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## anonymous 3 years ago find the intersections of the curve r=1+2sinθ r=1

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1. harsimran_hs4

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

2. ParthKohli

What?$1 + 2\sin(\theta) = 1$

3. anonymous

then?

4. ParthKohli

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

5. anonymous

$\Theta = 0$

6. ParthKohli

Yeah, there are many solutions

7. anonymous

how to find the intersection of the curve?

8. harsimran_hs4

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

9. ParthKohli

Indeed.

10. ParthKohli

But $$2\pi$$ is one too.

11. anonymous

what are the intersections of the curve?

12. anonymous

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

13. ParthKohli

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

14. ParthKohli

Wait, do you have polar coordinates?!

15. harsimran_hs4

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

16. anonymous

how about the other intersections??

17. ParthKohli

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

18. anonymous

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

19. harsimran_hs4

@ParthKohli i dont think (2npi, 1) will do because polar coordinates have limits 1. r is positive 2. theta lies b/w 0 to 2pi (so we dont talk of theta beyond them) point me out if i am wrong

20. ParthKohli

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

21. anonymous

$\theta \le 2\Pi ?$

22. harsimran_hs4

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

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