2:01 PM
@robjohn So, in the upper right-hand corner where it says "Maths general discussion" doesn't mean general discussion about Maths?
As a topic?

And in general, we do talk about Math. If it doesn't interrupt another conversation (which is common decency), a conversation about math is perfectly fine.

hello

Hi

@JohnSmith hi there

Same guy from yesterday (usr guy with the tangent question) -- couldn't log in on this computer for some reason
Would really appreciate some insight if anyone has some time

2:07 PM
@JohnSmith was he banned for some reason?

what is it?

No
QED: Basically, I have a function z=f(x,y) and an arbitrary point outside that function, (x0, y0, z0)
I an trying to find the tangent line, in the plane of z0, that connects to f(x,y) from my point

Oh, the question is not about the inability to log in, but the question from yesterday :-) doh!

in other words, it's like if I am sitting at a random point and I swing a large bar into a mountain parallel-style, and I want to know where the collision point is where it first makes contact
Yes

you can reduce it to a 2D problem immediately, then maybe lagrange multipliers

2:12 PM
I figure it's a 2d problem only, yes, but I'm completely clueless as to how this is set up
as it's basically operations within a cross-section

if your point is the origin, this seems like trying to find when $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{y}{x}$.
on the curve in the $z=0$ plane.

it's not the origin
technically 100,100 or so
here is a picture sorta illustrating what I am trying to find: i.imgur.com/ttxQh.png
this is all assuming a fixed level of z0 of course

@JohnSmith well, then $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{y-100}{x-100}$

for some reason that isn't rendering on my machine
whatever dollar-sign slash frac and slash mathrm d is

@JohnSmith You don't have the MathJax bookmark
There are three separate installation methods described on that page.
What browser are you using?

2:18 PM
Ah, there it goes. Rendering now
I'm not sure how to rectify dy/dx here (my calc is rusty)
but my function, f, consists of both x and y

What is the function?

it's technically quite large and complicated, but I do have its derivative (too large to paste here without making a mess)

so you'r really looking for a computer algorithm?

You were talking previously about a surface in $\mathbb{R}^3$. Are you still?
given by $z=f(x,y)$.

Yes, it's in R3, but I am "operating" within a particular slice z0
it's like sitting at a random spot outside a mountain and then clashing a long metal bar into said mountain and asking for the contact point
where the z coord of contact = your z coord

2:24 PM
@JohnSmith So $\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial y}}$

Not sure what that means; derivate f(x,y) with respect to x, and then divide the partial derivatives?

No, to get the slope of the curve in the $x-y$ plane, you compute the partials and divide.
then set that to $\frac{y-y_0}{x-x_0}$
and solve for $(x,y)$

okay, I'll definitely give this a shot
Thanks so much for the insight
(if I do this correctly I should get two solutions right?)
since there are technically two "tangent" connections
at either side of the circular shape?

So the final equation would be $\frac{y-y_0}{x-x_0}=-\frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial y}}$
There should then be two solutions to the equation.
@Bill: Hey Bill!
@Bill: Bye Bill! :-)

Like small kids scribble in park benches and whiteboards: Bill was here :-)

2:30 PM
Who is Bill?

@Srivatsan I just saw your comment, I'm up to chat about it if you have the time.

Bill Dubuque, Shoulders of Giants
52.5k 3 36 142
@Srivatsan Good morning!

Morning, robjohn.

@Srivatsan I have to go walk the dog. It has nothing to do with your arrival (or does it?) :-D

Up early?

2:34 PM
@Srivatsan No, I am actually a bit late to walk the dog.

@robjohn Nah, I figured that the dog is just an excuse long back. =)

@Srivatsan She would take exception to that comment, so I won't tell her :-)

Is there an easy way to do this in wolfram alpha?

bbl

when I calculate the full thing it doesn't fit in the input box
likely using NSolve

2:42 PM
@Srivatsan What do you think about these new "Chat Rules?"

"no solutions exist"
gahh!

@Skullpatrol Sorry, the time isn't right for me. WIll talk later.

the equation's right, too

@Srivatsan np

Agh, that's the second time this stupid laptop has frozen up on me in one hour!

2:47 PM
@ZhenLin Have you tried to reboot it?
@ZhenLin What is so special about tuna in Japan?

