Lesson 11: Implicit Differentiation (slides)

  • 1,672 views
Uploaded on

Using implicit differentiation we can treat relations which are not quite functions like they were functions. In particular, we can find the slopes of lines tangent to curves which are not graphs of …

Using implicit differentiation we can treat relations which are not quite functions like they were functions. In particular, we can find the slopes of lines tangent to curves which are not graphs of functions.

More in: Technology , Education
  • Full Name Full Name Comment goes here.
    Are you sure you want to
    Your message goes here
    Be the first to comment
    Be the first to like this
No Downloads

Views

Total Views
1,672
On Slideshare
0
From Embeds
0
Number of Embeds
2

Actions

Shares
Downloads
158
Comments
0
Likes
0

Embeds 0

No embeds

Report content

Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

Cancel
    No notes for slide

Transcript

  • 1. Sec on 2.6 Implicit Differen a on V63.0121.011: Calculus I Professor Ma hew Leingang New York University February 28, 2011.
  • 2. Music Selection “The Curse of Curves” by Cute is What We Aim For
  • 3. Announcements Quiz 2 in recita on this week. Covers §§1.5, 1.6, 2.1, 2.2 Midterm next week. Covers §§1.1–2.5
  • 4. Objectives Use implicit differenta on to find the deriva ve of a func on defined implicitly.
  • 5. Outline The big idea, by example Examples Basic Examples Ver cal and Horizontal Tangents Orthogonal Trajectories Chemistry The power rule for ra onal powers
  • 6. Motivating Example Problem y Find the slope of the line which is tangent to the curve . x x2 + y2 = 1 at the point (3/5, −4/5).
  • 7. Motivating Example Problem y Find the slope of the line which is tangent to the curve . x x2 + y2 = 1 at the point (3/5, −4/5).
  • 8. Motivating Example Problem y Find the slope of the line which is tangent to the curve . x x2 + y2 = 1 at the point (3/5, −4/5).
  • 9. Motivating Example, SolutionSolu on (Explicit) Isolate: √ y y = 1 − x =⇒ y = − 1 − x2 . 2 2 (Why the −?) . x
  • 10. Motivating Example, SolutionSolu on (Explicit) Isolate: √ y y = 1 − x =⇒ y = − 1 − x2 . 2 2 (Why the −?) Differen ate: . x dy −2x x =− √ =√ dx 2 1 − x2 1 − x2
  • 11. Motivating Example, SolutionSolu on (Explicit) Isolate: √ y y = 1 − x =⇒ y = − 1 − x2 . 2 2 (Why the −?) Differen ate: . x dy −2x x =− √ =√ dx 2 1 − x2 1 − x2 Evaluate: dy 3/5 3/5 3 =√ = = . dx x=3/5 1 − (3/5)2 4/5 4
  • 12. Motivating Example, SolutionSolu on (Explicit) Isolate: √ y y = 1 − x =⇒ y = − 1 − x2 . 2 2 (Why the −?) Differen ate: . x dy −2x x =− √ =√ dx 2 1 − x2 1 − x2 Evaluate: dy 3/5 3/5 3 =√ = = . dx x=3/5 1 − (3/5)2 4/5 4
  • 13. Motivating Example, SolutionSolu on (Explicit) Isolate: √ y y = 1 − x =⇒ y = − 1 − x2 . 2 2 (Why the −?) Differen ate: . x dy −2x x =− √ =√ dx 2 1 − x2 1 − x2 Evaluate: dy 3/5 3/5 3 =√ = = . dx x=3/5 1 − (3/5)2 4/5 4
  • 14. Motivating 2Example, another way 2 We know that x + y = 1 does not define y as a func on of x, but suppose it did. Suppose we had y = f(x), so that x2 + (f(x))2 = 1
  • 15. Motivating 2Example, another way 2 We know that x + y = 1 does not define y as a func on of x, but suppose it did. Suppose we had y = f(x), so that x2 + (f(x))2 = 1 We could differen ate this equa on to get 2x + 2f(x) · f′ (x) = 0
