polar coordinates...doing something wrong again...

(x,y)=(-4,-1)

find the polar coordiantes on -pi/2

Question

polar coordinates...doing something wrong again...

(x,y)=(-4,-1) 

find the polar coordiantes on -pi/2<theta<pi/2

r=sqrt(-4^2+-1^2)=sqrt 17 but webwork says it's wrong.  

same with (-2,0)
r = sqrt(-2^2+0^2)=2

and 

(0,-3)

r=sqrt(0+3^2)=3

now the theta's for those are right, but the theta for the last is wrong, too.  
theta=arctan (-3/0)
you can't divide by zero, so what do i do here?

anonymous · Answer

You've been restricted to angles that put you into quadrants 1 and 4, yet the points are mostly on the other side of the y axis. For the first one, try -sqrt(17) , it is flipped across the pole.

anonymous · Answer

Same problem with the second one. Try -2

anonymous · Answer

k. making the wrong r's negative worked.  how am i supposed to know when they are negative and when they are positive?  also, how do i do the theta for the last one?

anonymous · Answer

The last one...
Tangent is undefined in two places, pi/2 and -pi/2. Since your point is straight down, the angle is -pi/2.
I always graph the rectangular first so that I can find these little tricks when converting to polar.

anonymous · Answer

thanks.

anonymous · Answer

You're welcome.