THE UNIVERSITY OF MELBOURNE SCHOOL OF MATHEMATICS AND STATISTICS MAST10006 Calculus 2 Lecture Notes STUDENT NAME: EMAIL: This compilation has been made in accordance with the provisions of Part VB of the copyright act for the teaching purposes of the University. This booklet is for the use of students of the University of Melbourne enrolled in the subject MAST10006 Calculus 2. © School of Mathematics and Statistics, University of Melbourne, January 2019. These notes have been written by Christine Mangelsdorf, Anthony Morphett and Binzhou Xia. Reproduction of any part of this work other than that authorised by Australian Copyright Law without permission of the copyright owners is unlawful. Edition 11, July 2019. Table of Contents Section 0 – Notation used in MAST10006 Calculus 2 2 Section 1 – Limits, Continuity, Sequences and Series 9 Section 2 – Hyperbolic Functions 85 Section 3 – Complex Numbers 117 Section 4 – Integral Calculus 141 Section 5 – First Order Ordinary Differential Equations 183 Section 6 – Second Order Ordinary Differential Equations 249 Section 7 – Functions of Two Variables 309 1 / 391 Section 0 – Notation used in MAST10006 Calculus 2 Standard Abbreviations 1. such that or given that: | 2. therefore: ∴ 3. for all: ∀ 4. there exists: ∃ 5. equivalent to: ≡ 6. that is: i.e 7. approximate: ≈ 8. much smaller than: 2 / 391 Standard Notation for Sets of Numbers 1. natural numbers: N = {1, 2, 3, . . . } 2. integers: Z = {0,±1,±2, . . . } 3. rational numbers: Q = {mn | m,n ∈ Z,n , 0} 4. real numbers: R (rational numbers plus irrational numbers) 5. complex numbers: C = {x + iy | x, y ∈ R, i2 = −1} 6. R2 = {(x, y) | x, y ∈ R} (xy plane) 7. R3 = {(x, y, z) | x, y, z ∈ R} (3 dimensional space) 3 / 391 Standard Notation for Intervals 1. element of: ∈ so a ∈ X means “a is an element of the set X ” 2. open interval: (a, b) so x ∈ (0, 1) means “0 < x < 1” 3. closed interval: [a, b] so x ∈ [0, 1] means “0 ≤ x ≤ 1” 4. partial open and closed interval: (a, b] or [a, b) so x ∈ [0, 1) means “0 ≤ x < 1” 5. not including: \ so x ∈ R \ {0} means “x is any real number excluding 0”. Alternatively, we could write (−∞, 0) ∪ (0,∞) where ∪ means the ”union of the two intervals”. 4 / 391 More Standard Notation 1. natural logarithm: log x base 10 logarithm: log10 x Alternative notations for natural logarithms used in textbooks: loge x, ln x 2. inverse trigonometric functions: arcsin x, arctan x etc Alternative notations used in textbooks: sin−1 x, tan−1 x etc 3. implies: ⇒ so p⇒ q means “p implies q” 4. if and only if (iff): ⇔ (means both ⇐ and ⇒) so p⇔ q means “p implies q” AND “q implies p” 5. approaches: → so f (x)→ 1 as x→ 0 means “f (x) approaches 1 as x approaches 0” 5 / 391 Greek Alphabet α alpha ν nu β beta ξ xi γ gamma o omicron δ delta pi pi or ε epsilon ρ rho ζ zeta σ sigma η eta τ tau θ theta υ upsilon ι iota φ phi κ kappa χ chi λ lambda ψ psi µ mu ω omega 6 / 391 7 / 391 8 / 391 Section 1: Limits, Continuity, Sequences, Series Limits Let f be a real-valued function. We say that f has the limit L as x approaches a, lim x→a f (x) = L, if f (x) gets arbitrarily close to L whenever x is close enough to a but x , a. Note: If it exists, the limit L must be a unique finite real number. 9 / 391 Example 1.1: If f (x) = { 2x x , 1 4 x = 1 , evaluate limx→1 f (x). -3 -2 -1 1 2 3 x -1 1 2 3 4 f HxL Solution: Note: The limit of f as x approaches a does not depend on f (a). The limit can exist even if f is undefined at x = a. 10 / 391 Example 1.2: If f (x) = 1 x2 , evaluate lim x→0 f (x). x f HxL Solution: 11 / 391 Example 1.3: If f (x) = { 1 x < 0 2 x ≥ 0 , evaluate limx→0 f (x). x f HxL 2 1 Solution: 12 / 391 We can describe this behaviour in terms of one-sided limits. We write Theorem: lim x→a f (x) = L if and only if limx→a− f (x) = L and limx→a+ f (x) = L. Thus the limit exists if and only if the left and right hand limits exist and are equal. 13 / 391 Limit Laws Let f and g be real-valued functions and let c ∈ R be a constant. If lim x→a f (x) and limx→a g(x) exist, then 1. lim x→a [f (x) + g(x)] = limx→a f (x) + limx→a g(x). 2. lim x→a [cf (x)] = c limx→a f (x). 3. lim x→a [f (x)g(x)] = limx→a f (x) · limx→a g(x). 4. lim x→a [ f (x) g(x) ] = lim x→a f (x) lim x→a g(x) provided lim x→a g(x) , 0. 5. lim x→a c = c. 6. lim x→a x = a. 14 / 391 Example 1.4: Use the limit laws to evaluate lim x→2 x3 + 2x2 − 1 5 − 3x . Solution: 15 / 391 Limits as x Approaches Infinity We say that f has the limit L as x approaches positive infinity, lim x→∞ f (x) = L, if f (x) gets arbitrarily close to L whenever x is sufficiently large and positive. We say that f has the limit M as x approaches negative infinity: lim x→−∞ f (x) = M if f (x) gets arbitrarily close to M whenever x is sufficiently large and negative. Note: 1. L and M must be finite. 2. Limit laws (1)-(5) apply. 16 / 391 Example 1.5: Evaluate lim x→∞ e −x. x H L H0,1L Solution: 17 / 391 Evaluating Limits with Indeterminate Forms We say a function f (x) g(x) has indeterminate form 0 0 as x→ a if lim x→a f (x) = limx→a g(x) = 0. Example 1.6: Evaluate lim x→2 x2 − 4 x − 2 . Solution: 18 / 391 We say a function f (x) g(x) has indeterminate form ∞ ∞ as x→ a if f (x)→∞ and g(x)→∞. Example 1.7: Evaluate lim x→∞ 3x2 − 2x + 3 x2 + 4x + 4 . Solution: 19 / 391 We say a function f (x) − g(x) has indeterminate form ∞−∞ as x→ a if f (x)→∞ and g(x)→∞. Example 1.8: Evaluate lim x→∞ (√ x2 + 1 − x ) . Solution: 20 / 391 Sandwich Theorem: If g(x) ≤ f (x) ≤ h(x) when x is near a but x , a, and lim x→a g(x) = limx→a h(x) = L then lim x→a f (x) = L. x y hHxL gHxL f HxL a L 21 / 391 Note: 1. “x is near a but x , a” means that x lies in (b, a) ∪ (a, c) for some b < a < c. 2. The validity of Sandwich Theorem is based on the fact that g(x) ≤ f (x) ≤ h(x) ⇒|f (x) − L| ≤ |g(x) − L| + |h(x) − L| for all L. Can you prove this inequality or even the stronger conclusion that |f (x) − L| ≤ max{|g(x) − L|, |h(x) − L|}? 3. Sandwich Theorem also works for limits as x→∞. For example, if g(x) ≤ f (x) ≤ h(x) when x ∈ (c,∞) for some real number c, and lim x→∞ g(x) = limx→∞ h(x) = L, then lim x→∞ f (x) = L. The similar theorem holds for x→ −∞. 22 / 391 Example 1.9: Evaluate lim x→0 [ x2 sin (1 x )] . Solution: 23 / 391 Example 1.10: Evaluate lim x→0 [ x sin (1 x )] . Solution: 24 / 391 Continuity Let f be a real-valued function. The function f is continuous at x = a if lim x→a f (x) = f (a). 25 / 391 Example 1.11: Let f (x) = { 2x x , 1 4 x = 1. Is f continuous at x = 1? -3 -2 -1 1 2 3 x -1 1 2 3 4 f HxL Solution: 26 / 391 Example 1.12: Let f (x) = x2 − 4 x − 2 x , 2 4 x = 2. Is f continuous at x = 2? Solution: 27 / 391 28 / 391 At the endpoints of a domain, we cannot take both left and right hand limits, so we use the appropriate limit to test continuity. 1. A function f is left continuous (continuous from the left) at x = a if lim x→a− f (x) = f (a). 2. A function f is right continuous (continuous from the right) at x = a if lim x→a+ f (x) = f (a). 29 / 391 Example 1.13: Is f (x) = √ x continuous in its domain? x y 1 2 3 4 1 2 y = √ x Solution: 30 / 391 Let f and g be real-valued functions and let c ∈ R be a constant. Continuity Theorem 1: If the functions f and g are continuous at x = a, then the following functions are continuous at x = a: 1. f + g, 2. cf , 3. fg, 4. f g if g(a) , 0. Note: The theorem follows from limit laws. 31 / 391 Recall that ( g ◦ f )(x) = g( f (x)). Continuity Theorem 2: If f is continuous at x = a and g is continuous at x = f (a), then g ◦ f is continuous at x = a. Continuity Theorem 3: The following function types are continuous at every point in their domains: polynomials, trigonometric functions, exponentials, logarithms, nth root functions, hyperbolic f
