Solutions Introduction 0.3 v̄[0,2] = 4/2 = 2, v̄[1,4] = 15/3 = 5 and v̄[2,6] = 32/4 = 8. 0.5 The lower and upper bounds are 12.6 and 14.4 respectively. 1 1.23 y = −2x/k − 1 a) k = −2/3, b) impossible, c) impossible 1.25 a) b) y y 1 1 3 0.7 distance: 4c 0.9 The lower and upper estimates are 40 m and 50 m respectively. 0.11 v̄[t0 ,t] = (t3 − t30 )/(3(t − t0 )) = (t − t0 )(t2 + t0 t + t20 )/(3(t − t0 )) = (t2 + t0 t + t20 )/3. x 3 x d) y y 1 0.15 a) 0.728 ft, b) v̄ = 3.64 ft/s c) 0.615125 ft, v̄ = 12.3025 ft/s x -1 -1 c) 0.13 Since the velocity of A is greater than that of B throughout the interval from t = 0 to t = 1, and since the cars start from the same position, we may infer that the distance between them is increasing on the intervals [0, 0.25] and [0.75, 1]. Similarly, the distance is decreasing on [1, 1.25]. 3 1 3 x -1 -1 1.27 a) D(f ) = [1, 4], R(f ) = [0, 3], b) f (2) = 0, c) x = 1 Chapter 2 Chapter 1 1.1 {1, 2, 3, 4} and {1, 2} are subsets. 1.3 −1 and 1/2 1.11 (f /g)(x) = 1/(x2 + 1)2 and R(f /g) = (0, 1]. 1.13 2.3 limx→1 f (x) = 2.5 2.11 a) 1, b) 2, c) 2, d) 1, e) limx→0− f (x + 2) = limx→2− f (x) = 1, f ) limx→−1− f (x2 ) = limx→1+ f (x) = 2 2.13 a) y 12 10 8 y 6 2 4 2 1 x x 0 1 2 3 4 1.15 [1, 2) ∪ [2, 3] = [1, 3], [1, 2] r {2} = [1, 2), [1, 2] r {1, 2} = (1, 2), [0, 3] r [1, 3] = [0, 1), R r (−∞, 0) = [0, ∞), [1, 2] ∩ (3/2, 3) = (3/2, 2], [1, 2] ∩ [2, 3] = [2, 2], and (a, b) ∪ (b, a) = ((a + b − |a − b|)/2, (a + b + |a − b|)/2) 1.17 a) D(f ) = R r {2}√ b) f (3) = 0, f (−3) = − 3/5 c) f (f (1)) = f (−1) = −2/3 1.19 The equation y 2 = x2 is equivalent to the equation y = ±x. The latter equation does not define y as a function of x on R because it assigns every value x 6= 0 two distinct values of y. 1.21 f (x) = 4x + 1 and g(x) = −4x + 9. 0 0.5 1 2 3 4 b) R(f ) = [2, 11] r {3}, c) 3 2.15 If we define f (x) := 1 for x ≥ x0 and f (x) := −1 for x < x0 , then limx→x0 f (x) does not exist, but limx→x0 f (x)2 = 1. Consequently, the answer in general is “no.” 2.17 a) Proceeding as in Exercise 2.16, we find that limh→0 (g(1 + h) − g(1))/h = limh→0 (3 + h) = 3. b) limh→0 (g(t + h) − g(t))/h = limh→0 (2t + h + 1) = 2t + 1 Chapter 3 3.7 Let 1 > ε > 0 and limn→∞ xn = 2. Then there exists an N ∈ R such that |xn − 2| < ε/5 2 Solutions for all n > N . This shows that |xn + 2| ≤ |xn − 2| + 4 ≤ ε/5 + 4 < 5 for all n > N , and therefore, |x2n − 4| = |xn + 2||xn − 2| < 5ε/5 = ε for all n > N . Hence limn→∞ x2n = 4 and, by implication, limx→2 x2 = 4. 3.9 Assume that x0 is an accumulation point of D and let ε > 0. Then there exists a sequence (xn )∞ n=1 in D r {x0 } such that limn→∞ xn = x0 and, by implication, there exists an n such that 0 < |xn − x0 | < ε. Setting x := xn , it follows that x ∈ D and 0 < |x − x0 | < ε. Conversely, if for every ε > 0 there exists an x ∈ D such that 0 < |x − x0 | < ε then, in particular, for each n ∈ N there exists an xn ∈ D such that 0 < |xn − x0 | < 1/n. Since the latter inequality clearly shows that limn→∞ xn = x0 , we may infer that x0 is an accumulation point of D. 3.11 a) For ε > 1, we define δ := ε/7. If 0 < |x − 1| < δ, then |x| ≤ |x − 1| + 1 < 2, and therefore, |x3 − 1| = |x − 1||x2 + x + 1| ≤ |x − 1|(|x|2 + |x| + 1) < 7δ = ε. Consequently, limx→1 x3 = 1. b) For ε > 0, we set δ := ε. If 0 < |x − 2| < δ, then |(x2 − 4)/(x − 2) − 4| = |(x + 2) − 4| = |x − 2| < δ = ε. Thus, limx→2 (x2 − 4)/(x − 2) = 4. c) For ε > 0, we set δ := ε. If√0 < |x − √ 2| < δ for some√x ∈ [−1, √ ∞). Then | x + 1 − 3| = |x − 2|/| x + 1 + √ 3| < |x − √2| < δ = ε. Consequently, limx→2 x + 1 = 3. and similarly, √ √ n x − n x0 1 d √ n x = lim = √ n−1 . n x→x dx x − x0 0 n x0 x=x0 4.9 The line described by the equation y(x) = f 0 (x0 )(x − x0 ) + f (x0 ) has slope f 0 (x0 ) and passes through the point (x0 , f (x0 )) because y(x0 ) = f 0 (x0 )(x0 − x0 ) + f (x0 ) = 0 + f (x0 ) = f (x0 ). Thus, it must be the tangent line to the graph of f at (x0 , f (x0 )). Furthermore, for the function f (x) = x2 we find that (x + ∆x)2 − x2 = 2x, ∆x→0 ∆x f 0 (x) = lim and therefore, the tangent line to the graph of f at the point (2, 4) is described by the equation y(x) = 4(x − 2) + 4 = 4x − 4. 4.13 f ’(x) x 4.15 a) v̄[0,4] = 0 and v̄[1,4] = 4/3, b) y = 4(x − 1)/3 − 1 = 4x/3 − 7/3, c) y(x) = −1. d) It is positive for t ∈ (1, 2)∪(2, 4] and negative for t ∈ [0, 1). 4.17 tangent line: y(x) = 2(x − 2) + 3 = 4x − 1 Chapter 4 10 4.1 v(t) = 2ct 8 4.5 6 f(x) 4 (x + ∆x)3 − x3 f (x) = lim ∆x→0 ∆x 3x2 ∆x + 3x∆x2 + ∆x3 = lim ∆x→0 ∆x = lim (3x2 + 3x∆x + ∆x2 ) = 3x2 . 0 ∆x→0 4.7 √ √ 4 x − 4 x0 d √ 4 x = lim x→x0 dx x − x0 x=x0 1 = lim √ 3 √ 2 √ √ 4 2 √ x→x0 4 x + 4 x 4 x + 4 x √ x0 + 4 x0 3 0 1 = √ 3, 4 4 x0 2 x 0 -2 1 2 3 4.19 f (2.1) ≈ −0.85 and f (1.9) ≈ −1.15. √ 3 x x2 + 15 6.9 3.9708 4.21 Hence f 0 (64) ≈ 0.029. 7.0 4.0000 7.1 4.0291 4.23 a) TRUE because f (x) − f (2) = 0. x→2 x−2 f 0 (2) = lim b) POSSIBLY TRUE/POSSIBLY FALSE because the stated limit condition is satisfied for Solutions f (x) = 0 as well as for g(x) = 1, but only f assumes the value 0 at 2. c) TRUE because lim f (x) = f (2) + lim (f (x) − f (2)) x→2 x→2 f (x) − f (2) lim (x − 2) (x − 2) x→2 = f (2) + 0 · 0 = f (2). = f (2) + lim x→2 3 √ 6.5 tan(π/6)√ = cot(π/3) = 1/ 3, tan(π/3) = cot(π/6) = 3, √ tan(π/4) = cot(π/4) = 1, and cot(5π/6) = − 3. 6.7 The given equation cot(2α) = 1/2 implies that (cot2 (α) − 1)/(2 cot(α)) = 1/2, or equivalently, cot2 (α) − cot(α) − 1 =√0. Solving for cot(α) yields cot(α) = √ (1 + 5)/2. Consequently, tan(α) = 2/(1 + 5), and the√equation of the line is therefore y = −2x/(1 + 5) + 1. 6.9 a) y = −(x − 1)/3 + 2, b) y = 2x + 5, c) y = 3(x − 2)/2 − 1 4.25 f ’(x) g’(x) h’(x) 4.27 the instantaneous rate of change in the person’s weight 6.11 If the light ray were traveling downward as in Figure 6.2, it would be reflected in the direction of the line that connects the point of reflection with the origin. By symmetry, the reflected ray in Figure 6.8 must travel along the same line but in the opposite direction—not toward the origin but away from it. Consequently, the equation describing the path of the reflected ray is y = 2((3/2)2 − 1/4)x/3 = 4x/3. Chapter 5 Chapter 7 5.1 To construct y/x we mark two points D and C on the line L in Figure 5.1 such that DP = y and CP = 1. Furthermore, we mark a point B on the x-axis at distance x from P and draw a line segment from B to D. The point A is now obtained as the intersection of the x-axis with the line through C that runs parallel to BD. Since the triangles 4ACP and 4BDP are similar, it follows that x/y = BP /DP = AP /CP = AP . 7.9 a) yes, f (x0 ) = 0, b) limx→x0 f (x) = limx→x0 (x + x0 ) = 2x0 , c) f is continuous at x0 if and only if x0 = 0, because limx→x0 f (x) = f (x0 ) if and only if x0 = 0. d) Since f (x) = x + x0 for all x 6= x0 , it follows that f is differentiable at x0 if and only if x0 = 0. x x x 5.3 If x0 is the point where f assumes its minimum, then the difference quotient 4x0 e + 2e2 f (x0 + e) − f (x0 ) = = 4x0 + 2e e e must be zero for e = 0. Thus 4x0 = 0, and by implication, the minimal value of f is f (x0 ) = f (0) = −2. Chapter 6 6.1 tan(α) = a/b = (a/c)/(b/c) = sin(α)/ cos(α) and cot(α) = b/a = cos(α)/ sin(α). 6.3 Since the hypotenuse of a √ right triangle with two equal sides of length 1√is 2, √ it follows that sin(π/4) = cos(π/4) = 1/ 2 = 2/2. 7.11 a) TRUE because the assumption of continuity at x0 implies that x0 is contained in the domain of f . b) TRUE because f is continuous at x0 . c) Again TRUE because f is continuous at x0 . d) POSSIBLY TRUE/POSSIBLY FALSE. The function f : R → R, f (x) := |x − x0 | is continuous but not differentiable at x0 . 