Polar Graphing

How would I graph the straight line through the origin

theta=pi/4

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?

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.

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!

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.

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.

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

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.

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

edit: just figured out you can edit

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.

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

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

Is there a way to plot a point in polar?

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

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

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

@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 = 1
k = -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]π

h = 1
k = -2

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

@Jeffaschwarz

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

T(x,y) = arctan(y/x) + [0...4]π

1T(x,y) = π/4
1T(x,y) = 0