unctions. 32 / 391 Example 1.14: Let f (x) = log x + sin x√ x2 − 1 . For which values of x is f continuous? Solution: 33 / 391 34 / 391 Example 1.15: f (x) = x3 − cx + 8, x ≤ 1 x2 + 2cx + 2, x > 1. For which values of c is f continuous? Justify your answer. Solution: 35 / 391 36 / 391 Theorem: If f is continuous at b and lim x→a g(x) = b then lim x→a f [g(x)] = f [ lim x→a g(x) ] = f (b). Note: This theorem also holds for limits as x→∞, as long as b ∈ R is finite. 37 / 391 Example 1.16: Evaluate lim x→∞ sin (e −x) . Solution: 38 / 391 Differentiability Let f : R→ R be a real-valued function. The derivative of f at x = a is defined by f ′(a) = lim h→0 f (a + h) − f (a) h . The function f is differentiable at x = a if this limit exists. Geometrically, f is differentiable at x = a if the graph y = f (x) has a tangent line at x = a given by y − f (a) = f ′(a)(x − a) which gives a good approximation to the graph near x = a. Note: We can also define left differentiable and right differentiable. 39 / 391 x f HxL a Differentiable at x=a x f HxL Not differentiable at x=0 If f is differentiable at x = a, the linear approximation of f near x = a is given by f (x) ≈ f (a) + f ′(a)(x − a) Theorem: If f is differentiable at x = a, then f is continuous at x = a. 40 / 391 L’Hoˆpital’s Rule Let f and g be differentiable functions near x = a, and g′(x) , 0 at all points x near a with x , a. If lim x→a f (x) g(x) has the indeterminate form 0 0 or ∞ ∞ then lim x→a f (x) g(x) = lim x→a f ′(x) g′(x) if the limit involving the derivatives exists. Note: L’Hoˆpital’s Rule also holds when x approaches infinity. 41 / 391 Example 1.17: Evaluate lim x→0 sin x x . Solution: 42 / 391 Example 1.18: Evaluate lim x→∞ 3×2 − 2x + 3 x2 + 4x + 4 . Solution: 43 / 391 Example 1.19: Evaluate lim x→∞ ( x− 1 3 log x ) . (0 · ∞) Solution: 44 / 391 Aside – What is a limit really?* Recall our definition of limit: lim x→a f (x) = L if f (x) gets arbitrarily close to L whenever x is close enough to a but x , a. What do ‘arbitrarily close’ and ‘close enough’ mean? More formally, for any arbitrary positive real number ε, there is a positive real number δ such that |f (x) − L| < whenever |x − a| < δ. This formal definition of limit is covered in MAST20026 Real Analysis, but as a taster we do two examples. * Slides 45-47 are not examinable in MAST10006. 45 / 391 Aside - What is a limit really?* Example 1.20: Using the definition, prove that lim x→1 2x = 2 Solution: For an arbitrary positive real number ε, |f (x) − 2| = 2|x − 1| < ε if and only if |x − 1| < 1 2 ε = δ. This shows that |f (x) − 2| can be arbitrarily small whenever |x − 1| is small enough but not equal to 0. In other words, f (x) can be arbitrarily close to 2 whenever x is close enough to 1 but x , 1. Therefore, lim x→1 f (x) = 2. 46 / 391 Aside - What is a limit really?* Example 1.21: Sketch a proof of the limit law lim x→a [f (x) + g(x)] = limx→a f (x) + limx→a g(x) Solution: Suppose lim x→a f (x) = L and limx→a g(x) = M. For an arbitrary positive real number ε, to make |f (x) + g(x) − (L + M)| < ε we only need to make |f (x) − L| < ε 2 and |g(x) −M| < ε 2 . These will be satisfied whenever x is close enough to a but x , a since lim x→a f (x) = L and limx→a g(x) = M. Hence f (x) + g(x) can be arbitrarily close to L +M whenever x is close enough to a but x , a, which means that lim x→a [f (x) + g(x)] = L + M = limx→a f (x) + limx→a g(x). 47 / 391 48 / 391 Sequences A sequence is a function f : N→ R. It can be thought of as an ordered list of real numbers a1, a2, a3, a4, . . . , an . . . Thus, f (n) = an. The sequence is denoted by {an}, where an is the nth term. Example 1, 1 2 , 1 3 , 1 4 , 1 5 , . . . an = 1 n Example 1,−1, 1,−1, 1,−1, . . . an = (−1)n−1 49 / 391 The graph of a sequence {an} can be plotted on a set of axes with n on the x-axis and an on the y-axis. Example: an = 1n Example: an = (−1)n−1 2 4 n −1 1 an 1 3 5 2 4 n −1 1 an 1 3 5 50 / 391 Limits of Sequences A sequence {an} has the limit L if an approaches L as n approaches infinity. Note, that L must be finite. We write lim n→∞ an = L or an → L as n→∞. If the limit exists we say that the sequence converges. Otherwise, we say that the sequence diverges. 51 / 391 Example 1.22: Determine whether the following sequences converge or diverge: (a) {1 n } (b) { (−1)n−1 } (c) {n} Solution: 52 / 391 The only difference between lim n→∞ an = L and limx→∞ f (x) = L is that n is a natural number whereas x is a real number. Theorem: Let f : R→ R be a real function and {an} be a sequence of real numbers such that an = f (n). If lim x→∞ f (x) = L then limn→∞ an = L. This means that we can use the techniques for evaluating limits of functions to evaluate limits of sequences. Note: lim n→∞ an = L 6=⇒ limx→∞ f (x) = L. eg. an = sin(2pin), f (x) = sin(2pix). 53 / 391 Theorem: Let {an} and {bn} be sequences of real numbers and c ∈ R a constant. If lim n→∞ an and limn→∞ bn exist, then 1. lim n→∞ [an + bn] = limn→∞ an + limn→∞ bn. 2. lim n→∞ [can] = c limn→∞ an. 3. lim n→∞ [anbn] = limn→∞ an · limn→∞ bn. 4. lim n→∞ [an bn ] = lim n→∞ an lim n→∞ bn provided lim n→∞ bn , 0. 54 / 391 Sandwich Theorem: Let {an}, {bn} and {cn} be sequences of real numbers. If an ≤ cn ≤ bn for all n > N for some N, and lim n→∞ an = limn→∞ bn = L then lim n→∞ cn = L. 55 / 391 The Factorial Function The factorial function n! (n = 0, 1, 2, …) is defined by n! = n(n − 1)! , 0! = 1 or n! = n × (n − 1) × (n − 2) × … × 3 × 2 × 1 Therefore 1! = 1 2! = 2 × 1 = 2 3! = 3 × 2 × 1 = 6 4! = 4 × 3 × 2 × 1 = 24 Example (2n + 2)! = (2n + 2) × (2n + 1) × (2n) × (2n − 1) × … × 3 × 2 × 1 or (2n + 2)! = (2n + 2) × (2n + 1) × (2n)! 56 / 391 Standard Limits (1) lim n→∞ 1 np = 0 (p > 0) (2) lim n→∞ r n = 0 (|r| < 1) (3) lim n→∞ a 1 n = 1 (a > 0) (4) lim n→∞ n 1 n = 1 (5) lim n→∞ an n! = 0 (a ∈ R) (6) lim n→∞ logn np = 0 (p > 0) (7) lim n→∞ ( 1 + a n )n = ea (a ∈ R) (8) lim n→∞ np an = 0 (p ∈ R, a > 1) Note: Standard limits (1), (3), (4), (6), (7), (8) also hold for limits of real-valued functions as x→∞. Standard limit (2) also holds for x→∞ when 0 ≤ r < 1. 57 / 391 Example 1.23: Evaluate lim n→∞ [(n − 2 n )n + 4n2 3n ] . Solution: 58 / 391 Example 1.24: Find the limit of the sequence an = 3n + 2 4n + 2n , n ≥ 1. Solution: 59 / 391 Note: The order hierarchy can be used to help identify the largest term in an expression: logn np an n! 60 / 391 Example 1.25: Prove Standard Limit 6: lim n→∞ logn np = 0 (p > 0) Solution: Note: We must change to a continuous variable x ∈ R before applying L’Hoˆpital’s rule. 61 / 391 Example 1.26: Evaluate lim n→∞ [ log(3n − 2) − logn] . Solution: Note: We must change to a continuous variable x ∈ R before applying the continuity theorem. 62 / 391 Example 1.27: Evaluate lim n→∞ 1 + sin2 ( npi 3 ) √ n . Solution: 63 / 391 64 / 391 Adding Infinitely Many Numbers Starting with any sequence {an}, adding the an’s together in order gives a sequence {sn}: s1 = a1, s2 = a1 + a2, s3 = a1 + a2 + a3, … … … … The sequence of partial sums {sn} may or may not converge. If it does converge, we call S = lim n→∞ sn = limn→∞(a1 + a2 + . . . + an) the sum of the an’s. 65 / 391 Example 1.28: Find the sum S of an = (1 2 )n ,n ≥ 1. Solution: 66 / 391 Series The series with terms an is denoted by the sum ∞∑ n=1 an. If lim n→∞ sn exists, we say that the series converges. Otherwise we say that the series diverges. Example The sequence {n} = 1, 2, 3, 4, . . . The series ∞∑ n=1 n = 1 + 2 + 3 + 4 + . . . The sequence and series both diverge to infinity, so the sum does not exist. 67 / 391 Application: Decimals The decimal representation of a number is actually a series. Example The sequence { 1 10n } = 0.1, 0.01, 0.001, . . .. The series ∞∑ n=1 1 10n = 0.1 + 0.01