@robjohn Hi. Just dropped in quickly to see what was happening, sparked by the recent meta post. Did similarly on sci.math yesterday and it has really deteriorated now that most of the non-cranks have migrated here. Ah for the good ole days.

wolframalpha.com/input/?i=2*%282%5EN%5BZetaZero%5B2%5D%2C+50%5D%29-%282%5EN%5B1%2F2+%2B+I*Im%5BZetaZero%5B1%5D%5D%2F2%2C+50%5D%29-%282%5EN%5BZetaZero%5B1%5D%2C+50%5D%29
Oops that did not work

hello everyone

@BillDubuque, I liked that proof of left/right inverses for matrices using a polynomial that you wrote
Hi

3:02 PM
@QED Thanks. I thought about those matters far too much! (as you can see there)

hi QED

@QED Can you link to that proof?
Maybe I missed it.

hi srivatsan sir

Math cranks are pretty funny. EEE takes the cake, imo.

@MatsGranvik you wanted this: wolframalpha.com/input/…

3:04 PM
@Stom Hi Storm.

@MatsGranvik You need to replace * by %2A. However, Wolfram doesn't appear to like your input

haha hi

I can't find it

Can anyone explain why the setup designated earlier does not work?
Wolfram returns "no solutions" even though I know it does

@t.b. Yes, that is how I wanted it. I have been thinking about this because it gives 5 decimals in common with the the decimals of the first Riemann zeta zero.

3:07 PM
30

If $A$ and $B$ are square matrices such that $AB = I$ where $I$ is identity matrix. Show that $BA = I$. I do not understand anything more than the following. Elementary row operations. Linear dependence. Row reduced forms and their relations with the original matrix. If the entries of the mat...

@QED Beat me by a few secs. There are lots of nice proofs there.

@QED Oh, this one. We discussed this here of course. It's interesting that Bill could identify which proof you were referring to even with the sparse hint. =)

robojohn: I tried NSolve[{(y-100)/(x-100) = -(D(f,x))/(D(f,y)), f=(my huge disgusting function)},{x,y}] and it does not seem to work

@JohnSmith Could you try putting your "huge disgusting function" in PasteBin or some similar service? I'm now really curious as to how you obtained this function in the first place...

@Srivatsan As I said, I thought about it far too much. It's a FAQ in math forums at this level. Every time I revisit such questions I try to inject something fresh into them. Be sure follow the link to sci.math to the literature references if you want to understand it more generally.

3:13 PM
So, there's a really barebones question here, and it's exactly the same notation and essential problem as this much more fleshed-out question here. I wonder where it comes from.

If you graph it, it looks like a billow

What are the typical ranges for the independent variables? Where would the "guy with a big stick" usually be standing relative to this surface?
(FWIW: the function you have is actually relatively tame...)

for instance look at the contour here tinyurl.com/7qbvp9b

@BillDubuque Yes, your answers are in general thoughful -- and the ones QED mentioned are no exception. I try to follow your answers and subsequent pointers to the extent that I understand. :-) (And hi!)

So if you imagine the "guy" sitting in the air next to one of those contours, and he swings his "bar" into it, there is a particular contact point against the surface

3:24 PM
@JohnSmith here are the local max and min wolframalpha.com/input/?_=1325776823873&i=f%3d(3000-0.002*(x%5e2%2by%5e2%2bx*y)%2b16*(x%2by))*exp(-(0.00003*(x%5e2%2by%5e2)-0.0023*(x%2by)%2b0.3))&fp=1&incTime=true

I know how to get the local max and min; I'm trying to find a contact point given a fixed z. Thanks though

@BillDubuque, I read on wikipedia that you worked on macsyma!

@John Thanks, that's unfortunate that they cannot be edited, esp. since almost every link eventually rots.

3:28 PM
@QED I'm sure that some of his algorithms remain in use in current packages...
@JohnSmith So, the guy with the beam is standing on the surface itself?

no, in the air away from it
for some reason (y-y0)/(x-x0)=-(partial x)/(partial y) does not work

@JohnSmith You mean the guy is floating above the surface?

@JohnSmith Do you have two points along this bar?

why don't you phrase it as a 2D problem as I suggested
this removes a lot of confusion

@JohnSmith You have not only $\frac{y-y_0}{x-x_0}=-\frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial y}}$, but also $f(x,y)=z_0$
what is the $z_0$ you are using?

