Question

find the intersections of the curve
r=1+2sinθ
r=1

Answer 1

r = 1 is a circle with radius 1
we know r always has to positive
so tell me all the values of theta in (0, 2pi) such that r >= 0
1+ 2sin(theta) >= 0
sin(theta) >= -1/2

tell me the answer

Answer 2

What?\[1 + 2\sin(\theta) = 1\]

Answer 3

then?

Answer 4

\[2\sin(\theta) = 0\]Find the solutions to theta.

Answer 5

\[\Theta = 0\]

Answer 6

Yeah, there are many solutions

Answer 7

how to find the intersection of the curve?

Answer 8

in this system of coordinates theta can lie only between 0 to 2pi
so you can say theta = 0 is the required intersection

Answer 9

Indeed.

Answer 10

But \(2\pi\) is one too.

Answer 11

what are the intersections of the curve?

Answer 12

one of the point of intersection of the curve is (1,0), ?

Answer 13

If we can write the coordinates as \((\theta,r)\) then \((0,1)\) is the intersection.

Answer 14

Wait, do you have polar coordinates?!

Answer 15

@ParthKohli 2pi is just a depiction of one complete cycle
so 2pi coincides with zero itself

Answer 16

how about the other intersections??

Answer 17

\((2n\pi,1)\) where \(n \) is the set of all integers

Answer 18

the other intersection is ( 2\Pi, 1)?

Answer 19

@ParthKohli
i don`t think (2npi, 1) will do
because polar coordinates have limits
1. r is positive
2. theta lies b/w 0 to 2pi

(so we don`t talk of theta beyond them)
point me out if i am wrong

Answer 20

Yeah, you are right. So are we talking about polar coordinates?

Answer 21

\[\theta \le 2\Pi ?\]

Answer 22

yes i think these are polar

@yaho021 boundary conditions are always controversial
so either you include 0 (which is done always) or 2pi(which i have never seen till now)

0 will be safe side because after all 0 or 2pi depict same condition