Lesson 12: Implicit Differentiation

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A functional relationship between y and x can be made explicit. Even if the relation is not functional, however, we can assume the relation usually defines y as a function.

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Lesson 12: Implicit Differentiation

  1. 1. Section 2.6 Implicit Differentiation V63.0121, Calculus I February 24/25, 2009 Announcements Midterm in class March 4/5 ALEKS due Friday, 11:59pm . . Image credit: Telstar Logistics . . . . . .
  2. 2. Outline The big idea, by example Examples Vertical and Horizontal Tangents Chemistry The power rule for rational powers . . . . . .
  3. 3. y . Motivating Example Problem Find the slope of the line which is tangent to the curve . x . 2 2 x +y =1 at the point (3/5, −4/5). . . . . . .
  4. 4. y . Motivating Example Problem Find the slope of the line which is tangent to the curve . x . 2 2 x +y =1 at the point (3/5, −4/5). . . . . . .
  5. 5. y . Motivating Example Problem Find the slope of the line which is tangent to the curve . x . 2 2 x +y =1 at the point (3/5, −4/5). . . . . . . .
  6. 6. y . Motivating Example Problem Find the slope of the line which is tangent to the curve . x . 2 2 x +y =1 at the point (3/5, −4/5). . Solution (Explicit) √ Isolate: y2 = 1 − x2 =⇒ y = − 1 − x2 . (Why the −?) . . . . . .
  7. 7. y . Motivating Example Problem Find the slope of the line which is tangent to the curve . x . 2 2 x +y =1 at the point (3/5, −4/5). . Solution (Explicit) √ Isolate: y2 = 1 − x2 =⇒ y = − 1 − x2 . (Why the −?) −2x dy x =− √ =√ Differentiate: 2 1 − x2 1 − x2 dx . . . . . .
  8. 8. y . Motivating Example Problem Find the slope of the line which is tangent to the curve . x . 2 2 x +y =1 at the point (3/5, −4/5). . Solution (Explicit) √ Isolate: y2 = 1 − x2 =⇒ y = − 1 − x2 . (Why the −?) −2x dy x =− √ =√ Differentiate: 2 1 − x2 1 − x2 dx dy 3/5 3/5 3 =√ = =. Evaluate: 1 − (3/5) dx x=3/5 4/5 4 2 . . . . . .
  9. 9. y . Motivating Example Problem Find the slope of the line which is tangent to the curve . x . 2 2 x +y =1 at the point (3/5, −4/5). . Solution (Explicit) √ Isolate: y2 = 1 − x2 =⇒ y = − 1 − x2 . (Why the −?) −2x dy x =− √ =√ Differentiate: 2 1 − x2 1 − x2 dx dy 3/5 3/5 3 =√ = =. Evaluate: 1 − (3/5) dx x=3/5 4/5 4 2 . . . . . .
  10. 10. We know that x2 + y2 = 1 does not define y as a function of x, but suppose it did. Suppose we had y = f(x), so that x2 + (f(x))2 = 1 We could differentiate this equation to get 2x + 2f(x) · f′ (x) = 0 We could then solve to get x f′ (x) = − f(x) . . . . . .
  11. 11. The beautiful 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 function. So f(x) is defined “locally” . x . and is differentiable The chain rule then applies for this local . choice. . . . . . .
  12. 12. The beautiful 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 function. So f(x) is defined “locally” . x . and is differentiable The chain rule then applies for this local . choice. . . . . . .
  13. 13. The beautiful 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 function. So f(x) is defined “locally” . x . and is differentiable The chain rule then applies for this local . choice. l .ooks like a function . . . . . .
  14. 14. The beautiful 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 function. So f(x) is defined “locally” . x . and is differentiable The chain rule then applies for this local choice. . . . . . .
  15. 15. The beautiful 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 function. So f(x) is defined “locally” . x . and is differentiable The chain rule then applies for this local choice. . . . . . .
  16. 16. The beautiful 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 function. l .ooks like a function So f(x) is defined “locally” . x . and is differentiable The chain rule then applies for this local choice. . . . . . .
  17. 17. The beautiful 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 function. So f(x) is defined “locally” . . x . and is differentiable The chain rule then applies for this local choice. . . . . . .
  18. 18. The beautiful 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 function. So f(x) is defined “locally” . . x . and is differentiable The chain rule then applies for this local choice. . . . . . .
  19. 19. The beautiful 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 function. So f(x) is defined “locally” . . x . and is differentiable . The chain rule then does not look like a applies for this local function, but that’s choice. OK—there are only two points like this . . . . . .
  20. 20. Problem Find the slope of the line which is tangent to the curve x2 + y2 = 1 at the point (3/5, −4/5). Solution (Implicit, with Leibniz notation) Differentiate. Remember y is assumed to be a function of x: dy 2x + 2y = 0, dx dy Isolate : dx dy x =− . dx y Evaluate: dy 3/5 3 = =. dx ( 3 ,− 4 ) 4/5 4 5 5 . . . . . .
