How do I plot (1,1) (2,2) (3,3) and so on, on a Polar Plot?

I recently downloaded Mathematica and am extremely new to the software. To start off, I wanted to work with Polar Coordinates and Archimedean Spirals. I want to be able to graph a continuous spiral with the coordinates (r, theta)=(n, n) where n is a positive integer. Essentially I would be creating a spiral between the points (0,0) (1,1) (2,2) (3,3) and it continues on. I would like to have a graph with just the points as well as one with a line connecting the points.

Also, is there a way to specify different ranges?

Similar to the question posed on this Math Stack Exchange.

• F1 takes you to the docs. Search 'Polar'.
– Syed
Commented Aug 20, 2021 at 6:01

Another approach is as follows.

a = ListPolarPlot[{{0, 0}, {1, 1}, {2, 2}, {3, 3}},PlotStyle -> {Red, Thick}];
f = Interpolation[{{0, 0}, {1, 1}, {2, 2}, {3, 3}}];
b = PolarPlot[f[x], {x, 0, 3}];
Show[{a, b}]


• I'd like to add that PolarPlot[f[x], {x, 0, 10}] works too. Commented Aug 20, 2021 at 7:56
• a = ListPolarPlot[{{0, 0}, {1, 1.1}, {2.1, 2}, {3.3, 3}, {5, 5}, {8, 8.1}}, PlotStyle -> {Red, Thick}]; f = Interpolation[{{0, 0}, {1, 1.1}, {2.1, 2}, {3.3, 3}, {5, 5}, {8, 8.1}}]; b = PolarPlot[f[x], {x, 0, 13.5}]; Show[a, b, PlotRange -> All] Commented Aug 20, 2021 at 11:51
• a = ListPolarPlot[{{0, 0}, {1, 1.1}, {2.1, 2}, {3.3, 3}, {5, 5}, {8, 8.1}}, PlotStyle -> {Red, Thick}]; f = = Interpolation[{{0, 0}, {1, 1.1}, {2.1, 2}, {3.3, 3}, {5, 5}, {8, 8.1}}]; b = PolarPlot[f[x], {x, 0, 10}]; Show[a, b, PlotRange -> All]  works well. Everything has its limitations. Commented Aug 20, 2021 at 11:55

As i understand , you want this and this or this

ListPolarPlot[Table[{j, j}, {j, 0, 10}],
PlotStyle -> {Red, PointSize[.03]}]

ListPolarPlot[Table[{j, j}, {j, 0, 10}], Mesh -> {Range[0, 10]},
MeshStyle -> {Red, PointSize[.03]}, Joined -> True]

PolarPlot[th, {th, 0, 10}, Mesh -> {Range[0, 10]},
MeshStyle -> {Red, PointSize[.03]}]


Since you want to Archimedean Spirals,so we can set the form r == a*θ^(1/n) and then solve the equations.

polarpts = {{0, 0}, {1, 1}, {2, 2}, {3, 3}};
sol = Solve[
r == a*θ^(1/n) /. Thread[{r, θ} -> #] & /@ polarpts,
Reals];
ArchimedeanSpirals =
PolarPlot[a*θ^(1/n) /. sol[[1]], {θ, 0, 20},
MeshStyle -> Red];
Show[ArchimedeanSpirals, ListPolarPlot[polarpts, PlotStyle -> Red]]


Here we insist on using Archimedean Spirals.

polarpts = {{1, 1.1}, {2.1, 2}, {3.3, 3}, {5, 5}, {8, 8.1}};
f = NonlinearModelFit[Reverse /@ polarpts,
a*θ^(1/n), {a, n}, θ];
Show[PolarPlot[f[θ], {θ, 0, 20}],
ListPolarPlot[polarpts, PlotStyle -> Red]]


• The above does not work for polarpts = {{0, 0}, {1, 1.1}, {2, 2}, {3, 3}};, whereas Interpolation still does. Commented Aug 20, 2021 at 9:03
• @user64494 But Interpolation not always be Archimedean Spirals. The question is ambiguous. Commented Aug 20, 2021 at 9:11

Look for the function FromPolarCoordinates.

Unfortunately this function doesn't transform the Point {0,0}, That's why I used my own transformation:

r\[Theta] = Table[{i, i}, {i, 0, 10}]
xy = Map[#[[1]] {Cos[#[[2]]], Sin[#[[2]]]} &, r\[Theta]];
ListPlot[{xy, xy}, Joined -> {False, True}]


Hope it helps!

• f = Interpolation[{{0, 0}, {1, 1}, {2, 2}, {3, 3}}];PolarPlot[f[x], {x, 0, 3}] draws a spiral along these points. Commented Aug 20, 2021 at 6:50