Question

find the polar coordinates r>0 and r<0;
(-6 sqrt 2, -6 sqrt2)

Answer 1

for given (x,y) your polar coordinates are (r,\(\theta\))
r = \( \sqrt{x^2+y^2}\)
\( \theta \) = \( \arctan(\frac{y}{x})\)

Answer 2

so i got \[\sqrt{-12+\left( -12 \right)}\]

Answer 3

one of the 12 has to be positive but im not sure how to make it positive

Answer 4

\( (-6\sqrt{2})^2\)

Answer 5

= 72

Answer 6

oh 72+72=144
\[\sqrt{144}=\pm12\]

Answer 7

+12

Answer 8

the book as (12, 5pi/4) and (-12, pi/4); where did the pi/4 and 5pi/4 come from?

Answer 9

angle,
look at that theta.

Answer 10

also try to draw coordinate first.

Answer 11

Im a little confused how can i draw the coordinate if i only have 12 as an answer?