  21. 21. Summary y . If a relation is given between x and y, “Most of the time” “at most places” y can be assumed to . be a function of x . we may differentiate the relation as is dy Solving for does give the dx slope of the tangent line to the curve at a point on the curve. . . . . . .
  22. 22. Mnemonic Explicit Implicit y = f(x) F(x, y) = k . . Image credit: Walsh . . . . . .
  23. 23. Outline The big idea, by example Examples Vertical and Horizontal Tangents Chemistry The power rule for rational powers . . . . . .
  24. 24. Example Find the equation of the line tangent to the curve . y2 = x2 (x + 1) = x3 + x2 at the point (3, −6). . . . . . . .
  25. 25. Example Find the equation of the line tangent to the curve . y2 = x2 (x + 1) = x3 + x2 at the point (3, −6). . Solution Differentiating the expression implicitly with respect to x gives 3x2 + 2x dy dy 2y = 3x2 + 2x, so = , and dx dx 2y 3 · 32 + 2 · 3 dy 11 =− . = dx 2(−6) 4 (3,−6) . . . . . .
  26. 26. Example Find the equation of the line tangent to the curve . y2 = x2 (x + 1) = x3 + x2 at the point (3, −6). . Solution Differentiating the expression implicitly with respect to x gives 3x2 + 2x dy dy 2y = 3x2 + 2x, so = , and dx dx 2y 3 · 32 + 2 · 3 dy 11 =− . = dx 2(−6) 4 (3,−6) 11 Thus the equation of the tangent line is y + 6 = − (x − 3). 4 . . . . . .
  27. 27. Example Find the horizontal tangent lines to the same curve: y2 = x3 + x2 . . . . . .
  28. 28. Example Find the horizontal tangent lines to the same curve: y2 = x3 + x2 Solution . . . . . .
  29. 29. Example Find the horizontal tangent lines to the same curve: y2 = x3 + x2 Solution We solve for dy/dx = 0: 3x2 + 2x = 0 =⇒ 3x2 + 2x = 0 =⇒ x(3x + 2) = 0 2y . . . . . .
  30. 30. Example Find the horizontal tangent lines to the same curve: y2 = x3 + x2 Solution We solve for dy/dx = 0: 3x2 + 2x = 0 =⇒ 3x2 + 2x = 0 =⇒ x(3x + 2) = 0 2y The possible solution x = 0 leads to y = 0, which is not a smooth point of the function (the denominator in dy/dx becomes 0). . . . . . .
  31. 31. Example Find the horizontal tangent lines to the same curve: y2 = x3 + x2 Solution We solve for dy/dx = 0: 3x2 + 2x = 0 =⇒ 3x2 + 2x = 0 =⇒ x(3x + 2) = 0 2y The possible solution x = 0 leads to y = 0, which is not a smooth point of the function (the denominator in dy/dx becomes 0). The possible solution x = − 2 yields y = ± 3√3 . 2 3 . . . . . .
  32. 32. Example Find the vertical tangent lines to the same curve: y2 = x3 + x2 . . . . . .
  33. 33. Example Find the vertical tangent lines to the same curve: y2 = x3 + x2 Solution dx = 0. Tangent lines are vertical when dy Differentiating x implicitly as a function of y gives dx dx 2y = 3x2 + 2x , so dy dy dx 2y =2 3x + 2x dy . . . . . .
  34. 34. Example Find the vertical tangent lines to the same curve: y2 = x3 + x2 Solution dx = 0. Tangent lines are vertical when dy Differentiating x implicitly as a function of y gives dx dx 2y = 3x2 + 2x , so dy dy dx 2y =2 3x + 2x dy This is 0 only when y = 0. . . . . . .
  35. 35. Example Find the vertical tangent lines to the same curve: y2 = x3 + x2 Solution dx = 0. Tangent lines are vertical when dy Differentiating x implicitly as a function of y gives dx dx 2y = 3x2 + 2x , so dy dy dx 2y =2 3x + 2x dy This is 0 only when y = 0. We get the false solution x = 0 and the real solution x = −1. . . . . . .
  36. 36. 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: Scott Beale / Laughing Squid . . . . . .
  37. 37. . Definition The isothermic compressibility of a fluid is defined by dV 1 β=− dP V with temperature held constant. . Image credit: Neil Better . . . . . .
  38. 38. . Definition The isothermic compressibility of a fluid is defined by dV 1 β=− dP V with temperature held constant. The smaller the β, the “harder” the fluid. . Image credit: Neil Better . . . . . .
  39. 39. Example Find the isothermic compressibility of an ideal gas. . . . . . .