  • 16. Motivating 2Example, another way 2 We know that x + y = 1 does not define y as a func on of x, but suppose it did. Suppose we had y = f(x), so that x2 + (f(x))2 = 1 We could differen ate this equa on to get 2x + 2f(x) · f′ (x) = 0 We could then solve to get x f′ (x) = − f(x)
  • 17. Yes, we can! The beau ful fact (i.e., deep theorem) is that this works! y “Near” most points on the curve x2 + y2 = 1, the curve resembles the graph of a func on. . x
  • 18. Yes, we can! The beau ful fact (i.e., deep theorem) is that this works! y “Near” most points on the curve x2 + y2 = 1, the curve resembles the graph of a func on. . x
  • 19. Yes, we can! The beau ful fact (i.e., deep theorem) is that this works! y “Near” most points on the curve x2 + y2 = 1, the curve resembles the graph of a func on. . x looks like a func on
  • 20. Yes, we can! The beau ful fact (i.e., deep theorem) is that this works! y “Near” most points on the curve x2 + y2 = 1, the curve resembles the graph of a func on. . x
  • 21. Yes, we can! The beau ful fact (i.e., deep theorem) is that this works! y “Near” most points on the curve x2 + y2 = 1, the curve resembles the graph of a func on. . x
  • 22. Yes, we can! The beau ful fact (i.e., deep theorem) is that this works! y “Near” most points on the curve x2 + y2 = 1, the curve resembles the graph of a func on. looks like a func on . x
  • 23. Yes, we can! The beau ful fact (i.e., deep theorem) is that this works! y “Near” most points on the curve x2 + y2 = 1, the curve resembles the graph of a func on. . x
  • 24. Yes, we can! The beau ful fact (i.e., deep theorem) is that this works! y “Near” most points on the curve x2 + y2 = 1, the curve resembles the graph of a func on. . x
  • 25. Yes, we can! The beau ful fact (i.e., deep theorem) is that this works! y “Near” most points on the curve x2 + y2 = 1, the curve resembles the graph of a func on. . x does not look like a func on, but that’s OK—there are only two points like this
  • 26. Yes, we can! The beau ful fact (i.e., deep theorem) is that this works! y “Near” most points on the curve x2 + y2 = 1, the curve resembles the graph of a func on. So f(x) is defined “locally”, almost everywhere and is . x differen able looks like a func on
  • 27. Yes, we can! The beau ful fact (i.e., deep theorem) is that this works! y “Near” most points on the curve x2 + y2 = 1, the curve resembles the graph of a func on. So f(x) is defined “locally”, almost everywhere and is . x differen able The chain rule then applies for this local choice. looks like a func on
  • 28. Motivating Example, again, with Leibniz notation Problem Find the slope of the line which is tangent to the curve x2 + y2 = 1 at the point (3/5, −4/5).
  • 29. Motivating Example, again, with Leibniz notation Problem Find the slope of the line which is tangent to the curve x2 + y2 = 1 at the point (3/5, −4/5). Solu on dy Differen ate: 2x + 2y =0 dx
  • 30. Motivating Example, again, with Leibniz notation Problem Find the slope of the line which is tangent to the curve x2 + y2 = 1 at the point (3/5, −4/5). Solu on dy Differen ate: 2x + 2y = 0 dx Remember y is assumed to be a func on of x!
