Lesson 11: Implicit Differentiation (Section 21 handout)

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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.

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Lesson 11: Implicit Differentiation (Section 21 handout)

  1. 1. V63.0121.021, Calculus I Section 2.6 : Implicit Differentiation October 11, 2010 Notes Section 2.6 Implicit Differentiation V63.0121.021, Calculus I New York University October 11, 2010 Announcements Quiz 2 in recitation this week. Covers §§1.5, 1.6, 2.1, 2.2 Midterm next week. Covers §§1.1–2.5 Announcements Notes Quiz 2 in recitation this week. Covers §§1.5, 1.6, 2.1, 2.2 Midterm next week. Covers §§1.1–2.5 V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 2 / 34 Objectives Notes Use implicit differentation to find the derivative of a function defined implicitly. V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 3 / 34 1
  2. 2. V63.0121.021, Calculus I Section 2.6 : Implicit Differentiation October 11, 2010 Outline Notes The big idea, by example Examples Basic Examples Vertical and Horizontal Tangents Orthogonal Trajectories Chemistry The power rule for rational powers V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 4 / 34 Motivating Example y Notes Problem Find the slope of the line which is tangent to the curve x x2 + y2 = 1 at the point (3/5, −4/5). Solution (Explicit) Isolate: y 2 = 1 − x 2 =⇒ y = − 1 − x 2 . (Why the −?) dy −2x x Differentiate: =− √ =√ dx 2 1 − x2 1 − x2 dy 3/5 3/5 3 Evaluate: = = = . dx x=3/5 1 − (3/5)2 4/5 4 V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 5 / 34 Motivating Example, another way Notes We know that x 2 + y 2 = 1 does not define y as a function of x, but suppose it did. Suppose we had y = f (x), so that x 2 + (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) V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 6 / 34 2
  3. 3. V63.0121.021, Calculus I Section 2.6 : Implicit Differentiation October 11, 2010 Yes, we can! Notes The beautiful fact (i.e., deep theorem) is that this works! y “Near” most points on the curve x 2 + y 2 = 1, the curve resembles the graph of a function. looks like a function So f (x) is defined “locally”, x almost everywhere and is differentiable does not look like a The chain rule then applies function, but that’s for this local choice. OK—there are only two points like this looks like a function V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 7 / 34 Motivating Example, again, with Leibniz notation Notes Problem Find the slope of the line which is tangent to the curve x 2 + y 2 = 1 at the point (3/5, −4/5). Solution dy Differentiate: 2x + 2y =0 dx Remember y is assumed to be a function of x! dy x Isolate: =− . dx y dy 3/5 3 Evaluate: = = . dx ( 3 ,− 4 ) 4/5 4 5 5 V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 8 / 34 Summary Notes If a relation is given between x and y which isn’t a function: “Most of the time”, i.e., “at y most places” y can be assumed to be a function of x we may differentiate the relation x as is dy Solving for does give the dx slope of the tangent line to the curve at a point on the curve. V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 9 / 34 3
  4. 4. V63.0121.021, Calculus I Section 2.6 : Implicit Differentiation October 11, 2010 Outline Notes The big idea, by example Examples Basic Examples Vertical and Horizontal Tangents Orthogonal Trajectories Chemistry The power rule for rational powers V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 10 / 34 Another Example Notes Example Find y along the curve y 3 + 4xy = x 2 + 3. Solution Implicitly differentiating, we have 3y 2 y + 4(1 · y + x · y ) = 2x Solving for y gives 3y 2 y + 4xy = 2x − 4y (3y 2 + 4x)y = 2x − 4y 2x − 4y =⇒ y = 2 3y + 4x V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 11 / 34 Yet Another Example Notes Example Find y if y 5 + x 2 y 3 = 1 + y sin(x 2 ). Solution V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 12 / 34 4
  5. 5. V63.0121.021, Calculus I Section 2.6 : Implicit Differentiation October 11, 2010 Finding tangent lines with implicit differentitiation Notes Example Find the equation of the line tangent to the curve y 2 = x 2 (x + 1) = x 3 + x 2 at the point (3, −6). Solution V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 13 / 34 Recall: Line equation forms Notes 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. V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 14 / 34 Horizontal Tangent Lines Notes Example Find the horizontal tangent lines to the same curve: y 2 = x 3 + x 2 Solution We have to solve these two equations: 3x 2 + 2x = 0 y2 = x3 + x2 2y 1 [(x, y ) is on the curve] 2 [tangent line is horizontal] V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 15 / 34 5