7.13 g and h are continuous but f is not. 7.15 a) The statement is in general not true because it is for example not satisfied if f (x) := −1 for x ∈ [1, 2) and f (2) := 1. b) This statement is true because a continuous function cannot “jump” from negative to positive values and will pass through the x-axis somewhere in the interval from 1 to 2. c) The statement is in general not true because it fails, for example, for the function f (x) := |x − 3/2|. d) This statement is true because if f (1) and 4 Solutions f (2) had different signs, then the assumed continuity of f would imply that there exists a point z ∈ (1, 2) such that f (z) = 0. v’(t) v(t) v’(0)=0.28 7.17 Yes because if f were differentiable, then it would also be continuous (by Theorem 7.7). 7.19 t t y T=354.17 T=354.17 Figure S.1: the graphs of v 0 (t) and v(t). x 0 1 2 3 4 5 6 10.9 [0, 1] and [−1/2, 1/2] Chapter 8 8.5 Since q 0 (t) = 10.7 (f ◦ g)(4) = f (g(4)) = g(4)3 + g(4) + 1 = √ 3 √ 4 + 4 + 1 = 11 d (m(t)v(t)) = m 0 (t)v(t) + m(t)v 0 (t), dt equation (8.11) implies that m(t)v 0 (t) = F (t) − m 0 (t)v(t) + m 0 (t)(v(t) − u) = F (t) − m 0 (t)u. 8.7 a) f (x)g(x) b) (f (x + ∆x) − f (x))g(x) c) (f (x + ∆x) − f (x))(g(x + ∆x) − g(x)) d) f (x)(g(x + ∆x) − g(x)) e) f (x + ∆x)g(x) f ) (f (x + ∆x) − f (x))g(x + ∆x) 8.9 a) T = 85/0.24 s, and therefore, m(t) = −85t/(85/0.24) + 350 + 85 = −0.24t + 435 kg. b) m(t) 435 350 t T=354.17 c) m 0 (t) = −0.24 kg/s d) v 0 (t) = 85 · 50/(435 · 85/0.24 − 85t) = 50/(1812.5 − t) e) See Figure S.1. Chapter 10 10.3 (f 2 ) 0 = f 0 f + f f 0 = 2f f 0 and (f gh) 0 = (f g) 0 h+(f g)h 0 = (f 0 g +f g 0 )h+f gh 0 = f 0 gh+ f g 0 h + f gh 0 10.11 If f and g are continuous, then limx→x0 f (g(x)) = f (limx→x0 g(x)) = f (g(x0 )) for all x0 ∈ E (that are not isolated points of E). Thus, f ◦ g is continuous on E. 10.15 a) 4x3 − 6x2 , b) (3 − 3x2 )x2 + (3x − x3 + 2)2x, c) 600(3x3 − 4)599 9x2 , d) 3000(3x − 4)999 , e) 100(x2 +(x2 +1)100 )99 (2x+100(x2 +1)99 2x), f ) 20x(x2 + 1)9 (x − 2)10 + 10(x2 + 1)10 (x − 2)9 , g) ((1 − 4x)(3 − x) + (x − 2x2 + 7))/(3 − x)2 , h) (4x(x + 3) − (2x2 − 5))/(x + 3)2 , i) −12/x5 , j) (5x4 (x7 − 2) − 7x11 )/(x7 − 2)2 , −3 x−3 4 − x2 + 2x(x − 3) k) −2 , 4 − x2 (4 − x2 )2 l) 3x2 − 10x − 3/x2 10.17 a) y = 5(x − 1) + 2, b) y = −11(x − 2)/3 − 3, c) y = 0, d) y = 5(x − 4)/8 + 4 10.19 a) (f + g) 0 (2) = 1, (f g) 0 (2) = 7, (f /g) 0 (2) = 1, b) y = 7(x − 2) − 1 10.21 Since f (x)g(x) = x for all x ∈ R, it follows that f (0)g(0) = 0 and f 0 (0)g(0) + f (0)g 0 (0) = 1. While the former of these equations implies that either f (0) = 0 or g(0) = 0, the latter allows us to infer that either f (0) 6= 0 or g(0) 6= 0. Consequently, exactly one of the values f (0) and g(0) must be equal to zero and exactly one of them different from zero. 10.23 f20 = (f 0 ◦f1 )f 0 , f30 = (f 0 ◦f2 )(f 0 ◦f1 )f 0 , f40 = (f 0 ◦ f3 )(f 0 ◦ f2 )(f 0 ◦ f1 )f 0 , and in general fn0 = (f 0 ◦ fn−1 ) · · · (f 0 ◦ f1 )f 0 10.25 Since d/dx y 3 = 3y 2 y 0 , we may infer that the rate of change in y 3 is twelve times the rate Solutions 5 of change in y for y = 2 (because then 3 · y 2 = 12). Chapter 11 11.3 f assumes its global maximum at a and its global minimum at x1 . Furthermore, the set of all points at which f assumes a local maximum is {a, x4 } ∪ [x2 , x3 ), and similarly, f assumes a local minimum at all points in the set {x1 , b} ∪ (x2 , x3 ]. 2 1 -4 -3 -2 -1 x 0 1 2 3 Figure S.2: graph for Exercise 11.29a. 11.5 In solving for x the equation f 0 (x) = 3x2 − 6x − 9 = 0 we find the critical points of f to be at x = 3 and x = −1. 10 11.9 √ roots at x = 1, 3, 4, local maximum√ at x = (8 − 7)/3, local minimum at x = (8 + 7)/3 4 10 f(x) 7 6 5 4 3 f(x) 8 6 2 x f(x) -2 -1 0 1 2 3 4 Figure S.3: graph for Exercise 11.29b. 5 x 0 1 2 3 4 5 -5 -10 -15 11.11 d3 /dx3 (x + 1)4 = 24(x + 1), f 00 (x) = 20x3 + 30x4 , and f (4) (x) = 120x + 360x2 . 11.15 roots √ at x = 0, 1, 2, local maximum√at x = 1 − 1/ 3, local minimum at x = 1 + 1/ 3, point of inflection at x = 1 2 2x3 + x − 1 = 11.21 lim x→−∞ 3x3 − 4 3 x+1 11.23 lim+ 3 =∞ x→1 x − 1 11.25 a) concave up on (7/8, ∞), concave down on (−∞, 7/8), point p of inflectionpat x = 7/8 b) concave up on (− p 3/2, 0)∪( p 3/2, ∞), concave down on (−∞, − 3/2) ∪ (0,p 3/2), points of inflection at x = 0 and x = ± 3/2√ c) concave √ up on (−∞, −2/5 − 34/5) ∪ 34/5,√ ∞), concave down on (−2/5 − (−2/5 + √ 34/5, −2/5√+ 34/5), points of inflection at x = −2/5 ± 34/5 11.27 f is increasing on (500, ∞) and decreasing on (−∞, 500). 11.29 a) increasing on (−2, ∞), decreasing on (−∞, 2), local minima at all points in [−2, 1]; see Figure S.2 for the graph. b) increasing on (0, ∞), decreasing on (−∞, 0), local minimum at x = 0; see Figure S.3 for the graph. 11.31 a) maximum at x = −1, b) points of inflection at x = 0 and x = 1, c) f(x) x -3 -2 -1 1 2 3 11.33 a) [2, 4] ∪ [6, 8], b) [0, 2] ∪ [4, 6] ∪ [8, 10], c) [1, 3] ∪ [5, 7] ∪ [9, 10], d) [0, 1] ∪ [3, 4] ∪ [7, 9], e) x = 0, 2, 4, 6, 8, 10, f ) x = 1, 3, 5, 7, 9, g) global maximum at x = 0, global mimimum at x = 1 √ 11.35 a) √ increasing on ((14 − 3 3)/13, 1) ∪ 3)/13), decreasing on (−∞, −1) ∪ (1, (14 + 3 √ (−1, (14 − 3 3)/13), concave up on (−1, 1) ∪ (2, ∞), concave down on (−∞, −1) ∪ (1, 2), asymptotes: x = 1, x = −1, and y = 0; see Figure S.4 for the graph. b) decreasing on (−∞, −1) ∪ √ (1, ∞), increasing √ on (−1, 1), concave up on√(− 3, 0)√∪ ( 3, ∞), concave down on (−∞, − 3) ∪ (0, 3), asymptote: y = 0; see Figure S.5 for the graph. c) increasing on (0, 2) ∪ (2, ∞) decreasing on 6 Solutions the second derivative test implies that L assumes a minimum at is critical point. f(x) 12.7 length of the spiral: L ≈ 28055895 mm; storage capacity: 22444716000 bytes 5 12.9 infrared laser: L ≈ 5611179 mm; blue light laser: L ≈ 28055895 mm x -2 2 12.11 The area in dependence on x is A(x) = x(1√− x3 ), and the critical point of A is x0 = the maximal area is 1/ 3 4. Consequently, √ A(x0 ) = 3/(4 3 4). 4 -5 Figure S.4: graph for Exercise 11.35a. 12.13 dmin = 2 and P = (6/5, 8/5). √ 12.15 dmin = 50/7 and P = (−1, −1/7). 12.17 The distance from an arbitrary point (x, x2 + 2) on the parabola to the line is f(x) 0.5 d(x) = x -4 -2 2 4 -0.5 Figure S.5: graph for Exercise 11.35b. (−∞, 2) ∪ (−2, 0); concave up on (−2, 2), concave down on (−∞, −2) ∪ (2, ∞), asymptotes: x = 2, x = −2, and y = −1. f(x) x -2 Since d obviously assumes its minimal value at x0 = 1, it follows that the point on the parabola closest to the line is (1, 3). 12.19 Denoting by h and r the height and radius of the cylindrical can respectively, it follows that V = 355 = πr2 h. Consequently, the surface area is given by the equation A = 2πr2 + 2πrh = 2πr2 + 710 . r Setting the derivative of dA/dr equal to zero and solving for r yields r = p 3 355/(2π) ≈ 3.837 cm and, by implication, p h = 2 3 355/(2π) ≈ 7.674 cm. 5 -4 |x2 − 2x + 2| (x − 1)2 + 1 √ √ = . 5 5 2 4 -5 11.37 The largest slope is assumed at a critical point of y 0 . Solving therefore for x the equation y 00 = −6x + 6 = 0 yields x = 1, and since y 000 = −6 < 0, we may conclude that y 0 assumes its largest value at x = 1. 12.21 If we denote by x the distance along opposite shore from the point nearest to man to the point where he intends to leave water, then the total time it takes to get to car is 1 2p 2 x + 0.22 + (0.5 − x). T (x) = 3 6 Setting the derivative T 0 (x)√equal to zero and solving for x yields√x = 0.2/ √15, and the minimal time is T (0.2/ 15) = (2 15 + 5)/60 h. Chapter 12 Chapter 13 12.5 Since 13.5 z1 ≈ −6.47, z2 ≈ 0.54, and z3 ≈ 1.43. L 00 (x) = √ a2 b2 +p > 0, 2 2 +x b + (d − x)2 a2 the the the the 13.7 x0 ≈ 1.49937 and dmin ≈ 2.52796. 13.9 There is only one root at z ≈ 1.37880. Solutions 13.11 a) f(x) b) 1 1 3 0 -1 x c) f(x) 1 x 3 0 -1 0 f(x) 3 x -1 13.13 For x0 = 1 Newton’s method breaks down after one step, because x1 = 0 = f 0 (0) = f 0 (x1 ) and x2 = x1 −f (x1 )/f 0 (x1 ) is undefined. For x0 = 2 and x0 = 10, the corresponding sequences of values xn do not converge but alternate erratically between positive and negative values. 