# Cartesian points in polar coordinates.

1. Mar 18, 2006

### lakitu

Hey everyone, my lecture has given me this question, I am unsure where to start with it.

Express the Cartesian point (3, 3) in polar coordinates.

Do i need to use the sin and cos on my calc.

Any help would be very helpful

lakitu

2. Mar 18, 2006

### arildno

Instead of resorting to a calculator, draw the line segment from the origin to the point (3,3).
What is the angle this line segment makes with the positive x-axis?

3. Mar 18, 2006

### PPonte

lakitu, you could have done some research. That's one good thing I learned from this forum. I didn't learn yet polar coordinates and I think I can resolve this exercise by simply reading wikipedia's introduction on polar coordinates.

See- http://en.wikipedia.org/wiki/Polar_coordinates

4. Mar 18, 2006

### lakitu

your right, i guess i assumed it was a little tougher than it was :)

to arildno: Is the angle 45 deg ? would that make the answer (3,45)

kind regards lakitu

5. Mar 18, 2006

### Integral

Staff Emeritus
Look again at the radial component. How far is it from the origin to (3,3)?

6. Mar 18, 2006

### lakitu

im not sure what you mean? i can only think the distance is 6 if its not 3 what i originally believed :)

7. Mar 18, 2006

### lakitu

i read that the position of the point is defined by its direct distance from the origin (O) do you measure this with a ruler? I am just unsure :)

8. Mar 18, 2006

### PPonte

lakitu, see the introduction of wikipedia and the formula to determine the radial distance from the pole and think if it really is 6 or 3 or $$3\sqrt{2}$$.

9. Mar 18, 2006

### Integral

Staff Emeritus
Would you please show us the relationship between polar and cartesian coordinates.

10. Mar 18, 2006

### lakitu

i found this example in my text book, r = sqrt(x*x + y*y) a = atan(y / x) which would give me the distance of 4.24 for r (the origin to 3,3)

so would the answer be (4.24,45deg)?

i did read your recomendations but struggled to figure those out :)

am i on the right lines ?

11. Mar 18, 2006

### PPonte

Yes. That's right.
But you could use instead of the approximated 4.24 the precise r, which is $$3\sqrt{2}$$.

12. Mar 18, 2006

### lakitu

wow at last! I think i am going to have to change my username after this topic!

thanks

13. Mar 18, 2006

### lakitu

Could you explain to me how you work out that the precise r is $$3\sqrt{2}$$ ?

Thank you

14. Mar 18, 2006

### Integral

Staff Emeritus
What is the length of the hypotenus of a right triangle when both of the other sides have length 1?

15. Mar 18, 2006

### PPonte

lakitu, follow Integral's suggestion. I would have explained to you how do to it, but you would't learn as well as you will if you think for yourself.

16. Mar 18, 2006

### lakitu

i get it AC^ = AB^ + BC^ :)

17. Mar 18, 2006

### lakitu

no i stll dont get it:(

18. Mar 18, 2006

### Integral

Staff Emeritus
Why do you use the sides of some unknown triangle, when you have a number for the lengths of the sides?

19. Mar 19, 2006

### VietDao29

Uhmm, I suggest you reading your textbook again. There should be some chapter about the distance betweeen 2 points in Cartesian coordinate. The distance between 2 points P(xP, yP), and Q(xQ, yQ) is:
$$d = PQ = \sqrt{(x_P - x_Q) ^ 2 + (y_P - y_Q) ^ 2}$$.
Now apply this, adn see if you can work out $$r = 3 \sqrt{2}$$.
Remember that the origin O is (0, 0).
Can you go from here? :)

Last edited: Mar 19, 2006
20. Mar 20, 2006

### Markjdb

You were on the right track with this. If you have a point, (3, 3), you can use this theorem to work out the length of the hypotenuse, which is the distance between (3, 3) and the origin.