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11.2. DERIVATIVES OF DIFFERENTIAL FORMS 275 Four theorems, version 2 1. Conservative vector fields. Z C ∂f ∂x dx + ∂f ∂y dy + ∂f ∂z dz = f(q) −f(p) = Z ∂C f (C) 2. Green’s theorem. Z D ∂F2 ∂x −∂F1 ∂y  dx ∧dy = Z ∂D F1 dx + F2 dy (Gr) 3. Stokes’s theorem. Z S ∂F3 ∂y −∂F2 ∂z  dy ∧dz + ∂F1 ∂z −∂F3 ∂x  dz ∧dx + ∂F2 ...
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276 CHAPTER 11. WORKING WITH DIFFERENTIAL FORMS (a) For 0-forms: df = ∂f ∂x dx + ∂f ∂y dy + ∂f ∂z dz. (b) For 1-forms: d(F1 dx + F2 dy + F3 dz) = dF1 ∧dx + dF2 ∧dy + dF3 ∧dz. (c) For 2-forms: d(F1 dy ∧dz + F2 dz ∧dx + F3 dx ∧dy) = dF1 ∧dy ∧dz + dF2 ∧dz ∧dx + dF3 ∧dx ∧dy. In (b) and (c), F1, F2, and F3 are 0-forms, so t...
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11.2. DERIVATIVES OF DIFFERENTIAL FORMS 277 Next, we derive formulas for the derivatives of 1 and 2-forms. Example 11.3. For a 1-form ω = F1 dx + F2 dy + F3 dz, we find dω by calculating d(F1 dx), d(F2 dy), and d(F3 dz) separately. For instance, by definition: d(F1 dx) = dF1 ∧dx =
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278 CHAPTER 11. WORKING WITH DIFFERENTIAL FORMS Summing these calculations gives: d(F1 dy ∧dz + F2 dz ∧dx + F3 dx ∧dy) =
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11.3. A LOOK BACK AT THE THEOREMS OF MULTIVARIABLE CALCULUS 279 11.3 A look back at the theorems of multivariable calculus Take a moment to compare the formulas in Table 11.2 with the four theorems. It soon becomes apparent that, in each theorem, a differential form and its derivative appear. In this new framework, the...
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280 CHAPTER 11. WORKING WITH DIFFERENTIAL FORMS ψ: D →Rn as above, then the integral R M ζ is defined using substitution and the rules of forms to pull the integral back to an ordinary Riemann integral over the k-dimensional parameter domain D in Rk. (See the remarks before Exercise 4.14 at the end of the chapter for m...
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11.4. EXERCISES FOR CHAPTER 11 281 4.10. ω = 2xy3z4 dx + 3x2y2z4 dy + 4x2y3z3 dz 4.11. η = (x + 2y + 3z) dy ∧dz + exyz dz ∧dx + x4y5 dx ∧dy 4.12. Let U be an open set in R3. (a) Let f = f(x, y, z) be a 0-form (= real-valued function) on U. Find d(df). (b) Let ω = F1 dx + F2 dy + F3 dz be a 1-form on U. Find d(dω). (c) ...
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282 CHAPTER 11. WORKING WITH DIFFERENTIAL FORMS 4.14. Let α: R →R3 be given by α(t) = (cos t, sin t, t), where we think of R3 as xyz-space. (a) Find α∗(x2 + y2 + z2). (b) Show that α∗(dx) = −sin t dt. (c) Find α∗(dy) and α∗(dz). (d) Let ω be the 1-form ω = −y dx + x dy + z dz on R3. Find α∗(ω). 4.15. Let T : R2 →R2 be ...
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11.4. EXERCISES FOR CHAPTER 11 283 Differential forms are integrated over oriented domains, so, given a bounded subset D of R2, we first make a choice whether to assign it the positive orientation or the negative orientation. If this seems too haphazard, think of it as analogous to choosing whether, as a subset of R3, ...
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284 CHAPTER 11. WORKING WITH DIFFERENTIAL FORMS The remaining exercises call for speculation rather than proofs. You are asked to propose reasonable solutions consistent with patterns that have come before . It is considered a bonus if what you say is true. The correct answers are known and are covered in more advanced...
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Answers to selected exercises Section 1.1 Vector arithmetic 1.1. x + y = (5, −3, 9), 2x = (2, 4, 6), 2x −3y = (−10, 19, −12) 1.3. (−5, 3, −9) 1.5. (b) 1 3x + 2 3y (c) (−1, 1, 2) Section 1.2 Linear transformations 2.2. T(x + y) = T(x1 + x2, y1 + y2) = (x1 + y1, x2 + y2, 0) = (x1, x2, 0) + (y1, y2, 0) = T(x) + T(y) T(cx)...
