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    • The Mother of All Calculus Quizzes Louis A. Talman, Ph.D.Department of Mathematical & Computer Sciences Metropolitan State College of Denver February 22, 2008 1 / 56
    • Question 1: ADerivativeA Derivative IA Derivative IIA Derivative IIIA Derivative—TheIssueQuestion 2: IncreasingFunctionsQuestion 3: Concavity Question 1: A DerivativeQuestion 4: LocalMinimaQuestion 5: PolarArc-LengthQuestion 6: ImplicitDifferentiationQuestion 7: ImplicitFunctionsQuestion 8: ImproperIntegrals 2 / 56
    • A Derivative IQuestion 1: ADerivativeA Derivative IA Derivative IIA Derivative III Is it true that the function f given byA Derivative—TheIssueQuestion 2: Increasing x2 sin(1/x), when x = 0Functions f (x) = (1)Question 3: Concavity 0, when x = 0Question 4: LocalMinima is differentiable at x = 0? The differentiation rules giveQuestion 5: PolarArc-LengthQuestion 6: Implicit f ′ (x) = 2x sin(1/x) − cos(1/x), (2)DifferentiationQuestion 7: ImplicitFunctions and this is undefined when x = 0. What gives?Question 8: ImproperIntegrals 3 / 56
    • A Derivative IIQuestion 1: ADerivativeA Derivative IA Derivative IIA Derivative IIIA Derivative—TheIssue Here’s what the Product Rule actually says:Question 2: IncreasingFunctionsQuestion 3: Concavity If F (x) = u(x) · v(x), and if u′ (x0 ) and v ′ (x0 ) both exist, thenQuestion 4: LocalMinimaQuestion 5: Polar 1. F ′ (x0 ) exists, andArc-Length 2. is given byQuestion 6: ImplicitDifferentiationQuestion 7: Implicit F ′ (x0 ) = u′ (x0 ) · v(x0 ) + u(x0 ) · v ′ (x0 ). (3)FunctionsQuestion 8: ImproperIntegrals 4 / 56
    • A Derivative IIIQuestion 1: ADerivativeA Derivative IA Derivative IIA Derivative III For the functionA Derivative—TheIssueQuestion 2: Increasing x2 sin(1/x), when x = 0Functions f (x) = (4)Question 3: Concavity 0, when x = 0,Question 4: LocalMinima the difference quotient calculation gives usQuestion 5: PolarArc-Length ′ f (h) − f (0) h2 sin(1/h) − 0Question 6: Implicit f (0) = lim = lim (5)Differentiation h→0 h h→0 hQuestion 7: ImplicitFunctions = lim h sin(1/h) = 0.• (6)Question 8: Improper h→0Integrals 5 / 56
    • A Derivative—The IssueQuestion 1: ADerivativeA Derivative IA Derivative IIA Derivative IIIA Derivative—TheIssue The IssueQuestion 2: IncreasingFunctionsQuestion 3: Concavity When a theorem fails of applicability, that doesn’tQuestion 4: LocalMinima necessarily mean that no part of its conclusion can beQuestion 5: Polar true.Arc-LengthQuestion 6: ImplicitDifferentiationQuestion 7: ImplicitFunctionsQuestion 8: ImproperIntegrals 6 / 56
    • Question 1: ADerivativeQuestion 2: IncreasingFunctionsIncreasing Functions IIncreasing Functions IIIncreasing FunctionsIIIIncreasing FunctionsIV Question 2: IncreasingIncreasing Functions VIncreasing Functions FunctionsVIIncreasing FunctionsVIIIncreasingFunctions—The IssueQuestion 3: ConcavityQuestion 4: LocalMinimaQuestion 5: PolarArc-LengthQuestion 6: ImplicitDifferentiationQuestion 7: ImplicitFunctionsQuestion 8: ImproperIntegrals 7 / 56
    • Increasing Functions IQuestion 1: ADerivativeQuestion 2: IncreasingFunctionsIncreasing Functions IIncreasing Functions IIIncreasing FunctionsIIIIncreasing Functions How can we say that the function f (x) = x3 is increasing on theIVIncreasing Functions V interval [−1, 1], when f ′ (0) = 0 so that f isn’t increasing at 0?Increasing FunctionsVIIncreasing FunctionsVIIIncreasingFunctions—The IssueQuestion 3: ConcavityQuestion 4: LocalMinimaQuestion 5: PolarArc-LengthQuestion 6: ImplicitDifferentiationQuestion 7: ImplicitFunctionsQuestion 8: ImproperIntegrals 8 / 56
    • Increasing Functions IIQuestion 1: ADerivativeQuestion 2: IncreasingFunctions Def: f is increasing on a set A whenever u ∈ A, v ∈ A, and u < vIncreasing Functions IIncreasing Functions II implies f (u) < f (v).Increasing FunctionsIIIIncreasing Functions For the cubing function, we note that if u, v ∈ [−1, 1] with u < v thenIVIncreasing Functions V u − v < 0, whenceIncreasing FunctionsVIIncreasing FunctionsVII u3 − v 3 = (u − v)(u2 + uv + v 2 ) (7)Increasing √  Functions—The Issue 2  u+ v + v 3 2Question 3: Concavity = (u − v)  < 0, (8)Question 4: Local 2 2MinimaQuestion 5: PolarArc-Length so u3 < v 3 .•Question 6: ImplicitDifferentiationQuestion 7: ImplicitFunctionsQuestion 8: ImproperIntegrals 9 / 56
