# Cartesian points in polar coordinates.

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

## Answers and Replies

arildno
Homework Helper
Gold Member
Dearly Missed
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?

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

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

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

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

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 :)

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 $$3\sqrt{2}$$.

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

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 ?

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

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

thanks

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

Thank you

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

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.

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

no i stll dont get it:(

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

VietDao29
Homework Helper
lakitu said:
no i stll dont get it:(
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:
lakitu said:
i get it AC^ = AB^ + BC^ :)
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.

PPonte
lakitu, maybe you are not visualising well. Hope this image helps.