  40. 40. Example Find the isothermic compressibility of an ideal gas. Solution 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 So 1 dV 1 β=− · = V dP P Compressibility and pressure are inversely related. . . . . . .
  41. 41. Nonideal gasses Not that there’s anything wrong with that Example The van der Waals equation H .. makes fewer simplifications: O. ( ) . xygen . . H n2 P + a 2 (V − nb) = nRT, . V H .. O. H . ydrogen bonds . xygen where P is the pressure, V the H .. volume, T the temperature, n . the number of moles of the O. . xygen . . H gas, R a constant, a is a measure of attraction between H .. particles of the gas, and b a measure of particle size. . . . . . .
  42. 42. Nonideal gasses Not that there’s anything wrong with that Example The van der Waals equation makes fewer simplifications: ( ) n2 P + a 2 (V − nb) = nRT, V where P is the pressure, V the volume, T the temperature, n the number of moles of the gas, R a constant, a is a measure of attraction between particles of the gas, and b a measure of particle size. . . Image credit: Wikimedia Commons . . . . . .
  43. 43. Let’s find the compressibility of a van der Waals gas. Differentiating the van der Waals equation by treating V as a function of P gives ( ) ( ) an2 dV 2an2 dV + (V − bn) 1 − 3 P+ 2 = 0, dP V dP V . . . . . .
  44. 44. Let’s find the compressibility of a van der Waals gas. Differentiating the van der Waals equation by treating V as a function of P gives ( ) ( ) an2 dV 2an2 dV + (V − bn) 1 − 3 P+ 2 = 0, dP V dP V so V2 (V − nb) 1 dV β=− = 2abn3 − an2 V + PV3 V dP . . . . . .
  45. 45. Let’s find the compressibility of a van der Waals gas. Differentiating the van der Waals equation by treating V as a function of P gives ( ) ( ) an2 dV 2an2 dV + (V − bn) 1 − 3 P+ 2 = 0, dP V dP V so V2 (V − nb) 1 dV β=− = 2abn3 − an2 V + PV3 V dP What if a = b = 0? . . . . . .
  46. 46. Let’s find the compressibility of a van der Waals gas. Differentiating the van der Waals equation by treating V as a function of P gives ( ) ( ) an2 dV 2an2 dV + (V − bn) 1 − 3 P+ 2 = 0, dP V dP V so V2 (V − nb) 1 dV β=− = 2abn3 − an2 V + PV3 V dP What if a = b = 0? dβ Without taking the derivative, what is the sign of ? db . . . . . .
  47. 47. Let’s find the compressibility of a van der Waals gas. Differentiating the van der Waals equation by treating V as a function of P gives ( ) ( ) an2 dV 2an2 dV + (V − bn) 1 − 3 P+ 2 = 0, dP V dP V so V2 (V − nb) 1 dV β=− = 2abn3 − an2 V + PV3 V dP What if a = b = 0? dβ Without taking the derivative, what is the sign of ? db dβ Without taking the derivative, what is the sign of ? da . . . . . .
  48. 48. Nasty derivatives (2abn3 − an2 V + PV3 )(nV2 ) − (nbV2 − V3 )(2an3 ) dβ =− (2abn3 − an2 V + PV3 )2 db (2 ) nV3 an + PV2 = −( )2 < 0 PV3 + an2 (2bn − V) n2 (bn − V)(2bn − V)V2 dβ =( )2 > 0 da PV3 + an2 (2bn − V) (as long as V > 2nb, and it’s probably true that V ≫ 2nb). . . . . . .
  49. 49. Outline The big idea, by example Examples Vertical and Horizontal Tangents Chemistry The power rule for rational powers . . . . . .
  50. 50. Using implicit differentiation to find derivatives Example √ dy if y = x. Find dx . . . . . .
  51. 51. Using implicit differentiation to find derivatives Example √ dy if y = x. Find dx Solution √ If y = x, then y2 = x, so dy dy 1 1 = √. = 1 =⇒ = 2y dx dx 2y 2x . . . . . .
  52. 52. The power rule for rational numbers Example dy if y = xp/q , where p and q are integers. Find dx . . . . . .
  53. 53. The power rule for rational numbers Example dy if y = xp/q , where p and q are integers. Find dx Solution We have p xp−1 dy dy = · q−1 yq = xp =⇒ qyq−1 = pxp−1 =⇒ dx dx qy . . . . . .
  54. 54. The power rule for rational numbers Example dy if y = xp/q , where p and q are integers. Find dx Solution We have p xp−1 dy dy = · q−1 yq = xp =⇒ qyq−1 = pxp−1 =⇒ dx dx qy Now yq−1 = xp(q−1)/q = xp−p/q so xp−1 = xp−1−(p−p/q) = xp/q−1 yq−1 . . . . . .

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