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286 ANSWERS TO SELECTED EXERCISES Section 1.4 Matrix multiplication 4.1. AB = 3 1 1 7  , BA = 6 2 2 4  4.3. AB =   1 0 0 0 1 0 0 0 1  , BA =   1 0 0 0 1 0 0 0 1   4.5. AB = 21 0 0 14  , BA =   13 4 −5 4 5 6 −5 6 17   Section 1.5 The geometry of the dot product 5.1. (a) −3 (b) ∥x∥= √ 6, ∥y∥= 3 √ 2 (c) ...
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287 1.3. -1 0 1 2 3 4 5 1 2 3 4 5 1.5. 1.7. 1.9. α(x) = (x, x2), −1 ≤x ≤1 1.11. α(t) = (a cos t, −a sin t) , 0 ≤t ≤2π 1.13. α(t) = (1 + 4t, 2 + 5t, 3 + 6t) 1.15. α(t) = (1 −t, 0, t) 1.17. (a) v·(p−a) v·v (b) a + v·(p−a) v·v v 1.19. (b) No intersection (c) Intersection at (2, 1, 3) Section 2.2 Velocity, acceleration, sp...
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288 ANSWERS TO SELECTED EXERCISES 2.3. (a) v(t) = (−2 sin 2t, 2 cos 2t, 6t1/2) v(t) = 2√1 + 9t a(t) = (−4 cos 2t, −4 sin 2t, 3t−1/2) (b) 4 27
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289 Section 2.9 The classification of space curves 9.1. (a) √ 2 (b) 1 √ 2(−sin t, −cos t, 1) (c) (−cos t, sin t, 0) (d) −1 √ 2(sin t, cos t, 1) (e) 1 2 (f) −1 2 (g) a = 1, b = −1. Both α and β trace out the same helix but in opposite directions. The rotation F(x, y, z) = (x, −y, −z) about the x-axis by π transforms one...
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290 ANSWERS TO SELECTED EXERCISES Section 3.1 Graphs and level sets 1.1. (b) 0 1 2 3 4 5 6 -2 -1 0 1 2 -2 -1 0 1 2 (c) 1.3. (b) -2 -1 0 1 -2 -1 0 1 2 -2 -1 0 1 2 (c) 1.5. (b) 0 0.5 1 1.5 2 2.5 3 -2 -1 0 1 2 -2 -1 0 1 2
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291 (c) 1.7. (b) -1.5 -1 -0.5 0 0.5 1 1.5 -2 -1 0 1 2 -2 -1 0 1 2 (c) Section 3.2 More surfaces in R3 2.1. 2.3.
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292 ANSWERS TO SELECTED EXERCISES 2.5. The level set corresponding to c = 0 consists of the single point (0, 0, 0). Those for c = 1 and c = 2 are spheres centered at the origin of radius 1 and √ 2, respectively. 2.7. c = −1, 0, 1 (left, center, right) 2.9. c = −1, 0, 1 (left, center, right) Section 3.3 The equation of ...
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293 4.3. The point a = (1, 0) is in U, but no open ball centered at a stays within U. Thus U is not an open set. Section 3.5 Continuity 5.1. (a) If (x, y) ̸= (0, 0), then, in polar coordinates, f(x, y) = f(r cos θ, r sin θ) = (r cos θ)3(r sin θ) r2 = r2 cos3 θ sin θ, so |f(x, y)| = |r2 cos3 θ sin θ| ≤r2 = ∥(x, y)∥2. Th...
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294 ANSWERS TO SELECTED EXERCISES Section 3.8 Limits 8.1. For sums: Let L = limx→a f(x) and M = limx→a g(x), and consider the functions: ef(x) = ( f(x) if x ̸= a, L if x = a and eg(x) = ( g(x) if x ̸= a, M if x = a. By definition of limit, ef and eg are continuous at a, hence so is their sum. In other words, the functi...
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295 (b) ( 2x x2+y2 , 2y x2+y2 , 0) 1.17. (a) ∂f ∂x = 2x3 √ x4+y4 ∂f ∂y = 2y3 √ x4+y4 (b) ∂f ∂x(0, 0) = 0, ∂f ∂y (0, 0) = 0 Section 4.2 Conditions for differentiability 2.1. lim(x,y)→(0,0) f(x, y) = lim(x,y)→(0,0)(x2 −y2) = 0, while f(0, 0) = π, so f is not continuous at (0, 0), hence not differentiable there either. Se...
