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Finding polar coordinates of polar points

  1. Jun 8, 2012 #1
    1. The problem statement, all variables and given/known data
    Plot the Following points(given in polar coordinates). Find all the polar coordinates of each point.
    a. (2, pi/2)
    b. (2,0)
    c. (-2, pi/2)
    d. (-2,0)


    2. Relevant equations
    none


    3. The attempt at a solution
    I have plotted it on a graph but could someone explain to me how this answer is correct for a.?
    (2, pi/2 + 2npi) and (-2, pi/2 + (2n + 1)pi)
    where n is any integer.

    So what I have is there is a point on the positive x-axis on 2. An angle pi/2 since r=2 and theta = pi/2.
    The ray passes down from the positive y-axis to the negative y-axis. So the angle from the initial ray to the end of the ray on the negative y-axis is 3pi/2.


    *Sorry I didn't use Latex, I figured it out after finishing typing out the question.
     
  2. jcsd
  3. Jun 8, 2012 #2
    Once you have (2,π/2), then you should realize adding k2π to the angle works for any k an integer. Thus you have (2,π/2+k2π)

    But that does not describe all polar coordinates representing the point if we allow negative r values (more advanced classes typically do not). As you described, we have angle down of 3π/2, however to get the positive y value 2, we need [itex]r=-2[/itex].

    Thus we have (-2,3π/2). Then we can add k2π for any integer k and we have all the polar representations. Instead of writing 3π/2+k2π, the solution you described has π/2+(2k+1)π, which is the same thing.
     
  4. Jun 9, 2012 #3
    I see. Can you recommend any sites, videos, or readings that I can do to help me fully understand this concept or part in Polar Coordinates? I am taking summer courses now and I want to understand everything since its accelerated and I don't want it coming back to bite me when I have to take a cumulative final exam. I tried khanacademy but I couldn't get any videos that explain it. Maybe I'm not looking any harder. Any help is great!
    Also thanks for the reply!
     
  5. Jun 9, 2012 #4
    I did an internet search for "polar coordinate exercises", maybe someone else can point to videos, but I would first recommend just trying to apply what you know to exercises that you can find in your textbook or online.
     
  6. Jun 10, 2012 #5
    Ok thanks I will do that. I just have questions like about the Unit Circle. Its been awhile since I lasted used it and I forgot how to use it. What I mean is that I want to understand how to find the next angle of the unit circle like it starts at 0 and the next angle should be pi/6 if I remembered correctly. I want to be able to fully draw it out and also know what is sin, cos, tan, sec, csc, and cot of all of the angles in the unit circle. I am sure I need to find some Pre-Calculus materials to understand the concept.
     
  7. Jun 10, 2012 #6
    Then I have an exercise for you.
    1. For the 45-45-90 triangle, assume the hypotenuse (radius) is one, and find the lengths of the sides.
    2. For the 30-60-90 triangle, with hypotenuse one, find first the length of the shorter side, then the longer side.

    Hint: for the shorter side of the 30-60-90 triangle, compare to a 60-60-60 triangle.

    Then just remember that sin(theta)=y/r etc. So if r=1, like on the unit circle, you're life get's a little easier.



    I think this method will make it easier to remember a lot of trig.
     
  8. Jun 10, 2012 #7
    For your first question.
    So we have a 45-45-90 triangle.
    Hypotenuse is 1 or r=1 (radius)
    The equation I would use is: x^2 + y^2 = 1
    But both sides of the 45-45-90 triangle are the same "n" length.
    so I would make this equation: 2n^2 = 1
    square root both sides to get: sqroot(2)*(n) = 1
    divide by sqroot(2) and n = 1/sqroot(2)
    so both sides are equal to 1/sqroot(2)

    Thank you algebrat. This was a eye opener because I forgot what I learned in Pre-Calc but I will start on the 2nd one now.
     
  9. Jun 10, 2012 #8
    I am having trouble with the 2nd question you are asking me. I remembered there was a way to find two unknown sides given only the hypotenuse and the angles. Could you tell me if I am going in the right direction?
     
  10. Jun 10, 2012 #9
    How does the 30-60-90 triangle fit inside the 60-60-60 triangle. You'll have to draw it. Then, that long ago grade school geometry will tell you that if hypotenuse is 1, then the shorter side is 1/2. Then pythagoreans eqn will tell you the other side is root 3 over 2. Now you only have two triangles to memorize. And you'll remember how to get them if you forget the numbers. Since 3>1, the shorter side is the 1/2. Now you know sine of 30, 90, 120, 240, -60 et cetera. Try it. And try tan of 30, rationalize the denominator, and remember to recognize that number, one day someone could give you an inverse problem and you might find it useful.
     
  11. Jun 10, 2012 #10
    I see. Let me apply this to the triangle I drew and understand what you said/typed.
    I'm gonna try find a chart or something that has all angles and their sin, cos, and tan to study and memorize them.
     
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