13.15 (x0 , y0 ) ≈ (1.46557, 3.14790) 7 14.17 a) It is difficult to tell whether a left or right sum gives a lower or upper estimate because f is increasing on [0, 2] but decreasing on [2, 4]. b) We first determine an upper estimate for the area under the graph over the interval [0, 2] with a margin of error less than or equal to 0.05. Given the error estimate E ≤ |f (0) − f (2)| we choose 8/0.05 = 160 as our value for n so that ∆x = 2/160 = 1/80. The corresponding upper estimate for the area under the graph is represented by the following right sum: 160 X k 1 k 4− ≈ 5.358. 80 80 80 Chapter 14 P7 k k=−2 1/(k +3) = 49/20, k=2 2 = 253, P P 5 4 k k i=2 2 = 6·2 , n=1 2n = 30, n=−1 3 = 18. 14.3 P7 P3 14.5 n = 39.2/0.002 = 19600 14.7 f(x) upper estimate left sum f(x) lower estimate right sum x x a b 8 2 = ≤ 0.05, n n a b 14.9 Solving for n the inequality k=1 Now we can either repeat this process for the area under the graph over the interval [2, 4] or, in observing the graph to be symmetric with respect to the line x = 2, we can multiply the above result by 2 to infer that the total area under the graph is less than 10.72 with a margin of error less than 2 · 0.05 = 0.1. 14.19 Using the √ formula in Exercise 14.18 with f (x) = 1 − x2 , n = 100, and xk = −1 + p k∆x = −1 + k/50, we may infer that P100 ∆x2 + (f (xk ) − f (xk−1 ))2 ≈ 3.14 repk=1 resents an estimate for the length of the upper half of the circle described by the equation x2 + y 2 = 1. Consequently, an estimate for the entire circumference of the circle is C ≈ 2 · 3.14 = 6.28. E ≤ |f (1) − f (4)|3/n = 13.5/n ≤ 0.002 yields n ≥ 13.5/0.002 = 6750. 14.11 If s 0 (t) = v(t) = 5t2 , then s(t) = 5t3 /3 + C and therefore s(0) = C. Consequently, the total distance traveled is s(t) − s(0) = 5t3 /3. 14.13 The velocity of a segment of length 0.2 m at distance x from the center of rotation is approximately 2πx/2 = πx m/s, and the corresponding kinetic energy is 0.1(πx)2 /2 J. Consequently, the total kinetic energy is apP6 proximately equal to 0.1 k=1 (0.2 · πk)2 /2 = 91π 2 /500 J. 14.15 An P6upper estimate for the moment of inertia is k=1 0.1(0.2k)2 = 91/250. Chapter 15 15.1 8 Solutions Chapter 16 Pn−1 16.1 V ≈ k=0 π(R2 − x2k )(R + d)/n 16.3 With R = 0.02 m and d = 0.016 m we obtain: m = 1000π0.024 · 0.0362 /3 ≈ 0.0326 kg. P10 16.5 V ≈ k=1 8π(4−(−4+8k/10)2 /16)2 /10 ≈ 339.12 P20 16.7 V ≈ k=1 πk/16 = 13.125π 16.9 a) A(x) = 4x, P5 b) k=0 4k/9 = 20/3, P6 c) k=1 4k/9 = 28/3, P11 d) left sum: k=0 4k/36 = 22/3; P12 right sum: k=1 4k/36 = 26/3 16.11 Using aPright sum, we find the approxn 2 imation V ≈ k=1 16(4 − xk )/n. Setting for example n = 100 and xk = −2 + k/25 yields the numerical result V ≈ 42.66. Chapter 17 P4 2 17.3 left sum: k=1 (5xk − 5xk )/2 with xk = 1 + (k − 1)/2 P4 right sum: k=1 (5xk −5x2k )/2 with xk = 1+k/2 P4 2 midpoint sum: k=1 (5xk − 5xk )/2 with xk = 1 + (2k − 1)/4. 17.11 Proof of c): if a ≤ c ≤ b, then the equation in c) follows directly from the addition of areas between the graph of f and the x-axis over the intervals [a, c] and [c, b]. Given this observation, the equation can also be established for any other ordering of the boundaries of integration according to their magnitude. For instance, if b ≤ a ≤ c, then Z c Z b f (x) dx = f (x) dx − c b Z a Z c = f (x) dx + f (x) dx b =− a Z b f (x) dx + Z a or equivalently, Z b Z f (x) dx = a c f (x) dx + a c f (x) dx, a Z b f (x) dx c as desired. All other possible orderings can be handled in a similar fashion. Proof of d):R this follows R a from c) because Ra a f (x) dx = f (x) dx+ f (x) dx, and therea a a Ra fore, a f (x) dx = 0. 17.13 A rough estimate of the area under the graph shows that the correct value must have been 58.7. 17.15 a) 18 + 5 = 23, b) not enough information, c) 5 − 11 = −6, d) 5 + 11 = 16, e) not enough information, f ) 18 + 5 − 11 = 12 R 10 R 10 R 10 17.17 −5 g(u) du = −5 f (u) du + −5 2 du = 4 + 30 = 34. R0 R4 17.19 a) 0 f (x) dx = 1 + 7 = 8, 1 f (x) dx = R2 −2, and 1 f (x) dx = 1 − 2 = −1. b) If f (x) ≥ 0 for all x ∈ [1, 2], then −1 = R2 f (x) dx ≥ 0 which is obviously impossible. 1 c) If f (x) < 3.5 for all x ∈ [0, 4], then 7 = R4 R4 f (x) dx < 2 3.5 dx = 7 which again is false. 2 17.21 a) POSSIBLY TRUE/POSSIBLY FALSE: the statement is true for f (x) = 1 and g(x) = 2, and it is false for f (x) = arctan(x) and g(x) = 2. b) POSSIBLY TRUE/POSSIBLY FALSE: the statement is true for f (x) = 1 and g(x) = 2, and it is false for f (x) = − arctan(x) and g(x) = 2. c) TRUE: if f (x) ≤ g(x), then 0 ≤ g(x) − R1 f (x), and therefore, 0 ≤ 0 (g(x) − f (x)) dx = R1 R1 R1 g(x) dx − 0 f (x) dx. Hence 0 f (x) dx ≤ 0 R1 g(x) dx. 0 17.23 The distance is zero. Chapter 19 3 R3 19.7 14.12: A = 0 x3 dx = x4 /40 = 81/4, 5 R5 16.6: V = π 0 x dx = πx2 /20 = 25π/2, 1 R1 16.8: V = π 0 x6 dx = πx7 /70 = π/7, 2 R2 16.9: V = 0 4x dx = 2x2 0 = 8, 3 R3 16.10: V = π 0 x2 dx = πx3 /30 = 9π, R2 16.11: V = −2 4(4 − x2 ) dx 2 = 4(4x − x3 /3)−2 = 128/3. R2 2 19.9 −2 f 0 (x) dx = f (x)|−2 = 23 − ((−2)2 − 2 + 1) = 5 3 R3 19.11 a) 1 f1 (x) dx/2 = x2 /41 = 2, 3 R3 f (x) dx/2 = x3 /61 = 13/2, 1 2 3 R3 f (x) dx/2 = x4 /81 = 63/8, 1 R3 3 b) 1 fn (x) dx/2 = (3n+1 − 1)/(2n + 2) √ 19.13 a) Since the graph of 16 − x2 is symmetric withR respect to the y-axis, R 0 √ we may x√ 2 infer that 0 16 − t dt = −x 16 − t2 dt for all x ∈ [−4, 4]. Thus −f (−x) = Solutions R0 √ R −x √ 16 − t2 dt = −x 16 − t2 dt = f (x) as − 0 desired. b) f (0) = 0 p √ c) f 0 (±2) = 16 − (±2)2 = 12 19.15 To determine the boundaries of integration, we solve for x the equation 2x/a2 − x2 /a3 = 0. This yields x = 0 and x = 2a. Thus, the areaR enclosed by the parabola 2a and the x-axis is 0 (2x/a2 − x2 /a3 ) dx = 2a (x2 /a2 − x3 /(3a3 )) = 4/3. 0 19.17 a) x = 0, 4, 8, b) x = 2, 6, 10, c) x = 0, d) x = 2, e) x = 1, 3, 5, 7, 9 (Note: all answers are deduced from the observation that g 0 = f .) 19.19 f 0 (1) = 0 because f is constant. 19.21 We give a proof by contradiction: if f 0 (x) 0 all x ∈ were greater than or equal R xto g0 (x)+1R for x R, then f (x) − f (0) = 0 f (x) ≥ 0 (g 0 (x) + 1) dx = g(x) − g(0) + x, or equivalently f (x) ≥ g(x) − g(0) + f (0) + x for all x ∈ R. Thus, in contradiction to the assumption f ≤ g, it would follow that f (x) > g(x) for all x > g(0) − f (0). 19.23 Since f 00 (x) ≤ 0, it follows that f 0 is decreasing, and therefore, Z y Z x+y f 0 (t) dt ≤ f 0 (t) dt f (x + y) − f (x) = x 0 = f (y) − f (0) = f (y). Thus, f (x + y) ≤ f (x) + f (y) as desired. Chapter 20 20.1 sequence of cubes: 1, 8, 27, 64, 125, 216, . . . first differences: 7, 19, 37, 61, 91, . . . second differences: 12, 18, 24, 30, . . . third differences: 6, 6, 6, . . . fourth differences: 0, 0, . . . Furthermore, if a0 , a1 , . . . , an is a finite sequence such that a0 = 0, then the sum of the first differences is (a1 −a0 )+(a2 −a1 )+· · ·+(an −an−1 ) = (a1 +· · · +an )−(a0 +· · · +an−1 ) = an −a0 = an as desired. Chapter 21 21.5 The result follows by observing that 1 a 1 d 1 ln |ax + b| = · = . dx a a ax + b ax + b 21.7 a) D = (−∞, 2) ∪ (3, ∞), b) D = [−3, ∞), c) D = R r {4}, d) D = (2, ∞) 9 21.9 a) 3 ln |x| + C, b) ln(1 + x2 ) + C, c) −4 ln |1 − 5x|/5 + C, d) x + 4 ln |x − 3| + C, e) x/2 − 3 ln |2x + 3|/4 + C, f ) ln |x3 − 3| + C 21.11 y = x − 1 21.13 Since d/dx x ln(x) = ln(x) + 1, it follows that the slope of the normal line is −1/(ln(2) + 1). Consequently, the equation of the normal line is y = −(x − 2)/(ln(2) + 1) + 2 ln(2). 21.15 Since ln(x) is not defined for x < 0, the equation ln(x2 ) = 2 ln(x) cannot be valid for all x ∈ Rr{0}. The correct statement is “ln(x2 ) = 2 ln |x| for all x ∈ R r {0}.” Chapter 22 22.3 Since the number of elements in the domain is larger than the number of elements in the set of values, it follows that f assigns the same output value to at least two of the elements in the domain. Thus f is not invertible. 