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296 ANSWERS TO SELECTED EXERCISES (c) y = 1 3x2 −1 3 1 2 3 1 2 3 (d) y = −3 2 ln x 1 2 3 -1 1 2 3 4 5 Section 4.7 ∇f as normal vector 7.1. 2x −4y −z = −3 7.3. 4y + 3z = 7 7.5. α(t) = (1 + 14t, 2 + 5t, 3 −8t) Section 4.8 Higher-order partial derivatives 8.1. ∂2f ∂x2 = 12x2 −12xy + 6y2 ∂2f ∂y ∂x = −6x2 + 12xy −12y2 ∂2f ∂...
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297 10.11. y = x + 1 3 -1 1 2 3 4 1 2 3 4 Section 4.11 Classifying nondegenerate critical points 11.1. 1 −1 2(x + y)2 11.3. (a) h3 ∂3f ∂x3 + 3h2k ∂3f ∂x2 ∂y + 3hk2 ∂3f ∂x ∂y2 + k3 ∂3f ∂y3 (b) f(a) + ∂f ∂x(a) h + ∂f ∂y (a) k + 1 2
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298 ANSWERS TO SELECTED EXERCISES 1.11. (a) x y x y2 0.25 0.5 0.75 1 0.25 0.5 0.75 1 1.25 (b) Z 1 0 Z √x x f(x, y) dy  dx 1.13. (a) y = 1 - x2 -1 -0.5 0.5 1 0.25 0.5 0.75 1 1.25 (b) Z 1 0 Z √ 1−y2 −√ 1−y2 f(x, y) dx  dy 1.15. (a) y = 1/x 0 0.25 0.5 0.75 1 1.25 0 1 2 3 4 (b) Z 4 1 Z 1 1 y yexy dx  dy (c) e4 −4e ...
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299 Section 5.3 Interpretations of the double integral 3.1. 18 hundred birds 3.3. (a) 80 miles per hour (b) All points of R that lie on the circle x2 + y2 = 8 3, a quarter-circular arc 3.5. (2 3, 5 3) 3.7. q 2 3 Section 5.4 Parametrization of surfaces 4.1. (a) σ: D →R3, σ(x, y) = (x, y, p a2 −x2 −y2), where D is the di...
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300 ANSWERS TO SELECTED EXERCISES 6.7. a2 6.8. (a) (b) Z √ 2 0 Z q 1−y2 2 0 Z 2 q 1−y2 2 −z2 −2 q 1−y2 2 −z2 f(x, y, z) dx  dz  dy (c) Z 2 −2 Z q 1−x2 4 0 Z q 2−x2 2 −2z2 0 f(x, y, z) dy  dz  dx 6.10. (a) (b) Z 1 0 Z y 0 Z y z f(x, y, z) dx  dz  dy (c) Z 1 0 Z x 0 Z 1 x f(x, y, z) dy  dz  dx Section 6.1...
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301 2.7. (a)   ex+y ex+y −e−x cos y −e−x sin y −e−x sin y e−x cos y   (b) All entries of Df(x, y) are continuous on R2, so, by the C1 test, f is differentiable at every point of R2. 2.9. (a)   a b c d e f g h i   (= A) Section 6.3 The chain rule: a conceptual approach 3.1. (a) Df(s, t) =   1 −1 2e2s+3t 3e2s+3...
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302 ANSWERS TO SELECTED EXERCISES 1.6. 2π 1.8. (b) (1 −e−a2)π Section 7.3 Examples: linear changes of variables, symmetry 3.1. (a) D is the disk of radius 2 centered at (3, 2). (3,2) D 5 1 2 3 4 5 (b) D∗is the disk u2 + v2 ≤4 of radius 2 centered at the origin. (c) ZZ D∗(u + v + 5) du dv = 20π 3.4. (a) 150 (b) 100 3.7....
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303 1.3. -2 -1 0 1 2 -2 -1 0 1 2 1.5. -2 -1 0 1 2 -2 -1 0 1 2 1.7. (a) F(x, y) = 1 √ 4x2+1(−2x, 1) or its negative (b) F(x, y) = 1 √ 4x2+1(1, 2x) or its negative 1.9. (a) α′(t) =
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304 ANSWERS TO SELECTED EXERCISES Section 9.1 Definitions and examples 1.1. (a) -2 -1 0 1 2 -2 -1 0 1 2 (b) For instance, any line segment radiating directly away from the origin. If α(t) = (t, t), 0 ≤ t ≤1, then R C x dx + y dy = 1. (c) For instance, any circle centered at the origin. If α(t) = (cos t, sin t), 0 ≤t ≤2...
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305 (b) All positive p 3.13. 3 3.17. The vector field F = ∇(fg) is conservative with potential function fg, so R C ∇(fg) · ds = 0 for all piecewise smooth oriented closed curves C. By Exercise 1.19 in Chapter 4, ∇(fg) = f ∇g + g ∇f, so R C(f ∇g + g ∇f) · ds = 0. Hence R C(f ∇g) · ds = − R C(g ∇f) · ds. Section 9.4 Gree...