  • 31. Motivating Example, again, with Leibniz notation Problem Find the slope of the line which is tangent to the curve x2 + y2 = 1 at the point (3/5, −4/5). Solu on dy Differen ate: 2x + 2y = 0 dx Remember y is assumed to be a func on of x! dy x Isolate: =− . dx y
  • 32. Motivating Example, again, with Leibniz notation Problem Find the slope of the line which is tangent to the curve x2 + y2 = 1 at the point (3/5, −4/5). Solu on dy Differen ate: 2x + 2y = 0 dx Remember y is assumed to be a func on of x! dy x dy 3/5 3 Isolate: = − . Then evaluate: = = . dx y dx ( 3 ,− 4 ) 4/5 4 5 5
  • 33. Summary If a rela on is given between x and y which isn’t a func on: “Most of the me”, i.e., “at most places” y can be y assumed to be a func on of x we may differen ate the rela on as is . x dy Solving for does give the dx slope of the tangent line to the curve at a point on the curve.
  • 34. Outline The big idea, by example Examples Basic Examples Ver cal and Horizontal Tangents Orthogonal Trajectories Chemistry The power rule for ra onal powers
  • 35. Another Example Example Find y′ along the curve y3 + 4xy = x2 + 3.
  • 36. Another Example Example Find y′ along the curve y3 + 4xy = x2 + 3. Solu on Implicitly differen a ng, we have 3y2 y′ + 4(1 · y + x · y′ ) = 2x
  • 37. Another Example Example Find y′ along the curve y3 + 4xy = x2 + 3. Solu on Implicitly differen a ng, we have 3y2 y′ + 4(1 · y + x · y′ ) = 2x Solving for y′ gives 2x − 4y 3y2 y′ + 4xy′ = 2x − 4y =⇒ (3y2 + 4x)y′ = 2x − 4y =⇒ y′ = 3y2 + 4x
  • 38. Yet Another Example Example Find y′ if y5 + x2 y3 = 1 + y sin(x2 ).
  • 39. Yet Another Example Example Find y′ if y5 + x2 y3 = 1 + y sin(x2 ). Solu on Differen a ng implicitly: 5y4 y′ + (2x)y3 + x2 (3y2 y′ ) = y′ sin(x2 ) + y cos(x2 )(2x)
  • 40. Yet Another Example Example Find y′ if y5 + x2 y3 = 1 + y sin(x2 ). Solu on Differen a ng implicitly: 5y4 y′ + (2x)y3 + x2 (3y2 y′ ) = y′ sin(x2 ) + y cos(x2 )(2x) Collect all terms with y′ on one side and all terms without y′ on the other: 5y4 y′ + 3x2 y2 y′ − sin(x2 )y′ = −2xy3 + 2xy cos(x2 )
  • 41. Yet Another Example Example Find y′ if y5 + x2 y3 = 1 + y sin(x2 ). Solu on Collect all terms with y′ on one side and all terms without y′ on the other: 5y4 y′ + 3x2 y2 y′ − sin(x2 )y′ = −2xy3 + 2xy cos(x2 ) 2xy(cos x2 − y2 ) ′ Now factor and divide: y = 4 5y + 3x2 y2 − sin x2
  • 42. Finding tangents with implicit differentitiation Example Find the equa on of the line tangent to the curve . y2 = x2 (x + 1) = x3 + x2 at the point (3, −6).
  • 43. Solution Solu on dy 2 dy 3x2 + 2x Differen ate: 2y = 3x + 2x, so = , and dx dx 2y dy 3 · 32 + 2 · 3 33 11 = =− =− . dx (3,−6) 2(−6) 12 4
  • 44. Solution Solu on dy 2 dy 3x2 + 2x Differen ate: 2y = 3x + 2x, so = , and dx dx 2y dy 3 · 32 + 2 · 3 33 11 = =− =− . dx (3,−6) 2(−6) 12 4 11 Thus the equa on of the tangent line is y + 6 = − (x − 3). 4
  • 45. Finding tangents with implicit differentitiation Example Find the equa on of the line tangent to the curve . y2 = x2 (x + 1) = x3 + x2 at the point (3, −6).
  • 46. Recall: Line equation forms slope-intercept form y = mx + b where the slope is m and (0, b) is on the line. point-slope form y − y0 = m(x − x0 ) where the slope is m and (x0 , y0 ) is on the line.
