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Homework Help: Cartesian points in polar coordinates.

  1. Mar 18, 2006 #1
    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

  2. jcsd
  3. Mar 18, 2006 #2


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    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?
  4. Mar 18, 2006 #3
    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
  5. Mar 18, 2006 #4
    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
  6. Mar 18, 2006 #5


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    Look again at the radial component. How far is it from the origin to (3,3)?
  7. Mar 18, 2006 #6
    im not sure what you mean? i can only think the distance is 6 if its not 3 what i originally believed :)
  8. Mar 18, 2006 #7
    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 :)
  9. Mar 18, 2006 #8
    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 [tex]3\sqrt{2}[/tex].
  10. Mar 18, 2006 #9


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    Would you please show us the relationship between polar and cartesian coordinates.
  11. Mar 18, 2006 #10
    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 ?
  12. Mar 18, 2006 #11
    Yes. That's right. :approve:
    But you could use instead of the approximated 4.24 the precise r, which is [tex]3\sqrt{2}[/tex].
  13. Mar 18, 2006 #12
    wow at last! I think i am going to have to change my username after this topic!

  14. Mar 18, 2006 #13
    Could you explain to me how you work out that the precise r is [tex]3\sqrt{2}[/tex] ?

    Thank you
  15. Mar 18, 2006 #14


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    What is the length of the hypotenus of a right triangle when both of the other sides have length 1?
  16. Mar 18, 2006 #15
    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.
  17. Mar 18, 2006 #16
    i get it AC^ = AB^ + BC^ :)
  18. Mar 18, 2006 #17
    no i stll dont get it:(
  19. Mar 18, 2006 #18


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    Why do you use the sides of some unknown triangle, when you have a number for the lengths of the sides?
  20. Mar 19, 2006 #19


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    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:
    [tex]d = PQ = \sqrt{(x_P - x_Q) ^ 2 + (y_P - y_Q) ^ 2}[/tex].
    Now apply this, adn see if you can work out [tex]r = 3 \sqrt{2}[/tex].
    Remember that the origin O is (0, 0).
    Can you go from here? :)
    Last edited: Mar 19, 2006
  21. Mar 20, 2006 #20
    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.
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