22.11 ea /ab = eln(e a /eb ) = eln(e a )−ln(eb ) = ea−b 22.13 a) max. domain: D(f ) = R, domain of f −1 : D(f −1 ) = R, b) D(f ) = R, D(f −1 ) = R, c) D(f ) = R, not invertible, d) D(f ) = R r {0}, D(f −1 ) = R r {1}, e) D(f ) = [5/3, ∞), D(f −1 ) = [0, ∞), f ) D(f ) = R r {2}, D(f −1 ) = R r {1} 22.15 a) 2e2x , b) 3e3x−2 , c) 3e3x , d) 1, e) −1, 2 f ) 10xe5x , g) ex + xex , h) 2xe4x + 4x2 e4x , i) −4/(ex − e−x )2 , j) 2e2x /(e2x + 1)2 22.17 Since P 0 (t)/(2000 − P (t)) = 6/5, it follows that Z P 0 (t) dt − ln |2000 − P (t)| = 2000 − P (t) Z 6 6t = dt = + C. 5 5 Using the initial value P (0) = 100, we find that C = − ln(1900), and therefore 2000 − P (t) = ±1900e−6t/5 . The sign on the right-hand side must be positive, because otherwise, for t = 0, the values on the two sides do not match. Hence P (t) = 2000 − 1900e−6t/5 . 10 Solutions 22.19 a) b) y f(x) 1.0 y 8 6 12 4 8 0.5 x 0 1 x -2 c) -1 0 1 2 Figure S.6: graph for Exercise 22.23. d) y y 10 20 8 6 10 4 2 -2 -1 x 0 1 2 x -3 e) -2 -1 0 1 f) y y 8 12 6 8 4 4 2 -1 0 1 x -2 g) -1 0 1 2 h) y -1 y x 1 4 2 -2 3 -4 2 -6 1 -8 -2 0 -1 −bt 22.25 First we apply the exponential function to both sides of the given inequality to ob√ tain the equivalent inequality (x + y)/2 ≥ xy. (Note: this is permissible because the exponential function is increasing.) Since x and y are positive, the latter inequality in turn can be written in the form (x − y)2 ≥ 0, and the validity of the original equation has thus been established. Chapter 23 x -2 1 4 2 -1 x 0.0 23.3 proof of b): ax /ay = ex ln(a) /ey ln(a) = e(x−y) ln(a) = ax−y , proof of d): a0 = e0 ln(a) = e0 = 1, proof of f): (a/b)x = ex ln(a/b) = ex(ln(a)−ln(b)) = ex ln(a) /ex ln(b) = ax /ay 23.7 d/dx 3x−1 = 3x−1 ln(3), d/dx (1/2)3x = x x 3(1/2)3x ln(1/2), and d/dx 22 = 22 2x ln2 (2). x 0 1 2 −at 22.21 Since C (t) = K(−be +ae )/(a−b), it is easy to see that C has a critical point at t0 = ln(a/b)/(a − b). Furthermore, K b2 e−bt0 − a2 e−at0 a−b a2 b2 e−bt0 K 1 − 2 e−(a−b)t0 = a−b b 2 −bt0 b e a K = 1− a−b b < 0 (because a > b > 0). 23.9 a) y = −2x , b) y = 2−x , c) y = 10 − 2x , d) y = 2−(x−4) , e) y = −2−x , f ) y = −2−x . Graph of f : f(x) 8 6 4 C 00 (t0 ) = According to the second derivative test, we may thus conclude that C assumes its maximal value at t0 . 22.23 local minimum at x = e−1/2 , no local maximum, point of inflection at x = e−3/2 (see Figure S.6 for the graph of f ). 2 x -2 -1 0 1 2 3 23.11 a) Denoting by x the ticket price in dollars and by N (x) = N0 ax the number of spectators in dependence on the ticket price, the given information implies that N0 a25 = N (25) = 200 and N0 ax+5 = N (x + 5) = 9N (x)/10 = 9N0 ax /10. Solving for a and N0 yields a = (9/10)1/5 and N0 = 200(9/10)−5 . Thus N (x) = 200(9/10)(x−25)/5 . b) The revenue R(x) is given by the equation R(x) = xN (x) = 200x(9/10)(x−25)/5 . Solutions c) Since R 0 (x) = (1 + x ln(9/10)/5)N (x), it is easy to see that R assumes its maximal value at x = 5/ ln(10/9) ≈ $47.46. 23.13 a) smallest concentration: C(0) = 5000; largest concentration: C(4/ ln(2)) = 500(10 + 4 · 2−1/ ln(2) / ln(2)) = 5000 + 2000/(e ln(2)); b) t = 8/ ln(2) 23.15 It will take about 4.32 months, or approximately four months and 10 days. 23.17 Since the assumptions on f and g imply that limx→∞ 2f (x) = limx→∞ 3g(x) = ∞, we may apply L’Hôpital’s rule to conclude that 2f (x) 2f (x) f 0 (x) ln(2) ln(2) . = lim g(x) 0 = g(x) x→∞ 3 x→∞ 3 ln(3) g (x) ln(3) lim 23.19 If the current year with an income of $40, 000 is counted as the first year, and if payments into the retirement fund are always made at the end of each year then, at the end of n years, the amount of money in the fund will be 4000 n−1 X 1.06k 1.03n−1−k = k=0 4(1.06n − 1.03n ) . 0.00003 Setting n equal to 40 yields an amount of $936, 490.67, and the value in current dollars will be $936, 490.67/1.02540 = $348, 777.81. Chapter 24 24.3 proof of b): loga (x/y) = ln(x/y)/ ln(a) = ln(x)/ ln(a) − ln(y)/ ln(a) = loga (x) − loga (y), proof of d): loga (1) = ln(1)/ ln(a) = 0 24.5 The diagrams below for a = 1/4, 1/2, 2, 4 illustrate that loga (x) is decreasing for 0 < a < 1 and increasing for 1 < a. In fact, for 0 < a < 1 we have loga (x) = − log1/a (x), and the corresponding graphs are therefore reflections of each other across the x-axis. 2 log a(x) 11 24.7 d/dx xln(x) = xln(x) 2 ln(x)/x, √ √ 2 2−x 2 2−x ln(x2 −3)+ d/dx √(x −3) 2 = (x −3) e (− e−1 2x( 2 − x)/(x − 3)), d/dx x = ex 24.9 a) 3/((3x − 1) ln(2)), b) x1/ log2 (5)−1 / log2 (5), 2 c) (log5 (3x2 ))x (2x ln(log5 (3x2 ))) 2 +x/(log5 (3x ) ln(5))), d) x1/ ln(4)−1 / ln(4), e) −2, f ) 2x log10 (x2 ) + 2x/ ln(10), g) x4x (4 ln(x) + 4), h) 2x(6x)4x−1 +x2 (6x)4x−1 (4 ln(6x) + (4x − 1)/x), i) exe−1 + ex , j) (πxπ−1 log10 (πx) − xπ /(x ln(10)))/ log210 (πx) 24.11 Since fa (x) = log2 (x) + log2 (a), the graphs of fa for different values of a are vertical shifts of each other: f a(x) 4 a=4 3 a=2 2 a=1 1 0 x 1 2 3 -1 -2 24.13 Using 100 subdivisions along the interval R2 [1, a], we obtain: 1 log2 (x) dx P100 ≈ k=1 log2 (1 + k/100)/100 ≈ 0.562, R3 log (x) dx 1P 3 100 ≈ k=1 2 log3 (1 + 2k/100)/100 ≈ 1.190, R4 log (x) dx 1P 4 100 ≈ k=1 3 log4 (1 + 3k/100)/100 ≈ 1.851, R5 log (x) dx 1P 5 100 ≈ R e k=1 4 log5 (1 + 4k/100)/100 ≈ 2.535, ln(x) dx 1P 100 ≈ k=1 (e−1) ln(1+(e−1)k/100)/100 ≈ 1.009. 24.15 Since f 0 (x) = 0 and f (e) = f (1/e) = e, it follows that f (x) = e for all x ∈ (0, ∞) r {1}. a=2 1 0 -1 a=4 x 1 2 3 a = 1/4 a = 1/2 -2 Chapter 25 25.1 no answer 25.3 The trapezoid is easily seen to be composed of a rectangle of height h and width b and two right triangles of height h. Since the lengths of the base sides of the two triangles add up to a − b, it follows that the total area of the trapezoid is hb + h(a − b)/2 = h(a + b)/2. 12 Solutions Chapter 26 e) y 1.5 26.1 Dividing both sides of the equation cos2 (t) + sin2 (t) = 1 by cos2 (t) yields 1 + tan2 (t) = 1/ cos2 (t). The second identity is derived in a very similar fashion via a division by sin2 (t). 26.3 d/dt sin(3t2 ) = 6t cos(3t2 ), d/dt ecos(t) = − sin(t)ecos(t) , and d/dt cos3 (4t − 1) = −12 sin(4t − 1) cos2 (4t − 1). 26.5 a) sin(x) + x cos(x), b) 2 cos(x) sin(x), c) 3e3x / cos2 (e3x ), d) 4e4x sin(2x)+ 2e4x cos(2x), e) xcos(x) (cos(x)/x − sin(x) ln(x)), f ) 2x cos(3x) − 3x2 sin(3x), g) 2x cos(x2 ), h) −2−x ln(2) cos(5x) − 2−x 5 sin(5x), i) 2 sin(e4x ) + 8xe4x cos(e4x ), j) (4 cos(4x) − 3 sin(3x))/(sin(4x) + cos(3x)) 26.7 a) y x 6 -4 -2 2 4 -0.5 b) y 4 2 x -6 -4 -2 2 4 6 2 4 6 -2 c) y 1 x -6 -4 -2 -1 d) 1 -4 -2 2 4 2 4 6 -1.5 f) 1 -6 -4 -2 y x 6 -1 -2 Figure S.7: graphs for Exercise 26.7. 26.13 If x is measured in degrees, then limx→0 sin(x)/x = π/180 and d/dx sin(x) = π cos(x)/180. Chapter 27 0.5 -6 x -6 y x 6 -6 -1 For e) and f) see Figure S.7. 26.9 dy/dx = tan(x) − cot(x) 26.11 In solving for cos(θ) the equation Cb4 dT Cb4 cos(θ) = 4 2 − 4 2 , dθ R sin (θ) r sin (θ) we find that cos(θ) = (r/R)4 as desired. 27.3 sin(arcsin(−0.1)) = −0.1, arcsin(sin(7π/6)) = −π/6, arccos(cos(−π/6)) = π/6, and arctan(tan(π)) = 0. 27.5 Using the quotient rule and the trigonometric theorem of Pythagoras, we obtain √ √ 3s2 ( 3 − 2 cos(θ) + 3 cos2 (θ)) d2 A . = dθ2 2 sin3 (θ) √ √ Since the quadratic equation 3 − 2x + 3x2 = 0 is easily seen to have no solutions in R, it follows that d2 A/dθ2 > 0 and, by implication, A assumes √ a minimum at its critcal point θ = arccos(1/ 3). 