+ 0.001 + . . . = 0.11111111 . . . The sequence converges to 0 while the series converges to 1 9 . In General For a number x ∈ (0, 1) with decimal digits d1, d2, d3, d4, . . . x = 0.d1d2d3d4 . . . = ∞∑ n=1 dn 10n 68 / 391 Properties of Series Let ∞∑ n=1 an and ∞∑ n=1 bn be series, and c ∈ R\{0} a constant. If ∞∑ n=1 an and ∞∑ n=1 bn converge then 1. ∞∑ n=1 (an + bn) = ∞∑ n=1 an + ∞∑ n=1 bn converges. 2. ∞∑ n=1 (can) = c ∞∑ n=1 an converges. If ∞∑ n=1 an diverges then ∞∑ n=1 (can) diverges. Note: These follow from the properties of the limits of sequences. 69 / 391 Geometric Series A geometric series has the form ∞∑ n=0 arn = ∞∑ n=1 arn−1 = a + ar + ar2 + ar3 + . . . where a ∈ R\{0} and r ∈ R. The series converges if |r| < 1 and diverges if |r| ≥ 1. If |r| < 1, we have ∞∑ n=0 arn = a 1 − r . Note: This follows from the fact that n∑ k=0 ark = a(1 − rn+1) 1 − r for r , 1. 70 / 391 Example 1.29: What does the series ∞∑ n=0 (1 2 )n = 1 + 1 2 + (1 2 )2 + (1 2 )3 + . . . converge to? Solution: 71 / 391 Harmonic p Series A harmonic p series has the form ∞∑ n=1 1 np . The series converges if p > 1 and diverges if p ≤ 1. Example ∞∑ n=1 1 n2 converges BUT ∞∑ n=1 1 n diverges. 72 / 391 Divergence Test If lim n→∞ an , 0 then ∞∑ n=1 an diverges. Note: If lim n→∞ an = 0 then 1. ∞∑ n=1 an may converge or diverge. 2. The Divergence Test is not relevant, so we need to use another test to determine if ∞∑ n=1 an converges or diverges. 73 / 391 Example 1.30: Does the series ∞∑ n=1 n + 1 n converge? Solution: 74 / 391 Comparison Test Let ∞∑ n=1 an and ∞∑ n=1 bn be positive term series. 1. If an ≤ bn for all n and ∞∑ n=1 bn converges, then ∞∑ n=1 an converges. 2. If an ≥ bn for all n and ∞∑ n=1 bn diverges, then ∞∑ n=1 an diverges. To apply the comparison test we compare a given series to a harmonic p series or geometric series. 75 / 391 Example 1.31: Does ∞∑ n=1 7 2n2 + 4n + 3 converge or diverge? Solution: 76 / 391 Example 1.32: Does ∞∑ n=1 n2 + 4 n3 + 1 converge or diverge? Solution: 77 / 391 78 / 391 Ratio Test Let ∞∑ n=1 an be a positive term series and L = lim n→∞ an+1 an . 1. If L < 1, ∞∑ n=1 an converges. 2. If L > 1, ∞∑ n=1 an diverges. 3. If L = 1, the ratio test is inconclusive. The ratio test is useful if an contains an exponential or factorial function of n. 79 / 391 Example 1.33: Does ∞∑ n=1 10n n! converge or diverge? Solution: 80 / 391 Example 1.34: Does ∞∑ n=1 (2n)! n! n! converge or diverge? Solution: 81 / 391 82 / 391 83 / 391 84 / 391 Section 2: Hyperbolic Functions Even Functions A function f is an even function if f (x) = f (−x) Example f (x) = cos x and f (x) = x2 Odd Functions A function f is an odd function if f (x) = −f (−x) Example f (x) = sin x and f (x) = x3 x y −2 −1 1 2 1 2 3 4 y = x2 x y −2 −1 1 2 −8 −4 4 8 y = x3 85 / 391 We define the hyperbolic cosine function: cosh x = 1 2 ( ex + e−x ) , x ∈ R x y −3 −2 −1 1 2 3 −1 1 2 3 4 y = cosh(x) Properties 86 / 391 We define the hyperbolic sine function: sinh x = 1 2 ( ex − e−x) , x ∈ R x y −3 −2 −1 1 2 3 −4 −3 −2 −1 1 2 3 4 y = sinh(x) Properties 87 / 391 We define the hyperbolic tangent function: tanh x = sinh x cosh x = 1 2 (e x − e−x) 1 2 (e x + e−x) = ex − e−x ex + e−x , x ∈ R. x y −3 −2 −1 1 2 3 −2 −1 1 2 y = tanh(x) Properties 88 / 391 Why call them hyperbolic functions? Let x = cosh t and y = sinh t then 89 / 391 So (x, y) = (cosh t, sinh t) denotes a point on the hyperbola x2 − y2 = 1. Since x ≥ 1, the right hand branch of the hyperbola can be parametrised by x = cosh t, y = sinh t, t ∈ R. -4 -2 2 4 x -4 -2 2 4 y y=sinh t x=cosh t t t=0 t>0 t<0 x2-y2=1 90 / 391 Application: Catenary A flexible, heavy cable of uniform mass per length ρ and tension T at its lowest point has shape y = T ρg cosh (ρgx T ) where g is the acceleration due to gravity. x y Support Support 91 / 391 Example 2.1: If cosh x = 1312 and x < 0 find sinh x and tanh x. Solution: 92 / 391 93 / 391 Addition Formulae sinh(x + y) = sinh x cosh y + cosh x sinh y cosh(x + y) = cosh x cosh y + sinh x sinh y sinh(x − y) = sinh x cosh y − cosh x sinh y cosh(x − y) = cosh x cosh y − sinh x sinh y 94 / 391 Example 2.2: Prove the sinh(x + y) addition formula. Solution: 95 / 391 Double Angle Formulae sinh(2x) = 2 sinh x cosh x cosh(2x) = cosh2 x + sinh2 x cosh(2x) = 2cosh2 x − 1 cosh(2x) = 2sinh2 x + 1 These can be proved using the addition formulae. 96 / 391 Reciprocal Hyperbolic Functions We define the three reciprocal hyperbolic functions: sech x = 1 cosh x , x ∈ R x y −3 −2 −1 1 2 3 −0.5 0.5 1.0 y = sech(x) cosech x = 1 sinh x , x ∈ R \ {0} x y −3 −2 −1 1 2 3 −2 −1 1 2 y = cosech(x) 97 / 391 Reciprocal Hyperbolic Functions coth x = 1 tanh x = cosh x sinh x , x ∈ R \ {0} x y −3 −2 −1 1 2 3 −3 −2 −1 1 2 3 y = coth(x) 98 / 391 Basic Identities cosh2 x − sinh2 x = 1 coth2 x − 1 = cosech2 x 1 − tanh2 x = sech2 x 99 / 391 Derivatives of Hyperbolic Functions d dx (cosh x) = sinh x, x ∈ R d dx (sinh x) = cosh x, x ∈ R d dx (tanh x) = sech2 x, x ∈ R d dx (sech x) = − sech x tanh x, x ∈ R d dx (cosech x) = − cosech x coth x, x ∈ R \ {0} d dx (coth x) = − cosech2 x, x ∈ R \ {0} 100 / 391 Example 2.3: Prove that d (cosh x) dx = sinh x. Solution: 101 / 391 Example 2.4: Let y = √ sinh(6x), x > 0. Find dy dx . Solution: 102 / 391 Inverses of Hyperbolic Functions We define three inverse hyperbolic functions. 1. Inverse hyperbolic sine function: arcsinh x Since sinh x is a 1-1 function domain arcsinh x = range sinh x = R. range arcsinh x = domain sinh x = R. arcsinh(sinh x) = x, x ∈ R. sinh(arcsinh x) = x, x ∈ R. x y −4 −3 −2 −1 1 2 3 4 −3 −2 −1 1 2 3 y = arcsinh(x) y = sinh(x) 103 / 391 2. Inverse hyperbolic cosine function: arccosh x Restrict domain of cosh x to be [0,∞) to give a 1-1 function. Then domain arccosh x = range cosh x = [1,∞). range arccosh x = restricted domain cosh x = [0,∞). cosh(arccosh x) = x, x ≥ 1. arccosh(cosh x) = x, x ≥ 0. x y 1 2 3 4 5 1 2 3 4 5 y = arccosh(x) y = cosh(x) 104 / 391 3. Inverse hyperbolic tangent function: arctanh x Since tanh x is a 1-1 function domain arctanh x = range tanh x = (−1, 1). range arctanh x = domain tanh x = R. tanh(arctanh x) = x, −1 < x < 1. arctanh(tanh x) = x, x ∈ R. x y −4 −3 −2 −1 1 2 3 4 −3 −2 −1 1 2 3 y = arctanh(x) y = tanh(x) 105 / 391 The inverse hyperbolic functions can be expressed in terms of natural logarithms. arcsinh x = log ( x + √ x2 + 1 ) , x ∈ R arccosh x = log ( x + √ x2 − 1 ) , x ≥ 1 arctanh x = 1 2 log (1 + x 1 − x ) , −1 < x < 1 We can also define inverse reciprocal hyperbolic functions: • arcsech x (0 < x ≤ 1) • arccosech x (x , 0) • arccoth x (x < −1 or x > 1) 106 / 391 Example 2.5: Proof of arcsinh x relation. Solution: 107 / 391 108 / 391 Example 2.6: Simplify cosh(arctanh x) for −1 < x < 1. Solution: 109 / 391 110 / 391 111 / 391 112 / 391 Derivatives d dx (arcsinh x) = 1√ x2 + 1 (x ∈ R) d dx (arccosh x) = 1√ x2 − 1 (x > 1) d dx (arctanh x) = 1 1 − x2 (−1 < x < 1) Each formula is derived using implicit differentiation or by differentiating the logarithm definition of each function. 113 / 391 Example 2.7: Prove that d dx (arcsinh x) = 1√ x2 + 1 . Solution: 114 / 391 115 / 391 Example 2.8: Find d dx (arctanh(2x) cosh(3x)). Solution: 116 / 391 Section 3: Complex Numbers The Cartesian form of a complex number z ∈ C is z = x + iy where x, y ∈ R and • x = Re(z) is the real part of z, • y = Im(z) is the imaginary part of z, • i2 = −1. ReHzL ImHzL x r y z=x+iy Θ 117 / 391 The complex number can be written as z = r(cosθ + i sinθ) where • r = |z| = √ x2 + y2 • tanθ = y x Note: The angle θ is not unique – only defined up to multiples of 2pi. We choose θ such that −pi < θ ≤ pi and call this angle the principal argument of z. 118 / 391 The Complex Exponential We define the complex exponential using Euler’s formula
