Converting rectangular to polar

In summary: 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.)
  • #1
kElect
20
0

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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  • #2
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.
 
  • #3
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.
 
  • #4
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.
 
  • #5
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.
 
  • #6
I guess you could state your answer like this:
[itex]\left( \sqrt{5}, \arctan (-2) \right)[/itex]
(Fortunately, we can leave arctan (-2) as it is, ie. not add a multiple of pi, because θ is in Q IV.)
 

Related to Converting rectangular to polar

What is the process for converting rectangular coordinates to polar coordinates?

The process for converting rectangular coordinates (x, y) to polar coordinates (r, θ) involves using trigonometric functions to calculate the distance from the origin (r) and the angle (θ) between the point and the positive x-axis.

What are the formulas for converting rectangular coordinates to polar coordinates?

The formulas for converting rectangular coordinates to polar coordinates are:
r = √(x² + y²)
θ = tan⁻¹(y/x)

What are the advantages of using polar coordinates over rectangular coordinates?

Polar coordinates are useful for representing circular or symmetrical shapes, making them more suitable for certain mathematical problems. They also have a simpler form for representing complex numbers, making calculations easier.

Can you convert polar coordinates back to rectangular coordinates?

Yes, polar coordinates can be converted back to rectangular coordinates using the following formulas:
x = r cos(θ)
y = r sin(θ)

Are there any limitations to using polar coordinates?

One limitation of polar coordinates is that they can only represent points in the first quadrant of the Cartesian plane. Additionally, some equations may be more difficult to express in polar form.

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