27.7 d/dx arctan(x) = 1/(1 + tan2 (arctan(x))) = 1/(1 + x2 ) and d/dx arccot(x) = 1/(−1 − cot2 (arccot(x)) = −1/(1 + x2 ). √ 2 27.9 a) arcsin(x) √ + x/ 1 − x3x, 2 b) 2 arcsin(x)/ 1 − x , c)√3e /(1 + e6x ), 2 d) 4e4x arcsin(2x) + 2e4x / 1 − 4x √, arccos(x) e) x (arccos(x)/x√− ln(x)/ 1 − x2 ), f ) 2x arccos(3x) − 3x2 / 1 − 9x2 , √ − 2−x 10x/ 1 − 25x4 , g) −2−x ln(2) arccos(5x2 )√ 4x 4x h) 2 arcsin(e ) + 8xe / 1 − e8x , i) 0 27.11 a) 2 arctan(x) + C, b) 2 arcsin(x) + C, c) 5 ln(1 + x2 )/2 + C, d) √ 3 arctan(2x)/2 + C, e) arcsin(2x) + C, f ) 2 1 − x2 + C 27.13 Only the expressions in b) and d) are well defined. Solutions 27.15 Setting f (x) := arctan(x) + arctan(1/x), we find that 1 1 1 − · 2 =0 f (x) = 2 2 1+x 1 + 1/x x 0 for all x 6= 0. Consequently, there are constants C, D ∈ R such that f (x) = C for all x ∈ (0, ∞) and f (x) = D for all x ∈ (−∞, 0). In evaluating f at, for example, −1 and 1, we obtain C = arctan(1) + arctan(1) = π/2 and D = arctan(−1) + arctan(−1) = −π/2 as desired. 27.17 Since cos(x) > sin(x) for all x ∈ (−3π/4, π/4), it follows that f 0 (x) = cos(x) − sin(x) > 0 for all x ∈ [−π/4, π/4) and, by implication, f is strictly increasing. Furthermore, following the outline given in the hint, −1 (x) = arcsin(x2 − 1)/2 and we find that f√ (f −1 ) 0 (x) = x/ 2x2 − x4 . Chapter 28 28.1 δ must be between 0◦ and 42.5◦ . 28.3 As we trace the path of a light ray from the tip of the pencil to the observer (see Figure S.8), we notice that the refraction at the surface effects a decrease in the angle of observation and thus causes the pencil to appear bent. observer 13 29.9 Setting L := limx→∞ xa /bx , it follows that ln(L) = lim (a ln(x) − x ln(b)) x→∞ a ln(x) − ln(b) = −∞, = lim x x→∞ x because b > 1 and ln(b) > 0. Consequently, L = 0. 29.11 The limits are 3/2 and −1/2. 29.13 The limit is 1. 29.15 The limit is −1/2. 29.17 For k = 2, we find that 1 − cos(sin(x)) sin(sin(x)) cos(x) = lim x→0 xk 2x sin(sin(x)) cos(sin(x)) cos(x) 1 = lim = . = lim x→0 x→0 2x 2 2 lim x→0 29.19 continuous compounding: $200e0.06·10 ≈ $364.42, quarterly compounding: $200(1 + 0.06/4)4·10 ≈ $362.80 29.21 According to L’Hôpital’s rule, we may infer that ef (x) f 0 (x)ef (x) = lim = 2.346. x→∞ eg(x) x→∞ g 0 (x)eg(x) lim Consequently, f 0 (x) f 0 (x)ef (x) eg(x) = lim x→∞ g 0 (x) x→∞ g 0 (x)eg(x) ef (x) 2.346 = 1. = 2.346 lim angle of observation pencil Chapter 30 R2 R8 2 30.3 0 xe(2x ) dx = 0 eu /4 du = (e8 − 1)/4, R 2 2 x3 R8 u 3x e dx = 0 e du = e8 − 1 0 R π/4 R1 30.5 0 tan(x)/ cos2 (x) dx = 0 u du = 1/2 Figure S.8: a pencil immersed in water. 28.5 The minimal height is 0.5 tan(42.5◦ ) ≈ 0.46 miles. Chapter 29 29.1 f increases far more rapidly for large values of x than g. 29.7 The limits are 1 and 1/4. 30.7 H(x) = ln3 (x)/3 + C 30.9 increase in velocity: 20000 ln(2) mi/h 30.13 a) π 2 /32, b) − ln(3)/4, c) sin4 (x)/4 + C, d) −1/(2 cos2 (t)) + C, e) −1/ ln(ln(x)) + C, f ) − ln2 (x) + C, g) ee −e, h) arcsin2 (x−1)/2+C, i) 2/3, j) 8/15, k) ln | sin(x)|+C, l) − cot2 (x)/2−ln | sin(x)|+C m) 2(x +√2)5/2 /5 − 4(x +√2)3/2 /3 + C,√ n) (4 − 2 2)/3, o) 2(1 + x − ln(1 + x)) + C R5 R2 30.15 1 f (3x − 1) dx = 2 f (u)/3 du = 10/3 14 R π/2 30.17 Since 4 = 0 cos(x)f (2 sin(x)) dx = R2 R2 f (u)/2 du, we may infer that 0 f (u) du = 8. 0 30.19 v 0 (0) > 0 if and only if −g − 50m 0 (0)/m(0) > 0 if and only if |m 0 (0)| > g/50 30.21 a) arctan(x) + C, b) ln(1 + x2 )/2 + C, c) x − arctan(x) + C 30.23 a) F (x3 )/3 + C, b) F (ln(x)) + C, c) F (sin(x3 ))/3 + C Chapter 31 31.3 cosh2 (x) − sinh2 (x) = (ex + e−x )2 /4 − (ex − e−x )2 /4 = (e2x + 2 + e−2x − e2x + 2 − e−2x )/4 = 1. 31.5 x = cosh(y) if and only if 2x = ey + e−y if and only √ if 0 = (ey )2 − 2xey + 1 if and only if y e = x ± x2 − 1. √ 31.7 RSince A = xy − B = x x2 − 1 − B, x√ 2 B = 1 u − 1 du, and √ t = arcosh(x) = ln(x + x2 − 1), we may infer that p x2 d (A − B − t) = x2 − 1 + √ dx x2 − 1 √ p 1 + x/ x2 − 1 2 √ −2 x −1− x + x2 − 1 p x2 − 1 = − x2 − 1 + √ = 0. x2 − 1 Having thus shown A − B − t to be constant as a function of x, we may determine the value of this constant by setting x equal to 1. This yields A − B − t = 0 − 0 − 0 = 0. Consequently, t = A − B as desired. 31.11 The result easily follows from the following observation: √ √ (1 − 2)2 1 √ √ √ = = (1 − 2)2 . 2 2 (1 + 2) ((1 + 2)(1 − 2)) 31.17 In generalizing the calculation in Exercise 31.16, we find that h(x)+xh 0 (x) = g(h(x)), or equivalently, h 0 (x)/(g(h(x))−h(x)) = 1/x as desired. 31.19 In analogy to our calculation in Exercise 31.18, we differentiate both sides of the equation f (x) = h(x) − ax − b to conclude that h 0 (x) − a = f 0 (x) = g(h(x)), or equivalently, h 0 (x)/(g(h(x)) + a) = 1 as desired. Solutions R0√ R1 √ 31.21 a) 0 x 1 − x2 dx = − 1 u/2 du = 1/3, R1 √ R π/2 b) 0 x2 1 − x2 dx = 0 sin2 (t) cos2 (t) dt R π/2 R π/2 = 0 sin2 (2t)/4 dt = 0 (1 − cos(4t))/8 dt = π/16, R0 R1 √ √ c) 0 x/ 1 − x2 dx = − 1 1/(2 u) du = 1, R1 2 √ R π/2 2 d) 0 x / 1 − x2 dx = 0 sin (t) dt R π/2 = 0 (1 − cos(2t))/2 dt = π/4 31.23 a) In √ applying (31.12) with c = 0, d = 2, f (x) = 1/ 4 + x2 , and g(t) = 2 sinh(t), we obtain: Z 2 Z arsinh(1) √ 1 √ dx = dt = ln(1 + 2). 4 + x2 0 arsinh(0) √ b) Here we use (31.13) with f (x) = 1/ 4 + x2 , g(t) = 2 sinh(t), g −1 (x) = arsinh(x/2), and H(x) = x to infer that Z x 1 √ dx = arsinh + C. 2 4 + x2 31.25 a) sinh(x) cosh(y) + cosh(x) sinh(y) = (ex+y +ex−y −e−x+y −e−(x+y) +ex+y −ex−y + e−x+y − e−(x+y) )/4 = (ex+y − e−(x+y) )/2 = sinh(x + y). b) The proof is completely analogous to that in a). c) Setting y equal to x in a) yields sinh(2x) = 2 sinh(x) cosh(x). d) The first equality is an immediate consequence of the result in b) with x = y and the second and third inequalities are easily deduced from the hyperbolic theorem of Pythagoras. R1 √ 31.27 Given that 10 = 0 f ( 4 − x2 ) dx = R π/6 2f (2 cos(t)) cos(t) dt, we may infer that R0π/6 f (2 cos(t)) cos(t) dt = 5. 0 Chapter 32 32.1 Since a2 = b2 + c2 = 13, the equation of the ellipse is x2 /13 + y 2 /4 = 1. 32.3 In adopting essentially the same notation as in Figure 32.2, we denote by... • (x0 , y0 ) the point at which the light ray is reflected off the graph of the hyperbola. • α the angle between the incoming light ray and the tangent line to the hyperbola at (x0 , y0 ), which, according to the general law of reflection, is the same as the angle Solutions between this tangent line and the reflected light ray. • β the angle between the positive x-axis and the tangent line at (x0 , y0 ). Note: this choice of β represents a slight deviation from our use of notation in the case of the elliptic mirror, because in Figure 32.2 β denoted the angle between the tangent line and the negative x-axis. • γ the angle between the negative x-axis and (the extension of) the reflected light ray. • δ the angle between the positive x-axis and the incoming light ray. Since the upper branch of the hyperbola is the graph of the function bp 2 x − a2 , f (x) = y = a we may infer that bx0 b2 x0 tan(β) = f 0 (x0 ) = p 2 = 2 . a y0 a x0 − a2 Furthermore, elementary geometry shows that α + δ = β and α + β + γ = π. Hence 15 Chapter 33 33.1 The value of c will be greater for an adult because the larger body of the adult has a larger surface area and thus experiences a greater force of air resistance at a given speed. R1 R1 33.3 −1 x/(x2 +4x+4) dx = −1 1/(x+2) dx− R1 2/(x + 2)2 dx = ln(3) − 4/3 −1 √ R1 R1√ 33.5 1/(x2 −2) dx = 0 2/(4(x− 2)) dx− √ √ √ R1√ 0 2/(4(x + 2)) dx = ln( 2 − 1)/ 2. 0 g vmax − v(t) 1 = 2 t ln 33.7 − 2vmax vmax + v(t) vmax 2g vmax − v(t) =− t if and only if ln vmax + v(t) vmax vmax − v(t) = e−2gt/vmax . if and only if vmax + v(t) √ 33.11 Using the substitution u := x with 2u du = dx in each √ case, we obtain the following results: a) 2√ ln | x − 1| − ln(x) + C, √ b) 2 arctan( x − 1) + C, c) 4 − 4 ln( 2 + 1). 33.13 a) Rearranging the terms in the given identities concerning the preservation of the momentum and the (kinetic) energy yields γ = π − 2β + δ. Since the reflected light ray is described by the equation y = − tan(γ)(x − x0 ) + y0 , we need to prove that y0 = tan(γ), c − x0 or equivalently, y0 = tan(π − 2β + δ) c − x0 tan(δ)(1 − tan2 (β)) − 2 tan(β) = . 