    • Increasing Functions IIIQuestion 1: ADerivative Theorem: If f is continuous on [a, b] and increasing on a dense subset DQuestion 2: Increasing of [a, b], then f is increasing on [a, b].FunctionsIncreasing Functions IIncreasing Functions II Choose u, v ∈ [a, b], with u < v , and suppose that one or both of u, v doIncreasing FunctionsIII not lie in D . (Otherwise f (u) < f (v) and there is nothing to prove.)Increasing FunctionsIVIncreasing Functions V Select d0 ∈ (u, v) ∩ D . For each k ∈ N take αk−1 to be the midpoint ofIncreasing FunctionsVI (dk−1 , v)and choose dk ∈ (αk−1 , v) ∩ D.Increasing FunctionsVIIIncreasing Then d0 < d1 < · · · < dk < dk+1 < · · ·, with limk→∞ dk = v ,soFunctions—The Issue f (d0 ) < f (d1 ) < · · · < f (dk ) < f (dk+1 ) < · · ·, withQuestion 3: ConcavityQuestion 4: Local limk→∞ f (dk ) = f (v).MinimaQuestion 5: Polar It follows now that f (d0 ) < f (v).Arc-LengthQuestion 6: Implicit Similarly, f (u) < f (d0 ), and thus f (u) < f (v).•DifferentiationQuestion 7: ImplicitFunctionsQuestion 8: ImproperIntegrals 10 / 56
    • Increasing Functions IVQuestion 1: ADerivative It isn’t really clear what “increasing at 0” means. If g is given byQuestion 2: IncreasingFunctions x/2 + x2 sin(1/x) when x = 0,Increasing Functions I g(x) = (9)Increasing Functions IIIncreasing Functions 0 when x = 0,IIIIncreasing FunctionsIV then g ′ (0) = 1/2. But g isn’t increasing on any interval (−δ, δ).Increasing Functions VIncreasing FunctionsVIIncreasing FunctionsVIIIncreasingFunctions—The IssueQuestion 3: ConcavityQuestion 4: LocalMinimaQuestion 5: PolarArc-LengthQuestion 6: ImplicitDifferentiationQuestion 7: ImplicitFunctionsQuestion 8: ImproperIntegrals 11 / 56
    • Increasing Functions VQuestion 1: ADerivativeQuestion 2: IncreasingFunctionsIncreasing Functions I ′ d x 2 1 1 1Increasing Functions II g (0) = + x sin = +0= , (10)Increasing FunctionsIII dx x=0 2 x 2 2Increasing FunctionsIVIncreasing Functions V while x = 0 givesIncreasing FunctionsVIIncreasing Functions 1VII g ′ (x) = + 2x sin(1/x) − cos(1/x). (11)Increasing 2Functions—The IssueQuestion 3: Concavity So every interval (−δ, δ) contains sub-intervals on which f isQuestion 4: LocalMinima decreasing—even though f ′ (0) > 0.•Question 5: PolarArc-LengthQuestion 6: ImplicitDifferentiationQuestion 7: ImplicitFunctionsQuestion 8: ImproperIntegrals 12 / 56
    • Increasing Functions VIQuestion 1: ADerivativeQuestion 2: IncreasingFunctionsIncreasing Functions IIncreasing Functions II 0.06Increasing FunctionsIIIIncreasing FunctionsIV 0.04Increasing Functions VIncreasing FunctionsVI 0.02Increasing FunctionsVIIIncreasingFunctions—The IssueQuestion 3: Concavity 0.10 0.05 0.05 0.10Question 4: LocalMinima 0.02Question 5: PolarArc-LengthQuestion 6: Implicit 0.04DifferentiationQuestion 7: ImplicitFunctions 0.06Question 8: ImproperIntegrals 13 / 56
    • Increasing Functions VIIQuestion 1: ADerivativeQuestion 2: IncreasingFunctionsIncreasing Functions IIncreasing Functions II 0.06Increasing Functions y x 2 x2IIIIncreasing FunctionsIV 0.04Increasing Functions VIncreasing FunctionsVI 0.02Increasing FunctionsVIIIncreasingFunctions—The IssueQuestion 3: Concavity 0.10 0.05 0.05 0.10Question 4: LocalMinima 0.02Question 5: PolarArc-LengthQuestion 6: Implicit 0.04DifferentiationQuestion 7: Implicit y x 2 x2Functions 0.06Question 8: ImproperIntegrals 14 / 56
    • Increasing Functions—The IssueQuestion 1: ADerivativeQuestion 2: IncreasingFunctionsIncreasing Functions IIncreasing Functions IIIncreasing FunctionsIII The IssueIncreasing FunctionsIVIncreasing Functions VIncreasing Functions I The real problem here lies in failure to understandVIIncreasing FunctionsVII — the relationship between theorems andIncreasingFunctions—The Issue definitions, and, ultimately,Question 3: ConcavityQuestion 4: Local — ˆ the role of definition in mathematics.MinimaQuestion 5: PolarArc-LengthQuestion 6: ImplicitDifferentiationQuestion 7: ImplicitFunctionsQuestion 8: ImproperIntegrals 15 / 56
