5], t}], {k, 1, 100, 5}], {t, -Pi, Pi}, PlotRange -> All, PlotStyle -> Red, RegionFunction -> Function[{x, y}, (x - 3)^2 + y^2 < 9]] // Show[#, ContourPlot[(x - 3)^2 + y^2 == 9, {x, 0, 6}, {y, -3, 3}, ContourStyle -> Blue]] &?$endgroup$
– corey979
Feb 17 at 13:47
$begingroup$
@kglr nope. I editted my answer and showed what your solution is like.
$endgroup$
– Alper91
Feb 17 at 13:47
$begingroup$
@kglr nope. I editted my answer and showed what your solution is like.
$endgroup$
– Alper91
Feb 17 at 13:47
$begingroup$
@corey979 Yes! Yes! Exactly like that one and the sum of the curves' length within the circle
$endgroup$
– Alper91
Feb 17 at 13:49
$begingroup$
@corey979 Yes! Yes! Exactly like that one and the sum of the curves' length within the circle
$endgroup$
– Alper91
Feb 17 at 13:49
$begingroup$
@corey979 However, your solution is not the rotation case. It is the case where radial growth of the spiral increases.
$endgroup$
– Alper91
Feb 17 at 13:57
$begingroup$
@corey979 However, your solution is not the rotation case. It is the case where radial growth of the spiral increases.
$endgroup$
– Alper91
Feb 17 at 13:57
$begingroup$
@corey979 I am sorry corey what you provide was right, if I know where my limits are. And this is the code I wanted like you suggested: ParametricPlot[ Table[FromPolarCoordinates[{Exp[(t + 2 k Pi/100)*0.5], t}], {k, 1, 100, 5}], {t, -Pi, Pi}, PlotRange -> All, PlotStyle -> Red, RegionFunction -> Function[{x, y}, (x - (Exp[Pi*0.5] + 1)/2)^2 + y^2 < ((Exp[Pi*0.5] - 1)/2)^2]] // Show[#, ContourPlot[(x - (Exp[Pi*0.5] + 1)/2)^2 + y^2 == ((Exp[Pi*0.5] - 1)/2)^2, {x, 0, 6}, {y, -3, 3}, ContourStyle -> Blue]] & However, I need also sum of the lengths
$endgroup$
– Alper91
Feb 17 at 14:07
$begingroup$
@corey979 I am sorry corey what you provide was right, if I know where my limits are. And this is the code I wanted like you suggested: ParametricPlot[ Table[FromPolarCoordinates[{Exp[(t + 2 k Pi/100)*0.5], t}], {k, 1, 100, 5}], {t, -Pi, Pi}, PlotRange -> All, PlotStyle -> Red, RegionFunction -> Function[{x, y}, (x - (Exp[Pi*0.5] + 1)/2)^2 + y^2 < ((Exp[Pi*0.5] - 1)/2)^2]] // Show[#, ContourPlot[(x - (Exp[Pi*0.5] + 1)/2)^2 + y^2 == ((Exp[Pi*0.5] - 1)/2)^2, {x, 0, 6}, {y, -3, 3}, ContourStyle -> Blue]] & However, I need also sum of the lengths
$endgroup$
– Alper91
Feb 17 at 14:07
add a comment |
1 Answer
1
active
oldest
votes
8
$begingroup$
ctr = {3, 0};
radius = 3;
pp = PolarPlot[Evaluate@Table[Exp[(t + Pi*i/50)*0.5], {i, 1, 100}], {t, -Pi, Pi},
RegionFunction -> (Norm[{#, #2} - ctr] <= radius &)]
Total[ArcLength /@ Cases[pp, _Line, All]]
(* or Total[RegionMeasure /@ Cases[pp, _Line, All]] *)
314.511
If ypu have to use ParametricPlot, you can do
pp2 = ParametricPlot[Evaluate[Table[E^(((j*Pi)/50 + t)/2) { Cos[t], Sin[t]}, {j, 1, 100}]],
{t, -Pi, Pi},
RegionFunction -> (Norm[{#, #2} - ctr] <= radius &), PlotRange -> All]
Total[ArcLength /@ Cases[pp2, _Line, All]]
314.511
answered Feb 17 at 14:22
kglrkglr
185k10202421
$endgroup$
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
8
$begingroup$
ctr = {3, 0};
radius = 3;
pp = PolarPlot[Evaluate@Table[Exp[(t + Pi*i/50)*0.5], {i, 1, 100}], {t, -Pi, Pi},
RegionFunction -> (Norm[{#, #2} - ctr] <= radius &)]
Total[ArcLength /@ Cases[pp, _Line, All]]
(* or Total[RegionMeasure /@ Cases[pp, _Line, All]] *)
314.511
If ypu have to use ParametricPlot, you can do
pp2 = ParametricPlot[Evaluate[Table[E^(((j*Pi)/50 + t)/2) { Cos[t], Sin[t]}, {j, 1, 100}]],
{t, -Pi, Pi},
RegionFunction -> (Norm[{#, #2} - ctr] <= radius &), PlotRange -> All]
Total[ArcLength /@ Cases[pp2, _Line, All]]
314.511
answered Feb 17 at 14:22
kglrkglr
185k10202421
