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LordVictorius

Explain polar coords and give an example.

  • one year ago
  • one year ago

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  1. Ajk
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    With rectangular coords we typically represent locations as the intersection of lines on a horizontal/vertical axes. With polar coords, we take the magnitude of the x and y parts that we are used to on a cartesian coord system. so if we had a point (3,4), to find the magnitude we would just that the square root of (3^2 + 4^2). Just think of it like finding the hypotenuse if you were doing pythagorean theoram. What you now have, in polar, is rhe "r", which is just the length of the ray (line) from the origin (0,0). But, we also need something else called the argument, which is the angle (or theta) from the x-axis. If we had some point on the cartesian plane, like (1,1) then it would have an "r" of sqrt(2), and an argument of pi/4. I would write it like sqrt(2)[cos(pi/4) + sin(pi/4)]. You might also be able to write it like sqrt(2)*e^i*(pi/4).

    • one year ago
  2. LordVictorius
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    What is the Cartesian Plane?

    • one year ago
  3. Ajk
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    just the normal x,y coords that we are used to.

    • one year ago
  4. LordVictorius
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    Okay.

    • one year ago
  5. LordVictorius
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    Where did you get the r's from (1,1)?

    • one year ago
  6. Ajk
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    if I take the sqrt of 1^2 + 1^2, I have sqrt(2). Just think of it like finding the hypotenuse of a right triangle.

    • one year ago
  7. LordVictorius
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    So if this (1,1) was (x,y), then it is x^2+y^2?

    • one year ago
  8. LordVictorius
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    Or the sqrt(x^2+y^2)?

    • one year ago
  9. Ajk
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    with x,y coords we look for whee they intersect and that is the point. with polar, we have a ray that extends from the origen. the length of that ray is found just by the same way that you would find the hyp of a right triangle.

    • one year ago
  10. Ajk
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    sqrt(x^2 + y^2)

    • one year ago
  11. LordVictorius
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    Oh! Okay, thanks!

    • one year ago
  12. Ajk
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    remember, that just gives you the length of the ray from the origin. You still need to include an angle, or else we would have no idea where the endpoint of the ray is.

    • one year ago
  13. LordVictorius
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    Wait, so how do I get the angle?

    • one year ago
  14. Ajk
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    You will likely have to use your trig and the length of the x (cosine) and y (sine) parts to determine angle. If you have an angle that can be represented at y/x, then you also have something that can be represented as tangent theta, right? (because sin/cos). So once you have that, then you take the arctan (y/x) and that will give you the angle.

    • one year ago
  15. Ajk
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    so for the example I used above, if I have x = 1 and y = 1, then what I really have is y/x = 1/1 = 1. So I take tangent theta = 1, ----> theta = arctan (or tan inverse) (1), and that is pi/4 (or 45 degrees if you are in degree mode)

    • one year ago
  16. LordVictorius
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    Can you help me with this problem? I think it might be simpler to explain the steps. Use your grapher to determine which of the graphs matches the polar equation r = 3 sin 2θ.

    • one year ago
  17. Ajk
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    what is the theta?

    • one year ago
  18. LordVictorius
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    IT says 3 sin 2(theta)

    • one year ago
  19. Ajk
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    Without rectangular coords to convert to polar, or a polar coord to convert back to rectangular, I'm not really sure what to do with that. Where are the graphs the question is asking you to compare it to?

    • one year ago
  20. LordVictorius
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    The answer choices?

    • one year ago
  21. Ajk
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    yeah

    • one year ago
  22. LordVictorius
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    • one year ago
  23. Ajk
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    ah, ok. hold on, let me plot it.

    • one year ago
  24. Ajk
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    it is the third option

    • one year ago
  25. LordVictorius
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    How did you get that?

    • one year ago
  26. Ajk
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    do you have a graphing calculator? On mine, I switched the graph-type from rectangular to polar, then I just entered what you gave me: 3sin(2theta). It plotted the 3rd option.

    • one year ago
  27. LordVictorius
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    Oh, okay.

    • one year ago
  28. LordVictorius
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    Can you do another one? Find the rectangular coordinates of the point with the polar coordinates (8,3/2*pi ).

    • one year ago
  29. Ajk
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    My calculator isn't cooperating, but lets see if we can reason the answer. We know r=8, and we have a theta of 3/2*pi. So I think then that the point is on the y-axis below zero. We have a problem trying to use tangent right away, because cosine of (3/2*pi) = 0, that means tangent give a divide by zero error. Even so, we know the ray is length 8, so I think the rectangular coords are (0,-8)

    • one year ago
  30. LordVictorius
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    Okay,

    • one year ago
  31. LordVictorius
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    That's an answer! gj!