eiθ = cosθ + i sinθ for θ ∈ R. We can then write the polar form of a complex number as z = reiθ 119 / 391 Example 3.1: Write z = −1 + i in polar form. ReHzL ImHzL 1 2 1 -1+i Α Θ Solution: 120 / 391 Properties of the Complex Exponential 1. ei0 = 1 Proof: ei0 = cos 0 + i sin 0 = 1. 2. eiθeiφ = ei(θ+φ) Proof: eiθeiφ = (cosθ + i sinθ) ( cosφ + i sinφ ) = cosθ cosφ + i cosθ sinφ + i sinθ cosφ − sinθ sinφ = ( cosθ cosφ − sinθ sinφ ) + i ( cosθ sinφ + sinθ cosφ ) = cos ( θ + φ ) + i sin(θ + φ) = ei(θ+φ). 121 / 391 Products and Division in Polar Form If z = r1eiθ and w = r2eiφ then zw = r1r2ei(θ+φ) z w = r1 r2 ei(θ−φ) 122 / 391 Example 3.2: Using the complex exponential, simplify ( √ 3 − i)(1 + √3i) and √ 3 − i 1 + √ 3i . ReHzL ImHzL 3 1 2 1+ 3 i 3 -i 2 1 3Α1 Α2 123 / 391 Solution: 124 / 391 125 / 391 126 / 391 De Moivre’s Theorem: If z = reiθ and n is a positive integer then zn = ( reiθ )n = rneinθ 127 / 391 Example 3.3: Evaluate ( 1 + √ 3i )15 . Solution: 128 / 391 Exponential Form of sinθ and cosθ Now eiθ = cosθ + i sinθ (1) ⇒ e−iθ = cos(−θ) + i sin(−θ) ⇒ e−iθ = cosθ − i sinθ (2) Equation (1) + (2) gives eiθ + e−iθ = 2 cosθ ⇒ cosθ = 1 2 ( eiθ + e−iθ ) 129 / 391 Equation (1) − (2) gives eiθ − e−iθ = 2i sinθ ⇒ sinθ = 1 2i ( eiθ − e−iθ ) Note: These formulae give a connection between the hyperbolic and trigonometric functions. cosh(iθ) = 1 2 ( eiθ + e−iθ ) = cosθ sinh(iθ) = 1 2 ( eiθ − e−iθ ) = i sinθ 130 / 391 Example 3.4: Express sin5 θ in terms of the functions sin(nθ) for integers n. Solution: 131 / 391 Differentiation via the Complex Exponential If z = x + yi where x, y ∈ R then we define ez = ex+iy = ex eiy = ex(cos y + i sin y). Derivatives of functions from R to C are defined similarly as those from R to R. Differentiation to functions from R to C is also linear and follows the product law. Show that d dt ( ekt ) = kekt when k = a + bi ∈ C. d dt [ e(a+bi)t ] = d dt [ eateibt ] 132 / 391 = d dt [ eat (cos(bt) + i sin(bt)) ] = aeat [cos(bt) + i sin(bt)] + eat [−b sin(bt) + bi cos(bt)] = aeat [cos(bt) + i sin(bt)] + eat [ bi2 sin(bt) + bi cos(bt) ] = aeat [cos(bt) + i sin(bt)] + bieat [cos(bt) + i sin(bt)] = (a + bi)eat [cos(bt) + i sin(bt)] = (a + bi)eateibt = (a + bi)e(a+ib)t. 133 / 391 Example 3.5: Find d56 dt56 ( e−t cos t ) . Solution: 134 / 391 135 / 391 Note: Example 3.5 also gives the answer to d56 dt56 ( e−t sin t ) . 136 / 391 Integration via the Complex Exponential Since d dx ( ekx ) = k ekx if k = a + bi (a, b ∈ R) , then ∫ k ekx dx = ekx + C ⇒ ∫ ekx dx = 1 k ekx + D 137 / 391 Example 3.6: Evaluate ∫ e3x sin(2x) dx. Solution: 138 / 391 139 / 391 Note: Example 3.6 also gives the answer to ∫ e3x cos(2x) dx. 140 / 391 Section 4: Integral Calculus Derivative Substitutions To evaluate ∫ f [g(x)]g′(x)dx put u = g(x)⇒ du dx = g′(x). Then ∫ f [g(x)]g′(x)dx = ∫ f (u) du dx dx = ∫ f (u) du 141 / 391 Example 4.1: Evaluate ∫ (6x2 + 10) sinh(x3 + 5x − 2)dx. Solution: 142 / 391 Example 4.2: Evaluate ∫ sech2(3x) 10 + 2 tanh(3x) dx. Solution: 143 / 391 Trigonometric and Hyperbolic Substitutions We can use trigonometric and hyperbolic substitutions to integrate expressions containing √ a2 − x2, √ a2 + x2, √ x2 − a2, where a is a positive real number. Method: Put x = g(θ). Then∫ f (x) dx = ∫ f [g(θ)]g′(θ) dθ 144 / 391 Integrand Substitution √ a2 − x2, 1√ a2 − x2 , ( a2 − x2 ) 3 2 etc. x = a sinθ or x = a cosθ √ a2 + x2, 1√ a2 + x2 , ( a2 + x2 )− 32 etc. x = a sinhθ √ x2 − a2, 1√ x2 − a2 , ( x2 − a2 ) 5 2 etc. x = a coshθ 1 a2 + x2 x = a tanθ 145 / 391 Example 4.3: Evaluate ∫ 1√ x2 + 25 dx using a substitution. Solution: 146 / 391 147 / 391 Example 4.4: Evaluate ∫ 1 x2 + 2 dx using a substitution. Solution: 148 / 391 149 / 391 Example 4.5: Evaluate ∫ √ 9 − 4x2 dx if |x| ≤ 32 . Solution: 150 / 391 151 / 391 Example 4.6: Evaluate ∫ ( x2 − 1 ) 3 2 dx if x ≥ 1. Solution: 152 / 391 Powers of Hyperbolic Functions Consider the integral:∫ sinhm x coshn x dx where m,n are integers (≥ 0). • If m or n is odd, create a “derivative” substitution by rewriting one of the odd power terms using identities. • If m and n are even, use double angle formulae. 153 / 391 Example 4.7: Evaluate ∫ sinh4 θ dθ. Solution: 154 / 391 155 / 391 Finish Example 4.6: 156 / 391 Example 4.8: Evaluate ∫ sinh5 x cosh6 x dx. Solution: 157 / 391 158 / 391 Example 4.9: Evaluate I = ∫ sinh5 x cosh7 x dx. Solution: 159 / 391 160 / 391 Partial Fractions Let f (x) and g(x) be polynomials, then f (x) −→ degree n g(x) −→ degree d can be written as the sum of partial fractions. Case 1: n < d 1. Factorise g over the real numbers. 2. Write down partial fraction expansion. 3. Find unknown coefficients A,A1,A2, . . . ,Ar,B,B1,B2, . . . ,Br. 161 / 391 Denominator Factor Partial Fraction Expansion (x − a) A x − a (x − a)r A1 x − a + A2 (x − a)2 + · · · + Ar (x − a)r (x2 + bx + c) Ax + B x2 + bx + c (x2 + bx + c)r A1x+B1x2+bx+c + A2x+B2 (x2+bx+c)2 + · · · + Arx+Br(x2+bx+c)r 162 / 391 Example 4.10: Evaluate ∫ 4 x2(x + 2) dx (x , 0,−2). Solution: 163 / 391 164 / 391 165 / 391 Example 4.11: Evaluate ∫ 4x (x2 + 4)(x − 2) dx (x , 2). Solution: 166 / 391 167 / 391 168 / 391 Example 4.12: Evaluate ∫ 2x4 + 3x2 (x2 + 1)2(x2 + 2) dx. Solution: 169 / 391 170 / 391 171 / 391 Note: In general, for a positive integer n if we put x = tanθ then∫ 1 (x2 + 1)n dx = ∫ cos2n−2 θ dθ. 172 / 391 Case 2: n ≥ d Use long division, then apply case 1. Example 4.13: Find∫ 5x4 + 13x3 + 6x2 + 4 x3 + 2x2 dx (x , 0,−2). Solution: 173 / 391 174 / 391 Integration by Parts The product rule for differentiation is d dx (uv) = du dx v + u dv dx Integrate ∫ d dx (uv) dx = ∫ ( du dx v + u dv dx ) dx ⇒ uv = ∫ du dx v dx + ∫ u dv dx dx ⇒ ∫ u dv dx dx = uv − ∫ v du dx dx 175 / 391 Example 4.14: Evaluate ∫ x2 log x dx (x > 0). Solution: 176 / 391 Example 4.15: Evaluate ∫ xe5x dx. Solution: 177 / 391 Example 4.16: Evaluate ∫ log x dx (x > 0). Solution: 178 / 391 Note: This technique can also be used to integrate inverse trigonometric functions and inverse hyperbolic functions. 