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306 ANSWERS TO SELECTED EXERCISES Section 10.4 Curl fields 4.1. ∇· F = 3, not 0, so F is not a curl field. 4.2. It’s a curl field with G(x, y, z) = (1 2z2, 1 2x2, 1 2y2), for example. Section 10.5 Gauss’s theorem 5.1. 384 5 π 5.3. 104 3 π 5.7. 225 −32π Section 10.6 The inverse square field 6.2. 11 6.4. It’s the total m...
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Index acceleration, 28 add zero, 32, 78 affine function, 84 annulus, 223 arclength, 29, 30 area as double integral, 131 as line integral, 219 of graph z = f(x, y), 152 of parallelogram, 15, 36, 175 of sphere, 143, 238 of surface, 142 of surface of revolution, 152 average value, 132, 144 basis, 5 binormal, 38 boundary o...
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308 INDEX dot product, 7, 21 eigenvalue, 120 eigenvector, 120 first-order approximation, 83, 86, 91, 92, 95, 159, 161, 174, 283 flux, 235 Frenet vectors, 39 Frenet-Serret formulas, 42 Fubini’s theorem, 129 fundamental theorem of calculus, 279 Gauss’s law, 259, 270 Gauss’s theorem, 255, 273, 274, 279 global maximum/mini...
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INDEX 309 of surface, 134 partial derivative, 85 path, 25 piecewise smooth curve, 210 surface, 241, 249, 252 plane as span, 13 equation of, 61 polar coordinates, 69, 136, 176 potential energy, 230 potential function, 211, 212 principal normal, 33 product rules, 32, 41 pullback, 178, 210, 240, 261, 264, 281, 282 related...
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Elementary Calculus 0 v2 0 g v2 0 2g Michael Corral
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Elementary Calculus Michael Corral Schoolcraft College
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About the author: Michael Corral is an Adjunct Faculty member of the Department of Mathematics at School- craft College. He received a B.A. in Mathematics from the University of California, Berkeley, and received an M.A. in Mathematics and an M.S. in Industrial & Operations Engineering from the University of Michigan. ...
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Preface This book covers calculus of a single variable. It is suitable for a year-long (or two-semester) course, normally known as Calculus I and II in the United States. The prerequisites are high school or college algebra, geometry and trigonometry. The book is designed for students in engineering, physics, mathemati...
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iv PREFACE describing A, B and C would be “Easy”, “Moderate” and “Challenging”, respectively. However, many of the B exercises are easy and not all the C exercises are difficult. Appendix A provides answers and hints to many of the odd-numbered and some of the even-numbered exercises. A few exercises require the student...
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Contents Preface iii 1 The Derivative 1 1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.2 The Derivative: Limit Approach . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 1.3 The Derivative: Infinitesimal Approach . . . . . . . . . . . . . . . . . . . . . . . ...
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102 4.3 Numerical Approximation of Roots of Functions . . . . . . . . . . . . . . . . . . 109 4.4 The Mean Value Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118 5 The Integral 124 5.1 The Indefinite Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124 v
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vi CONTENTS 5.2 The Definite Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132 5.3 The Fundamental Theorem of Calculus . . . . . . . . . . . . . . . . . . . . . . . 140 5.4 Integration by Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146 5.5 Improper Integral...
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. . . . . . . . . . . . . . . . . . . . . . . 252 8.2 Average Value of a Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 258 8.3 Arc Length and Curvature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263 8.4 Surfaces and Solids of Revolution . . . . . . . . . . . . . . . . . . . . ....
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The Greek Alphabet Letters Name Letters Name Letters Name A α alpha I ι iota P ρ rho B β beta K κ kappa Σ σ sigma Γ γ gamma Λ λ lambda T τ tau ∆ δ delta M µ mu Υ υ upsilon E ǫ epsilon N ν nu Φ φ phi Z ζ zeta Ξ ξ xi X χ chi H η eta O o omicron Ψ ψ psi Θ θ theta Π π pi Ω ω omega Mathematical Notation Symbol Meaning Examp...
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CHAPTER 1 The Derivative 1.1 Introduction Calculus can be thought of as the analysis of curved shapes.1 Its development grew out of attempts to solve physical problems. For example, suppose that an object at rest 100 ft above the ground is dropped. Ignoring air resistance and wind, the object will fall straight down un...
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2 Chapter 1 • The Derivative §1.1 First, the object travels 100 ft in 2.5 seconds, so its average speed in that time is distance traveled time elapsed = 100 ft 2.5 seconds = 40 ft/s, and its average velocity in that time is change in position change in time = final position −initial position end time −start time = 0 ft ...