  • 47. Horizontal Tangent Lines Example Find the horizontal tangent lines to the same curve: y2 = x3 + x2
  • 48. Horizontal Tangent Lines Example Find the horizontal tangent lines to the same curve: y2 = x3 + x2 Solu on We have to solve these two equa ons: y2 = x3 + x2 [(x, y) is on the curve] 3x2 + 2x = 0 [tangent line is horizontal] 2y
  • 49. Solution, continued Solving the second equa on gives 3x2 + 2x = 0 =⇒ 3x2 + 2x = 0 =⇒ x(3x + 2) = 0 2y (as long as y ̸= 0). So x = 0 or 3x + 2 = 0.
  • 50. Solution, continued Solving the second equa on gives 3x2 + 2x = 0 =⇒ 3x2 + 2x = 0 =⇒ x(3x + 2) = 0 2y (as long as y ̸= 0). So x = 0 or 3x + 2 = 0.
  • 51. Solution, continued Solving the second equa on gives 3x2 + 2x = 0 =⇒ 3x2 + 2x = 0 =⇒ x(3x + 2) = 0 2y (as long as y ̸= 0). So x = 0 or 3x + 2 = 0. Subs tu ng x = 0 into the first equa on gives y2 = 03 + 02 = 0 =⇒ y = 0 which we’ve disallowed. So no horizontal tangents down that road.
  • 52. Solution, continued Subs tu ng x = −2/3 into the first equa on gives ( )3 ( )2 2 2 4 y = − 2 + − = 3 3 27 √ 4 2 =⇒ y = ± =± √ , 27 3 3 so there are two horizontal tangents.
  • 53. Horizontal Tangents ( ) − 3 , 3√3 2 2 . ( ) − 2 , − 3√3 3 2
  • 54. Horizontal Tangents ( ) − 3 , 3√3 2 2 . ( ) − 2 , − 3√3 3 2 node
  • 55. ExampleFind the ver cal tangent lines to the same curve: y2 = x3 + x2
  • 56. ExampleFind the ver cal tangent lines to the same curve: y2 = x3 + x2Solu on dx Tangent lines are ver cal when = 0. dy
  • 57. ExampleFind the ver cal tangent lines to the same curve: y2 = x3 + x2Solu on dx Tangent lines are ver cal when = 0. dy Differen a ng x implicitly as a func on of y gives dx dx dx 2y 2y = 3x2 + 2x , so = 2 (no ce this is the dy dy dy 3x + 2x reciprocal of dy/dx).
  • 58. ExampleFind the ver cal tangent lines to the same curve: y2 = x3 + x2Solu on dx Tangent lines are ver cal when = 0. dy Differen a ng x implicitly as a func on of y gives dx dx dx 2y 2y = 3x2 + 2x , so = 2 (no ce this is the dy dy dy 3x + 2x reciprocal of dy/dx). We must solve y2 = x3 + x2 [(x, y) is on the curve] and 2y = 0 [tangent line is ver cal] 3x2 + 2x
  • 59. Solution, continued Solving the second equa on gives 2y = 0 =⇒ 2y = 0 =⇒ y = 0 3x2 + 2x (as long as 3x2 + 2x ̸= 0).
  • 60. Solution, continued Solving the second equa on gives 2y = 0 =⇒ 2y = 0 =⇒ y = 0 3x2 + 2x (as long as 3x2 + 2x ̸= 0). Subs tu ng y = 0 into the first equa on gives 0 = x3 + x2 = x2 (x + 1) So x = 0 or x = −1.
  • 61. Solution, continued Solving the second equa on gives 2y = 0 =⇒ 2y = 0 =⇒ y = 0 3x2 + 2x (as long as 3x2 + 2x ̸= 0). Subs tu ng y = 0 into the first equa on gives 0 = x3 + x2 = x2 (x + 1) So x = 0 or x = −1. x = 0 is not allowed by the first equa on, but x = −1 is.
  • 62. Tangents ( ) − 3 , 3√3 2 2 (−1, 0) . ( ) − 3 , − 3√3 node 2 2
  • 63. Examples Example Show that the families of curves xy = c, x2 − y2 = k are orthogonal, that is, they intersect at right angles.
  • 64. Orthogonal Families of Curves y xy =xy = c 1x2 − y2 = k . x
  • 65. Orthogonal Families of Curves y xy y = x = 1 2xy = cx2 − y2 = k . x
  • 66. Orthogonal Families of Curves y xy = 1 xy y = = 2 x 3xy = cx2 − y2 = k . x