179 / 391 Example 4.17: Evaluate ∫ e3x sin(2x) dx. Solution: 180 / 391 181 / 391 182 / 391 Section 5: First Order Differential Equations Ordinary Differential Equations (1) An equation of the form f ( x, y, dy dx , d2y dx2 , . . . , dny dxn ) = 0 is an ordinary differential equation (o.d.e) of order n. Example 5.1: What order is 3 d4y dx4 = ( dy dx )2 + 2x2y? Solution: 183 / 391 (2) A solution of an o.d.e is a function y that satisfies the o.d.e for all x in some interval. Example 5.2: Verify that y(x) = x2 + 2 x is a solution of dy dx + y x = 3x for all x ∈ R \ {0}. Solution: 184 / 391 First Order O.D.E’s The general form of a first order o.d.e is dy dx = f (x, y). Example 5.3: Solve dy dx = x3. Solution: This is the general solution where c ∈ R is an arbitrary constant. Each value of c corresponds to a different solution of the o.d.e. 185 / 391 -4 -2 2 x -4 -2 2 4 6 y c=0 c=-2 c=2 c=4 186 / 391 Initial value problem for a first order o.d.e Solve dy dx = f (x, y) subject to the condition y(x0) = y0. Example 5.4: Solve dy dx = x3 given y(0) = 2. Solution: 187 / 391 Separable O.D.E’S A separable first order o.d.e has the form: dy dx =M(x)N(y), (M(x) , 0, N(y) , 0) To solve use separation of variables dy dx =M(x)N(y) ⇒ 1N(y) dy dx =M(x) ⇒ ∫ 1 N(y) dy dx dx = ∫ M(x)dx ⇒ ∫ 1 N(y)dy = ∫ M(x) dx 188 / 391 Example 5.5: Solve dy dx = y 1 + x (x , −1). Solution: 189 / 391 190 / 391 -6 -4 -2 2 4 x -6 -4 -2 2 4 6 y c=1 c=2 c=-1 c=-2 H-1,0L 191 / 391 192 / 391 Example 5.6: Solve dy dx = 1 2y √ 1 − x2 (−1 < x < 1, y , 0) if y(0) = 3. Solution: 193 / 391 194 / 391 x y c=9 c=4 c=1 c=0 c=9 c=4 c=1 c=0 y= arcsin x + c y=- arcsin x + c x=-1 x=1 * Solution 195 / 391 Linear First Order O.D.E’s Example 5.7: Solve x dy dx + y = ex. Solution: 196 / 391 A linear first order o.d.e has the form: dy dx +P(x)y = Q(x)
To solve: Multiply o.d.e by I(x) I(x)dy dx +P(x)I(x)y = Q(x)I(x) If the left side can be written as the derivative of y(x)I(x), then d dx [ y(x)I(x)] = Q(x)I(x) which can be solved by integrating with respect to x. 197 / 391 Aim: Find an integrating factor I so the left side will be the derivative of yI. Then d dx ( yI) ≡ Idy dx +PIy ⇒ dy dx I + ydI dx = Idy dx +PIy ⇒ ydI dx = PIy To solve for all y ⇒ dI dx = PI (separable) 198 / 391 ⇒ 1I dI dx = P ⇒ ∫ 1 IdI = ∫ Pdx ⇒ log |I| = ∫ P dx + c ⇒ |I| = e ∫ Pdx+c = e ∫ Pdx · ec ⇒ I = ±ec︸︷︷︸ constant ·e ∫ Pdx 199 / 391 So one integrating factor is I(x) = e ∫ Pdx Note: Since we only need one integrating factor I, we can neglect the ‘+c’ and modulus signs when calculating I. 200 / 391 Example 5.8: Find the general solution of dy dx + y x = sin x (x , 0). Solution: 201 / 391 202 / 391 203 / 391 x y c=0 c=1 c=-2 c=-2 c=1 -2 2 -2 2 y(x) = − cos x + 1 x sin x + c x 204 / 391 Example 5.9: Solve 1 2 dy dx − xy = x if y(0) = −3. Solution: 205 / 391 206 / 391 Note: 207 / 391 x y c=1 H0,1L c=2 c=-2 c=-1 H0,-2L y = −1 + cex2 208 / 391 Other First Order O.D.E’s Sometimes it is possible to make a substitution to reduce a general first order o.d.e to a separable or linear o.d.e. • A homogeneous type o.d.e has the form dy dx = f (y x ) Substituting u = yx reduces the o.d.e to a separable o.d.e. • Bernoulli’s equation has the form dy dx + P(x)y = Q(x)yn Substituting u = y1−n reduces the o.d.e to a linear o.d.e. 209 / 391 Example 5.10: Solve the homogeneous type differential equation dy dx = y x + cos2 (y x ) ( −pi2 < yx < pi2 ) by substituting u = y x . Solution: 210 / 391 211 / 391 212 / 391 Example 5.11: Solve the Bernoulli equation dy dx + y = e3xy4 (y , 0) by substituting u = y−3. Solution: 213 / 391 214 / 391 215 / 391 Population Models Malthus (Doomsday) model Rate of growth is proportional to the population p at time t. dp dt ∝ p ⇒ dp dt = kp (separable/linear) where k is a constant of proportionality representing net births per unit population per unit time. If the initial population is p(0) = p0, then the solution is p(t) = p0ekt 216 / 391 Note: The Doomsday model is unrealistic since if • k > 0 − unbounded exponential growth • k < 0 − population dies out • k = 0 − population stays constant 217 / 391 Equilibrium Solutions 1. An equilibrium solution is a solution that does not change with time. i.e. dx dt = 0 ⇒ x(t) = x0 2. Stable equilibrium – solutions that start nearby move closer as t increases. t x x0 Stable 218 / 391 3. Unstable equilibrium – solutions that start nearby move further away as t increases. t x x0 Unstable 219 / 391 4. Semistable equilibrium – on one side of x0 solutions that start nearby move closer as t increases whereas on the other side of x0 solutions move further away as t increases. t x x0 Semistable t x x0 Semistable 220 / 391 5. Phase plots: If dx dt = f (x), a plot of dx dt as a function of x will give the equilibrium solutions and the behaviour of solutions close by. 221 / 391 Doomsday model with harvesting. Remove some of the population at a constant rate. dp dt = kp − h, h > 0. Example 5.12: dp dt = 3p − 2 (k = 3, h = 2) Solution: • Equilibrium solutions 222 / 391 • Phase plot p dp dt 23 -2 dp dt >0 dp dt <0 dies out! 223 / 391 t p Unstablep(0)=2/3 Note: Solving dp dt = 3p − 2 with p(0) = p0 gives p(t) = 23 + ( p0 − 23 ) e3t which agrees with predicted behaviour. 224 / 391 Logistic model. Include “competition” term in Malthus’ model since overcrowding, disease, lack of food and natural resources will cause more deaths. dp dt = kp − k a p2 = kp ( 1 − p a ) ↗︷ ︸︸ ︷ net birth rate ↖︷ ︸︸ ︷ competition term where a > 0 is the carrying capacity. 225 / 391 Example 5.13: dp dt = p ( 1 − p 4 ) (k = 1, a = 4) Solution: • Equilibrium solutions • Phase plot p dp dt 1 0 2 4 dp dt >0 dp dt <0 226 / 391 t p pH0L=4 pH0L=0 227 / 391 • Exact solution dp dt = p 4 ( 4 − p) (separable) 228 / 391 Suppose p(0) = 1 229 / 391 t p 1 4 limiting growth initial exponential growth Note: Logistic model accurately describes • population in a limited space (e.g. bacteria culture). • population of USA from 1790-1950. 230 / 391 Logistic model with harvesting. Remove some of the population at constant rate: dp dt = kp ( 1 − p a ) − h, h > 0, a > 0 Example 5.14: dp dt = p ( 1 − p 4 ) − 3 4 ( a = 4, k = 1, h = 3 4 ) Solution: 231 / 391 • Phase plot p dp dt 1 4 – 3 4 1 3 dp dt >0 dp dt <0 dp dt <0 dies out! 232 / 391 tp pH0L=3 pH0L=1 dies out! 233 / 391 Find the time taken until the population dies out if p(0) = 1 2 . dp dt = −1 4 (p − 3)(p − 1) (separable) 234 / 391 235 / 391 Mixing Problems Example 5.15: Effluent (pollutant concentration 2g/m3) flows into a pond (volume 1000m3, initially 100g pollutant) at a rate of 10m3/min. The pollutant mixes quickly and uniformly with pond water and flows out of pond at a rate of 10m3/min. Find the concentration of pollutant in the pond at any time. Solution: 236 / 391 Let x be the amount (grams) of pollutant in pond at time t minutes. Then C = x V is the concentration of pollutant in pond (grams/m3), where V is the volume of the pond (m3) at time t. dx dt = rate pollutant flows in - rate pollutant flows out 237 / 391 238 / 391 x dx dt 20 2000 dx dt >0 dx dt <0 239 / 391 Expect solution (for all initial conditions) to look like t (min) x (grams) xH0L=2000 Stable 240 / 391 • Exact solution dx dt + x 100 = 20 (Linear/Separable) 241 / 391 tHminL conc. Hgm3L 2 0.1 242 / 391 Definitions 1. Transient terms: terms decaying to 0 as t→∞. 2. Steady state terms: terms NOT decaying to 0 as t→∞. The solution for the concentration can be classified as follows. 243 / 391 Example 5.16: Find the concentration of pollutant in pond if input flow rate is decreased to 5m3/min. Solution: 244 / 391 Let V be volume in pond (m3) at time t minutes. 