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Introduction • Section 1.1 3 ∆s ∆t = s(t+∆t) −s(t) ∆t = −16(t+∆t)2 + 100 −(−16t2 + 100) ∆t = −16t2 −32t∆t −16(∆t)2 + 100 + 16t2 −100 ∆t = −32t∆t −16(∆t)2 ∆t = ✚✚ ∆t(−32t −16∆t) ✚✚ ∆t = −32t −16∆t , Now let the interval [t, t+∆t] get smaller and smaller indefinitely—that is, let ∆t get closer and closer to 0. Then the av...
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4 Chapter 1 • The Derivative §1.1 Finding the area inside curved regions is another type of problem that calculus can solve. The basic idea is to use simpler regions—rectangles—whose areas are known, then use those to approximate the area inside the curved region. One such method is to draw more and more rectangles of ...
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Introduction • Section 1.1 5 N = the set of all natural numbers, i.e. the set of nonnegative integers: 0,1,2,3,4,... Z = the set of all integers: 0,±1,±2,±3,±4,... Q = the set of all rational numbers m n , where m and n are integers, with n ̸= 0 R = the set of all real numbers Note that N ⊂Z ⊂Q ⊂R. The set of real numb...
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6 Chapter 1 • The Derivative §1.1 Exercises A For Exercises 1-4, suppose that an object moves in a straight line such that its position s after time t is the given function s = s(t). Find the instantaneous velocity of the object at a general time t ≥0. You should mimic the earlier example for the instantaneous velocity...
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Introduction • Section 1.1 7 C 7. What is the flaw in the following “proof” that π = 4?: d = 1 Step 1: Draw a square around a circle of diameter d = 1. The circumference of the circle is thus πd = π, and the perimeter of the square is 4. d = 1 Step 2: Remove corners from the square as shown in the picture on the right, ...
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8 Chapter 1 • The Derivative §1.2 1.2 The Derivative: Limit Approach The following definition generalizes the example from the previous section (concerning instan- taneous velocity) to a general function f (x): The derivative of a real-valued function f (x), denoted by f ′(x), is f ′(x) = lim ∆x→0 ∆f ∆x = lim ∆x→0 f (x+...
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The Derivative: Limit Approach • Section 1.2 9 Notice in the above example that replacing ∆x by 0 was unnecessary when taking the limit, since the ratio f (x+∆x) −f (x) ∆x simplified to 0 before taking the limit, and the limit of 0 is 0 regard- less of what ∆x approaches. In fact, the answer—namely, f ′(x) = 0 for all x...
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10 Chapter 1 • The Derivative §1.2 Linear functions have a constant derivative—the constant being the slope of the line. The converse turns out to be true: a function with a constant derivative must be a linear function.12 What types of functions do not have constant derivatives? The previous section discussed such a f...
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The Derivative: Limit Approach • Section 1.2 11 The problem with using the limit definition to find the derivative of a curved function is that the calculations require more work, as the above example shows. As the functions become more complicated those calculations can become difficult or even impossible. And though lim...
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12 Chapter 1 • The Derivative §1.2 Another formulation is to set h = w−x in formula (1.4), which yields f ′(x) = lim h→0 f (x+ h) −f (x) h = lim w−x→0 f (x+(w−x)) −f (x) w −x , so that f ′(x) = lim w→x f (w) −f (x) w −x (1.5) since w−x approaches 0 if and only if w approaches x. Another formulation replaces h by −h: f ...
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The Derivative: Limit Approach • Section 1.2 13 As an example of using these different formulations, recall that a function f is even if f (−x) = f (x) for all x in the domain of f , and f is odd if f (−x) = −f (x) for all x in its domain. For example, x2, x4, and cos x are even functions; x, x3, and sin x are odd func...
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14 Chapter 1 • The Derivative §1.2 If the derivative f ′(x) exists then f is differentiable at x. A differentiable function is one that is differentiable at every point in its domain. For example, f (x) = x is a differentiable function, but f (x) = |x| is not differentiable at x = 0. The act of calculating a derivative...
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The Derivative: Infinitesimal Approach • Section 1.3 15 1.3 The Derivative: Infinitesimal Approach Traditionally a function f of a variable x is written as y = f (x). The dependent variable y is considered a function of the independent variable x. This allows taking the derivative of y with respect to x, i.e. the derivat...
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16 Chapter 1 • The Derivative §1.3 This might seem like a strange notion, but it really is not all that different from the limit notion where, say, you let ∆x approach 0 but not necessarily let it equal 0.18 As for the square of a nonzero infinitesimal being 0, think of how a calculator handles the squares of small numb...
