Calculus of Polar Curves

Slope of a Tangent Line:

If r = f(q), then x = f(q) cos q and y = f(q) sin q

dx/dq = f ’(q) cos q  - f(q) sin q and dy/dq  = f ‘(q) sin q + f(q)cos q

dy/dx = dy/dq
            dx/dq

Therefore,
    dy  =  f ‘(q) sin q + f(q) cos q
    dx      f ‘(q) cos q – f(q) sin q

Examples:
(a)  Find the slope of r = cos 2q at q = 0, p/2, -p/2, and p

     x = cos 2q cos q
     y = cos 2q sin q

    dy = -2 sin 2q cos q - cos 2q sin q
    dx    -2 sin 2q sin q + cos 2q cos q

    At q = 0,       dy/dx = 0/1    horizontal tangent
    At q = p/2,   dy/dx = 1/0    vertical tangent
    At q = -p/2,  dy/dx = 1/0   vertical tangent
    At q = p,      dy/dx = 0/-1 horizontal tangent
 

(b)  Find the horizontal and vertical tangents to the cardioid r = 1 +cos q, 0 < q < 2p

        x = (1 + cos q) cos q
        y = (1 + cos q) sin q

        dy = (1 + cos q)(-sin q) + (-sin q)(cos q)
        dx    (1 + cos q) cos q  +  (-sin q) sin q

Area of Polar Curves:

The area of the region between the origin and the curve r = f(q) , a < q < b, is

            b
A = 1/2 ò r2 dq
          a

Examples:

(a)  Find the area inside the outer loop and outside the inner loop of the limacon r = 2 cos q + 1

       Graph this in your calculator for 0 < q < 2p and watch the calculator draw the curve.
       The curve starts at (1,0) and ends at (0,2p).  The top of the outer loop is formed from q = 0 to q = 2p/3 and
       from q = 4p/3 to q = 2p.  The inner loop is formed from q = 2p/3 to q = 4p/3.  The area we are looking for is

             2p/3                                       2p                                          4p/3
        1/2 ò   (2 cos q + 1)2 dq   +   1/2 ò   (2 cos q + 1)2 dq  -   1/2  ò   (2 cos q + 1)2 dq  =  8.338
             0                                          4p/3                                         2p/3

(b)  Find the area shared by the circles r = 1 and r = 2 sin q

       Graph this in your calculator for 0 < q < 2p and watch the calculator draw the curve.
       The first curve starts at (1,0) and ends at (0,2p).  The second curve starts at (0,0) and ends
       at (p,0).  The intersection points of the two curves can be found by solving

       2 sin q = 1
          sin q = 1/2
              q = p/6,

       The radii in the area shared by the circles hit r = 2 sin q for 0 < q < p/6   and  for
        5p/6 < q < p.

               2p                                        4p/3
        1/2 ò   (2 cos q + 1)2 dq   -   1/2   ò   (2 cos q + 1)2 dq  =  8.881
             0                                           2p/3

Length of a Polar Curve:

If r = f(q) has a continuous first derivative for a < q < b and if the point P(r,q) traces the curve exactly once, then the length of the curve is

                      b
               L =  ò \/ ( r2 + (dr/dq)2 ) dq
                    a

Examples:

(1)  Find the length of the parabolic segment r = 6/(1 + cos q), 0 < q < p/2

                      p/2
               L =  ò \/ ( (6 / (1 + cos q))2 + (-6 / (1 + cos q)2 )2 ) dq   =  7.248
                     0

(2)  Find the length of the curve r = cos3(q/3), 0 < q <  p/4

                      p/2
               A =  ò \/ (cos3(q / 3))2 +( (-sin(q/3)cos(q / 3))2 )2) dq   =  1.384
                     0

Surface Area of Revolution:

Revolution about x-axis (y > 0)

                       b
               A =  ò 2pr sinq \/ ( r2 + (dr/dq)2 ) dq
                    a

Revolution about y-axis (x > 0)

                       b
               A =  ò 2pr cosq \/ ( r2 + (dr/dq)2 ) dq
                    a

Example:

Find the area of the surface generated by revolving

               r = \/(cos 2q)  ,   0 <q < p/4

   about the y-axis.