245 / 391 Note: There are no equilibrium solutions for x. 246 / 391 247 / 391 tHminL conc. Hgm3L 2 0.1 200 248 / 391 Section 6: Second Order Differential Equations A second order o.d.e has the form F ( x, y, dy dx , d2y dx2 ) = 0 The general form of a linear second order o.d.e is d2y dx2 +P(x)dy dx + Q(x)y = R(x) • If R(x) = 0, the o.d.e is homogeneous (H). • If R(x) , 0, the o.d.e is inhomogeneous (IH). Note: A homogeneous linear o.d.e is different to a homogeneous type first order o.d.e. 249 / 391 Initial value problem for a second order o.d.e Solve d2y dx2 +P(x)dy dx + Q(x)y = R(x) subject to the conditions y(x0) = y0 and y′(x0) = y1. Boundary value problem for a second order o.d.e Solve d2y dx2 +P(x)dy dx + Q(x)y = R(x) subject to the conditions y(a) = y0 and y(b) = y1. 250 / 391 Homogeneous 2nd Order Linear O.D.E’s Theorem: The general solution of y” +P(x)y′ + Q(x)y = 0 is the function y given by y(x) = c1y1(x) + c2y2(x) where • y1, y2 are two linearly independent solutions of the homogeneous o.d.e, • c1, c2 ∈ R are arbitrary constants. 251 / 391 Definition: Two functions y1 and y2 are linearly independent if c1y1(x) + c2y2(x) = 0 ⇒ c1 = c2 = 0. Example 6.1: Are y1(x) = x2, y2(x) = 2x2 linearly independent? Solution: Example 6.2: Are y1(x) = e2x, y2(x) = xe2x linearly independent? Solution: 252 / 391 Homogeneous 2nd Order Linear O.D.E’s with Constant Coefficients General form: ay” + by′ + cy = 0 where a, b, c are constants. To solve for y(x): Try y(x) = eλx ⇒ y′(x) = λeλx, y”(x) = λ2eλx so ( aλ2 + bλ + c ) eλx︸︷︷︸ ,0 = 0 ⇒ aλ2 + bλ + c = 0 Characteristic Equation 253 / 391 ⇒ λ = −b ± √ b2 − 4ac 2a Case 1: b2 − 4ac > 0 • 2 distinct real values λ1, λ2 • 2 linearly independent solutions eλ1x, eλ2x • General Solution: y(x) = Aeλ1x + Beλ2x 254 / 391 Example 6.3: Solve y” + 7y′ + 12y = 0 for y(x). Solution: 255 / 391 256 / 391 Case 2: b2 − 4ac = 0 • 1 real value λ = −b 2a • 1 solution is eλx • 2nd linearly independent solution is xeλx (found using variation of parameters — not in syllabus). • General Solution: y(x) = Aeλx + Bxeλx We now verify that xeλx is a solution: If y(x
) = xeλx, then 257 / 391 y′(x) = (λx + 1) eλx, y”(x) = ( λ2x + 2λ ) eλx. So ay” + by′ + cy = a ( λ2x + 2λ ) eλx + b (λx + 1) eλx + cxeλx = xeλx ( aλ2 + bλ + c ) ︸ ︷︷ ︸ =0 + (2λa + b)︸ ︷︷ ︸ =0 eλx = 0 So y(x) = xeλx is a solution. 258 / 391 Example 6.4: Solve y” + 2y′ + y = 0 for y(x). Solution: 259 / 391 260 / 391 Case 3: b2 − 4ac < 0 • 2 complex conjugate values λ1 = α + iβ, λ2 = α − iβ • 2 complex linearly independent solutions e(α+iβ)x, e(α−iβ)x • General Solution: y(x) = C1e(α+iβ)x + C2e(α−iβ)x where C1,C2 ∈ C = C1eαx ( cos(βx) + i sin(βx) ) + C2eαx ( cos(βx) − i sin(βx)) = (C1 + C2)︸ ︷︷ ︸ A eαx cos(βx) + (C1i − C2i)︸ ︷︷ ︸ B eαx sin(βx) 261 / 391 Put A = C1 + C2 and B = (C1 − C2)i. If C1 = C2, then A,B ∈ R. Note: Imposing real conditions on the o.d.e will always lead to real coefficients A and B. • 2 real linearly independent solutions eαx cos(βx), eαx sin(βx) Real General Solution: y(x) = Aeαx cos(βx) + Beαx sin(βx) 262 / 391 Example 6.5: Solve y”− 4y′ + 13y = 0 for y(x) if y(0) = −1 and y′(0) = 2. Solution: 263 / 391 264 / 391 This solution has the form y(x) = e2x (a cosθ + b sinθ). In general, a cosθ + b sinθ = √ a2 + b2 ( a√ a2 + b2 cosθ + b√ a2 + b2 sinθ ) = √ a2 + b2 ( cosφ cosθ + sinφ sinθ ) = √ a2 + b2 cos(θ − φ). 265 / 391 Hence we can rewrite the solution from Example 6.5 as This form is sometimes preferable for graphing or further manipulation. 266 / 391 Inhomogeneous 2nd Order Linear O.D.E’s Theorem: The general solution of y” +P(x)y′ + Q(x)y = R(x) is the function y given by y(x) = yH (x) + yP(x) where • yH (x) = c1y1(x) + c2y2(x) is the general solution of the homogeneous o.d.e (called the homogeneous solution, GS(H)), • yP(x) is a solution of the inhomogeneous o.d.e (called a particular solution, PS(IH)), 267 / 391 Inhomogeneous 2nd Order Linear O.D.E’s with Constant Coefficients General form: ay” + by′ + cy = R(x) where a, b, c are constants. Example 6.6: Solve y” + 2y′ − 8y = R(x) where (a) R(x) = 1 − 8x2 (b) R(x) = e3x (c) R(x) = 85 cos x (d) R(x) = 3 − 24x2 + 7e3x. 268 / 391 Solution: Step 1: Find the general solution of y” + 2y′ − 8y = 0. 269 / 391 Step 2: Find a particular solution of y” + 2y′ − 8y = R(x). (a) R(x) = 1 − 8x2 : y” + 2y′ − 8y = 1 − 8x2 270 / 391 271 / 391 (b) R(x) = e3x : y” + 2y′ − 8y = e3x( e3x is NOT part of GS(H) ) 272 / 391 273 / 391 (c) R(x) = 85 cos x : y” + 2y′ − 8y = 85 cos x 274 / 391 275 / 391 Superposition of Particular Solutions Theorem: A particular solution of ay” + by′ + cy = c1R1(x) + c2R2(x) is yP(x) = c1y1(x) + c2y2(x) where • y1(x) is a particular solution of ay” + by′ + cy = R1(x), • y2(x) is a particular solution of ay” + by′ + cy = R2(x), • a, b, c, c1, c2 are constants. 276 / 391 Example 6.6 (d): R(x) = 3 − 24x2 + 7e3x. Solution: 277 / 391 Example 6.7: Solve y” − y = ex. Solution: GS(H) : yH (x) = Aex + Be−x 278 / 391 Example 6.8: Solve y” + 2y′ + y = e−x. Solution: GS(H) : yH (x) = (A + Bx) e−x 279 / 391 280 / 391 Example 6.9: Solve y” + 49y = 28 sin(7t). Solution: GS(H) : yH (t) = A cos(7t) + B sin(7t) 281 / 391 282 / 391 Springs - Free Vibrations An object (mass m kg) stretches a spring (natural length L m) hanging from a fixed support by s m. 283 / 391 The forces are: • gravitational force = mg (g = 9.8 m/s2) • restoring force in spring (from Hooke’s Law) T = k · extension (k > 0)0 spring constant At equilibrium, forces balance so: 284 / 391 Suppose the mass is set in motion. Let y be the displacement of the object from the equilibrium position (y = 0) at any time t. Assume • downward direction is positive • spring is stretched below equilibrium • mass is moving down (so damping is upwards) 285 / 391 Extra force: • damping force is proportional to velocity R = βy˙ (β ≥ 0)0 damping constant Using Newton’s Law (F = ma) 286 / 391 To solve, try y(t) = eλt ⇒ mλ2 + βλ + k = 0 ⇒ λ = −β ± √ β2 − 4mk 2m • If β = 0 : λ = ±ib simple harmonic motion • If 0 < β < 2√mk : λ = a ± ib underdamped, weak damping • If β = 2√mk : λ = a, a critical damping • If β > 2√mk : λ = a, b overdamped, strong damping 287 / 391 288 / 391 Example 6.10: A 4049 kg mass stretches a spring hanging from a fixed support by 0.2m. The mass is released from the equilibrium position with a downward velocity of 3m/s. Find the position of the mass y below equilibrium at any time t, if the damping constant β is: (a) 0 (b) 16049 (c) 80 7 (d) 2000 49 Solution: 289 / 391 290 / 391 (a) β = 0 : y¨ + 49y = 0 291 / 391 t y Simple harmonic motion 3 7 – 3 7 Π 7 2 Π 7 3 Π 7 Equil. 292 / 391 (b) β = 16049 : y¨ + 4y˙ + 49y = 0 293 / 391 t y Weak damping 1 5 – 1 5 21 Equil. 294 / 391 (c) β = 807 : y¨ + 14y˙ + 49y = 0 295 / 391 t y critical damping 0.15 0.1 0.05 0.5 1 Equil. 296 / 391 (d) β = 200049 : y¨ + 50y˙ + 49y = 0 297 / 391 t y strong damping 0.06 0.03 2 4 Equil 298 / 391 Springs – Forced Vibrations If an external downwards force f is applied to the spring-mass system at time t, the forces acting on the mass are: 299 / 391 Example 6.11: Apply an external downwards force f (t) = 1607 sin(7t) in Example 6.10. Solution: 300 / 391 (a) β = 807 : y¨ + 14y˙ + 49y = 28 sin(7t) GS(IH) : y(t) = (A + Bt) e−7t − 27 cos(7t) 301 / 391 t y GSHHL contributes oscillatory motion due to PSHIHL 0.4 -0.4 1 2 3 Equil. 302 / 391 (b) β = 0 : y¨ + 49y = 28 sin(7t) GS(IH) : y(t) = A cos(7t) + B sin(7t) − 2t cos(7t) 303 / 391 t y Resonance 10 -10 5 -5 2 4 6 Equil Definition Resonance: Resonance occurs when the external force f has the same form as one of the terms in the GS(H). If β = 0, then the PS(IH) will grow without bound as t→∞. 