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The Derivative: Infinitesimal Approach • Section 1.3 17 You might have noticed that the above example did not involve limits, and that the derivative 2x represents a real number (i.e. no infinitesimals appear in the final answer); this will always be the case. Infinitesimals possess another useful property: Microstraightne...
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18 Chapter 1 • The Derivative §1.3 The Microstraightness Property can be extended to smooth curves—that is, curves without sharp edges or cusps. For example, circles and ellipses are smooth, but polygons are not. The properties of infinitesimals can be applied to determine the derivatives of the sine and cosine function...
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The Derivative: Infinitesimal Approach • Section 1.3 19 A similar argument (left as an exercise) using the cosine addition formula shows: d dx (cos x) = −sin x One of the intermediate results proved here bears closer examination. Namely, sin dx = dx for an infinitesimal angle dx measured in radians. At first, it might see...
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20 Chapter 1 • The Derivative §1.3 Notice also that the value of a function at an infinitesimal may itself be an infinitesimal (e.g. sin dx = dx) or a real number (e.g. cos dx = 1). For a differentiable function f (x), df dx = f ′(x) and so multiplying both sides by dx yields the important relation: df = f ′(x) dx (1.9) ...
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Derivatives of Sums, Products and Quotients • Section 1.4 21 1.4 Derivatives of Sums, Products and Quotients So far the derivatives of only a few simple functions have been calculated. The following rules will make it easier to calculate derivatives of more functions: Rules for Derivatives: Suppose that f and g are dif...
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22 Chapter 1 • The Derivative §1.4 Note that by the Product Rule, in general the derivative of a product is not the product of the derivatives. That is, d(f ·g) dx ̸= df dx · dg dx . This should be obvious from some earlier examples. For instance, if f (x) = x and g(x) = 1 then (f · g)(x) = x·1 = x so that d(f ·g) dx =...
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Derivatives of Sums, Products and Quotients • Section 1.4 23 Example 1.5 Use the Quotient Rule to show that d dx (tan x) = sec2 x. Solution: Since tan x = sin x cos x then: d dx (tan x) = d dx µsin x cos x ¶ = (cos x)· d dx (sin x) −(sin x)· d dx (cos x) cos2 x = (cos x)·(cos x) −(sin x)·(−sin x) cos2 x = cos2 x + sin2...
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24 Chapter 1 • The Derivative §1.4 For n ≥1 differentiable functions f1, ..., fn and constants c1, ..., cn: d dx (c1f1 + ··· + cn fn) = c1 df1 dx + ··· + cn dfn dx (1.10) Note that the above formula includes differences, by using negative constants. The formula also shows that differentiation is a linear operation, whi...
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Derivatives of Sums, Products and Quotients • Section 1.4 25 2. Assuming P(n) is true for some n ≥0, show that P(n+1) is true. Assuming that d dx (xn) = n xn−1, show that d dx ¡ xn+1¢ = (n +1) x(n+1)−1 = (n +1) xn. It was shown in Section 1.2 that d dx (x) = 1, so: d dx ¡ xn+1¢ = d dx ¡ x· xn¢ = x· d dx ¡ xn¢ + xn · d ...
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26 Chapter 1 • The Derivative §1.4 A way to remember the Power Rule is: bring the exponent down in front of the variable then reduce the variable’s original exponent by 1. This works even for negative exponents. Example 1.8 Find the derivative of f (t) = 3t100 − 2 t100 . Solution: Differentiate term by term: df dt = d ...
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The Chain Rule • Section 1.5 27 1.5 The Chain Rule From what has been discussed so far it might be tempting to think that the derivative of a function like sin 2x is simply cos 2x, since the derivative of sin x is cos x. It turns out that is not correct: d dx (sin 2x) = d dx (2 sin x cos x) (by the double-angle formula...
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28 Chapter 1 • The Derivative §1.5 The Chain Rule should make sense intuitively. For example, if df du = 4 then that means f is increasing 4 times as fast as u, and if du dx = 3 then u is increasing 3 times as fast as x, so overall f should be increasing 12 = 4·3 times as fast as x, exactly as the Chain Rule says. Exam...
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The Chain Rule • Section 1.5 29 Recall that the composition f ◦g of two functions f and g is defined as (f ◦g)(x) = f (g(x)). Using prime notation the Chain Rule can be written as: Chain Rule: If g is a differentiable function of x, and f is a differentiable function on the range of g, then f ◦g is a differentiable func...
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30 Chapter 1 • The Derivative §1.5 Exercises A For Exercises 1-18, find the derivative of the given function. 1. f (x) = (1 −5x)4 2. f (x) = 5(x3 + x −1)4 3. f (x) = p 1 −2x 4. f (x) = (1 −x2) 3 2 5. f (x) = px x + 1 6. f (x) = px + 1 px −1 7. f (t) = µ1 −t 1 + t ¶4 8. f (x) = µ x2 + 1 x −1 ¶6 9. f (x) = sin2 x 10. f (x...
