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harsimran_hs4
 one year ago
Best ResponseYou've already chosen the best response.0r = 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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

harsimran_hs4
 one year ago
Best 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

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

harsimran_hs4
 one year ago
Best ResponseYou've already chosen the best response.0yes 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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