3:37 PM
Yes
but I still get no solutions even when I run the thing in Wolfram that way

wolfram alpha adds a level of indirection, when it doesn't give you what you want you don't know if you stated the problem in a way it doesn't interpret

Well, Alpha isn't always the best for multivariates; you don't have an actual computing environment on you?

Is there a font of mine broken or does this post contain some weird characters?

@tb I'm seeing cruft, too.

pheew

3:43 PM
@J.M. Here on such rotted link from my comment in Pete L. Clark's post in that thread bit.ly/Shift1-1notOnto If your replace "www" by "groups" it resolves fine.

@tb That character is $\unicode{x2592}$ called "medium shade" in unicode
They are probably using some old OS and copy-pasting weird encodings.

@robjohn not "cruft?"

@Skullpatrol It could be called cruft perhaps...

@robjohn Here I see a big fat black square. On main a greyish pixel-y rectangle.

I definitely gave it something it can interpret

3:46 PM
@BillDubuque Interesting; I think complaints should be directed at Google for this; you'd think they could do redirects...

@Srivatsan I'm happy to hear that you found those posts helpful. Sometimes I never know since my posts often get few votes since they appear later than the FGITWs. If there is something that you don't understand then please do ask questions since it will help me to clarify the post, which will help many other readers. It's difficult to "go backward" and remember what the conceptual difficulties were when one was learning topics.

I give up; this is frustrating me to no end
there's no reason why Wolfram should be telling me there's no solution

@JohnSmith Again, it might be because of Alpha's own limitations. You have no access to any computing environment?

I don't know what you mean by computing environment

@J.M. The problem is that the short syntax I used was never "officially" supported. If I had a good contact at google perhaps we could get a fix. But it seems that google doesn't care about usenet groups generally, since the support has gone downhill as of late.

3:49 PM
I don't have Mathematica or Maple if that's what you mean

@tb yeah, I think that MathJax's choice of font is different. If you zoom in on the square it shows some white spots.

is there a way I could synthesize the result with a programming language?

@BillDubuque, I didn't see the connection with the euclidean algorithm

@JohnSmith Yes, that's what I meant; rather unfortunate...

Costs too much

3:50 PM
you can try some free software
maxima or gp

What QED said.

@QED yes, I was one of the core Macsyma developers when I was an MIT undergad. I didn't get much sleep during that phase of my life.

obviously you need to learn to use it
@BillDubuque, that's so cool! Did you meet Gosper? ..that would explain all the proofs using telescopes

@QED Did you know Gosper poked his head in math.SE once?

@BillDubuque or the support was never really that good. It has steadily gone downhill since Google took the archives from deja.news

3:53 PM
@JM, no I never heard that

All they're crazy about now is Plus, Plus, Plus... ech.

@QED Gosper had already left for Stanford by the time I arrived at MIT, but I met him later, and worked with him (remotely) on Macsyma.

poor guy
he couldn't delete his account

@BillDubuque Is that the Gosper of Soddy-Gosper, or some younger Gosper?

> Now how do I "drain away" from this little Utopia?

3:54 PM
I may be mathematically illiterate nowadays but at least I can lay claim to Stanford/MIT admission XD

this new style of website where you can't delete anything is a bit weird

@QED All your mistakes preserved for eternity :-)

@QED Do you mean that you still don't see the connection with the Euclidean algorithm?

@BillDubuque, yes for the matrix AB = BA thing
you just wrote it as a passing remark though

JM: Is there an iterative approach I could take, perhaps? Like finding the equation of the line through (x0,y0) at z0 and see where it best intersects the function at a particular range if I have a rough idea where it should intersect?

3:57 PM
Is there a standard etiquette for two roughly-equivalent answers being posted at the same time here? Matt E and I have basically the same answer here:math.stackexchange.com/questions/96644/quadratic-reciprocity
His was in a minute earlier -- should I delete mine?

@robjohn Indeed, it seems that google was more interested in using usenet newgroups to gather users for its own "groups" than it was in supporting usenet newsgroups and their archives.

it's okay just to leave it @CamMcLeman

@BillDubuque It's too bad. I noticed it most when the search facility (this is Google!) was broken most of the time for usenet groups.