  • 67. Orthogonal Families of Curves y xy = 1 xy y = = 2 x 3xy = cx2 − y2 = k . x 1 − = xy
  • 68. Orthogonal Families of Curves y xy = 1 xy y = = 2 x 3xy = cx2 − y2 = k . x − 1 = − 2 xy y = x
  • 69. Orthogonal Families of Curves y xy = 1 xy y = = 2 x 3xy = cx2 − y2 = k . x = − 1 xy = − − 2 xy y = 3 x
  • 70. Orthogonal Families of Curves y xy = 1 xy y = = 2 x 3 x2 − y2 = 1xy = cx2 − y2 = k . x = − 1 xy = − − 2 xy y = 3 x
  • 71. Orthogonal Families of Curves y xy = 1 xy y = = 2 x 3 x2 − y2 = 2 x −y =1xy = cx2 − y2 = k . x 2 = − 1 xy = − − 2 xy y = 3 2 x
  • 72. xy = c x2 − y2 = k 2 2 x2 − y2 = 3 x2 − y2 = 2 x −y =1 x . y xy y = xxy = − = − 1 xy y = − 2 xy = 1 3 = 2 3 x Orthogonal Families of Curves
  • 73. Orthogonal Families of Curves y xy = 1 xy y = = 2 x 3 x2 − y2 = 3 x2 − y2 = 2 x −y =1xy = cx2 − y2 = k . x 2 = − 1 xy = − − 2 x2 − y2 = −1 xy y = 3 2 x
  • 74. Orthogonal Families of Curves y xy = 1 xy y = = 2 x 3 x2 − y2 = 3 x2 − y2 = 2 x −y =1xy = cx2 − y2 = k . x 2 = − 1 xy = − − 2 x2 − y2 = −1 xy y = 3 2 x2 − y2 = −2 x
  • 75. Orthogonal Families of Curves y xy = 1 xy y = = 2 x 3 x2 − y2 = 3 x2 − y2 = 2 x −y =1xy = cx2 − y2 = k . x 2 = − 1 xy = − − 2 x2 − y2 = −1 xy y = 3 2 x2 − y2 = −2 x2 − y2 = −3 x
  • 76. Examples Example Show that the families of curves xy = c, x2 − y2 = k are orthogonal, that is, they intersect at right angles. Solu on y In the first curve, y + xy′ = 0 =⇒ y′ = − x
  • 77. Examples Example Show that the families of curves xy = c, x2 − y2 = k are orthogonal, that is, they intersect at right angles. Solu on y In the first curve, y + xy′ = 0 =⇒ y′ = − x ′ ′ x In the second curve, 2x − 2yy = 0 =⇒ y = y
  • 78. Examples Example Show that the families of curves xy = c, x2 − y2 = k are orthogonal, that is, they intersect at right angles. Solu on y In the first curve, y + xy′ = 0 =⇒ y′ = − x ′ ′ x In the second curve, 2x − 2yy = 0 =⇒ y = y The product is −1, so the tangent lines are perpendicular wherever they intersect.
  • 79. Ideal gases The ideal gas law relates temperature, pressure, and volume of a gas: PV = nRT (R is a constant, n is the amount of gas in moles) .Image credit: Sco Beale / Laughing Squid
  • 80. Compressibility Defini on The isothermic compressibility of a fluid is defined by dV 1 β=− dP V
  • 81. Compressibility Defini on The isothermic compressibility of a fluid is defined by dV 1 β=− dP V Approximately we have ∆V dV ∆V ≈ = −βV =⇒ ≈ −β∆P ∆P dP V The smaller the β, the “harder” the fluid.
  • 82. Compressibility of an ideal gas Example Find the isothermic compressibility of an ideal gas.
  • 83. Compressibility of an ideal gas Example Find the isothermic compressibility of an ideal gas. Solu on If PV = k (n is constant for our purposes, T is constant because of the word isothermic, and R really is constant), then dP dV dV V ·V+P = 0 =⇒ =− dP dP dP P 1 dV 1 So β = − · = . Compressibility and pressure are inversely V dP P related.
  • 84. Nonideal gassesNot that there’s anything wrong with thatExample H.The van der Waals equa on makesfewer simplifica ons: . . Oxygen H ( ) H. n2 . P + a 2 (V − nb) = nRT, Oxygen Hydrogen bonds V H.where a is a measure of a rac on . . Oxygen Hbetween par cles of the gas, and b ameasure of par cle size. H.