    • Question 1: ADerivativeQuestion 2: IncreasingFunctionsQuestion 3: ConcavityConcavity IConcavity IIConcavity—The MoralQuestion 4: Local Question 3: ConcavityMinimaQuestion 5: PolarArc-LengthQuestion 6: ImplicitDifferentiationQuestion 7: ImplicitFunctionsQuestion 8: ImproperIntegrals 16 / 56
    • Concavity IQuestion 1: ADerivativeQuestion 2: IncreasingFunctionsQuestion 3: ConcavityConcavity IConcavity IIConcavity—The Moral If y = 6x2 − x4 , then y ′′ = 12 − 12x2 , and this is positive exactlyQuestion 4: LocalMinima when −1 < x < 1. Where is the curve concave upward?Question 5: PolarArc-LengthQuestion 6: ImplicitDifferentiationQuestion 7: Implicit Is the answer “(−1, 1)”, or is it “[−1, 1]”?FunctionsQuestion 8: ImproperIntegrals 17 / 56
    • Concavity IIQuestion 1: ADerivative This is trickier than the last question. f is concave upward on an interval IQuestion 2: Increasing provided:FunctionsQuestion 3: Concavity I f ′′ (x) > 0 when x ∈ I . (G. L. Bradley & K. J. Smith, 1999; S. K. Stein, 1977)Concavity IConcavity II I f ′ is an increasing function on I . (R. Larson, R. Hostetler & B. H. Edwards,Concavity—The Moral 2007; J. Stewart, 2005)Question 4: LocalMinima I The tangent line at each point of the curve lies (locally) below the curve in I .Question 5: PolarArc-Length (C. H. Edwards & D. E. Penney, 2008; M. P. Fobes & R. B. Smyth, 1963)Question 6: ImplicitDifferentiation I f [(1 − λ)x1 ] + f (λx2 ) ≤ (1 − λ)f (x1 ) + λf (x2 ) when x1 , x2 ∈ I andQuestion 7: ImplicitFunctions 0 < λ < 1. (G. B. Thomas, Jr., 1972)Question 8: ImproperIntegrals I {(x, y) : x ∈ I ⇒ y ≥ f (x)} is a convex set. (R. P. Agnew, 1962) 18 / 56
    • Concavity—The MoralQuestion 1: ADerivativeQuestion 2: IncreasingFunctionsQuestion 3: ConcavityConcavity IConcavity II The Moral?Concavity—The MoralQuestion 4: LocalMinimaQuestion 5: PolarArc-Length Read your author’s definitions.Question 6: ImplicitDifferentiationQuestion 7: ImplicitFunctionsQuestion 8: ImproperIntegrals 19 / 56
    • Question 1: ADerivativeQuestion 2: IncreasingFunctionsQuestion 3: ConcavityQuestion 4: LocalMinimaLocal Minima ILocal Minima II Question 4: Local MinimaLocal Minima IIILocal Minima IVLocal Minima—TheIssueQuestion 5: PolarArc-LengthQuestion 6: ImplicitDifferentiationQuestion 7: ImplicitFunctionsQuestion 8: ImproperIntegrals 20 / 56
    • Local Minima IQuestion 1: ADerivativeQuestion 2: IncreasingFunctionsQuestion 3: ConcavityQuestion 4: LocalMinimaLocal Minima ILocal Minima II If a smooth function f has a local minimum at x = x0 , must there beLocal Minima III δ > 0 so that f ′ (x) ≤ 0 on (x0 − δ, x0 ) but f ′ (x) ≥ 0 onLocal Minima IVLocal Minima—The (x0 , x0 + δ)?IssueQuestion 5: PolarArc-LengthQuestion 6: ImplicitDifferentiationQuestion 7: ImplicitFunctionsQuestion 8: ImproperIntegrals 21 / 56
    • Local Minima IIQuestion 1: ADerivativeQuestion 2: IncreasingFunctions The First Derivative Test, of course, says:Question 3: ConcavityQuestion 4: Local If there is a δ > 0 such that f ′ is negative on (x0 − δ, x0 ) and positive onMinimaLocal Minima I (x0 , x0 + δ), then f has a local minimum at x = x0 .Local Minima IILocal Minima IIILocal Minima IVLocal Minima—The And here’s a counter example to the converse:IssueQuestion 5: PolarArc-Length 4x4 − 3x4 cos(1/x), when x = 0;Question 6: Implicit f (x) = (12)Differentiation 0, when x = 0.Question 7: ImplicitFunctionsQuestion 8: ImproperIntegrals 22 / 56
    • Local Minima IIIQuestion 1: ADerivativeQuestion 2: IncreasingFunctionsQuestion 3: Concavity 0.0015Question 4: LocalMinimaLocal Minima ILocal Minima IILocal Minima IIILocal Minima IVLocal Minima—The 0.0010IssueQuestion 5: PolarArc-LengthQuestion 6: ImplicitDifferentiationQuestion 7: ImplicitFunctions 0.0005Question 8: ImproperIntegrals 0.2 0.1 0.1 0.2 23 / 56
    • Local Minima IVQuestion 1: ADerivativeQuestion 2: IncreasingFunctionsQuestion 3: Concavity 0.0015Question 4: LocalMinima y 7x4Local Minima ILocal Minima IILocal Minima IIILocal Minima IVLocal Minima—The 0.0010IssueQuestion 5: PolarArc-LengthQuestion 6: ImplicitDifferentiationQuestion 7: ImplicitFunctions 0.0005Question 8: ImproperIntegrals y x4 0.2 0.1 0.1 0.2 24 / 56