@Cam I wouldn't worry too much about it. But it would be helpful to readers to add a remark, e.g. "Note to readers: this is essentially the same as Matt's answer, except...." Even if you think they are equivalent, they may not prove equivalent to students, since one "trivial" difference might make the answer more comprehensible to a student. What's trivial to experts may not be so to students.
3

@JohnSmith I'd do iterative methods as a last resort, TBH. The function really isn't that complicated.

4:04 PM
Huzzah. I have returned for now.

@AsafKaragila The Return of the King!

Huzzah indeed; we've a party...

@BillDubuque "What's trivial to experts may not be so to students." because students can't see the forest for the trees.

@robjohn Yes, the search has gone seriously downhill. It seems that the many of the servers have latency issues, so the search results often timeout (the DB search is apparently distributed across many servers). I often cannot find old posts that I know are in the archive. Dejanews never had that problem.

JM: Well if the iterative approach gets the job done, I'd rather go that route, no?

4:06 PM
@BillDubuque I take it that the circuitous route of using site:groups.google.com on the Google search engine (gasp!) doesn't work either...
@JohnSmith Sure, but iterative methods need a starting point. They operate according to GIGO. What's your starting point?

GIGO?

"garbage in, garbage out"

@JohnSmith: did you say what $z_0$ was?

Okay, thanks all. The conventions seem to differ slightly between MO and MSE, so I like to check.

Arbitrary, can be anything as long as it's within the range of the peak

4:09 PM
@robjohn The Return of the Jedi, if anything.

@JohnSmith That can be a bit problematic. "Arbitrary" isn't necessarily "good".

@BillDubuque: one thing that usenet did for me, was to improve my ascii-art skills :-) Here, I've been improving my speed TeX skills.

@J.M. I tried all the obvious permutations long ago with no success. The good thing about the groups search engine is that you can search on various fields, e.g. I usually use author:dubuque when I want to search only my posts.

@JM Do you have a recipe for cookies?
Something nice and fun.

@AsafKaragila What kind? Would you have oatmeal on you, for instance?

4:11 PM
Hmmm... No I don't have that around.

@BillDubuque That sucks...

@JM: I finally got the seal of approval from Didier. He has removed his downvote. I wonder what typos thinks are in my answer.

@robjohn Ha! Luckily for you it didn't also improve your flaming skills. Many of us had to make great efforts to tone it down when moving here after hanging out in the wild, wild west for far too long. This is like a church compared to sci.math.

@JM Well duh [read: dough]... :-)

4:13 PM
@BillDubuque I tried very hard not to flame. It just seems so unproductive, and I figure if I don't do it in real life, I shouldn't do it online.

@AsafKaragila Well, I'm not quite sure what you can put in your cookies... chocolate chips? Nuts?

@BillDubuque But I do agree that sci.math was a bar room brawl compared to here.
6

No flame is good. @BillDubuque: Hello Bill! May I ask what the picture in your avatar means?

@JM Probably nothing. I am going to buy some juice later, so I can fetch stuff for that as well.

@AsafKaragila Except for the coloring (yech), this is more or less what I use.
@robjohn Heh, "brawl". :D "I was expecting a faculty room and ended up in a truck stop!"

4:17 PM
@Tim Google "modular group" for images to see what my avatar means.

@BillDubuque That's the Klein invariant, isn't it?

Hmmm.

@AsafKaragila I think we should remove our most recent exchange on main...

Probably.

OH! I think I know why Wolfram doesn't work!
technically NSolve[{(y-100)/(x-100) = -(D(f,x))/(D(f,y)), f=(my huge disgusting function)},{x,y}] assumes the tangent is a true tangent, right? As opposed to a best-approximation

4:19 PM
@JohnSmith Have you tried sacrificing a virgin on the altar in the hour of the witch on the night of the bloody moon?

I did: I neutralized my last comment.

@AsafKaragila That doesn't work; I've tried that years ago... :D

Yep. In return, I got SARS

@JohnSmith That shouldn't make too much difference unless your derivatives go wild.

@JohnSmith Then you did it wrong...

4:20 PM
rob: Is there a way to make it approximate?

I really think you should slice your surface...

@JM Was the moon in the scorpio?

I don't know how to slice my surface

@JM You're suggesting he cut himself?? W|A not working is not a reason for self mutilation...

@AsafKaragila No, it was in Gemini...

4:21 PM
@Asaf: there still is a comment from you that doesn't make sense...

