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Polar co-ordinates

  1. Sep 20, 2011 #1
    1. The problem statement, all variables and given/known data


    A point has coordinates (−5, 3*square root of 7).
    What is the polar coordinates of this point?

    (in the form a,b)


    2. Relevant equations


    x= rcos theta
    y=rsin theta

    3. The attempt at a solution

    Using phythagoras thrown

    -5=x=rcos theta (eq 1)
    3*square root 7=Y=rsin theta (eq 2)

    r=-5/cos theta
    r=3*square root 7 / sin theta
     
  2. jcsd
  3. Sep 20, 2011 #2

    Mark44

    Staff: Mentor

    I don't see that you have actually used the Pythagoras Theorem. Isn't r2 = x2 + y2?

    It might be helpful to draw a right triangle, with base x, altitude y, and hypotenuse r.
     
  4. Sep 20, 2011 #3
    Try to find the value of r by eliminating [itex]\theta[/itex] using sin[itex]^{2}[/itex][itex]\theta[/itex] + cos[itex]^{2}[/itex][itex]\theta[/itex] = 1.
     
  5. Sep 20, 2011 #4
    (-5)^2=r^2cos^2 theta (eq 1)
    25 = r^2cos^2
    r^2 = 25 / cos^2
    r^2 = 25 / (1 - sin ^2 theta)
    r^2 * (1 - sin ^2 theta) = 25
    (1 - sin ^2 theta) = 25 / r^2
    1 - (25 / r^2) = sin ^2 theta

    (3)^2* ( 7) =Y=r^2sin^2 theta (eq 2)
    9*7 = r^2sin^2 theta
    64 = r^2 (sin^2 theta)
    r^2 = 64 / sin^2 theta
    r^2* sin^2 theta = 64
    sin^2 theta = 64 /r^2

    1 - (25 / r^2) = 64 /r^2
    r^2 - (25*r^2 / r^2) = 64
    r^2 - (25) = 64
    r^2 = 64-25
    r^2 = 39
    r = =/- (square root of 39)

    Polar co-ordinates of the point?
     
  6. Sep 20, 2011 #5

    verty

    User Avatar
    Homework Helper

    x = r cos(theta)
    y = r sin(theta)

    hmm, I know x and y but not r and theta. I have two unknowns. I need to eliminate r or theta. How can I do that?

    if I can eliminate r, I can find theta.
    if I can eliminate theta, I can find r.

    ------------------------------------------------------------
     
  7. Sep 20, 2011 #6

    Mark44

    Staff: Mentor

    You're really going at it the hard way.
    9*7 != 64.
    Add 25 to both sides. Also, as noted above, 64 isn't the right number.
    If you redo your calculations for r, you will have one of the coordinates. As already recommended, draw a right triangle with base -5 (i.e., to the left of the origin), altitude 3*sqrt(7). You can find the hypotenuse as r = [itex]\sqrt{(-5)^2 + (3\sqrt{7})^2}[/itex]
     
  8. Sep 23, 2011 #7
    Need answer in the form

    (1 , pie/3)
     
  9. Sep 23, 2011 #8
    r = [itex]\sqrt{(-5)^2 + (3\sqrt{7})^2}[/itex]

    r = [itex]\sqrt{(25 + (21}[/itex]

    r = [itex]\sqrt{(41}[/itex]

    r = 6.40

    But still need polar point in the form (f, g/pie)
     
  10. Sep 23, 2011 #9

    Mark44

    Staff: Mentor

    [itex](3 \sqrt{7})^2 \neq 21[/itex]
    Your continuing algebra errors will make completing this problem very difficult.
    If you haven't drawn a coordinate system showing the point in rectangular coordinates, you should do so. The x-coordinate and y-coordinate of the point determine a right triangle. What quadrant is this triangle in?

    Label the base, altitude, and hypotenuse of this triangle with the values of x, y, and r. Then use one of the trig functions to find the angle that the hypotenuse makes with the base. To get the value of theta for your polar point, you will need to take into account which quadrant the triangle is in.
     
  11. Sep 23, 2011 #10

    Mark44

    Staff: Mentor

    I meant to mention this before. The name of this Greek letter - [itex]\pi[/itex] - is pi, not pie.
     
  12. Sep 26, 2011 #11
    (3√7)^2 = 63


    √ (25 + 63) =

    =√88

    = √(4 x 22)

    =2√22

    = 9.38


    x co-ordinate = -5
    y co-ordinate = 3√7 = 7.94

    cos (theta) = -5 / 3√7
    theta = 129 degrees

    pi = 180 degrees

    129 degrees = 0.0243 radians
     
  13. Sep 26, 2011 #12

    Mark44

    Staff: Mentor

    Up to here, you're OK, although the numbers 9.38 and 7.97 are only rough approximations to 2√22 and 3√7.
    No. The cosine of an angle in a right triangle is the adjacent side divided by the hypotenuse.
    How did you get this? 129 degrees is a little over 2 radians.
     
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