1. Parameterization

2. Section 2 Parametric Differentiation

3. Theorem Let x = f(t), y = g(t) and dx/dt is nonzero, then dy/dx = (dy/dt) / (dx/dt) ; provided the given derivatives exist

4. Examples

5. Example 1 Let x = 4sint, y = 3cost. Find: 1. dy/dx and d 2 y/dx 2 2. dy/dx and d 2 y/dx 2 at t = π /4

6. Solution dx/dt = 4cost, dy/dt = -3sint → dy/dx = (dy/dt) / (dx/dt) = - 3sint / 4cost = -(3/4)tant d(dy/dx)/dt = -(3/4)sec 2 t d 2 y/dx 2 = d(dy/dx )/dx= [d(dy/dx)/dt] / [dx/dt] = [-(3/4)sec 2 t ] / 4cost = (-3/16) sec 3 t (dy/dx)( π /4) = -(3/4)tan( π /4) = -(3/4) (d 2 y/dx 2 )( π /4) = (-3/16) sec 3 ( π /4) = -(3/16)(√2) 3 = - 3√2/8

7. Example 2 Let x = 4sint, y = 3cost. Find: dy/dx and d 2 y/dx 2 at the point (0, -3 )

8. Solution. Let x = 4sint, y = 3cost. First we find any value of t corresponding to the point (0, -3 ). It is clear that one such value is t = π. Why? (dy/dx)( π ) = -(3/4)tan( π ) = 0 (d 2 y/dx 2 )( π ) = (-3/16) sec 3 ( π ) = -(3/16)(-1) 3 = 3/16

9. Example 3 Let x = -3cos2t, y = 2sin3t. Determine at which point (points): 1. The graph has a horizontal tangent. 2. The graph has a vertical tangent

10. Solution a. We have: dx/dt = 6sin2t, dy/dt = 6cos3t The graph has a horizontal tangent at a point at which dy/dx=0, or equivalently dy/dt = 0 and dx/dt; ≠0, that's at which 6cos3t = 0 but 6sin2t ≠ 0 Why?

11. { t: 6cos3t = 0 and 6sin2t ≠ 0 } = {t: 3t = (2n+1) π /2 and 2t ≠ n π, nЄZ } = {t: t = (2n+1) π /6 and t ≠ n π /2, nЄZ } = {…,-9 π /6, -7π /6, -5π /6, -3π /6, -π /6, π /6, 3 π /6, 5π /6, 7π /6, 9π /6,…} - {..., -2 π, -3π /2, -π, - π /2, 0, π /2, π, 3π /2, 2 π,……} = {…….., -7π /6, -5π /6, -π /6, π /6, 5π /6, 7π /6, ……..…}

12. { (x,y)=(-3cos2t, 2sin3t): 6cos3t = 0 and 6sin2t ≠ 0 } = { …., ( -3cos (-7π /3 ), 2sin (-7π /2 ) ), ( -3cos (-5π /3 ), 2sin (-5π /2 ) ), ( -3cos (-π /3 ), 2sin (-π /2 ) ), ( -3cos (π /3 ), 2sin (π /2 ) ), ( -3cos (5π /3 ), 2sin (5π /2 ) ), ( -3cos (7π /3 ), 2sin (7π /2 ) ), ……..…} = { …., ( -3cos (-7π /3 ), 2sin (-7π /2 ) ), ( -3cos (-5π /3 ), 2sin (-5π /2 ) ), ( -3cos (-π /3 ), 2sin (-π /2 ) ), ( -3cos (π /3 ), 2sin (π /2 ) ), ( -3cos (5π /3 ), 2sin (5π /2 ) ), ( -3cos (7π /3 ), 2sin (7π /2 ) ), ……..…} = { ( -3/2, 2 ), ( -3/2, -2) }

13. Thus the graph has horizontal asymptotes at the points ( -3/2, 2 ) and ( -3/2, -2 )

14. b. We have: dx/dt = 6sin2t, dy/dt = 6cos3t The graph has a vertical tangent at a point at which dy/dx does not exist; that's at which dx/dt = 6sin2t = 0 but dy/dt = 6cos3t ≠ 0

15. { t: 6sin2t = 0 and 6cos3t ≠ 0 } = {t: 2t = n π and 3t ≠(2n+1) π /2 } = {t: t = n π /2 and t ≠ (2n+1) π /6 } = {..., -2 π, -3π /2, -π, - π /2, 0, π /2, π, 3π /2, 2 π,……} - {…,-9 π /6, -7π /6, -5π /6, -3π /6, -π /6, π /6, 3 π /6, 5π /6, 7π /6, 9π /6,…} = {…….., -2π, -π, 0, π, 2π, ……..…} {(x,y): 6sin2t = 0 and 6cos3t ≠ 0 } = { ….., ( -3cos (-4π), 2sin (-6π) ), ( -3cos (-2π), 2sin (-3π) ), ( -3cos (0), 2sin (0) ), ( -3cos (2π), 2sin (3π) ), ( -3cos (4π), 2sin (6π) ), ……..…} = { ( -3 0 ) }

16. { (x,y)=(-3cos2t, 2sin3t): 6cos3t = 0 and 6sin2t ≠ 0 } = { …., ( -3cos (-7π /3 ), 2sin (-7π /2 ) ), ( -3cos (-5π /3 ), 2sin (-5π /2 ) ), ( -3cos (-π /3 ), 2sin (-π /2 ) ), ( -3cos (π /3 ), 2sin (π /2 ) ), ( -3cos (5π /3 ), 2sin (5π /2 ) ), ( -3cos (7π /3 ), 2sin (7π /2 ) ), ……..…} = { …., ( -3cos (-7π /3 ), 2sin (-7π /2 ) ), ( -3cos (-5π /3 ), 2sin (-5π /2 ) ), ( -3cos (-π /3 ), 2sin (-π /2 ) ), ( -3cos (π /3 ), 2sin (π /2 ) ), ( -3cos (5π /3 ), 2sin (5π /2 ) ), ( -3cos (7π /3 ), 2sin (7π /2 ) ), ……..…} = { ( -3/2, 2 ), ( -3/2, -2) } Thus the graph has a vertical asymptote at the points ( - 3, 0)

17. Homework 1. Find dy/dx and d 2 y/dx 2 at t=t 0, for each of the following cases: a. x= e 3t, y=e -4t, where t 0 =0 b. x=cos2t, y=sin2t, where t 0 = π /6 2. Let x=cos2t, y=sin2t Find dy/dx and d 2 y/dx 2 at (1/2, √3/2)

18. 3. Let x= e 3t, y=2e -4t Find dy/dx and d 2 y/dx 2 at (e 3, 2e 4 ) 4. Let x= e 3t, y=2e -4t Find dy/dx and d 2 y/dx 2 at (1, 2)