1 − tan2 (β) + 2 tan(β) tan(δ) Using the fact that tan(δ) = y0 /(c + x0 ), the term on the right-hand side of the equation above is easily seen to be equal to b4 x2 2b2 x0 y0 1 − 4 02 − 2 c + x0 a y0 a y0 . b4 x20 2b2 x0 1− 4 2 + 2 a y0 a (c + x0 ) Finally, in performing almost exactly the same steps as in the calculation on p.248 (with the identity b2 = c2 − a2 in place of b2 = a2 − c2 ), we arrive at the desired conclusion that tan(γ) is indeed equal to y0 /(c − x0 ). M (v − v0 ) = m(u0 − u), M (v 2 − v02 ) = m(u20 − u2 ). Since v 2 − v02 = (v − v0 )(v + v0 ) and u20 − u2 = (u0 + u)(u0 − u), we may infer that M (v − v0 ) = m(u0 − u), v + v0 = u0 + u. In solving these equations for v and u, we find that 2mu0 (M − m)v0 + , M +m M +m 2M v0 (m − M )u0 + . u= m+M m+M v= b) If u0 = 0, then the change in momentum is M (v−v0 ) = M (M − m)v0 2mM v0 −M v0 = − . M +m m+M c) Since the number of collisions N (t) is certainly proportional to the distance traveled, and since the distance traveled is approximately equal to v(t)∆t, it follows that there is a constant of proportionality λ such that N (t) = λv(t)∆t. 16 Solutions d) Since the total change in momentum over the time span from t to t + ∆t is approximately equal to the number of collisions multiplied by the change in momentum effected by each of them, we may use the results of b) (with v(t) in place of v0 ) and c) to infer that illustration for Rn m−1 f (x) dx ≥ Pn k=m f (k): f(x) 2mM v(t)N (t) p(t + ∆t) − p(t) ≈− ∆t (m + M )∆t 2mM λ v(t)2 . =− m+M F (t) ≈ In defining c to be equal to 2mM λ/(m + M ) we arrive at the desired conclusion: F (t) ≈ −cv(t)2 . 33.15 a) v(t) = 2 tan(2t + arctan(2)) − 2 = (6 tan(2t) + 2)/(1 − 2 tan(2t)), 2 b) v(t) = 4/(2 − et ) − 3 Chapter 34 34.1 3/8 34.3 210 1 9 9 9 9 130 34.5 9 + + + = 0 1 2 3 2 512 20 X 32 1 ≈ 0.89 34.9 32 k 2 k=12 550 X 1 1000 34.11 1000 ≈ 0.9986 k 2 k=450 3 1 4 9 34.13 · 4 = 4 2 2 32 365! ≈ 0.507 34.15 1 − 36523 342! Chapter 35 35.3 R n+1 Pn illustration for k=m f (k) ≥ m f (x) dx: f(x) x m n+1 x m-1 n R 35.13 arcsin(x) dx√ R = x arcsin(x) − R x/ √ 1 − x2 dx = x arcsin(x) + √1/(2 u) du = x arcsin(x) + 1 − x2 + C Qn n 35.15 k=2 k k = e k=2 k ln(k) 2 ≤ e(n+1) (ln(n+1)−1/2)/2−ln(4)+1 2 2 = (n + 1)(n+1) /2 e−(n+1) /4 e− ln(4) e1 2 √ (n+1) −(n+1)2 /4 = (1/4) n + 1 e R1 35.17 a) 0 arctan(x) dx R1 1 = x arctan(x)|0 − 0 x/(1 + x2 ) dx = 1 (x arctan(x) − ln(1 + x2 )/2)0 =R π/4 − ln(2)/2, R b) x sinh(x) dx = x cosh(x) − cosh(x) dx = xRcosh(x) − sinh(x) + C, R c) x3 ln(x) dx = x4 ln(x)/4 − x3 /4 dx = x4 ln(x)/4 − x4 /16 + C, R2 R4 2 d) 0 x3 ex dx = 0 ueu /2 du R 4 4 = ueu /2|0 − 0 eu /2 du = (3e4 + 1)/2, R 2 R e) ln (x) dx = x ln2 (x) − 2 ln(x) dx = xR ln2√(x) − 2x ln(x) + 2x + C, f ) x x + 1 dx R √ √ 3 3 = 2x x + 1 /3 − 2 x + 1 /3 dx √ √ 3 5 = 2x x + 1 /3 − 4 x + 1 /15 + C Pn Qn 1/k = k=4 ln(k)/k 35.19 a) ln k=4 k Rn ≤ 3 ln(x)/x dx = ln2 (n)/2 − ln2 (3)/2, and Qn √ ln(n) √ ln(3) / 3 . Furtherefore, k=4 k 1/k ≤ n R n+1 Pn thermore, ln(x)/x dx = k=4 ln(k)/k ≥ 4 ln2 (n + 1)/2 − ln2 (4)/2 and, by implication, √ Qn ln(n+1) 1/k n + 1 P /2ln(4) . k=4 kQ ≥ n n k ln(k) = k=1 ln2 (k) b) ln R Pn k=1 n = k=2 ln2 (k) ≥ 1 ln2 (x) dx 2 (n) − 2n ln(n) + 2n − 2, and therefore, = Qnn ln ln(k) n ln(n) 2n 2 2n k ≥ e /(e k=1 n ). Qn n 2 c) Since ln k=1 (k + 1) R n+1 Pn 2 = k=1 ln(k + 1) ≤ 1 ln(x2 + 1) dx R n+1 n+1 − 2x2 /(1 + x2 ) dx = x ln(x2 + 1) P 1 1 Solutions = (n + 1) ln(n2 + 2n + 2) − ln(2) − 2n +2 arctan(n + 1) − 2 arctan(1) and arctan(n + 1) < π/2, we may infer that Qn 2 (k + 1) k=1 ≤ (n2 + 2n + 2)n+1 e2 arctan(n+1) /(2e2n+π/2 ) ≤ (n2 + 2n + 2)n+1 eπ/2 /(2e2n ) < 5(n2 + 2n + 2)n+1 /(2e2n ). R n+1 Pn d) ln(n + 1) = 1 1/x dx ≤ k=1 1/k R Pn n = 1 + k=2 1/k ≤ 1 + 1 1/x dx = 1 + ln(n). e) With ∆x := π/(2n) and xk := k∆x it follows that R Pn π/2 1 = 0 sin(x) dx k=1 sin(xk )∆x, and Pn≤ therefore, 2n/π ≤ k=1 sin(πk/(2n)). 35.21 a) Since f (a) = a and f (b) = b if and only if f −1 (a) = a and f −1 (b) = b, the reRb sult of Exercise 35.20 implies that a f (x) dx + R f −1 (b) Rb R b −1 f (x) dx = f −1 (a) f (t) dt + a f −1 (x) dx = a b xf −1 (x)a = b2 − a2 as desired. b) Here the given assumptions allow us to infer that f −1 (b) = a and f −1 (a) = b. Hence b R b −1 R f −1 (b) f (x) dx = xf −1 (x)a − f −1 (a) f (t) dt = a Rb Ra 0 − b f (x) dx = a f (x) dx. 35.23 The correct answer is c). 17 Rn ≥ n ln(n)/2 + ln(2) + 2 x ln(x) dx = n ln(n)/2 − ln(2) + n2 ln(n)/2 −n2 /4 + 1 or, equivalently, Qn √ n2 +n √ n2 −4 k 4 n / 2 e . k=1 k ≥ Pn 37.9 a) k=3 arctan(k + 1) ≤ Rarctan(n + 1)/2 + arctan(4)/2 n + 3 arctan(x + 1) dx = (n + 3/2) arctan(n + 1) − 7 arctan(4)/2 + ln((n2 + 2n + 2)/17)/2, R n+1/2 Pn arctan(x + 1) dx k=3 arctan(k + 1) ≥ 5/2 = (n + 3/2) arctan(n + 3/2) − 7 arctan(7/2)/2 + ln((4n2 + 12n + 13)/53)/2, R n+1/2 Pn b) k=5 k/(k 2 + 1) ≤ 9/2 x/(x2 + 1) dx = ln((4n2 + 4n + 5)/85)/2 P n k/(k 2 + 1) ≥ 5/52 + n/(2(n2 + 1)) Rk=5 n + 5 x/(x2 + 1) dx = 5/52 + n/(2(n2 + 1)) + ln((n2 + 1)/26)/2 Qn ln(k) 37.11 a) Since ln k=4 k Pn = R k=4 ln2 (k) ≤ ln2 (n)/2 + ln2 (4)/2 n + 4 ln2 (x) dx = (n + 1/2) ln2 (n) −2n ln(n) + 2n − 7 ln(2) ln(4) + 8 ln(4) − 8, we may infer that n Y k ln(k) ≤ k=4 Chapter 36 1 34 6 1 33 6 1 32 6 36.3 2 · 4 + 3· 3 + 4· 2 4 4 2 4 4 3 4 4 4 ≈ 0.46 Chapter 37 37.5 Trap(n1 , f, −2, 0) = P20 −1/20 + k=0 ((−2 + k/10)3 /8 + 1)/10 ≈ 1.499, Mid(n2 , f, 0, 3) P30 = k=1 ((−1 + (2k − 1)/20)3 /8 + 1)/10 ≈ 5.530 37.7 Since the function f (x) = x ln(x) is easily seenQ to be concave may Qnup, kwe Pn infer that n k k k ln = ln = k=1 k=2 k=2 k ln(k) R n+1/2 x ln(x) dx ≤ 3/2 n+1/2 2 = (x ln(x)/2 − x2 /4)3/2 2 = (n + 1/2)2 ln(n + 1/2)/2 − (n + 1/2) Qn /4 k −9 ln(3/2)/8 + 9/16, and therefore, k=1 k p (n+1/2)2 9/16 9/8 2 9/8 ≤ n + 1/2 e 2 / e(n+1/2) /43 p √ 9p (n+1/2)2 √ 9 √ (n+1/2)2 . n + 1/2 / 83 4e = 8 2 e Using now a trapezoid approximation for the Qn k lower estimate, we see that ln k=1 k n(n+1/2) ln(n) e2n−8 . 47 ln(2)−8 n2n Furthermore, R n+1/2 2 Pn 2 ln (x) dx k=4 ln (k) ≥ 7/2 = (n + 1/2) ln2 (n + 1/2) −2(n + 1/2) ln(n + 1/2) + 2(n + 1/2) −7 ln2 (7/2)/2 + 7 ln(7/2) − 7 and, by implication, n Y k ln(k) ≥ k=4 (n + 1/2)(n+1/2)(ln(n+1/2)−2) . e6−2n (7/2)7 ln(7/2)/2−7 Q n ek b) In this case we have ln k=1 e R n+1/2 x Pn √ = k=1 ek ≤ 1/2 e dx = e(en − 1), and √ Qn k n therefore, k=1 ee ≤ e e(e −1) . The estimate is R obtained as follows: Pn lower n k n e ≥ e /2 + e/2 + 1 ex dx = 3en /2 − e/2 k=1 Qn √ 3en −e k and, by implication, k=1 ee ≥ e . √ √ Qn n 1/(k k) 1/(k k) 37.13 k=1 e = e k=1 ≥ √ √ 5 √ (4−1/n)/√n 5/2−(2−1/(2n))/ n e = e / e P Chapter 38 38.3 R∞ 0 ∞ 2−x dx = −2−x / ln(2)|0 = 1/ ln(2) 18 Solutions 38.5 Since Z ∞ 2 Chapter 39 1 dx = x lnα (x) Z ∞ ln(2) 1 du, uα 39.5 The verification is completely analogous to the one in Exercise 39.2. (j) we may infer that the integral is convergent for α > 1 and divergent for α ≤ 1. R1 √ 3 R1 √ 3 38.17 0 x/ x2 + x dx ≤ 0 x/ x dx R1 = 0 1/x3/2 dx < ∞ (convergent), R1 2 R1 x /(x4 + x5 + x6 ) dx ≥ 0 x2 /(3x4 ) dx 0R 1 1/(3x2 ) dx = ∞ (divergent), = R1 0 √ R1 1/ x2 + x dx ≤ 0 1/x1/2 dx < ∞ (conver0 gent), R 1/2 √ R 1/2 1/ x3 − x6 dx ≥ 0 1/x3/2 dx = ∞ (di0 vergent), √ R2 R1 √ 3 (x + 1)/ 3 x4 + x6 dx ≥ 0 1/ 2x4 dx 0R 1 ≥ 0 1/(2x4/3 ) dx = ∞ (divergent) 38.19 a) 1, b) 1/2, c) ln(3)/2, d) π/2 38.21 a) Since f is strictly decreasing, it follows Rb Rb that c f (x) dx ≥ c f (b) dx = f (b)(b − c), and Rb therefore, bf (b) ≤ cf (b) + c f (x) dx. b) Since the limit of f (x) as x tends to ∞ is assumed to be zero, it follows that Rlimb→∞ bf (b) R∞ ∞ ≤ c limb→∞ f (b) + c f (x) dx = c f (x) dx. c) According R ∞ 0 ≤ limb→∞ bf (b) R ∞to b), we have ≤ limc→∞ c f (x) dx = 0 f (x) dx− R∞ R∞ Rc limc→∞ 0 f (x) dx = 0 f (x) dx − 0 f (x) dx = 0, and therefore limb→∞ bf (b) = 0 as desired. d) f is invertible because it is assumed to be strictly decreasing. Furthermore, since f (x) is positive and limx→∞ f (x) = 0, we may infer that limx→0+ f −1 (x) = limx→∞ f −1 (f (x)) = limx→∞ x = ∞. e) Using R a the result of Exercise R a35.20, it follows that 0 f −1 (x) dx = limc→0+ c f −1 (x) = a R f −1 (a) limc→0+ xf −1 (x) − limc→0+ −1 f (t) dt = c f (c) 39.7 For 0 ≤ j ≤ n we have Pn,x0 (x0 ) Pn = k=j f (k) (x0 )(x − x0 )k−j /(k − j)! x=x0 = f (j) (x0 ) as desired. Pn 39.11 Q2n (x) = k=0 (−1)k x2k /(2k)! 