@JM Explains everything.
@tb No there isn't!

I mean even when I slap in f=arbitrary point, it still says no solutions
I think I need to somehow make it find the best fit instead of solve

@AsafKaragila Can't argue with that :)

@JM Last night the moon was in Taurus. I know because it was almost over the Pleiades

@JohnSmith One way to take slices of your surface is to rotate your independent variables...

4:22 PM
@tb Yes you can, this is room 12B

@robjohn I don't flame in real life either, but if you hang out long enough in a forum like sci.math the cranks and charlatans can get to you. E.g. see some of Tim Golden's threads, which seriously riled even Arturo.

@robjohn Interesting. (And I like the Pleiades myself...)

@AsafKaragila No I can't. You win for the third time today...

@JM We were at my uncle's last night looking at the Moon and Jupiter.

@tb You'll have your day, don't worry...

4:24 PM
is there a better way to ask it to solve lhs=rhs where rhs=value, solve for x,y?

@tb I think that you need to see an argument specialist. I know an expert.

errr, f=value

@JohnSmith That's a bit difficult in general; especially since you have stuff inside and outside an exponential function...

@AsafKaragila I was thinking of this, before :) // don't say "no you weren't"

4:25 PM
@tb Obviously. :-)

@robjohn That isn't false coloring, is it? (Still, pretty...)

@JM Nope, that is as true as I know of.

is there a way to do it in GP?

@JM Did you think the Pleiades was a different color?
However, I have much better equipment now so I should do another image and expose longer to get more of the nebulosity.

4:29 PM
@robjohn Maybe it's because I haven't seen 'em that bright... (I do live a few latitudes down from you.)

@JM Oh, you won't see them with that intensity of color with the naked eye, even with a telescope. The low-light receptors (rods) are not very sensitive to color.

@robjohn Oh, that explains it; I can't take pics... :)

The camera can pick up the color as if the intensity of light were greatly magnified.
@BillDubuque Riling Arturo sounds to be a pretty hard thing to do. I don't know if I've actually been in too many "discussions" with him here or on sci.math, but from what I've seen, he is pretty level headed.

I can confirm that the answer is correct with the partial equation
But for whatever reason Wolfram isn't showing it

@JohnSmith Alpha is not the end-all in equation solving

4:36 PM
I just downloaded GP but don't know how to synthesize my input\

@JohnSmith A reading of the manual certainly helps...

I am reading it, but it's not listing what I am after precisely
I'm trying to get an approximation, not an exact tangent

@JohnSmith Sorry, what is GP?

@robjohn Speaking of bar rooms, should you find yourself in BeanTown please do look me up. I owe you a few Bezout brews!

@robjohn Speaking of debts, you owe me a few whatnots. I know you took them from the whatnot jar!!

4:43 PM
Tim: Trying to find a good mathematical computation environment to use
I can't afford Mathematica or Maple

@Tim PARI/GP.

Trying Maple Syrup. Put it on your harddrive and add water. I heard it makes Maple appear on your computer.

@AsafKaragila Speaking of whatnot, I asked a question about it here, inspired by seeing it being often used in this chatroom

@AsafKaragila Sheesh, I thought you had that miracle of the perpetual whatnot jar going on...

They said "whatnot" would give you some clues as to the speaker/writer's social background, about which I have no clue.

4:45 PM
@JM I do. I'm also keeping track about where my whatnots are going and what they are doing!

@Tim Oh, just like the peculiar uses of "doubt"?

@BillDubuque :-) My son is going to college in Chicago, but if we decide to make our way further east, I'd like to visit Princeton once more and see Boston. I've never been to Boston, but I have a friend here who went to MIT. He tells me horror stories about the traffic there.

@JM Where is the use of "doubt"?
@JM I didn't use doubt often. But I remember I saw that post at meta.

@AsafKaragila I guess I can't get anything by you, especially if it's from the whatnot jar.

Sage
@robjohn Well... it's the jar, and whatnot :-)

4:48 PM
@Tim Yeah, John could try Sage if GP's not his groove...

I don't see how "whatnot" can give you any clue about people outside the USA.

Puck Frinceton

@JM I didn't use "doubt" often. But I remember I saw that post at meta.

@JohnSmith That spoonerism almost sounds like a name...

@JM So is GP preferred over Sage? Is it better than Sage in most aspects?