    • Local Minima—The IssueQuestion 1: ADerivativeQuestion 2: IncreasingFunctionsQuestion 3: Concavity The issue?Question 4: LocalMinimaLocal Minima ILocal Minima IILocal Minima IIILocal Minima IVLocal Minima—TheIssue Confusion of a theorem with its converse,Question 5: PolarArc-Length among other things.Question 6: ImplicitDifferentiationQuestion 7: ImplicitFunctionsQuestion 8: ImproperIntegrals 25 / 56
    • Question 1: ADerivativeQuestion 2: IncreasingFunctionsQuestion 3: ConcavityQuestion 4: LocalMinimaQuestion 5: PolarArc-Length Question 5: Polar Arc-LengthPolar Arc-Length IPolar Arc-Length IIPolar Arc-length IIIPolar Arc-length IVQuestion 6: ImplicitDifferentiationQuestion 7: ImplicitFunctionsQuestion 8: ImproperIntegrals 26 / 56
    • Polar Arc-Length IQuestion 1: ADerivativeQuestion 2: IncreasingFunctions Why don’t we approach arc-length in polar coordinates the way weQuestion 3: Concavity do in cartesian coordinates?Question 4: LocalMinimaQuestion 5: PolarArc-LengthPolar Arc-Length I In cartesian coordinates:Polar Arc-Length IIPolar Arc-length IIIPolar Arc-length IV s = lim (xk − xk−1 )2 + [f (xk ) − f (xk−1 )]2 (13)Question 6: ImplicitDifferentiation = lim (xk − xk−1 )2 + [f ′ (ξk )]2 (xk − xk−1 )2 (by MVT) (14)Question 7: ImplicitFunctionsQuestion 8: Improper = lim 1 + [f ′ (ξk )]2 (xk − xk−1 ) (15)Integrals b = 1 + [f ′ (x)]2 dx (16) a 27 / 56
    • Polar Arc-Length IIQuestion 1: ADerivative When r = f (θ) in polar coordinates (so that f (θk ) = rk ), the Law of CosinesQuestion 2: Increasing gives:FunctionsQuestion 3: Concavity 2 2 s = lim rk + rk−1 − 2rk rk−1 cos(θk − θk−1 ) (17)Question 4: LocalMinimaQuestion 5: Polar = lim (rk − rk−1 )2 + 2(1 − cos ∆θk )rk rk−1 (18)Arc-LengthPolar Arc-Length I 1 − cos ∆θkPolar Arc-Length II = lim [f ′ (ξk )]2 + 2f (θk )f (θk−1 ) 2 ∆θk . (19)Polar Arc-length III (∆θk )Polar Arc-length IVQuestion 6: Implicit This is not a Riemann sum. . . and I see no way to fudge it into one.DifferentiationQuestion 7: Implicit The fact thatFunctions 1 − cos t 1Question 8: Improper lim = (20)Integrals t→0+ t2 2 is very suggestive—though not particularly helpful. 28 / 56
    • Polar Arc-length IIIQuestion 1: ADerivative Duhamel’s Theorem (Standard Model)Question 2: IncreasingFunctionsQuestion 3: Concavity Theorem1 : Let f be a continuous function of three variables onQuestion 4: LocalMinima [a, b] × [a, b] × [a, b]. If P = {x0 , x1 , . . . , xn }, whereQuestion 5: Polar a = x0 < x1 < x2 < · · · < xn−1 < xn = b, is a partition of [a, b], withArc-Length xk−1 ≤ ξk , ηk , ζk ≤ xk for k = 1, 2, . . . , n, then for every ǫ > 0 there is aPolar Arc-Length IPolar Arc-Length II δ > 0 such that whenever P < δ it follows thatPolar Arc-length IIIPolar Arc-length IV n bQuestion 6: Implicit f (ξk , ηk , ζk )(xk − xk−1 ) − f (t, t, t) dt < ǫ. (21)Differentiation k=1 aQuestion 7: ImplicitFunctionsQuestion 8: ImproperIntegrals 1 Adapted from Advanced Calculus, David V. Widder, Second Edition, Prentice-Hall, 1961, and reprinted by Dover, 1989; p 174. 29 / 56
    • Polar Arc-length IVQuestion 1: ADerivative Duhamel’s Theorem (Deluxe Model)Question 2: IncreasingFunctionsQuestion 3: Concavity Theorem: Let η > 0, and suppose that F is a continuous function fromQuestion 4: Local [a, b] × [a, b] × [a, b] × [0, η] to R. To each partitionMinimaQuestion 5: Polar P = {x0 , x1 , . . . , xn }, where a = x0 < x1 < · · · < xn = b, and toArc-Length each choice of triples of numbers ξk , ηk , ζk ∈ [xk−1 , xk ], k = 1, . . . , n,Polar Arc-Length IPolar Arc-Length II we associate the sumPolar Arc-length IIIPolar Arc-length IV nQuestion 6: Implicit S(F, [a, b], P, {(ξk , ηk , ζk )}n ) = k=1 F (ξk , ηk , ζk , ∆xk ) ∆xk .DifferentiationQuestion 7: Implicit k=1FunctionsQuestion 8: Improper If ǫ > 0, there is a δ > 0 such that P < δ impliesIntegrals b S(F, [a, b], P, {(ξk , ηk , ζk )}n ) k=1 − F (t, t, t, 0) dt < ǫ. a 30 / 56