39.15 ln(1/2) ≈ P7,1 (1/2) P7 k+1 (1/2)k /k ≈ −0.692 and, sim= k=1 (−1) ilarly ln(3/2) ≈ P7,1 (3/2) ≈ 0.405, ln(2) ≈ P7,1 (2) ≈ 0.760 and ln(3) ≈ P7,1 (3) ≈ 12.69. Using a calculator, we notice that the errors in the approximation are markedly smaller for ln(1/2) and ln(3/2) than for ln(2) and ln(3). 39.17 Using the result of Exercise 39.8, it follows that P7,3 (x) = f (x). Pn 39.19 a) Pn,2 = k=0 4 lnk (2)(x − 2)k /k!, P(x) n b) Qn (x) = Pk=1 xk /(k − 1)!, n c) Pn,1 (x) =P k=0 (k + 1)(−1)k (x − 1)k , n d) Qn (x) = k=0 xk 39.21 Here we use the formula cos2 (x) = (cos(2x) + 1)/2 to infer that Q2n (x) = 1 + P n k 2k−1 2k x /(2k)!. k=1 (−1) 2 Chapter 40 40.3 Setting f (x) := arctan(x), the mean value theorem allows us to infer that | arctan(x) − arctan(x0 )| = |f (x) − f (x0 )| = |f 0 (t)(x − x0 )| = |x − x0 |/(1 + t2 ) ≤ |x − x0 | as desired. 40.11 With f (x) := sin(x) we obtain: limx→0 sin(x)/x = limx→0 (Q1 (x) + f 00 (t)x2 /2)/x = limx→0 (1 − sin(t)x/2) = 1. af −1 (a) − limb→∞ f (b)f −1 (f (b)) Rb + limb→∞ f −1 (a) f (t) dt = − limb→∞ bf (b) Rb R∞ + limb→∞ 0 f (t) dt = 0 f (x) dx. R∞ R∞ 38.23 a) e 1/(x ln2 (x)) dx = 1 1/u2 du = 1, R∞ R∞ x b) 0 e−e ex dx = 1 e−u du = 1/e, c) divergent, d) divergent 40.13 Since 211 /11! ≈ 0.51 · 10−4 , it follows that the error in the approximation sin(2) ≈ Q10 (2) ≈ 0.90935 is less that 10−4 . 38.25 Since the inequality ln(f (x))/x ≥ 2.416 can be written in the form (x) ≤ e−2.416x , R ∞ 1/f −2.416x dx is obviand since the integral 1 e ously convergent, Theorem 38.7 allows us to inR∞ fer that the integral 1 1/f (x) dx is convergent as well. f (t) = g 0 (t) = 40.17 In applying theR mean value theorem to x the function g(x) := a f (u) du, we may conclude that there exists a t in (a, b) such that 1 g(b) − g(a) = b−a b−a Z b f (x) dx. a 40.19 p(x) = x2 = f (x). In general, for f (x) = Solutions αx2 + βx + γ we find that f (b) − f 0 (a)(b − a) − f (a) (x − a)2 (b − a)2 + f 0 (a)(x − a) + f (a) p(x) = α(b + a) + β − (2αa + β) (x − a)2 b−a + (2αa + β)(x − a) + αa2 + βa + γ = 19 for all sufficiently large values of n, and the comparison test, as stated in Theorem 41.7, P∞ therefore allows us to infer that the series k=0 ak is convergent. c) Setting an := 1, we easily observe that the given condition does not guarantee the series to be convergent. d) This condition guarantees the convergence of the series by the ratio test. = α(x − a)2 + 2αa(x − a) + βx + αa2 + γ = αx2 + βx + γ = f (x). In other words, the second order secant polynomial of a second order polynomial is the polynomial itself. Chapter 41 41.17 If 0 < |q| < 1, then ln |q| < 0 and, by implication, limn→∞ |q|n = limn→∞ en ln |q| = limx→−∞ ex = 0. Hence limn→∞ q n = 0 whenever −1 < q < 1 (the case q = 0 is trivial). Furthermore, for q = 1 we have limn→∞ q n = limn→∞ 1 = 1, and for q > 1, it follows that limn→∞ q n = limn→∞ en ln(q) = limx→∞ ex = ∞. Finally, if q ≤ −1, then the limit of q n does not exist because q n ≥ 1 for even values of n and q n ≤ −1 for odd values of n. P∞ 41.19 k=2 1/5k = 1/(1−1/5)−1−1/5 = 1/20 41.25 Since limn→∞ |an+1 /an | = limn→∞ (n + 1)!nn /(n!(n + 1)n+1 ) = limn→∞ nn /(n + 1)n = limn→∞ 1/(1 + 1/n)n = 1/e < 1,Pthe ratio test allows us to infer that k the series ∞ k=1 k!/k is convergent. P∞ P∞ 41.27 k=m q k = q m k=0 q k = q m /(1 − q) 41.29 Since the ratio of the side lengths of two successive equilateral triangles that are cut out is obviously 1/2, it follows that the total area of the remaining figure is √ √ ∞ k 3 3X 1 3 = A= − . 4 4 2 6 √ k=1 41.31 a) 6/5, b) 1/90, c) 1, d) 1/(e3 − 1), e) 1/24 P∞ 41.33 a) Since the series k=1 1/k is divergent, and since limn→∞ 1/n = 0, it follows that the given condition does not guarantee the series to be convergent. b) The condition in b) implies that |an | < 1/n2 Chapter 42 42.3 The convergence at x = x0 is obvious. If x 6= x0 , then the assumption λ = ∞ in conjunction with (42.1) implies that p limn→∞ n |an (x − x0 )n | = ∞. Consequently, Theorem P∞ 41.20b allows us to infer that the series k=0 ak (x − x0 )k is divergent. 42.5 radius of convergence: √ r = 1/λ = 1/ limn→∞ n 3n = 1/3; interval of convergence: (2/3, 4/3) 42.7 Since limn→∞ |an+1 (x − x0 )n+1 /an (x − x0 )n | = λ|x−x0 | for all x 6= x0 , Theorem P∞ 41.23 allows us to conclude that the series k=0 ak (x − x0 )k is convergent whenever λ|x − x0 | < 1 and divergent whenever λ|x−x0 | > 1. In other words, the series is convergent for x ∈ (x0 − 1/λ, x0 + 1/λ) and divergent for x ∈ R r [x0 − 1/λ, x0 + 1/λ]. By implication, the radius of convergence is r = 1/λ. 42.9 radius of convergence: r = limn→∞ (n + 1)n+1 n!/(nn (n + 1)!) = limn→∞ (1 + 1/n)n = e; interval of convergence: (−2 − e, −2 + e) 42.11 According to L’Hôpital’s rule, we have limx→∞ ln(x1/x ) = limx→∞ ln(x)/x = limx→∞ √1/x = 0 and, by implication, limn→∞ n n = e0 = 1. This proves that the two series have thepsame radius of convergence because limn→∞ n |nan | = p √ n limn→∞ n n · limn→∞ |an |. 42.21 a) and c) are power series but b) and d) √ are not. In b) it is the term x that is inconsistent with the general representation√of a power series, and in d) it is the exponent k because √ k is not always an integer. 3 42.23 − 3x2 + 5 for x ∈ R, P∞a) x2k+1 /(2k + 1)! for x ∈ R, b) P k=0 x ∞ c) Pk=0 (−1)k x4k+5 /(2k + 1)! for x ∈ R, ∞ d) k=0 (−1)k 22k x2k+1 /(2k + 1)! for x ∈ R, 20 Solutions P∞ e) P k=0 (−1)k 22k x2k /(2k)! for x ∈ R, ∞ f ) k=0 (−8)k x3k for x ∈ (−1/2, √ 1/2), √ P∞ −k−2 2k g) Pk=0 2 x for x ∈ (− 2, 2), ∞ h)P k=3 k(k − 1)(k − 2)xk−3 /6 for x ∈ (−1, 1), ∞ i) Pk=0 22k x2k+1 /(2k + 1)! for x ∈ R, ∞ j) Pk=0 (−1)k xk+2 for x ∈ (−1, 1), ∞ k) P k=0 (−1)k x6k+2 /(2k + 1)! for x ∈ R, ∞ l) k=0 (−2)k xk for x ∈ (−1/2, 1/2) 42.25 a) The interval of convergence is √ n n = (−1, 1) in each case because lim n→∞ √ √ n limn→∞ n2 = limn→∞ n 1 = 1 (see also Exercise 42.11). P∞ b) For x = 1 the corresponding series k=1 1/k is divergent by the integral test. For x = −1 we have ∞ X (−1)k 1 1 1 − + ± ... k 2 3 4 k=1 ∞ ∞ X X 1 1 1 − =− , = 2k 2k − 1 2k(2k − 1) b) Using the results in a), we may infer that ∞ X 1 d 2 = = k(k − 1)xk−2 (1 − x)3 dx (1 − x)2 k=2 = ∞ X ∞ X k 2 xk−2 − k=2 kxk−2 , k=2 and therefore, ∞ X k 2 xk = x + k=1 ∞ X k 2 xk k=2 ∞ ∞ k=2 k=1 X X 2x2 2x2 k =x+ + kx = + kxk 3 3 (1 − x) (1 − x) 2x2 x x2 + x = + = . (1 − x)3 (1 − x)2 (1 − x)3 = −1 + k=1 k=1 this series is Rconvergent because ∞ 1/(2x(2x − 1)) dx ≤ 1 1/(2x − 1)2 dx = 1 1/2 < ∞. c) Here the convergence at x = ±1 follows from the criterion stated in Exercise 41.8 in conjunction with the Rintegral test because ∞ |(±1)n /n2 | = 1/n2 and 1 1/x2 dx < ∞. d) The series is divergent at x = ±1 by the sequence test because (±1)n does not approach zero as n tends to ∞. R −x3 dx 42.27 P∞ a) e k 3k+1 /(k!(3k + C, = R k=0 (−1) x P + 1)) k x /(k!k) + C, b) R ex /x dx = ln(x) + ∞ k=1 c) Parctan(x2 )/x dx ∞ = k=0 (−1)k x4k+2 /((2k + 1)(4k + 2)) + C Substituting 1/3 for x yields (1/9 + 1/3)/(8/27) = 3/2. b) Using the formula in Theorem 42.6, we obtain 1 limn→∞ |an+1 /an | α(α − 1) · · · (α − n + 1)(n + 1)! = lim n→∞ α(α − 1) · · · (α − n)n! n+1 = 1. = lim n→∞ α − n r= c) Given the definition of g in a), it follows that (1 + x)g 0 (x) = ∞ X α(α − 1) · · · (α − k + 1) k=1 + (k − 1)! k=1 =α+ it follows that k=1 x kx = (1 − x)2 + k! ∞ X α(α − 1) · · · (α − k + 1)k k=1 =α 1+ for all x ∈ (−1, 1). Substituting 1/2 for x yields P ∞ k 2 k=1 k/2 = (1/2)/(1 − 1/2) = 2. k! = αg(x), k! xk xk ∞ X α(α − 1) · · · (α − k + 1) k=1 xk−1 xk ∞ X α(α − 1) · · · (α − k + 1)(α − k) k=1 k (k − 1)! ∞ X α(α − 1) · · · (α − k + 1) k=1 ∞ X k 2 /3k = f (k) (x) = α(α − 1) · · · (α − k + 1)(1 + x)α . 