304 / 391 RLC series electric circuit An RLC series electric circuit is an electric circuit with 4 components connected sequentially in a loop: Circuits such as this are common in radio communications. 305 / 391 RLC series electric circuit The circuit components are 306 / 391 RLC series electric circuit Let q(t) be the charge on the capacitor (measured in Coulomb) at time t seconds. The charge satisfies the second-order ODE L d2q dt2 + R dq dt + q C = V This equation has the same form as the equation of motion for a spring with external driving force, and can exhibit all the same solution types as the spring system. 307 / 391 308 / 391 Section 7: Functions of Two Variables Example The temperature T at a point on the Earth’s surface at a given time depends on the latitude x and the longitude y. We think of T being a function of the variables x, y and write T = f (x, y). In general A function of two variables is a mapping f that assigns a unique real number z = f (x, y) to each pair of real numbers (x, y) in some subset D of the xy plane R2. We also write f : D→ R where D is called the domain of f . Example If f (x, y) = x2 + y3 then f (2, 1) = 4 + 1 = 5. 309 / 391 We can represent the function f by its graph in R3. The graph of f is: { (x, y, z) : (x, y) ∈ D and z = f (x, y) } . This is a surface lying directly above the domain D. The x and y axes lie in the horizontal plane and the z axis is vertical. 310 / 391 Equations of a Plane The Cartesian equation of a plane has the form ax + by + cz = d where a, b, c, d are real constants. n = ai + bj + ck is a normal vector to the plane. In fact, the plane passing through a point (x0, y0, z0) with a normal vector (a, b, c) consists of the points (x, y, z) such that (a, b, c) is perpendicular to (x − x0, y − y0, z − z0) and thus has equation a(x − x0) + b(y − y0) + c(z − z0) = 0, that is, ax + by + cz = ax0 + by0 + cz0. 311 / 391 312 / 391 Example 7.1: The plane 4x + 3y + z = 2 can be written as z = 2 − 4x − 3y, so is the graph of the function f : R2 → R given by f (x, y) = 2 − 4x − 3y. Sketch the plane. Solution: 313 / 391 314 / 391 Level Curves A curve on the surface z = f (x, y) for which z is a constant is a contour. The same curve drawn in the xy plane is a level curve. So a level curve of f has the form {
(x, y) : f (x, y) = c} where c ∈ R is a constant. 315 / 391 Sketching Functions of Two Variables The key steps in drawing a graph of a function of two variables z = f (x, y) are: 1. Draw the x, y, z axes. For right handed axes: the positive x axis is towards you, the positive y axis points to the right, and the positive z axis points upward. 2. Draw the y − z cross section. 3. Draw some level curves and their contours. 4. Draw the x − z cross section. 5. Label any x, y, z intercepts and key points. 316 / 391 Example 7.2: Find the level curves of z = √ 1 − x2 − y2. Hence identify the surface and sketch it. Solution: 317 / 391 x y c=1 c=0 H0,1L H0,-1L H-1,0L H1,0L Level curves 318 / 391 Consider cross sections (slices) to help sketch graph. x z H-1,0L H1,0L H0,1L y z H-1,0L H1,0L H0,1L 319 / 391 Surface is a hemisphere radius 1, centre at (0, 0, 0) for z ≥ 0. 320 / 391 Example 7.3: Sketch the graph of z = 4×2 + y2. Solution: x y c=0 c=4 H0,2L H0,-2L H-1,0L H1,0L Level curves 321 / 391 Consider cross sections (slices) to help sketch graph. x z -1 1 4 z=4×2 y z -1 1 4 z=y2 322 / 391 The surface is an elliptic paraboloid (parabolic bowl). 323 / 391 Example 7.4: Sketch the graph of z = √ 4×2 + y2. Solution: x y c=0 c=2 H0,2L H0,-2L H-1,0L H1,0L Level curves 324 / 391 Cross sections x z -1 1 2 z=2ÈxÈ y z -1 1 2 z=ÈyÈ 325 / 391 The surface is an elliptic cone. 326 / 391 Limits Let f : R2 → R be a real-valued function. We say f has the limit L as (x, y) approaches (x0, y0) lim (x,y)→(x0,y0) f (x, y) = L if when (x, y) approaches (x0, y0) along ANY path in the domain, f (x, y) gets arbitrarily close to L. Note: 1 L must be finite. 2 The limit can exist if f is undefined at (x0, y0). 3 The usual limit laws apply. 327 / 391 Continuity Let f : R2 → R be a real-valued function. f is continuous at (x, y) = (x0, y0) if lim (x,y)→(x0,y0) f (x, y) = f (x0, y0) Note: The continuity theorems for functions of one variable can be generalised to functions of two variables. 328 / 391 Example 7.5: Let f (x, y) = x2 + y2. For which values of x and y is f continuous? Solution: 329 / 391 Example 7.6: Evaluate lim (x,y)→(2,1) log(1 + 2×2 + 3y2). Solution: 330 / 391 First Order Partial Derivatives Let f : R2 → R be a real-valued function. The first order partial derivatives of f with respect to the variables x and y are defined by the limits: fx = ∂f ∂x = lim h→0 f (x + h, y) − f (x, y) h fy = ∂f ∂y = lim h→0 f (x, y + h) − f (x, y) h Note: • ∂f ∂x measures the rate of change of f with respect to x when y is held constant. • ∂f ∂y measures the rate of change of f with respect to y when x is held constant. 331 / 391 Geometric Interpretation of ∂f ∂x and ∂f ∂y 332 / 391 Let C1 be the curve where the vertical plane y = y0 intersects the surface. Then ∂f ∂x ∣∣∣∣ (x0,y0) gives the slope of the tangent to C1 at (x0, y0, z0). Let C2 be the curve where the vertical plane x = x0 intersects the surface. The ∂f ∂y ∣∣∣∣ (x0,y0) gives the slope of the tangent to C2 at (x0, y0, z0). • T1 and T2 are the tangent lines to C1 and C2. 333 / 391 Example 7.7: Let f (x, y) = xy2. Find ∂f ∂y from first principles. Solution: 334 / 391 Example 7.8: Let f (x, y) = 3x3y2 + 3xy4. Find ∂f ∂x and ∂f ∂y . Solution: 335 / 391 Example 7.9: Let f (x, y) = y log x + x tanh(3y). Find fx, fy at (1, 0). Solution: 336 / 391 Tangent Planes and Differentiability Let f : R2 → R be a real-valued function. We say that f is differentiable at (x0, y0) if the tangent lines to all curves on the surface z = f (x, y) passing through (x0, y0, z0) form a plane, called the tangent plane. This holds if fx and fy exist and are continuous near (x0, y0). 337 / 391 The tangent line T1 has equation (y = y0 fixed): z − z0 = ∂f∂x ∣∣∣∣ (x0,y0) (x − x0) The tangent line T2 has equation (x = x0 fixed): z − z0 = ∂f∂y ∣∣∣∣ (x0,y0) (y − y0) Since a plane passing through (x0, y0, z0) has the form z − z0 = α(x − x0) + β(y − y0) the tangent plane has equation z − z0 = ∂f∂x ∣∣∣∣ (x0,y0) (x − x0) + ∂f∂y ∣∣∣∣ (x0,y0) (y − y0). 338 / 391 Example 7.10: Find the equation of the tangent plane to the surface z = f (x, y) = 2×2 + y2 at (1, 1, 3). Solution: 339 / 391 Linear Approximations If f is differentiable at (x0, y0), we can approximate z = f (x, y) by its tangent plane at (x0, y0, z0). This linear approximation of f near (x0, y0) is: f (x, y) ≈ f (x0, y0) + ∂f∂x ∣∣∣∣ (x0,y0) (x − x0) + ∂f∂y ∣∣∣∣ (x0,y0) (y − y0) Let ∆x = x − x0, ∆y = y − y0, ∆f = z − z0 = f (x, y) − f (x0, y0). Then the approximate change in f near (x0, y0), for given small changes in x and y, is: ∆f ≈ ∂f∂x ∣∣∣∣ (x0,y0) ∆x + ∂f∂y ∣∣∣∣ (x0,y0) ∆y 340 / 391 Example 7.11: Let z = f (x, y) = x2 + 3xy − y2. If x changes from 2 to 2.05 and y changes from 3 to 2.96, estimate the change in z. Solution: 341 / 391 Note: The actual change in f is ∆f = f (2.05, 2.96) − f (2, 3) = 13.6449 − 13 = 0.6449 342 / 391 Example 7.12: Find the linear approximation of f (x, y) = xexy at (1, 0). Hence, approximate f (1.1,−0.1). Solution: 343 / 391 Note: The actual value is (1.1)e−0.11 ≈ 0.98542 344 / 391 Second Order Partial Derivatives Let f : R2 → R be a real-valued function. The second order partial derivatives of f with respect to x and y are defined by: • fxx = ( fx ) x = ∂ ∂x ( ∂f ∂x ) = ∂2f ∂x2 • fyy = ( fy ) y = ∂ ∂y ( ∂f ∂y ) = ∂2f ∂y2 • fxy = ( fx ) y = ∂ ∂y ( ∂f ∂x ) = ∂2f ∂y∂x • fyx = ( fy ) x = ∂ ∂x ( ∂f ∂y ) = ∂2f ∂x∂y 345 / 391 Theorem: If the second order partial derivatives of f exist and are continuous then fxy = fyx. 346 / 391 Example 7.13: Find the second order partial derivatives of f : R2 → R given by f (x, y) = x sin(x + 2y). Solution: 347 / 391 Note: fxy = fyx as expected since trigonometric functions and polynomials are continuous for all (x, y) ∈ R2. 