4:50 PM
lol

@robjohn Actually Boston traffic is not that bad compared to other major cities.

GP appears to be command line
and really fidgety

@Tim I wouldn't say "preferred". I've only used PARI/GP thrice, and Sage once. What I'm saying is that if you still don't feel comfy after a few hours of poking around an environment, you could try switching...

@JM Oh, I remember you like Mathematica the most.

@BillDubuque He was mainly a pedestrian/bicyclist, so he viewed things from a pedestrian vs car standpoint. Perhaps that has improved since he was there.

4:54 PM
@Tim It's more "grew on me" than "like"... :D To be fair, I still use MATLAB from time to time, and I check other environments every so often.

@robjohn Ah, that makes sense. Many of the natives do lack road manners.

@AsafKaragila For example, they are most likely from Israel?

@Tim No...

hi $\forall$

It's going to take a lot longer than a few hours to get comfortable

5:00 PM
must say that the latex support for site makes me want to type more and more :)

@QED Okay, I was exaggerating a bit with "few"...

5:36 PM
@robjohn: Congrats for the "Aaaaahh..." !
(and the well-done)

@tb Thanks. I think I displayed that mostly it was a matter of not being detailed enough.
@tb Didier pointed out a missing ' and suggested changing $x$ to $x(u)$ in a few places, which I did as it makes the fact that $x$ is being considered a function of $u$ clearer.

@robjohn Yes, certainly. Sorry for not following up on my promise from yesterday. It is a bit sad that the neat computation is now a bit hidden in clutter... Anyway: all is well that ends well :)

@tb at least the downvote is removed :-)

@robjohn I saw that and you got the "nice answer" on top of it...

@tb yes, and another upvote, too (Didier?)

5:44 PM
if I am to use the arclength formula do I have to use parametrics to split up my z function?

@JohnSmith are you parameterizing the curve at $z=z_0$?

yes

@robjohn I would guess so. (doesn't a "well done" amount to +1?) All I can say is that he voted twice today.

@tb the upvote came just a few minutes ago, and he acquiesced to my answer an hour or two before.
@tb how do you know he voted twice? is that a trusted power?

@robjohn no, scroll down to the bottom of a user profile you see total votes then a vote count day/week/month (or the other way around)

5:48 PM
basically finding arclength from (x1,y1) to (x2,y2) of a function z where z=value

@tb ah, never looked down there. interesting.

@tb hmm. I thought that I had voted more than once this month, but perhaps I have been just too busy irl. Gotta get back in the program :-)
@JohnSmith what curve are you following? the curve $(x,y)$ where $f(x,y)=z_0$?

6:04 PM
yes

@JohnSmith Is that for part of the problem?

6:20 PM
not really but it's good info to know
(but yes for the same problem)

Hi.

6:37 PM
hello

zzz

fine
how are you

I am OK.
Found an interesting graph theory book: math.jussieu.fr/~jabondy/books/gtwa/gtwa.html

oh yeah

6:49 PM
does anyone know hiow to calculate arclength for z=f(x,y) for given bounds of x, y, and a z=z0?

7:16 PM
@JohnSmith This came up yesterday in here. You might want to try the main site. The involvement of z seems superficial. You have an implicit equation f(x, y) = 0 for a curve in the plane and you want to calculate arclength along it. To me it seems like you need to find parametrizations for the curve. I don't know of a general procedure for doing this. In examples it probably isn't so hard. But it sounds like you want something general. That would be an interesting question to ask.
@JohnSmith This theorem can give you an answer to "Onto which axis can I project the curve (in a small neighborhood) to get a parametrization?" but it's not so clear to me how you can find this neighborhood.
The proof uses the inverse function theorem, and the proof of that which I remember relies on this. Which is some iterative process. Maybe you could explore that?

i think i figured out how to get the solution
trying to convert some stuff to polar coordinates
god why does every single math paper i look at assume origin 0,0

7:34 PM
Because you can always shift?

not easily
like if my arbitrary origin is at point px, py and my radius is R, landing on point x,y, and I sweep counterclockwise theta degrees -- being able to translate that into cartesian coords isn't simply rcostheta rsintheta etc

@JohnSmith as Dylan mentioned (px + (R cos theta), py + (R sin theta))

7:52 PM
Mmmmm... stew... so good...