1. III.TRACING OF PARAMETRIC CURVES To trace a cartesian curve defined by the parametric equations x = f(  ), y = g(  ), we use the following properties. If either x or y is a periodic function of  with period T then trace the curve in one period, say for   [0, T].

2. Symmetry: (i) About X-axis if f(-  ) = f(  ), g(-  ) = - g(  ) (ii) About Y-axis if f(-  ) = - f(  ), g(-  ) = g(  ) (iii) About Opposite Quadrants if f(-  ) = f(  ), g(-  ) = g(  ) f(-  ) = f(  ), g(-  ) = g(  )

3. Region : Find the region for  for which x and y are well defined. Tabulation: Tabulate the values of x and y for selected values of . Derivatives : Find dy/dx defined by

4. The tangents to the curve at different points can be determined by evaluating dy/dx at that point. If dy/dx >0, in an interval then the curve increases at that interval. If dy/dx < 0, then the curve decreases. If dy/dx = 0, the curve has a stationary point.

5. Example : Trace the curve x = a(  - sin  ), y = a(1-cos  ); 0 ≤  ≤ 2  ;a >0. Note that f(-  ) = -a(  - sin  )= - f(  ); g(-  ) = a(1-cos  ) = g(  ). g(-  ) = a(1-cos  ) = g(  ). Therefore the curve is symmetric about the y-axis. Therefore the curve is symmetric about the y-axis. Also y is a periodic function of  with period 2 . It is sufficient to trace the curve for   [0, 2  ]. For   [0, 2  ], x and y are well defined. Note that y ≥ 0. Entire curve lies above the Y-axis.

6. 0 2222 x =a(  - sin  ) y = a(1-cos  ) dy/dx

7. At  = 0, dy/dx = . Tangent to the curve at  = 0 is perpendicular to x-axis. At  = , dy/dx = 0. Tangent to the curve is parallel to x-axis at  = . At  =2 , dy/dx = . Tangent to the curve is again perpendicular to x-axis at  = 2 .

8. For 0 0. Therefore the function is increasing in this interval. Therefore the function is increasing in this interval. For  <  <2 , dy/dx <0. Therefore the function is decreasing in this interval.

9. X Y (a ,0)(2a ,0) (0,0) (a , 2a) The curve is called a cycloid

10. 2. Trace the following curve with explanation. x = a(  + sin  ), y = a(1 - cos  ); -  ≤  ≤  ; a >0. x = a(  + sin  ), y = a(1 - cos  ); -  ≤  ≤  ; a >0. Note that f(-  ) = -a(  + sin  )= - f(  ), g(-  ) = a(1 - cos  ) = g(  ). Therefore the curve is symmetric about the y-axis. Also y is a periodic function of  with period 2 . It is sufficient to trace the curve for   [- ,  ]. For   [- ,  ], x and y are well defined. Note that y ≥ 0. Entire curve lies above the Y-axis.

11.  -  0  x =a(  + sin  ) y = a(1-cos  ) dy/dx

12. For 0 0. Therefore the function is increasing in this interval. Therefore the function is increasing in this interval. For -  <  <0, dy/dx <0. Therefore the function is decreasing in this interval.

13. X Y (a ,0) (-a ,0) (0,0) (a , 2a)(-a , 2a)