Polar Coordinates! Please help!

[gabespeep]

I am not sure if I am doing this correctly and I was hoping to get some guidence in the right direction if I am doing this wrong. I have to take the rectangular coordinates of (-5,2) and translate them into polar coordinates.
So far I have found r:

r^2=(-5)^2+(2)^2
r^2=25+4
r=√29


Θ=π+tan^-1(2/-5), since the coordinates fall in the second quadrant.
≈ -18.66
Polar Coordinates≈(√29,-18.66) ? I dont know if this is correct or not? If not what am I doing wrong so that I will know for the future how I am supposed to solve this type of problem. Any help is much appreciated!

 

[Yogi]

r is correct. You said the point is in the second quadrant but if the starting point is 0 degrees, your answer is clockwise 18.66 degrees (in the 4th quadrant).

I would draw a picture, then don't worry about the sign of -2/5 too much, just tan^-1(2/5)=21.8 degrees tells you the smaller angle of the right triangle which should lead you to the answer. Hint: if its in quadrant II the theta portion of the answer should be between 90 and 180). After you understand it doing it like that a couple times then go to your formula, where I am guessing the values of n can be 90, 180, 270, 360 in which you pick the one to get you to the correct quadrant. I don't know how you got 18.66.

 

[pka]

I am not sure if I am doing this correctly and I was hoping to get some guidence in the right direction if I am doing this wrong. I have to take the rectangular coordinates of (-5,2) and translate them into polar coordinates.
So far I have found r:
r^2=(-5)^2+(2)^2
r^2=25+4
r=√29
Θ=π+tan^-1(2/-5), since the coordinates fall in the second quadrant.

Actually you are correct as for as I quoted above.
$\displaystyle \phi = \arctan \left( {\frac{2}{{ - 5}}} \right) + \pi$
Your mistake is trying to convert to 'degrees'. You should leave it as number.

 

[gabespeep]

I got -18.66 by having my calculator on the DEG mode. When its on the GRAD mode I get the same answer as you did. I still don't fully understand what I am supposed to do though. I know how to find r, and I know that the coordinates of (-5,2) fall into the second quadrant. My book says to use the formula Θ=Pi +tan^-1(y/x) for points that fall in quad 2 or 3, and Θ=tan^-1(y/x) for points that fall into quads 1 or 4.... I am so lost on this second part, even when I draw it out. This is the first day of me attempting to learn the polar coordinates.

 

[pka]

I got -18.66 by having my calculator on the DEG mode. When its on the GRAD mode I get the same answer as you did. I still don't fully understand what I am supposed to do though. I know how to find r, and I know that the coordinates of (-5,2) fall into the second quadrant. My book says to use the formula Θ=Pi +tan^-1(y/x) for points that fall in quad 2 or 3, and Θ=tan^-1(y/x) for points that fall into quads 1 or 4.... I am so lost on this second part, even when I draw it out. This is the first day of me attempting to learn the polar coordinates.

Well your book is correct about $\displaystyle \theta$. But again the mistake is using degrees. Most current mathematical practices have no place for degrees! In fact, the default setting on most calculators and computer algebra systems is ths so-called radian mode. But in this context, radian just means number.

 

[mmm4444bot]

When [my calculator is] on the GRAD mode I get the same answer as you did. I still don't fully understand what I am supposed to do though.

You are supposed to use radian mode, when working with polar coordinates.

The formulas are stated for angles measured in radians.

In other words, if the terminal ray of an angle in standard position has rotated from 0 to 2.76 radians, then measuring outward along that ray √29 units from the origin puts you at the point (-5,2).

Also, you understand that -2/5 is not an angle measurement when it appears in arctan(-2/5), yes? (Just wondering :cool

 

[gabespeep]

Thank you for taking the time to help me!

Ok! Thank you very much for that information because I did not know that, and it was very helpful!