$endgroup$
add a comment |
8
$begingroup$
ctr = {3, 0};
radius = 3;
pp = PolarPlot[Evaluate@Table[Exp[(t + Pi*i/50)*0.5], {i, 1, 100}], {t, -Pi, Pi},
RegionFunction -> (Norm[{#, #2} - ctr] <= radius &)]
Total[ArcLength /@ Cases[pp, _Line, All]]
(* or Total[RegionMeasure /@ Cases[pp, _Line, All]] *)
314.511
If ypu have to use ParametricPlot, you can do
pp2 = ParametricPlot[Evaluate[Table[E^(((j*Pi)/50 + t)/2) { Cos[t], Sin[t]}, {j, 1, 100}]],
{t, -Pi, Pi},
RegionFunction -> (Norm[{#, #2} - ctr] <= radius &), PlotRange -> All]
Total[ArcLength /@ Cases[pp2, _Line, All]]
314.511
answered Feb 17 at 14:22
kglrkglr
185k10202421
$endgroup$
add a comment |
8
8
8
$begingroup$
ctr = {3, 0};
radius = 3;
pp = PolarPlot[Evaluate@Table[Exp[(t + Pi*i/50)*0.5], {i, 1, 100}], {t, -Pi, Pi},
RegionFunction -> (Norm[{#, #2} - ctr] <= radius &)]
Total[ArcLength /@ Cases[pp, _Line, All]]
(* or Total[RegionMeasure /@ Cases[pp, _Line, All]] *)
314.511
If ypu have to use ParametricPlot, you can do
pp2 = ParametricPlot[Evaluate[Table[E^(((j*Pi)/50 + t)/2) { Cos[t], Sin[t]}, {j, 1, 100}]],
{t, -Pi, Pi},
RegionFunction -> (Norm[{#, #2} - ctr] <= radius &), PlotRange -> All]
Total[ArcLength /@ Cases[pp2, _Line, All]]
314.511
answered Feb 17 at 14:22
kglrkglr
185k10202421
$endgroup$
ctr = {3, 0};
radius = 3;
pp = PolarPlot[Evaluate@Table[Exp[(t + Pi*i/50)*0.5], {i, 1, 100}], {t, -Pi, Pi},
RegionFunction -> (Norm[{#, #2} - ctr] <= radius &)]
Total[ArcLength /@ Cases[pp, _Line, All]]
(* or Total[RegionMeasure /@ Cases[pp, _Line, All]] *)
314.511
If ypu have to use ParametricPlot, you can do
pp2 = ParametricPlot[Evaluate[Table[E^(((j*Pi)/50 + t)/2) { Cos[t], Sin[t]}, {j, 1, 100}]],
{t, -Pi, Pi},
RegionFunction -> (Norm[{#, #2} - ctr] <= radius &), PlotRange -> All]
Total[ArcLength /@ Cases[pp2, _Line, All]]
314.511
answered Feb 17 at 14:22
kglrkglr
185k10202421
edited Feb 17 at 14:30
answered Feb 17 at 14:22
kglrkglr
185k10202421
answered Feb 17 at 14:22
kglrkglr
185k10202421
answered Feb 17 at 14:22
kglrkglr
185k10202421
185k10202421
add a comment |
add a comment |
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1
$begingroup$
Something like this:
ParametricPlot[ Table[FromPolarCoordinates[{Exp[(t + 2 k Pi/100)*0.5], t}], {k, 1, 100, 5}], {t, -Pi, Pi}, PlotRange -> All, PlotStyle -> Red, RegionFunction -> Function[{x, y}, (x - 3)^2 + y^2 < 9]] // Show[#, ContourPlot[(x - 3)^2 + y^2 == 9, {x, 0, 6}, {y, -3, 3}, ContourStyle -> Blue]] &?$endgroup$
– corey979
Feb 17 at 13:47
$begingroup$
@kglr nope. I editted my answer and showed what your solution is like.
$endgroup$
– Alper91
Feb 17 at 13:47
$begingroup$
@corey979 Yes! Yes! Exactly like that one and the sum of the curves' length within the circle
$endgroup$
– Alper91
Feb 17 at 13:49
$begingroup$
@corey979 However, your solution is not the rotation case. It is the case where radial growth of the spiral increases.
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– Alper91
Feb 17 at 13:57
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@corey979 I am sorry corey what you provide was right, if I know where my limits are. And this is the code I wanted like you suggested: ParametricPlot[ Table[FromPolarCoordinates[{Exp[(t + 2 k Pi/100)*0.5], t}], {k, 1, 100, 5}], {t, -Pi, Pi}, PlotRange -> All, PlotStyle -> Red, RegionFunction -> Function[{x, y}, (x - (Exp[Pi*0.5] + 1)/2)^2 + y^2 < ((Exp[Pi*0.5] - 1)/2)^2]] // Show[#, ContourPlot[(x - (Exp[Pi*0.5] + 1)/2)^2 + y^2 == ((Exp[Pi*0.5] - 1)/2)^2, {x, 0, 6}, {y, -3, 3}, ContourStyle -> Blue]] & However, I need also sum of the lengths
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– Alper91
Feb 17 at 14:07