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yaho021

  • one year ago

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

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  1. harsimran_hs4
    • one year ago
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    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
    • one year ago
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    What?\[1 + 2\sin(\theta) = 1\]

  3. yaho021
    • one year ago
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    then?

  4. ParthKohli
    • one year ago
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    \[2\sin(\theta) = 0\]Find the solutions to theta.

  5. yaho021
    • one year ago
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    \[\Theta = 0\]

  6. ParthKohli
    • one year ago
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    Yeah, there are many solutions

  7. yaho021
    • one year ago
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    how to find the intersection of the curve?

  8. harsimran_hs4
    • one year ago
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    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
    • one year ago
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    Indeed.

  10. ParthKohli
    • one year ago
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    But \(2\pi\) is one too.

  11. yaho021
    • one year ago
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    what are the intersections of the curve?

  12. yaho021
    • one year ago
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    one of the point of intersection of the curve is (1,0), ?

  13. ParthKohli
    • one year ago
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    If we can write the coordinates as \((\theta,r)\) then \((0,1)\) is the intersection.

  14. ParthKohli
    • one year ago
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    Wait, do you have polar coordinates?!

  15. harsimran_hs4
    • one year ago
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    @ParthKohli 2pi is just a depiction of one complete cycle so 2pi coincides with zero itself

  16. yaho021
    • one year ago
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    how about the other intersections??

  17. ParthKohli
    • one year ago
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    \((2n\pi,1)\) where \(n \) is the set of all integers

  18. yaho021
    • one year ago
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    the other intersection is ( 2\Pi, 1)?

  19. harsimran_hs4
    • one year ago
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    @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

  20. ParthKohli
    • one year ago
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    Yeah, you are right. So are we talking about polar coordinates?

  21. yaho021
    • one year ago
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    \[\theta \le 2\Pi ?\]

  22. harsimran_hs4
    • one year ago
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    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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