    • Question 1: ADerivativeQuestion 2: IncreasingFunctionsQuestion 3: ConcavityQuestion 4: LocalMinimaQuestion 5: Polar Question 6: ImplicitArc-LengthQuestion 6: Implicit DifferentiationDifferentiationImplicit Differentiation IImplicit DifferentiationIIImplicit DifferentiationIIIImplicit DifferentiationIVImplicit DifferentiationVImplicit DifferentiationVIImplicit DifferentiationVIIImplicitDifferentiation—TheIssueQuestion 7: ImplicitFunctions 31 / 56
    • Implicit Differentiation IQuestion 1: ADerivative y 2 (2 − x) Given the problem “Find y ′ when 2 2+1 = 1,”Question 2: IncreasingFunctions x +yQuestion 3: Concavity I Æthelbert differentiated both sides (correctly), solved (correctly),Question 4: Local ′ y(x2 − y 2 − 4x − 1)Minima and got yÆ = 3 − 2x2 + x − 2) .Question 5: Polar 2(xArc-LengthQuestion 6: ImplicitDifferentiation I Brunhilde multiplied through by x2 + y 2 + 1 (correctly) before ¨Implicit Differentiation IImplicit Differentiation she differentiated (correctly), and when she solved (correctly),II ′ 2x + y 2Implicit Differentiation she got yB = .IIIImplicit Differentiation 2y(1 − x)IVImplicit DifferentiationVImplicit DifferentiationVIImplicit Differentiation Who was wrong?VIIImplicitDifferentiation—TheIssueQuestion 7: ImplicitFunctions 32 / 56
    • Implicit Differentiation II Here are the slope fields2 :Question 1: ADerivativeQuestion 2: IncreasingFunctions 2 2Question 3: ConcavityQuestion 4: LocalMinima 1 1Question 5: PolarArc-LengthQuestion 6: ImplicitDifferentiation 2 1 1 2 2 1 1 2Implicit Differentiation IImplicit DifferentiationII 1 1Implicit DifferentiationIIIImplicit DifferentiationIV 2 2Implicit DifferentiationVImplicit Differentiation ′ ′VI yÆ yBImplicit DifferentiationVIIImplicitDifferentiation—The 2Issue My thanks to Prof. Diane Davis for the ideas that underlie the Mathematica code I usedQuestion 7: Implicit to generate these slope fields.Functions 33 / 56
    • Implicit Differentiation IIIQuestion 1: ADerivative Pick a point on the curve—say (−1, 1):Question 2: IncreasingFunctionsQuestion 3: Concavity y 2 (2 − x) 12 (3) I = = 1, so (−1, 1) is on the curve.Question 4: LocalMinima x2 + y 2 + 1 (−1,1) 3Question 5: PolarArc-LengthQuestion 6: Implicit ′ y(x2 − y 2 − 4x − 1) 1 · (3) 1Differentiation I yÆ = = =− .Implicit Differentiation I (−1,1) 2(x3 − 2x2 + x − 2) (−1,1) 2 · (−6) 4Implicit DifferentiationIIImplicit DifferentiationIII ′ 2x + y 2 −2 + 1 1Implicit Differentiation I yB = = =− .IVImplicit Differentiation (−1,1) 2y(1 − x) (−1,1) 2·1·2 4VImplicit DifferentiationVIImplicit DifferentiationVII The issue goes away.ImplicitDifferentiation—TheIssueQuestion 7: ImplicitFunctions 34 / 56
    • Implicit Differentiation IVQuestion 1: ADerivativeQuestion 2: IncreasingFunctionsQuestion 3: Concavity Let y be defined implicitly as a function of x byQuestion 4: LocalMinima F (x, y)Question 5: Polar = H(x, y). (22)Arc-Length G(x, y)Question 6: ImplicitDifferentiation ThenImplicit Differentiation IImplicit DifferentiationII ′ Fx G − F Gx − G2 HxImplicit Differentiation yÆ = − , and (23)IIIImplicit Differentiation Fy G − F Gy − G2 HyIVImplicit Differentiation ′ Fx − Gx H − GHxV yB = − . (24)Implicit Differentiation Fy − Gy H − GHyVIImplicit DifferentiationVIIImplicit It is easy to use (22) to reduce(23) to (24).Differentiation—TheIssueQuestion 7: ImplicitFunctions 35 / 56
    • Implicit Differentiation VQuestion 1: ADerivative Exercise: Assume thatQuestion 2: IncreasingFunctions y 2 (2 − x) 2 + y2 + 1 = 1, (25)Question 3: Concavity xQuestion 4: LocalMinima and show—without using the analysis just given—how to reduceQuestion 5: PolarArc-LengthQuestion 6: Implicit ′y(x2 − y 2 − 4x − 1)Differentiation yÆ = (26)Implicit Differentiation I 2(x3 − 2x2 + x − 2)Implicit DifferentiationIIImplicit DifferentiationIII toImplicit DifferentiationIVImplicit Differentiation ′ 2x + y 2V yB = . (27)Implicit DifferentiationVI 2y(1 − x)Implicit DifferentiationVIIImplicitDifferentiation—TheIssueQuestion 7: ImplicitFunctions 36 / 56
    • Implicit Differentiation VIQuestion 1: ADerivative 2Question 2: IncreasingFunctionsQuestion 3: ConcavityQuestion 4: LocalMinima 1Question 5: PolarArc-LengthQuestion 6: ImplicitDifferentiationImplicit Differentiation IImplicit DifferentiationII 2 1 1 2Implicit DifferentiationIIIImplicit DifferentiationIVImplicit DifferentiationVImplicit Differentiation 1VIImplicit DifferentiationVIIImplicitDifferentiation—TheIssueQuestion 7: Implicit 2Functions 37 / 56