  6. 6. V63.0121.021, Calculus I Section 2.6 : Implicit Differentiation October 11, 2010 Solution, continued Notes Solving the second equation gives 3x 2 + 2x = 0 =⇒ 3x 2 + 2x = 0 =⇒ x(3x + 2) = 0 2y (as long as y = 0). So x = 0 or 3x + 2 = 0. Substituting x = 0 into the first equation gives y 2 = 03 + 02 = 0 =⇒ y = 0 which we’ve disallowed. So no horizontal tangents down that road. Substituting x = −2/3 into the first equation gives 3 2 2 2 4 2 y2 = − + − = =⇒ y = ± √ , 3 3 27 3 3 so there are two horizontal tangents. V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 16 / 34 Tangents Notes − 2 , 3√3 3 2 (−1, 0) − 2 , − 3√3 3 2 node V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 17 / 34 Example Find the vertical tangent lines to the same curve: y 2 = x 3 + x 2 Notes Solution V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 18 / 34 6
  7. 7. V63.0121.021, Calculus I Section 2.6 : Implicit Differentiation October 11, 2010 Solution, continued Notes V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 19 / 34 Examples Notes Example Show that the families of curves xy = c x2 − y2 = k are orthogonal, that is, they intersect at right angles. Solution In the first curve, y y + xy = 0 =⇒ y = − x In the second curve, x 2x − 2yy = 0 = =⇒ y = y The product is −1, so the tangent lines are perpendicular wherever they intersect. V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 21 / 34 Orthogonal Families of Curves Notes y xy xy = 3 = xy 2 = x2 − y2 = 3 x −y =2 x2 − y2 = 1 1 xy = c 2 x2 − y2 = k x 1 − 2 − 2 = x 2 − y 2 = −1 − 3 xy = x 2 − y 2 = −2 xy = x 2 − y 2 = −3 xy V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 22 / 34 7
  8. 8. V63.0121.021, Calculus I Section 2.6 : Implicit Differentiation October 11, 2010 Examples Notes Example Show that the families of curves xy = c x2 − y2 = k are orthogonal, that is, they intersect at right angles. Solution In the first curve, y y + xy = 0 =⇒ y = − x In the second curve, x 2x − 2yy = 0 = =⇒ y = y The product is −1, so the tangent lines are perpendicular wherever they intersect. V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 23 / 34 Ideal gases Notes 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 V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 25 / 34 Compressibility Notes Definition The isothermic compressibility of a fluid is defined by dV 1 β=− dP V with temperature held constant. Approximately we have ∆V dV ∆V ≈ = −βV =⇒ ≈ −β∆P ∆P dP V The smaller the β, the “harder” the fluid. V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 26 / 34 8
  9. 9. V63.0121.021, Calculus I Section 2.6 : Implicit Differentiation October 11, 2010 Compressibility of an ideal gas Notes 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. V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 27 / 34 Nonideal gasses Not that there’s anything wrong with that Notes Example The van der Waals equation H makes fewer simplifications: Oxygen H n2 P +a 2 (V − nb) = nRT , V H Oxygen Hydrogen bonds where P is the pressure, V the H volume, T the temperature, n the number of moles of the gas, Oxygen H R a constant, a is a measure of attraction between particles of H the gas, and b a measure of particle size. V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 28 / 34 Compressibility of a van der Waals gas Notes Differentiating the van der Waals equation by treating V as a function of P gives an2 dV 2an2 dV P+ + (V − bn) 1 − = 0, V2 dP V 3 dP so 1 dV V 2 (V − nb) β=− = V dP 2abn3 − an2 V + PV 3 Question 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 V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 29 / 34 9
  10. 10. V63.0121.021, Calculus I Section 2.6 : Implicit Differentiation October 11, 2010 Nasty derivatives Notes dβ (2abn3 − an2 V + PV 3 )(nV 2 ) − (nbV 2 − V 3 )(2an3 ) =− db (2abn3 − an2 V + PV 3 )2 nV 3 an2 + PV 2 =− <0 (PV 3 + an2 (2bn − V ))2 dβ n2 (bn − V )(2bn − V )V 2 = >0 da (PV 3 + an2 (2bn − V ))2 (as long as V > 2nb, and it’s probably true that V 2nb). V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 30 / 34 Outline Notes The big idea, by example Examples Basic Examples Vertical and Horizontal Tangents Orthogonal Trajectories Chemistry The power rule for rational powers V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 31 / 34 Using implicit differentiation to find derivatives Notes Example dy √ Find if y = x. dx Solution √ If y = x, then y 2 = x, so dy dy 1 1 2y = 1 =⇒ = = √ . dx dx 2y 2 x V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 32 / 34 10
  11. 11. V63.0121.021, Calculus I Section 2.6 : Implicit Differentiation October 11, 2010 The power rule for rational powers Notes Theorem p If y = x p/q , where p and q are integers, then y = x p/q−1 . q Proof. V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 33 / 34 Summary Notes Implicit Differentiation allows us to pretend that a relation describes a function, since it does, locally, “almost everywhere.” The Power Rule was established for powers which are rational numbers. V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 34 / 34 Notes 11

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