Cartesian points in polar coordinates.

  • Thread starter lakitu
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  • #1
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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

  • #2
arildno
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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?
 
  • #3
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
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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
Integral
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Look again at the radial component. How far is it from the origin to (3,3)?
 
  • #6
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im not sure what you mean? i can only think the distance is 6 if its not 3 what i originally believed :)
 
  • #7
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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
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 [tex]3\sqrt{2}[/tex].
 
  • #9
Integral
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Would you please show us the relationship between polar and cartesian coordinates.
 
  • #10
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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
PPonte
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].
 
  • #12
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wow at last! I think i am going to have to change my username after this topic!

thanks
 
  • #13
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Could you explain to me how you work out that the precise r is [tex]3\sqrt{2}[/tex] ?

Thank you
 
  • #14
Integral
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What is the length of the hypotenus of a right triangle when both of the other sides have length 1?
 
  • #15
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
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i get it AC^ = AB^ + BC^ :)
 
  • #17
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no i stll dont get it:(
 
  • #18
Integral
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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?
 
  • #19
VietDao29
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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:
[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:
  • #20
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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.
 
  • #21
PPonte
lakitu, maybe you are not visualising well. Hope this image helps.

http://img72.imageshack.us/img72/2572/radial7nd.gif [Broken]
 
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