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The Chain Rule • Section 1.5 31 21. In an internal combustion engine, as a piston moves downward the connecting rod rotates the crank in the clockwise direction, as shown in Figure 1.5.1 below.28 connecting rod l A B O crank a s θ piston Figure 1.5.1 The point A can only move vertically, causing the point B to move aro...
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32 Chapter 1 • The Derivative §1.6 1.6 Higher Order Derivatives The derivative f ′(x) of a differentiable function f (x) can be thought of as a function in its own right, and if it is differentiable then its derivative—denoted by f ′′(x)—is the second derivative of f (x) (the first derivative being f ′(x)). Likewise, th...
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Higher Order Derivatives • Section 1.6 33 A natural question to ask is: what do higher order derivatives represent? Recall that the first derivative f ′(x) represents the instantaneous rate of change of a function f (x) at the value x. So the second derivative f ′′(x) represents the instantaneous rate of change of the f...
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34 Chapter 1 • The Derivative §1.6 v > 0 a < 0 v < 0 a < 0 v = 0 t = 3.47 s = 0 s = 2 t = 0 Notice in Example 1.14 that the acceleration of the ball is not only constant but also negative. To see why this makes sense, first consider the case where the ball is moving upward. The ball has an initial upward velocity of 34 ...
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Higher Order Derivatives • Section 1.6 35 For example: 1! = 1 3! = 1·2·3 = 6 2! = 1·2 = 2 4! = 1·2·3·4 = 24 By convention 0! is defined to be 1. The following statement can be proved using induction: dn dxn (xn) = n! for all integers n ≥0 Thus, dn+1 dxn+1 ¡ xn¢ = d dx µ dn dxn ¡ xn¢¶ = d dx (n!) = 0 for all integers n ≥...
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36 Chapter 1 • The Derivative §1.6 12. Find the general expression for the n-th derivative of f (x) = 1 ax+b for all constants a and b (a ̸= 0). 13. Show that the function y = Acos(ωt+φ) + Bsin(ωt+φ) satisfies the differential equation d2 y dt2 + ω2y = 0 for all constants A, B, ω, and φ. 14. If s(t) represents the posit...
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CHAPTER 2 Derivatives of Common Functions 2.1 Inverse Functions The derivatives calculated in the previous chapter were mostly for polynomials and a few trigonometric functions. This chapter will show how to find the derivatives of other types of functions, beginning in this section with inverse functions. The idea here...
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38 Chapter 2 • Derivatives of Common Functions §2.1 y x y = f (x) (a) f is one-to-one y x y = f (x) (b) f is not one-to-one Figure 2.1.3 Horizontal rule for one-to-one functions If a function f is one-to-one on its domain, then f has an inverse function, denoted by f −1, such that y = f (x) if and only if f −1(y) = x. ...
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Inverse Functions • Section 2.1 39 Since y is a function of x, dy dx will be in terms of x and hence 1 dy dx will be in terms of x. However, since (by invertibility) x is a function of y, dx dy would normally be in terms of y, not x, so that the two sides of the equation dx dy = 1 dy dx are not in the same terms! One w...
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40 Chapter 2 • Derivatives of Common Functions §2.1 To obtain a formula in prime notation for the derivative of an inverse function, notice that for all x in the domain of an invertible differentiable function f , f −1(f (x)) = x ⇒ d dx ¡ f −1(f (x)) ¢ = d dx (x) ⇒ ¡ f −1¢′ (f (x)) · f ′(x) = 1 by the Chain Rule, and h...
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Trigonometric Functions and Their Inverses • Section 2.2 41 2.2 Trigonometric Functions and Their Inverses The graphs of the six trigonometric functions are shown in Figure 2.2.1: x y 0 1 −1 π 2 π 3π 2 2π (a) y = sin x x y 0 1 −1 π 2 π 3π 2 2π (b) y = cos x x y 0 −π 2 π 2 (c) y = tan x x y 0 1 −1 π 2 π 3π 2 2π (d) y = ...
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42 Chapter 2 • Derivatives of Common Functions §2.2 For example, y = sin x is one-to-one over the interval £ −π 2 , π 2 ¤ , as shown in Figure 2.2.2 below: x y 0 −1 1 π 2 π −π 2 −π y = sin x Figure 2.2.2 y = sin x is one-to-one with x restricted to £ −π 2 , π 2 ¤ Similarly, recall that cos x is one-to-one over [0,π], t...