    • Implicit Differentiation VIIQuestion 1: ADerivative The implicit differentiation technique is justified by theQuestion 2: IncreasingFunctions Implicit Function Theorem: Let f be a smooth real-valued functionQuestion 3: ConcavityQuestion 4: Local defined on an open subset D of R, and let (x0 , y0 ) be a solution of theMinima equation f (x, y) = 0. If fy (x0 , y0 ) = 0, there are positive numbers, ǫQuestion 5: PolarArc-Length and δ , and a smooth function ϕ : (x0 − δ, x0 + δ) → (y0 − ǫ, y0 + ǫ)Question 6: ImplicitDifferentiation such that for each x ∈ (x0 − δ, x0 + δ), y = ϕ(x) is the only solution ofImplicit Differentiation I f (x, y) = 0 lying in (y0 − ǫ, y0 + ǫ). Moreover, for eachImplicit DifferentiationII x ∈ (x0 − δ, x0 + δ),Implicit DifferentiationIIIImplicit Differentiation ′ fx [x, ϕ(x)]IVImplicit Differentiation ϕ (x) = − . (28)V fy [x, ϕ(x)]Implicit DifferentiationVIImplicit DifferentiationVIIImplicitDifferentiation—TheIssueQuestion 7: ImplicitFunctions 38 / 56
    • Implicit Differentiation—The IssueQuestion 1: ADerivativeQuestion 2: IncreasingFunctions The IssueQuestion 3: ConcavityQuestion 4: LocalMinimaQuestion 5: Polar Our textbook problems encourage students (andArc-LengthQuestion 6: Implicit teachers) to think about these problemsDifferentiationImplicit Differentiation IImplicit DifferentiationII I globally instead of locally, andImplicit DifferentiationIIIImplicit DifferentiationIVImplicit Differentiation I without considering the hypotheses needed to justifyVImplicit Differentiation what they are doing.VIImplicit DifferentiationVIIImplicitDifferentiation—TheIssueQuestion 7: ImplicitFunctions 39 / 56
    • Question 1: ADerivativeQuestion 2: IncreasingFunctionsQuestion 3: ConcavityQuestion 4: LocalMinimaQuestion 5: PolarArc-Length Question 7: Implicit FunctionsQuestion 6: ImplicitDifferentiationQuestion 7: ImplicitFunctionsImplicit Functions IImplicit Functions IIImplicit Functions IIIImplicit Functions IVImplicit Functions VImplicit Functions VIImplicitFunctions—TheIssuesQuestion 8: ImproperIntegrals 40 / 56
    • Implicit Functions IQuestion 1: ADerivativeQuestion 2: IncreasingFunctionsQuestion 3: ConcavityQuestion 4: LocalMinima When I apply implicit differentiation to the equationQuestion 5: PolarArc-Length (x2 + y 2 )2 = x2 − y 2 to find y ′ , I getQuestion 6: ImplicitDifferentiation ′ x(2y 2 + 2x2 − 1)Question 7: Implicit y =− 2 + 2x2 + 1) , (29)Functions y(2yImplicit Functions IImplicit Functions IIImplicit Functions III which gives the indeterminate form 0/0 at the origin. Can I use limitsImplicit Functions IV to find the slope of the line tangent to this curve at the origin? How?Implicit Functions VImplicit Functions VIImplicitFunctions—TheIssuesQuestion 8: ImproperIntegrals 41 / 56
    • Implicit Functions IIQuestion 1: A yDerivativeQuestion 2: IncreasingFunctions 1.0Question 3: ConcavityQuestion 4: LocalMinimaQuestion 5: Polar 0.5Arc-LengthQuestion 6: ImplicitDifferentiationQuestion 7: ImplicitFunctions 0.0 xImplicit Functions IImplicit Functions IIImplicit Functions IIIImplicit Functions IVImplicit Functions V 0.5Implicit Functions VIImplicitFunctions—TheIssuesQuestion 8: Improper 1.0Integrals 1.0 0.5 0.0 0.5 1.0 42 / 56
    • Implicit Functions IIIQuestion 1: ADerivativeQuestion 2: IncreasingFunctionsQuestion 3: ConcavityQuestion 4: LocalMinimaQuestion 5: Polar The surplus of tangent lines at (0, 0) results from the fact that there is noArc-Length open rectangle, centered at (0, 0), whose intersection with the curve is theQuestion 6: ImplicitDifferentiation graph of a function.Question 7: ImplicitFunctionsImplicit Functions I But the conclusion of the Implicit Function Theorem asserts that there isImplicit Functions II such a rectangle. Because the conclusion is false, the IFT must not applyImplicit Functions IIIImplicit Functions IV to this function at the origin.Implicit Functions VImplicit Functions VIImplicitFunctions—TheIssuesQuestion 8: ImproperIntegrals 43 / 56
