Advanced Calculus(MATH 104)-Polar Coordinates
Harish Bhandari
           Department of Mathematics
              Kathmandu University
                Dhulikhel, Nepal
                   Lecture 5
               August 17, 2019
Relations of spherical coordinates
          Harish Bhandari     Coordinates systems
Relations of spherical coordinates
  Spherical coordinates and rectangular coordinates
  Spherical coordinates and Cylindrical coordinates
           Harish Bhandari           Coordinates systems
Example
   1. Find the spherical coordinate equation for
        a. x 2 +p
                y 2 + (z − 1)2 = 1
        b. z = x 2 + y 2
                                              √
   2. Find the spherical coordinates for (0, 2 3, −2)
          Harish Bhandari           Coordinates systems
Conics in polar coordinates
           PF = r
           FB = rcosθ
           PD = k − FB
               = k − rcosθ
           PF
     Also,     = e (Constant)
           PD
           PF = ePD
             r = e(k − rcosθ)
                                          e = 1 : Parabola
                                          e < 1 : Ellipse
                                          e > 1 : Hyperbola
           Harish Bhandari      Coordinates systems
Conics in polar coordinates
         Harish Bhandari      Coordinates systems
Conics in polar coordinates
  Example
   1. Find the equation of the conic with one focus at origin:
        a. e = 1, x = 2
        b. e = 51 , y = −10
                                                                 25
   2. Find the equation of directrix of the parabola: r =     10+10cosθ
  Ellipse with eccentricity e and semi-major axis a
                                                     a
                                                  k=   − ae
                                                     e
                                                ke = a(1 − e 2 )
                                   ∴ Polar equation of ellipse with
                                   eccentricity e and semi-major axis
                                   a is
                                                  a(1 − e 2 )
                                              r=
                                                  1 + ecosθ
           Harish Bhandari            Coordinates systems
Example
                 400
  Sketch r =   16+8sinθ
          Harish Bhandari   Coordinates systems
Equations of Line
                                   Example
                                                       √
                                   If θ0 = π4 , r0 =        2, we
                                   obtain that
 From right angled triangle                                    π  √
 OP0 P,                                             rcos(θ − ) = 2
                                            π              π4 √
   OP0 = OPcos(∠POP0 )             rcosθcos      + rsinθsin      = 2
                                             4                4
    r0 = rcos(θ − θ0 )                       1           1        √
                                            √ rcosθ + √ rsinθ = 2
 where P0 (r0 , θ0 ) is the foot              2           2
 of perpendicular of the line                              x +y =2
 L and r0 ≥ 0
            Harish Bhandari           Coordinates systems
Equations of Circle
                                         b. If the circle’s center
                                            lies on +ve x-axis
   From Cosine law,
                                            r = 2acosθ (θ0 = 0)
   a2 = r02 +r 2 −2r0 rcos(θ−θ0 )        c. If the circle’s center
                                            lies on +ve y-axis 
   ,                                        r = 2asinθ θ0 = π2
       a. If the circle passes
          through the origin,
          then r0 = a, so
            r = 2acos(θ − θ0 )
             Harish Bhandari        Coordinates systems
Some Exercises
  Lines
                     π
                            √
                                                             2π
                                                               
     a. rcos θ −     4 =         2          b. rcos θ −       3    =3
                     3π                                      π
                                                              
     c. rcos θ +    1 4 =                   d. rcos θ +   =2 3
        √      √                               √
     e. 2x + 2y = 6                         f. 3x − y = 1
     g. y = −5                              h. x = −4
                                     Sketch the parabolas and ellipses:
  Find a polar equation for
  each conic section:
           Harish Bhandari             Coordinates systems