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Trigonometric Functions and Their Inverses • Section 2.2 43 x y 0 π 2 −π 2 1 −1 (a) y = csc−1 x x y 0 π π 2 1 −1 (b) y = sec−1 x x y 0 π π 2 (c) y = cot−1 x Figure 2.2.4 Graphs of csc−1 x, sec−1 x, cot−1 x The derivatives of the six inverse trigonometric functions are: d dx (sin−1 x) = 1 p 1−x2 (for |x| < 1) d dx (csc−...
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44 Chapter 2 • Derivatives of Common Functions §2.2 The proofs of the derivative formulas for the remaining inverse trigonometric functions are similar, and are left as exercises. Example 2.3 Find the derivative of the function y = 3 tan(π−2x). Solution: By the Chain Rule with u = π−2x, the derivative of y = 3 tan(π−2x...
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The Exponential and Natural Logarithm Functions • Section 2.3 45 2.3 The Exponential and Natural Logarithm Functions Functions of the form ax, where the exponent x varies, are called exponential functions. Unless otherwise noted, assume that a > 0 (0x is just 0, and (−1)1/2 is not a real number). You already know how a...
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46 Chapter 2 • Derivatives of Common Functions §2.3 The limit in this definition means that as x becomes larger—approaching infinity (∞)—the values of ¡ 1 + 1 x ¢x approach a number, denoted by e. More decimal places for e can be obtained by making x sufficiently large.4 For example, when x = 5×106 the value is 2.71828155...
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The Exponential and Natural Logarithm Functions • Section 2.3 47 The function ex is often referred to simply as the exponential function, even though there are obviously many exponential functions. What makes the base e so special? Take y = Aekt to represent the amount of some physical quantity at time t, for some cons...
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48 Chapter 2 • Derivatives of Common Functions §2.3 From algebra you should be familiar with the following properties of the natural logarithm, along with their equivalent properties in terms of the exponential function:7 ln(ab) = ln a + ln b ea · eb = ea+b ln ³a b ´ = ln a −ln b ea eb = ea−b ln ab = b ln a ¡ ea¢b = ea...
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The Exponential and Natural Logarithm Functions • Section 2.3 49 Combine that result with the derivative d dx (ln x) = 1 x for x > 0 to get: d dx (ln|x|) = 1 x Logarithmic Differentiation For some functions it is easier to differentiate the natural logarithm of the function first and then solve for the derivative of the...
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50 Chapter 2 • Derivatives of Common Functions §2.3 Radioactive Decay A classic example of the differential equation dy dt = ky is the case of exponential decay of a radioactive substance, often referred to simply as radioactive decay. In this case the general solution y = Aekt represents the amount of the substance at...
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The Exponential and Natural Logarithm Functions • Section 2.3 51 Then the half-life tH is: tH = −ln 2 k = − ln 2 1 6 ln 0.6 = 8.14 hours Note in the above example that the given time t = 6 was used for finding the constant k and then the half-life tH. For the converse problem—given the half-life find the time required fo...
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52 Chapter 2 • Derivatives of Common Functions §2.3 Exercises A For Exercises 1-12, find the derivative of the given function. 1. y = e2x 2. y = xex2 3. y = e−x −ex 4. y = esin x 5. y = 1 + ex 1 −ex 6. y = 1 1 + e−2x 7. y = eex 8. y = e2 ln x 9. y = ln(3x) 10. y = ln(x2 + 2x + 1)4 11. y = ¡ ln(tan x2) ¢3 12. y = ln(ex +...
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General Exponential and Logarithmic Functions • Section 2.4 53 2.4 General Exponential and Logarithmic Functions For a general exponential function y = ax, with a > 0, use logarithmic differentiation to find its derivative: ln y = ln ¡ ax¢ = x ln a d dx (ln y) = d dx (x ln a) = ln a y′ y = ln a ⇒ y′ = y·ln a Thus, the d...
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54 Chapter 2 • Derivatives of Common Functions §2.4 x y 0 1 a > 1 a < 1 Figure 2.4.1 y = ax x y 0 1 a > 1 a < 1 Figure 2.4.2 y = loga x Hence, for any a > 0 with a ̸= 1 the function f (x) = ax is one-to-one, so it has an inverse function, called the base a logarithm and denoted by f −1(x) = loga x. It is often spoken a...
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General Exponential and Logarithmic Functions • Section 2.4 55 d dx ¡ loga x ¢ = 1 x ln a In general, when taking the logarithm of a function u = u(x): d dx ¡ loga u ¢ = 1 u ln a · du dx = u′ u ln a Example 2.14 Find the derivative of y = log2(cos 4x). Solution: This is the case where a = 2, so: dy dx = 1 (cos 4x)(ln 2...
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