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Mathematics 14 Online
OpenStudy (dugalde):

Can you help me find the polar coordinates of each rectangular coordinates (4,0) (0,3) (2,2)

OpenStudy (jdoe0001):

notice, they're just x,y coordinate values thus, plot them in a cartesian plane grid get the distance, that'd be "r" get the angle, that'd be \(\theta\) then plug the values for the polar coordinates using that :)

OpenStudy (jdoe0001):

usually you'd need to use the tangent function to get the angle but in this case, is too simple

OpenStudy (dugalde):

so will the first one be (4,0 degrees)

OpenStudy (jdoe0001):

well.... plot here |dw:1431733421540:dw|

OpenStudy (dugalde):

|dw:1431732726359:dw|

OpenStudy (jdoe0001):

yeap... so r = 4 and angle is 0 so 4, 0 is the polar

OpenStudy (dugalde):

what equation would I use

OpenStudy (jdoe0001):

|dw:1431733663462:dw| notice, the angle is 90 degrees and the distance is 3

OpenStudy (jdoe0001):

hmm what equation? for.... well.. that was the 1st one, (4,0) in rectangular is the same in polar

OpenStudy (dugalde):

ok Next one would be (3,90 degrees)

OpenStudy (jdoe0001):

0, 3 has an r = 3, and an angle of \(\frac{\pi }{2}\)

OpenStudy (jdoe0001):

yeap or \(\bf (3,\frac{\pi }{2})\)

OpenStudy (dugalde):

next one would be (2sqrt(2), 45 degrees)

OpenStudy (jdoe0001):

now... 2.2.... well.. hmmm yeap, that's correct \(\bf r=\sqrt{x^2+y^2}\implies r=\sqrt{2^2+2^2}\implies r=\sqrt{8}\implies r=2\sqrt{2} \\ \quad \\ \theta=tan^{-1}\left( \cfrac{y}{x} \right)\implies \theta=tan^{-1}\left( \cfrac{\cancel{2}}{\cancel{2}} \right)\)

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