# Polar Graphing

To change the graph paper to a polar grid, start by clicking on the wrench in the upper right hand corner of the graph paper to open the Graph Settings menu. To graph in polar form, write your equation using r and theta notation: Polar graphs are automatically set to a domain of [0,12pi]. It's possible to restrict this domain using restrictions. However, it's not currently possible to use a larger domain. Currently, graphing polar coordinates isn't supported - but it's on our list! You can find more helpful example graphs by clicking on the three horizontal lines in the upper left hand corner of the calculator.

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• Poloinc

How would I graph the straight line through the origin

theta=pi/4

• Troy Alexander

I do not see the option of polar functions with negative values for theta, even when I set specific ranges. Will this be changed/updated?

• bej ugo

Nice app! Similar to Troy's comment, I'm trying to graph the polar equation r=3.1cos(-1.1*theta). This program will only graph about 60% of it, b/c it won't let us extend the upper limit above about 12pi (and we need to let theta run from 0 all the way to 20pi in order to see this entire curve.)

I hope the creators of this site will program in more flexibility with extending restrictions so that we can graph beautiful graphs like this one.

Thanks, again, for such a nice program.

• G. W.

Why does the spiral created by r=theta stop at r=37 and some change? Shouldn't it continue? Is this some kind of convention? Thanks for the fantastic site!

• Leonardo Cisija

a quick temporary workaround for the negative theta range problem for those who need it is to replace theta with "theta + a2pi" or "theta + 360a" where "a" is a scaling factor. in the slider for a (or whichever letter you choose), write something like "a = [-50, -49, -48,..., 50]" to set your range in multiples of 2pi. it's basically drawing a bunch of different graphs of different ranges that look like one graph, which means a bigger range will take longer, but my computer can barely open ms word yet handles 600 items on that list pretty well.

here's an example with a log function, which decays pretty fast. notice i've changed a2pi to a12pi because that's where the function usually cuts off for me that way it saves time. • Wojciech Gościcki

But is there a way for a spiral created by r = θ (or r = θ + a12π) to have less than 6 turns?

Edit: Never mind, range restriction works fine.

Edited by Wojciech Gościcki
• Flexagontnt

Is it possible to do implicit polar equations, like r-1=sin(theta)? If so, how?

• Leonardo Cisija

no, but theres always a work around. define these:
R(x,y)=sqrt(x^2+y^2)
T(x,y)=arctan(y/x)+[0,...,50]pi

now use R(x,y) and T(x,y) instead of r and theta. the example you gave can just be rewritten for r though.

Edited by Leonardo Cisija
• Leonardo Cisija

*R(x,y)=sqrt(x^2+y^2)

edit: just figured out you can edit

Edited by Leonardo Cisija
• Aalia Khazi

Hi, how do I graph polar coordinates on this? I can't seem to get the degree function.

Edit: Ohhh, I see it's not supported.

Edited by Aalia Khazi
• Leonardo Cisija

you can use a similar work-around. define these

X=r*cos(theta)

Y=r*sin(theta)

then just plot it with the values for r and theta plugged-in. here's an example: https://www.desmos.com/calculator/arlph9kxks

• Pomona CH

Leonardo's fix for range worked well! I couldn't run it with 601 elements as Desmos semi-crashed and started glitching really bad, but with elements from 0 to 60 it let me expand the range enough to see the center of this graph pretty well: https://www.desmos.com/calculator/izkovapy1m

• Mikaelarhertel

Is there a way to plot a point in polar?

• Leonardo Cisija

Plotting points in polar isn't technically supported yet, but you can use the work-around I described above:

Define these

`X(r,θ)=r*cos(θ)Y(r,θ)=r*sin(θ)`

then just plot it with the values for r and theta plugged-in. Here's an example: https://www.desmos.com/calculator/arlph9kxks

Edited by Leonardo Cisija
• 22mh0280

how can I translate the polar graph across the x and y axis?

• Jeffaschwarz

I can't graph theta=pi/4 and theta=0.

• Leonardo Cisija

@22mh0280

The easiest way to translate polar graphs across x and y is by writing them parametrically.

`R(x,y) = sqrt(x^2+y^2)T(x,y) = arctan(y/x) + [0...10]π`

(The 10 can be any number higher than 2 depending on how many times you wan the graph to go around).

The, something like r = 2*θ can now be written as

`1R(x,y) = 2*T(x,y)`

(The 1 has to be there before the R so desmos doesn't think you're just defining another function)

To translate, just offset the x's and y's. Here it is for 1 right and 2 down:

`1R(x-1,y+2) = 2*T(x-1,y+2)`

Or you can make h and k sliders:

`1R(x-h,y-k) = 2*T(x-h,y-k)h = 1k = -2`

Even just the original definitions:

`R(x,y) = sqrt((x-h)^2+(y-k)^2)T(x,y) = arctan((y-k)/(x-h)) + [0...10]πh = 1k = -21R(x,y) = 2T(x,y)`

• Leonardo Cisija

@Jeffaschwarz

Desmos can't normally do polars solved for theta, so just do it parametrically.

`T(x,y) = arctan(y/x) + [0...4]π1T(x,y) = π/41T(x,y) = 0`