  • 85. Nonideal gassesNot that there’s anything wrong with thatExampleThe van der Waals equa on makesfewer simplifica ons: ( ) n2 . P + a 2 (V − nb) = nRT, Vwhere a is a measure of a rac onbetween par cles of the gas, and b ameasure of par cle size.
  • 86. Compressibility of a van der Waals gas Differen a ng the van der Waals equa on by trea ng V as a func on of P gives ( ) ( ) an2 dV 2an2 dV P+ 2 + (V − bn) 1 − 3 = 0, V dP V dP
  • 87. Compressibility of a van der Waals gas Differen a ng the van der Waals equa on by trea ng V as a func on of P gives ( ) ( ) an2 dV 2an2 dV P+ 2 + (V − bn) 1 − 3 = 0, V dP V dP so 1 dV V2 (V − nb) β=− = V dP 2abn3 − an2 V + PV3
  • 88. Nonideal compressibility,continued 1 dV V2 (V − nb) β=− = V dP 2abn3 − an2 V + PV3 Ques on
  • 89. Nonideal compressibility,continued 1 dV V2 (V − nb) β=− = V dP 2abn3 − an2 V + PV3 Ques on What if a = b = 0?
  • 90. Nonideal compressibility,continued 1 dV V2 (V − nb) β=− = V dP 2abn3 − an2 V + PV3 Ques on What if a = b = 0? dβ Without taking the deriva ve, what is the sign of ? db
  • 91. Nonideal compressibility,continued 1 dV V2 (V − nb) β=− = V dP 2abn3 − an2 V + PV3 Ques on What if a = b = 0? dβ Without taking the deriva ve, what is the sign of ? db dβ Without taking the deriva ve, what is the sign of ? da
  • 92. Nasty derivatives Answer We get the old (ideal) compressibility We have ( ) dβ nV3 an2 + PV2 = −( )2 < 0 db PV3 + an2 (2bn − V) dβ n2 (bn − V)(2bn − V)V2 We have =( ) > 0 (as long as da 3 PV + an 2 (2bn − V) 2 V > 2nb, and it’s probably true that V ≫ 2nb).
  • 93. Outline The big idea, by example Examples Basic Examples Ver cal and Horizontal Tangents Orthogonal Trajectories Chemistry The power rule for ra onal powers
  • 94. Using implicit differentiation tofind derivatives Example dy √ Find if y = x. dx
  • 95. Using implicit differentiation tofind derivatives Example dy √ Find if y = x. dx Solu on √ If y = x, then y2 = x, so dy dy 1 1 2y = 1 =⇒ = = √ . dx dx 2y 2 x
  • 96. The power rule for rational powers Theorem p If y = xp/q , where p and q are integers, then y′ = xp/q−1 . q
  • 97. The power rule for rational powers Theorem p If y = xp/q , where p and q are integers, then y′ = xp/q−1 . q Proof. First, raise both sides to the qth power: y = xp/q =⇒ yq = xp
  • 98. The power rule for rational powers Theorem p If y = xp/q , where p and q are integers, then y′ = xp/q−1 . q Proof. First, raise both sides to the qth power: y = xp/q =⇒ yq = xp Now, differen ate implicitly: q−1 dy dy p xp−1 qy = px p−1 =⇒ = · dx dx q yq−1
  • 99. The power rule for rational powers Theorem p If y = xp/q , where p and q are integers, then y′ = xp/q−1 . q Proof. Now, differen ate implicitly: q−1 dy dy p xp−1 qy = px p−1 =⇒ = · dx dx q yq−1
  • 100. The power rule for rational powers Theorem p If y = xp/q , where p and q are integers, then y′ = xp/q−1 . q Proof. Simplify: yq−1 = x(p/q)(q−1) = xp−p/q so xp−1 xp−1 q−1 = p−p/q = xp−1−(p−p/q) = xp/q−1 y x
  • 101. Summary Using implicit differen a on we can treat rela ons which are not quite func ons like they were func ons. In par cular, we can find the slopes of lines tangent to curves which are not graphs of func ons.