348 / 391 Chain Rule 1. If z = f (x, y) and x = g(t), y = h(t) are differentiable functions, then z = f (g(t), h(t)) is a function of t, and dz dt = ∂z ∂x dx dt + ∂z ∂y dy dt 349 / 391 Example 7.14: If z = x2 − y2, x = sin t, y = cos t. Find dz dt at t = pi 6 . Solution: 350 / 391 351 / 391 2. If z = f (x, y) and x = g(s, t), y = h(s, t) are differentiable functions, then z is a function of s and t with ∂z ∂s = ∂z ∂x ∂x ∂s + ∂z ∂y ∂y ∂s ∂z ∂t = ∂z ∂x ∂x ∂t + ∂z ∂y ∂y ∂t 352 / 391 Example 7.15: If z = ex sinh y, x = st2, y = s2t. Find ∂z ∂s and ∂z ∂t . Solution: 353 / 391 Directional Derivatives Let uˆ = (u1,u2) be a unit vector in the xy-plane (so u21 + u 2 2 = 1). The rate of change of f at P0 = (x0, y0) in the direction uˆ is the directional derivative Duˆf ∣∣∣ P0 . Geometrically this represents the slope of the surface z = f (x, y) above the point P0 in the direction uˆ. 354 / 391 The straight line starting at P0 = (x0, y0) with velocity uˆ = (u1,u2) has parametric equations: x = x0 + tu1, y = y0 + tu2. Hence, Duˆf ∣∣∣ P0 = rate of change of f along the straight line at t = 0 = value of d dt f (x0 + tu1, y0 + tu2) at t = 0 = fx(x0, y0)x′(0) + fy(x0, y0)y′(0) by the chain rule = fx(x0, y0)u1 + fy(x0, y0)u2. We can also write this as a dot product Duˆf ∣∣∣ P0 = ( ∂f ∂x ∣∣∣∣∣ P0 , ∂f ∂y ∣∣∣∣∣ P0 ) · (u1,u2). 355 / 391 Gradient Vectors If f : R2 → R is a differentiable function, we can define the gradient of f to be the vector grad f = ∇f = ∂f ∂x i + ∂f ∂y j = ( ∂f ∂x , ∂f ∂y ) Then the directional derivative of f at the point P0 in the direction uˆ is the dot product Duˆf ∣∣∣ P0 = ∇f ∣∣∣ P0 · uˆ 356 / 391 Example 7.16: Find the directional derivative of f (x, y) = xey at (2, 0) in the direction from (2, 0) towards (1 2 , 2 ) . Solution: • direction uˆ 357 / 391 358 / 391 Example 7.17: Find the directional derivative of f (x, y) = arcsin ( x y ) at (1, 2) in the direction pi 4 anticlockwise from the positive x axis. Solution: • direction u
ˆ 359 / 391 360 / 391 361 / 391 Properties of ∇f and Duˆf The directional derivative of f is Duˆf = ∇f · uˆ = |∇f | |uˆ| cosθ = |∇f | cosθ where θ is the angle between ∇f and uˆ, and |v| denotes the length of a vector v. 362 / 391 So for fixed ∇f : • Duˆf is maximum when cosθ = 1 so θ = 0 ⇒ f increases most rapidly along ∇f . • Duˆf is minimum when cosθ = −1 so θ = pi ⇒ f decreases most rapidly along −∇f . 363 / 391 • Duˆf = 0 when cosθ = 0 so θ = pi2 and ∇f ⊥ uˆ. But Duˆf = 0, whenever uˆ is tangent to a level curve of f (where f = constant). ⇒ ∇f ⊥ level curves of f 364 / 391 Example 7.18: Let f (x, y) = 4×2 + y2. (a) Find ∇f at (1, 0) and (0, 2). (b) Show that ∇f is perpendicular to the level curves, by sketching ∇f at these points and the level curves of f . Solution: 365 / 391 (b) 0,2 1,0 4j 8i Level curves of f x,y y x 366 / 391 Example 7.19: In what direction does f (x, y) = xey (a) increase (b) decrease most rapidly at (2, 0)? Express direction as a unit vector. Solution: From Example 7.16 ∇f (2, 0) = i + 2j 367 / 391 368 / 391 Stationary Points A stationary point of f is a point (x0, y0) at which ∇f = 0 So ∂f ∂x = 0 and ∂f ∂y = 0 simultaneously at (x0, y0). Geometrically, this means that the tangent plane to the graph z = f (x, y) at (x0, y0) is horizontal, i.e. parallel to the xy-plane. 369 / 391 Three important types of stationary points are 370 / 391 A function f has a 1. local maximum at (x0, y0) if f (x, y) ≤ f (x0, y0) for all (x, y) in an open disk centred at (x0, y0), 2. local minimum at (x0, y0) if f (x, y) ≥ f (x0, y0) for all (x, y) in an open disk centred at (x0, y0), 3. saddle point at (x0, y0) if (x0, y0) is a stationary point, and there are points near (x0, y0) with f (x, y) > f (x0, y0) and other points near (x0, y0) with f (x, y) < f (x0, y0). 371 / 391 Any local maximum or minimum of f will occur at a critical point (x0, y0) such that 1. ∇f (x0, y0) = 0 or 2. ∂f ∂x and/or ∂f ∂y do not exist at (x0, y0). z = √ x2 + y2. Minimum at (0, 0) BUT ∇f does not exist at (0, 0). 372 / 391 Second Derivative Test If ∇f (x0, y0) = 0 and the second partial derivatives of f are continuous on an open disk centred at (x0, y0), consider the Hessian function H(x, y) = fxxfyy − (fxy)2 evaluated at (x0, y0). Then (x0, y0) is a 1. local minimum if H(x0, y0) > 0 and fxx(x0, y0) > 0. 2. local maximum if H(x0, y0) > 0 and fxx(x0, y0) < 0. 3. saddle point if H(x0, y0) < 0. Note: Test is inconclusive if H(x0, y0) = 0. 373 / 391 Example 7.20: Find and classify the stationary points of f : R2 → R given by f (x, y) = x3 + y3 + 3x2 − 3y2 − 8. Solution: 374 / 391 375 / 391 376 / 391 Example 7.21: Find and classify the stationary points of f : R2 → R given by f (x, y) = y sin x. Solution: 377 / 391 378 / 391 Partial Integration Let f : R2 → R be a continuous function over a domain D in R2. The partial indefinite integrals of f with respect to the first and second variables (say x and y) are denoted by:∫ f (x, y) dx and ∫ f (x, y) dy. • ∫ f (x, y) dx is evaluated by holding y fixed and integrating with respect to x. • ∫ f (x, y) dy is evaluated by holding x fixed and integrating with respect to y. 379 / 391 Example 7.22: Evaluate ∫ (3x2y + 12y2x3) dx. Solution: Note: 380 / 391 Example 7.23: Evaluate ∫ 1 0 (3x2y + 12y2x3) dy. Solution: 381 / 391 Double Integrals Let f : R2 → R be a continuous function over a domain D in R2. We can evaluate the double integral:" D f (x, y) dA = " D f (x, y) dx dy " D f (x, y) dA is the volume under the surface z = f (x, y) that lies above the domain D in the xy plane, if f (x, y) ≥ 0 in D. 382 / 391 Δx Δy Volume of thin rod = (Area base)︸ ︷︷ ︸ ‖ · (height)︸ ︷︷ ︸ ‖ ∆x∆y f (x, y) 383 / 391 The double integral is defined as the limit of sums of the volumes of the rods:" D f (x, y) dA = " D f (x, y) dx dy = lim ∆x→0 lim∆y→0 n∑ i=1 [ f (x, y)∆x∆y ] i Note: If f (x, y) = 1 then " D dA = " D dx dy gives the area of the domain D. 384 / 391 Double Integrals Over Rectangular Domains Definitions 1. R = [a, b] × [c, d] is a rectangular domain defined by a ≤ x ≤ b, c ≤ y ≤ d. 2. ∫ d c ∫ b a f (x, y) dx dy = ∫ d c [∫ b a f (x, y) dx ] dy means integrate with respect to x first and then integrate with respect to y. 385 / 391 Fubini’s Theorem: Let f : R2 → R be a continuous function over the domain R = [a, b] × [c, d]. Then" R f (x, y)dA = ∫ d c ∫ b a f (x, y) dx dy = ∫ b a ∫ d c f (x, y) dy dx So order of integration is not important. 386 / 391 Example 7.24: Evaluate " R (x2 + y2) dx dy if R = [−1, 1] × [0, 1]. Solution: 387 / 391 388 / 391 Note: As expected, the order of integration is not important since polynomials are continuous for all (x, y) ∈ R2. 389 / 391 Example 7.25: Using double integrals, find the volume of the wedge shown below. Solution: 390 / 391 391 / 391 欢迎咨询51作业君
