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harsimran_hs4 Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.1
What?\[1 + 2\sin(\theta) = 1\]
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.1
\[2\sin(\theta) = 0\]Find the solutions to theta.
 one year ago

yaho021 Group TitleBest ResponseYou've already chosen the best response.0
\[\Theta = 0\]
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.1
Yeah, there are many solutions
 one year ago

yaho021 Group TitleBest ResponseYou've already chosen the best response.0
how to find the intersection of the curve?
 one year ago

harsimran_hs4 Group TitleBest ResponseYou've already chosen the best response.0
in this system of coordinates theta can lie only between 0 to 2pi so you can say theta = 0 is the required intersection
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.1
Indeed.
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.1
But \(2\pi\) is one too.
 one year ago

yaho021 Group TitleBest ResponseYou've already chosen the best response.0
what are the intersections of the curve?
 one year ago

yaho021 Group TitleBest ResponseYou've already chosen the best response.0
one of the point of intersection of the curve is (1,0), ?
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.1
If we can write the coordinates as \((\theta,r)\) then \((0,1)\) is the intersection.
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.1
Wait, do you have polar coordinates?!
 one year ago

harsimran_hs4 Group TitleBest ResponseYou've already chosen the best response.0
@ParthKohli 2pi is just a depiction of one complete cycle so 2pi coincides with zero itself
 one year ago

yaho021 Group TitleBest ResponseYou've already chosen the best response.0
how about the other intersections??
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.1
\((2n\pi,1)\) where \(n \) is the set of all integers
 one year ago

yaho021 Group TitleBest ResponseYou've already chosen the best response.0
the other intersection is ( 2\Pi, 1)?
 one year ago

harsimran_hs4 Group TitleBest ResponseYou've already chosen the best response.0
@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
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.1
Yeah, you are right. So are we talking about polar coordinates?
 one year ago

yaho021 Group TitleBest ResponseYou've already chosen the best response.0
\[\theta \le 2\Pi ?\]
 one year ago

harsimran_hs4 Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago
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