    • one year ago
  32. LordVictorius
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    Wait, can I try one, and tell me if it's correct/

    • one year ago
  33. Ajk
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    Sure

    • one year ago
  34. LordVictorius
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    Find the rectangular coordinates of the point with the polar coordinates . This is the question.

    • one year ago
  35. LordVictorius
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    (-3, 5pi/8)

    • one year ago
  36. LordVictorius
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    Those are the coords.

    • one year ago
  37. LordVictorius
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    So r=-3, right?

    • one year ago
  38. LordVictorius
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    Agh, I'm confused.

    • one year ago
  39. Ajk
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    r = -3, but remember that r is sqrt(x^2 + y^2). We get the negative sign because 5/8*pi is in quadrant II, which means the cosine (x) part is a negative number.

    • one year ago
  40. LordVictorius
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    So I will need a unit circle, right?

    • one year ago
  41. Ajk
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    That is probably what I would do, look at tan(5/8pi) first

    • one year ago
  42. LordVictorius
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    It's not on the unit circle.

    • one year ago
  43. Ajk
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    if you divide pi (the top half of the circle) into 8, you can see that 5/8pi is just a little past pi/2

    • one year ago
  44. LordVictorius
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    So pi/2?

    • one year ago
  45. LordVictorius
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    0,1

    • one year ago
  46. Ajk
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    what I did to find x was 3cos(5/8*pi), and to find y was 3sin(5/8*pi). I had to round off a bit, but I got x = 1.15 and y = -2.78. Just to doublecheck, I did sqrt((1.15)^2 + (-.2.78)^2) and it did come back as 3.

    • one year ago
  47. Ajk
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    Since 5/8pi is in quadrant 2, I know the x value must be negative, and the y value will be positive.

    • one year ago
  48. LordVictorius
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    Can you tell me the equation you useed>

    • one year ago
  49. Ajk
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    I only used what you gave me. So, my reasoning went like this: first, locate what quadrant 5/8*pi is located. It is in Q2, so we are dealing with a negative x and a positive y. Then, to find x, I just entered 3cos(5/8*pi) into my calculator and got -1.15 (rounded). To find y, I entered 3sin(5/8*pi) into my calculator, and it gave me 2.78 (rounded). So with that I now have my x and y values to plot.

    • one year ago
  50. LordVictorius
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    So for r, you used x(sin or cos) of y, right?

    • one year ago
  51. Ajk
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    Yes, but I just thought of something. If that r is -3, then I should have used it in my computations for x and y, which would have switched the signs of both x and y. I think we need to change my answer to 1.15 and -2.78, sorry about that.

    • one year ago
  52. LordVictorius
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    So x =-2.78?

    • one year ago
  53. Ajk
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    nope, x = 1.15, and y = -2.78. Cosine gives us the x part, and Sine gives us the y part.

    • one year ago
  54. LordVictorius
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    Wait, I thought x was negative.

    • one year ago
  55. Ajk
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    it was, but look at that r. r = -3, so we have to include that in our calculation. If we have some negative value for x and multiply by -3, the answer will turn positive, right?

    • one year ago
  56. LordVictorius
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    Oh, okay.

    • one year ago
  57. LordVictorius
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    But I don't have that answer choice.

    • one year ago
  58. Ajk
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    same for y. It would be positive, but times -3 it turns negative.

    • one year ago
  59. Ajk
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    what are the options?

    • one year ago
  60. LordVictorius
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    I only have (-3/2,3/2) (3sqrt(3)/2),-3/2) (-3/2, 3sqrt(3)/2) and 3/2, -3/2

    • one year ago
  61. LordVictorius
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    Hello?

    • one year ago
  62. LordVictorius
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    I need to leave in half an hour, with the assignment done and Istill dont understand polar coords. I'm osrry, but I need a brief descriptiion that wil help get my assignment done.

    • one year ago
  63. LordVictorius
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    @satellite73 Please help!

    • one year ago
  64. Ajk
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    It looks like there is probably a much easier/better way to do this problem than how I've been working on it. You might be better off spending that 30 minutes reviewing the textbook or getting help from someone better at this than I am.

    • one year ago
  65. LordVictorius
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    Soeey Ajk, maybe someome else can explani it better.

    • one year ago
  66. LordVictorius
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    Thanks anyway.

    • one year ago
  67. LordVictorius
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    :(

    • one year ago
  68. LordVictorius
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    @cwrw238

    • one year ago
  69. LordVictorius
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    @dpaInc

    • one year ago
  70. Ajk
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    I'd be interested to see an explanation, if someone else helps. I'll check back and hopefully learn something too.

    • one year ago
  71. LordVictorius
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    @satellite73

    • one year ago
  72. LordVictorius
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    @dpaInc

    • one year ago
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