    • Implicit Functions IVQuestion 1: ADerivativeQuestion 2: IncreasingFunctionsQuestion 3: ConcavityQuestion 4: LocalMinimaQuestion 5: Polar Actually, we should have known that the IFT doesn’t apply:Arc-LengthQuestion 6: Implicit Putting F (x, y) = (x2 + y 2 )2 − x2 + y 2 , we haveDifferentiationQuestion 7: ImplicitFunctionsImplicit Functions I Fy (0, 0) = 4(x2 + y 2 )y + 2y = 0, (30)Implicit Functions II (0,0)Implicit Functions IIIImplicit Functions IVImplicit Functions V so that one of the hypotheses of the IFT fails.Implicit Functions VIImplicitFunctions—TheIssuesQuestion 8: ImproperIntegrals 44 / 56
    • Implicit Functions VQuestion 1: ADerivativeQuestion 2: IncreasingFunctionsQuestion 3: Concavity We could find the slope of either branch of the curve by using the implicitQuestion 4: LocalMinima derivative if we were to solve, algebraically, for y in terms of x and thenQuestion 5: Polar replace y with the solution throughout the implicit differentiation expressionArc-LengthQuestion 6: Implicit for y ′ —and take the limit as we approach the origin.DifferentiationQuestion 7: Implicit That’s nice. . . except that the whole point of implicit differentiation is toFunctionsImplicit Functions I cirvumvent the necessity of solving for y in terms of x. . .Implicit Functions IIImplicit Functions IIIImplicit Functions IVImplicit Functions VImplicit Functions VIImplicitFunctions—TheIssuesQuestion 8: ImproperIntegrals 45 / 56
    • Implicit Functions VIQuestion 1: ADerivativeQuestion 2: IncreasingFunctionsQuestion 3: ConcavityQuestion 4: LocalMinimaQuestion 5: Polar If we absolutely must have the slope of a branch of the curve as it passesArc-Length through the origin, the best option is probably to re-parametrize. In thisQuestion 6: ImplicitDifferentiation case, polar coordinates work nicely. They give us the equationQuestion 7: ImplicitFunctions r2 = cos 2θ for our curve, and it’s easy to see from this that the slopes ofImplicit Functions I the two tangent lines are ±1.Implicit Functions IIImplicit Functions IIIImplicit Functions IVImplicit Functions VImplicit Functions VIImplicitFunctions—TheIssuesQuestion 8: ImproperIntegrals 46 / 56
    • Implicit Functions—The IssuesQuestion 1: ADerivativeQuestion 2: IncreasingFunctionsQuestion 3: Concavity Two Issues:Question 4: LocalMinimaQuestion 5: PolarArc-Length I Hypotheses, hypotheses, hypotheses!Question 6: ImplicitDifferentiationQuestion 7: ImplicitFunctions I In this case, 0/0 isn’t an “indeterminate form”— it’sImplicit Functions IImplicit Functions II undefined!Implicit Functions IIIImplicit Functions IVImplicit Functions VImplicit Functions VIImplicitFunctions—TheIssuesQuestion 8: ImproperIntegrals 47 / 56
    • Question 1: ADerivativeQuestion 2: IncreasingFunctionsQuestion 3: ConcavityQuestion 4: LocalMinimaQuestion 5: PolarArc-Length Question 8: Improper IntegralsQuestion 6: ImplicitDifferentiationQuestion 7: ImplicitFunctionsQuestion 8: ImproperIntegralsImproper Integrals IImproper Integrals IIImproper Integrals IIIImproper Integrals IVImproper Integrals VImproper Integrals VIImproperIntegrals—The Moral 48 / 56
    • Improper Integrals IQuestion 1: ADerivative Why don’t we useQuestion 2: Increasing 1 tFunctions 2x 2xQuestion 3: Concavity 2 dx = lim 2 dx (31) −1 1−x t→1− −t 1−xQuestion 4: LocalMinima tQuestion 5: Polar = lim ln(1 − x2 ) (32)Arc-Length t→1− −tQuestion 6: ImplicitDifferentiation = lim {ln(1 − t2 ) − ln[1 − (−t)2 )]} = 0 (33)Question 7: Implicit t→1−FunctionsQuestion 8: Improper as the elementary-calculus definition for that improper integral?IntegralsImproper Integrals IImproper Integrals IIImproper Integrals III It would make freshman life so much easier.Improper Integrals IVImproper Integrals VImproper Integrals VIImproperIntegrals—The Moral 49 / 56
