Question

Help with Polar coordinates

Answer 1

Convert the rectangular coordinate (-3, 4) to polar coordinates.

Answer 2

So tan^-1(4/3)= 53.13 degrees, but our points are in the 2nd quadrant would we add 180?

Answer 3

So we need R and theta

\[\large r = \sqrt{x^2 + y^2}\]
\[\large \theta = \tan^{-1}(\frac{y}{x})\]

Answer 4

http://www.teacherschoice.com.au/maths_library/coordinates/polar_-_rectangular_conversion.htm

Answer 5

r=sqrt(-3)^2+(4)^2
r=sqrt(9+16)
r=sqrt25
r=5

Answer 6

Right, your R = 5

As far as the angle...remember its actually \(\large tan^{-1}(\frac{4}{-3})\) so you would have -53.13 degrees right?

Answer 7

x=r cos t and y=r sin t put find valve of r by square and add by divide find angle polar coordinate is (r,t)

Answer 8

Indeed, but I was confused with this example in my text

Answer 9

Oh wait nvm I understand now negative / negative = positive *facepalm*

Answer 10

Yeah lol dont worry about it :D

Answer 11

so in this case would we still add 180 to get back to the second quadrant?

Answer 12

Correct!

Answer 13

180+-53.13=126.87 degrees which almost looks like

Answer 14

Yeah good enough! lol

So we have

\[\large (5, 126.87^\circ)\]

Answer 15

Indeedly It almost went smoothly besides my failure to recognize 4/-3

Answer 16

Lol no problem, mistakes happen!