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Parameterization

Parameterization. Section 2. Parametric Differentiation. 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. Examples. Example 1. Let x = 4sint, y = 3cost. Find: 1. dy/dx and d 2 y/dx 2

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Parameterization

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  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 d2y/dx2 2. dy/dx and d2y/dx2 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)sec2t d2y/dx2 = d(dy/dx )/dx= [d(dy/dx)/dt] / [dx/dt] = [-(3/4)sec2t ] / 4cost = (-3/16) sec3t (dy/dx)( π/4) = -(3/4)tan( π/4) = -(3/4) (d2y/dx2 )( π/4) = (-3/16) sec3( π/4) = -(3/16)(√2) 3 = - 3√2/8

  7. Example 2 Let x = 4sint, y = 3cost. Find: dy/dx and d2y/dx2 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 (d2y/dx2 )( π) = (-3/16) sec3( 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. Solutiona. 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 ≠ nπ , nЄZ }= {t: t = (2n+1)π/6 and t ≠ nπ/2 , nЄZ }= {…,-9π/6, -7π/6 , -5π/6, -3π/6 , -π/6, π/6 , 3π/6, 5π/6 , 7π/6,9π/6 ,…}- {... , -2π , -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 = nπ and 3t ≠(2n+1)π/2 }= {t: t = nπ/2 and t ≠ (2n+1)π/6 }= {... , -2π , -3π/2 , -π , -π/2 , 0 ,π/2 , π ,3π/2 , 2 π,……} - {…,-9π/6, -7π/6 , -5π/6, -3π/6 , -π/6, π/6 , 3π/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 andd2y/dx2 at t=t0, for each of the following cases: a. x= e3t, y=e-4t , where t0=0 b. x=cos2t, y=sin2t, where t0= π/6 2. Let x=cos2t, y=sin2t Find dy/dx and d2y/dx2 at (1/2 , √3/2)

  18. 3. Let x= e3t, y=2e-4t Find dy/dx and d2y/dx2 at (e3, 2e4) 4. Let x= e3t, y=2e-4t Find dy/dx and d2y/dx2 at (1, 2)

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