    • Improper Integrals IIQuestion 1: ADerivativeQuestion 2: IncreasingFunctionsQuestion 3: ConcavityQuestion 4: LocalMinima The calculation we have just examined gives something called the “CauchyQuestion 5: Polar Principal Value” (CPV) of the improper integral. The CPV is writtenArc-LengthQuestion 6: Implicit 1Differentiation 2xQuestion 7: Implicit PV 2 dx.Functions −1 1−xQuestion 8: ImproperIntegralsImproper Integrals IImproper Integrals IIImproper Integrals IIIImproper Integrals IVImproper Integrals VImproper Integrals VIImproperIntegrals—The Moral 50 / 56
    • Improper Integrals IIIQuestion 1: ADerivativeQuestion 2: IncreasingFunctionsQuestion 3: ConcavityQuestion 4: Local Why not use the CPV in elementary Calculus?MinimaQuestion 5: PolarArc-Length The short answer:Question 6: ImplicitDifferentiation Using Cauchy Principal Values would break the equationQuestion 7: ImplicitFunctions 1 ξ 1Question 8: Improper 2x 2x 2xIntegrals 2 dx = 2 dx + 2 dx. (34)Improper Integrals I −1 1−x −1 1−x ξ 1−xImproper Integrals IIImproper Integrals IIIImproper Integrals IVImproper Integrals VImproper Integrals VIImproperIntegrals—The Moral 51 / 56
    • Improper Integrals IVQuestion 1: ADerivative The long answer:Question 2: IncreasingFunctions Choose B , with |B| > 1. Let P be the polynomial function given byQuestion 3: ConcavityQuestion 4: Local P (u) = (u − 1)(u + 1)(u + 2B − 1)(u + 2B + 1), (35)MinimaQuestion 5: PolarArc-Length and putQuestion 6: ImplicitDifferentiation P ′ (u)Question 7: Implicit f (u) = − . (36)Functions P (u)Question 8: ImproperIntegralsImproper Integrals I Then f is continuous in (−1, 1), and a tedious calculation shows thatImproper Integrals IIImproper Integrals III 1Improper Integrals IV B−1Improper Integrals V PV f (u) du = ln . (37)Improper Integrals VI −1 B+1ImproperIntegrals—The Moral 52 / 56
    • Improper Integrals VQuestion 1: ADerivativeQuestion 2: IncreasingFunctions Now put g(u) = (u2 + 2Bu − 1)/(2B), and note that g(−1) = −1Question 3: ConcavityQuestion 4: Local while g(1) = 1. Putting x = g(u), we ought therefore to be able to writeMinimaQuestion 5: Polar 1 1Arc-Length PV F (x) dx = PV F [g(u)]g ′ (u) du, (38)Question 6: ImplicitDifferentiation −1 −1Question 7: ImplicitFunctions where F (x) = 2x/(1 − x2 ).Question 8: ImproperIntegralsImproper Integrals I However, the CPV on the left side of (38) is zero, as we have seen; theImproper Integrals II integrand on the right side of (38) turns out to be the integrand of (37),Improper Integrals III B−1Improper Integrals IV above, and so that CPV is ln . As it happens, the single value thisImproper Integrals V B+1Improper Integrals VI latter quantity cannot assume is zero.ImproperIntegrals—The Moral 53 / 56
    • Improper Integrals VIQuestion 1: ADerivativeQuestion 2: IncreasingFunctionsQuestion 3: ConcavityQuestion 4: LocalMinimaQuestion 5: PolarArc-Length A very good reason for not using Cauchy Principal Values inQuestion 6: Implicit elementary calculus is that they break the Substitution Theorem forDifferentiation Definite Integrals—which is too valuable a theorem to give up. . . inQuestion 7: ImplicitFunctions elementary calculus.Question 8: ImproperIntegralsImproper Integrals IImproper Integrals IIImproper Integrals IIIImproper Integrals IVImproper Integrals VImproper Integrals VIImproperIntegrals—The Moral 54 / 56
    • Improper Integrals—The MoralQuestion 1: ADerivativeQuestion 2: IncreasingFunctionsQuestion 3: ConcavityQuestion 4: Local MoralMinimaQuestion 5: PolarArc-Length The definitions we adopt condition the theorems we canQuestion 6: ImplicitDifferentiation prove. We have chosen (not necessarily consciously orQuestion 7: ImplicitFunctions with full knowledge) the definitions we use in elementaryQuestion 8: Improper calculus to support the tools we want students to be ableIntegralsImproper Integrals I to use.Improper Integrals IIImproper Integrals IIIImproper Integrals IVImproper Integrals VImproper Integrals VIImproperIntegrals—The Moral 55 / 56
    • Question 1: ADerivativeQuestion 2: IncreasingFunctionsQuestion 3: ConcavityQuestion 4: LocalMinimaQuestion 5: PolarArc-LengthQuestion 6: ImplicitDifferentiation The EndQuestion 7: ImplicitFunctionsQuestion 8: ImproperIntegralsImproper Integrals IImproper Integrals IIImproper Integrals IIIImproper Integrals IVImproper Integrals VImproper Integrals VIImproperIntegrals—The Moral 56 / 56