Question

How to draw a semi-circle in polar coordinates without using conditions?
Center:0,0 Radius:1

Answer 1

@Callisto @trollwillrule @hartnn @ganeshie8 @dan815 @Loser66 @Diyadiya @Preetha

Answer 2

@derpface @ranga

Answer 3

from wherever you like to wherever you like+pi

Answer 4

Standard way is to go anticlockwise direction

Answer 5

from positive X axis

Answer 6

Polar coordinates

Answer 7

someone?

Answer 8

\(r = 1\)
\(0 <= \theta <= \pi\)

Answer 9

\(r = \sqrt{\csc^2\theta - r^2\cot^2 \theta}\)

im sure ur looking for something more profound hmm

Answer 10

@ikram002p @eliassaab @atlas will have more ideas

Answer 11

second one doesnt work, it gives full circle :|

Answer 12

:/

Answer 13

do you need an equation for a semi circle

Answer 14

yep

Answer 15

Do you know the equation of a semi circle in cartesian equation?

Answer 16

nope :P

Answer 17

@atlas here? :O

Answer 18

The semi circle in cartesian is simpler, since when you look at:

r^2=x^2+y^2 if you want to graph it in terms of say, y it'll look like:

\[y=\pm \sqrt{r^2-x^2}\]

However, you have to pick if you want to graph + or - of this to see it will only give you half of your circle.

I don't think that playing around with that trying to convert it to polar coordinates will help you, I think it really is just as simple as restricting your domain between 0 and pi or something like that. No reason to over complicate this.

Answer 19

But I do not want conditions :/

Answer 20

I don't want objects to fall and break when I let go of them, but I can't wish gravity away!

Answer 21

But I'll try to help figure this out, because maybe it exists.

So from what we see of this graph:

r=1
0<theta<pi

That's a unit circle's upper semi circle.

Now what happens when theta is between pi and 2pi? It can be seen as the negative radius possibly, until we hit 2pi and go to 3pi where it's normal again. So we have some level of alternation going on here.

Any ideas?

Answer 22

Crazy suggestion: TJ style :3
\[\Large r = \frac{|\sin(\theta)|+\sin(\theta)}{|\sin(\theta)|+\sin(\theta)}\]
It's 1 for every \(\theta\) between 0 and \(\pi\) and undefined elsewhere, right? :D

Answer 23

Haha yeah that's clever. I was thinking something along the lines of introducing a rounding function and having

r=(-1)^mathfloor(theta/pi)

Answer 24

Both of which fall under "overcomplicating things" right?
Who asks these questions?!

Answer 25

Well I'm ok with this, it's kind of fun.

Answer 26

Well, I've done my part (sort of) so where the heck are you Lau???
@kc_kennylau XD

Answer 27

Some day in a few years someone will ask me how to graph a semicircle in polar coordinates without explicitly restricting the domain and I will valiantly write out a couple sine's and children with be astounded.

Answer 28

And I'll chuckle and say restricting the domain is easier XD
Are you a maths professor by any chance? :>

Answer 29

Nope just a university student. I don't think we'll ever encounter this question ever again.

Answer 30

Ahh right.
What year/how old are you?
(nice boat btw... meant to say that ages ago XD)

Answer 31

I'm 23, I'll be graduating this spring actually with a Chemistry major with a math and spanish minor. I think math is probably one of the most interesting and misunderstood subjects and wouldn't mind trying to become a math teacher for highschool or something in a few years. It's a nice sailboat, but honestly I think catamarans are better. =D What about you, are you a university student or how do you come about doing math here?

Answer 32

O.O
That's a lot
I'm a university student, hopefully graduating with my BS Mathematics this year.

Encountered this site after a weird (and extremely rare) Omegle conversation with a girl about my age about two years back, where things were pleasant and I ended up helping her with her maths homework... when it ended, I decided I would kill time by helping others online with maths... hey presto, it led me here XD

Answer 33

Yeah not as exciting of a story for me, I was in calculus and never bought the textbook because I figured the subject hadn't changed in a long time (ironically), so I used Paul's Online Notes as my textbook basically. They have a link to OpenStudy on the side, but I rarely asked any questions, I started out by just answering peoples questions in biology about mitosis and meiosis and that sort of thing since I hadn't discovered how awesome math and physics were.

I'm trying to teach myself Java, but now I'm procrastinating here so I better get off of here lol.

Answer 34

This is sort of my 'break' if that makes any sense... lol
But yeah, I did start to cut down OS time... not that it made much difference as it just increased my gaming time XD

Answer 35

@terenzreignz @Kainui lolz when did this question become a platform for you two to discuss about your future xD

Answer 36

@terenzreignz Wow that function is amazing :D

Answer 37

A slightly more general version would be:

\[r=\frac{ |\sin(\theta + \phi)| + \sin(\theta + \phi)}{ |\sin(\theta + \phi)| + \sin(\theta + \phi) }\]

Where phi is the amount in radians that you want to rotate your semicircle, so if you choose pi, then you'd get the other half.

Another alternative might be to replace sine with cosine, negative sine, or negative cosine rather than worry about adding a angle inside, but that will only get you to the four obvious semicircles.

Answer 38

ok
u know
1=x^2+y^2 drow a circl in rectangle coordinate wid center:0,0
nw
1=root x^2+y^2 drow a semi circl in rectangle coordinate wid center:0,0
the same idea of polar
asume
x=sin theta
y=cos theta

sin theta ^2 +cos theta^2 = 1 the same of first equation but in polar coordinate wid center:0,0
nw
root sin theta ^2 +cos theta^2 = 1 is semi circle
so
r^2=1 is a circle wid radus 1 center 0,0 (unit circle)
r =1 is a semi circl radus 1 center 0,0