Converting rectangular to polar

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To convert the rectangular coordinates (1, -2) to polar form, the radius r is calculated using the formula r = sqrt(x^2 + y^2), resulting in r = sqrt(5). The angle theta is determined using tan(theta) = y/x, which gives tan(theta) = -2. The angle is in the fourth quadrant, and while a calculator can provide a decimal value, it can also be expressed as theta = arctan(-2). The final polar coordinates are (sqrt(5), arctan(-2)), with the understanding that theta does not need further adjustments for quadrant placement.
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Homework Statement


How do you convert the rectangular coordinate points (1, -2) to polar form?

note: rectangular is (x,y) polar is (r, theta)

Homework Equations


r^2 = x^2 + y^2 , x = rcos(theta) , y = rsin(theta) , tan(theta) = y/x

The Attempt at a Solution


So basically, I tried getting it to polar form by first finding the radius. This part was easy since all I had to do was to plug it into the first equation.

r^2 = (1)^2 + (-2)^2
= sqrt(5)

Next, I tried getting theta by getting tan(theta).

tan(theta) = -2/1

Here is where the problem came in. When I tried putting this onto the unit circle, I didn't get any recognizable triangles(such as the 45-45 right triangle or the 30-60-90 right triangle).

So how would I find theta?(without using calculator)
 
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kElect said:

Homework Statement


How do you convert the rectangular coordinate points (1, -2) to polar form?

note: rectangular is (x,y) polar is (r, theta)


Homework Equations


r^2 = x^2 + y^2 , x = rcos(theta) , y = rsin(theta) , tan(theta) = y/x


The Attempt at a Solution


So basically, I tried getting it to polar form by first finding the radius. This part was easy since all I had to do was to plug it into the first equation.

r^2 = (1)^2 + (-2)^2
= sqrt(5)

Next, I tried getting theta by getting tan(theta).

tan(theta) = -2/1

Here is where the problem came in. When I tried putting this onto the unit circle, I didn't get any recognizable triangles(such as the 45-45 right triangle or the 30-60-90 right triangle).

So how would I find theta?(without using calculator)

You wouldn't. You could express it is Arctan(-2) since that is in the 4th quadrant but you will need a calculator or tables for a decimal answer.
 
LCKurtz said:
You wouldn't. You could express it is Arctan(-2) since that is in the 4th quadrant but you will need a calculator or tables for a decimal answer.

even using a calculator my answer comes out to be -63.435 as theta whereas the answer is -1.107.
 
kElect said:
even using a calculator my answer comes out to be -63.435 as theta whereas the answer is -1.107.
That's because -63.435 is in degrees whereas -1.107 is in radians.
 
eumyang said:
That's because -63.435 is in degrees whereas -1.107 is in radians.

oh ty.

hmm, strange. professor said calculator wasn't necessary.
 
I guess you could state your answer like this:
\left( \sqrt{5}, \arctan (-2) \right)
(Fortunately, we can leave arctan (-2) as it is, ie. not add a multiple of pi, because θ is in Q IV.)
 
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