42.29 a) Since ∞ k=1 42.31 a) Here we only need to observe that the kth derivative of f is and R∞ X 1 d 1 = = kxk−1 , (1 − x)2 dx 1 − x P∞ x k ! Solutions or equivalently, α g 0 (x) = for all x ∈ (−1, 1). g(x) 1+x d) Integrating both sides of the resulting equation in c) with the substitution u = g(x) on the left yields ln |g(x)| = α ln |1 + x| + C = α ln(1 + x) + C. Setting x equal to zero, we find that 0 = ln(1) = ln |g(0)| = ln(1) + C = C, and therefore, g(x) = ±(1 + x)α . The sign on the right-hand side of this equation must be positive because g(0) = 1 > 0. Thus, we have shown that g(x) = (1 + x)α for all x ∈ (−1, 1) as desired. e) If α = n, then the product α(α − 1) · · · (α − k + 1) is zero for all k > n. Given the definition of n k as stated in Chapter 34, the result in d) therefore allows us to infer that n x n n (x + y) = y 1 + y ! n X n(n − 1) · · · (n − k + 1) xk n 1+ =y k! yk k=1 n n k X X n x n k n−k n x y =y = . k yk k k=0 k=0 21 and the general solution is y = x/(ln(x2 ) + C). b) The equation p is separable, and the general solution is y = ± − ln(C − 3x2 ) 43.17 y = C|x|−b/a 43.19 a) 7 ln |y| − 3y = x + C, 2 b) y = 2 arctan(Cesin(x) ), f ) y = 2 + Cex /2+x 43.21 b) y/x + 2 ln |y/x − 1| = ln |x| + C, c) y = x tan(ln |x| + C) Chapter 44 44.1 The derivative of Q(t) as given in (44.5) is −Q0 e−t/RC /RC = −Q(t)/RC. Thus Q 0 (t) + Q(t)/RC = 0. 44.3 Using the result of Exercise 44.2 and setting C := y0 eA(x0 ) , it follows that y(x) = y0 eA(x0 )−A(x) = Ce−A(x) is a solution of the differential equation in (44.10). Furthermore, y(x0 ) = y0 eA(x0 )−A(x0 ) = y0 as desired. 44.5 y 0 (x) + a(x)y(x) = f (x)eA(x) e−A(x) Z − a(x) f (x)eA(x) dx + D e−A(x) Z + a(x) f (x)eA(x) dx + D e−A(x) = f (x) as desired. 44.7 y(x) = Chapter 43 43.9p solution of y 2 y 0 − 2xy 2 y 0 − 1 = 0: 3 y = C − 3 ln |1 − 2x|/2; solution of y 0 = xy − xy 0 : y = Cex /(1 + x) 2 0 3 43.11 p solution of y y = x , y(1) = 0: y = 3 3(x4 − 1)/4; solution of sin(y)y 0 = cos(x), y(π/6) = π: y = arccos(− sin(x) − 1/2) 43.13 Since z 0 /(z 2 −2z) = z 0 /(2z−4)−z 0 /(2z), it follows that ln |z − 2|/2 − ln |z|/2 = ln |x| + D or, equivalently, (z − 2)/z = ±e2D x2 . Thus, in defining C to be equal to ±e2D , we find that z(1 − Cx2 ) = 2, and therefore, z = 2/(1 − Cx2 ) as desired. 43.15 a) The equation is homogeneous because (xy−2y 2 )/x2 = y/x−2(y/x)2 = z−2z 2 = F (z), R x 1 tet 2 /2 2 dt + 2e1/2 e−x /2 2 = 1 + e(1−x )/2 Verification: since y 0 /(1−y) = x, it follows that − ln |1 − y| = x2 /2 + C, and the initial condition y(1) = 2 implies that C = −1/2. Hence y − 1 = 2 2 e(1−x )/2 , or equivalently, y = 1 + e(1−x )/2 as desired. 44.9 Using integration by parts, it follows that Z t 0 t sin(ωτ )eτ /RC dt = RC sin(ωτ )eτ /RC − ωRC 0 Z t cos(ωτ )eτ /RC dτ 0 t = RC sin(ωt)et/RC − ωR2 C 2 cos(ωτ )eτ /RC 2 2 −ω R C 2 0 Z t τ /RC sin(ωτ )e 0 dτ, 22 Solutions and therefore, Z t sin(ωτ )eτ /RC dt 0 = RC(et/RC (sin(ωt) − ωRC cos(ωt)) + ωRC) . 1 + ω 2 R2 C 2 Multiplying both sides by E0 /R yields the desired result in (44.16). 44.11 a) y(x) = x2 (3x + D)e−x , b) y(x) = ln(x) + D/ ln(x), c) y(x) = − cos(x) + (sin(x) + D)/x 44.13 Introducing the substitution u = 1/y, it follows that y 0 = −u 0 /u2 , and the corresponding differential equation in u is u 0 −3u/x = −x2 . Solving for u yields u(x) = (− ln(x)+D)x3 , and therefore, y(x) = 1/((− ln(x) + D)x3 ). √ 44.15√a) Q(t) = (cos(10t) + 35 sin(10t))/72 − e−10t/ 35 /72. √ b) Since Q(t) = sin(10t + α)/12 − e−10t/ 35 /72 for α = arcsin(1/6), it follows that the phase shift is arcsin(1/6)/10. √ √ c) I(t) = 5 cos(10t+α)/6+5e−10t/ 35 /(36 35) and Is (t) = 5 cos(10t + α)/6. d) Is(t) 1.0 0.5 t 0.0 0.5 1.0 1.5 -0.5 -1.0 44.17 A sin(t + α) = 2 sin(t) √ − 3 cos(t) for A = √ 13 and α = − arcsin(3/ 13), and √ A cos(t + β) = 2 sin(t)√− 3 cos(t) for A = 13 and β = π + arcsin(2/ 13). 44.19 a) y = x + C, b) 4 ln |1 + 3y|/9 − y/3 = x2 /2 + C, c) y = Cex , d) y = −1/(x + C), e) y = x/(1 + Cx), f ) y = x arcsin(Cx), g) y = e2x + Dex , h) y = x2 /2 + C p 2πk for some in|z| = n |w| and nφ = θ +p teger k. This shows that n |w|ei(θ+2πk)/n is an nth root of z for all integers k. Furthermore, the roots corresponding to values k between 0 and n − 1 are distinct because the corresponding angles (θ + 2πk)/n are distinct in the complex plane, and if j is an integer outside the range from 0 to n − 1, then a division of j by n with a remainder k ∈ {0, . . . , n − 1} allows us to infer the existence of an integer p such that j = pn + k and, by implication, e(θ+2πj)/n = e(θ+2πk)/n+2πp = e(θ+2πk)/n . The latter observation shows that there are no nth roots other than those corresponding to values k ∈ {0, . . . , n − 1}. iπ/4 )(z − ei3π/4 45.21 z 4 + 1 = (z − e√ √ )(z − i5π/4 i7π/4 2 )(z −e ) = (z − 2z +1)(z 2 + 2z +1) e 45.23 real part: 5/17; imaginary part: 14/17 45.25 x5 −32 = (x−2)(x2 −4 cos(2π/5)+4)(x2 − 4 cos(4π/5) + 4) 45.27 x4 + x2 + 1 = (x2 − ei2π/3 )(x2 − ei4π/3 ) = (x2 − x + 1)(x2 + x + 1) 45.29 ei7π/20 , ei(7π/20+2π/5) , ei(7π/20+4π/5) , ei(7π/20+6π/5) , ei(7π/20+8π/5) 45.31 r = 5, θ = π + arctan(3/4) 45.33 The √ real and imaginary parts are 512/243 and 512 3/243 respectively, because √ 10 √ (1 − i/ 3)10 = 2ei5π/6 / 3 = 210 ei50π/6 /35√= 1024ei2π/6 /243 = 1024(1/2 + i 3/2)/243. 45.35 Since f1 f2 = u1 u2 −v1 v2 +i(u1 v2 +v1 u2 ), we may apply the product rule (for ordinary real-valued functions) to infer that (f1 f2 ) 0 = (u1 u2 − v1 v2 ) 0 + i(u1 v2 + v1 u2 ) 0 = u10 u2 + u1 u20 − v10 v2 − v1 v20 +i(u10 v2 + u1 v20 + v10 u2 + v1 u20 ) = (u10 + iv10 )(u2 + iv2 ) + (u1 + iv2 )(u20 + iv20 ) = f10 f2 + f1 f20 as desired. Chapter 46 Chapter 45 45.1 z1 +z2 = 4−2i, z1 z2 = 7−4i, and z1 −z2 = −4i. p √ 2 2 45.3 √ z z̄ = a + b + i(ba − ab) 2 2 = a + b = |z| 45.11 z 4 + 2z 2 + 1 = (z + i)2 (z − i)2 45.19 If z = |z|eiφ is an nth root of w = |w|eiθ , then |w|eiθ = z n = |z|n einφ . Consequently, 46.5 According to Theorem 46.4, the assumption that y1 and y2 are solutions of (46.7) implies that the scalar multiples C1 y1 and C2 y2 are solutions of (46.7) as well. Since Theorem 46.4 also asserts the sum of any two solutions to be a solution, we may infer that C1 y1 + C2 y2 , too, is a solution of (46.7). 46.7 x = (1 · 5 + 2 · 3)/(2 · 5 + 3 · 1) = 11/13 and x = (2 · 2 − 1 · 1)/(2 · 5 + 3 · 1) = 3/13. Solutions Verification: 2x − 3y = 22/13 − 9/13 = 1 and x + 5y = 11/13 + 15/13 = 2. 46.15 Since Z x0 −f (t)y2 (t) C1 (x0 ) = dt = 0, Wy1 ,y2 (t) x Z x0 0 f (t)y1 (t) C2 (x0 ) = dt = 0, W y1 ,y2 (t) x0 it follows that y0 (x0 ) = 0. To verify the second equality, we apply the first fundamental theorem of calculus to infer that f (x)y2 (x) f (x)y1 (x) and C20 (x) = . C10 (x) = − Wy1 ,y2 (x) Wy1 ,y2 (x) This yields y00 (x0 ) = C10 (x0 )y1 (x0 ) + C1 (x0 )y10 (x0 ) + C20 (x0 )y2 (x0 ) + C2 (x0 )y20 (x0 ) =− f (x0 )y2 (x0 )y1 (x0 ) f (x0 )y1 (x0 )y2 (x0 ) + Wy1 ,y2 (x0 ) Wy1 ,y2 (x0 ) = 0. −5x 46.21 a) y(x) = C1 e√ + C2 xe−5x , √ 3x 11x) + C2 e3x√sin( 11x), b) y(x) = C1 e cos( √ −(4+ 18)x c) y(x) = C1 e + C2 e−(4− 18)x R 2 2 2 2 46.23 a) y2 (x) = ex e2x /e2x dx = xex , 2 and the general solution is y(x) = C1 ex + 2 C2 xex . R b) y2 (x) = e2xp e−x−ln |2x+1|/2 /e4x dx R = e2x e−5x / |2x + 1| dx, and the general solution is y(x) = C1 e2x + C2 y2 (x). 46.25 The assumptions y1 (x0 ) = y2 (x0 ) and y10 (x0 ) = y20 (x0 ) imply that y1 and y2 are solutions of the same initial value problem. Since the solution of an initial value problem is unique (by Theorem 46.2), we may infer that y1 (x) = y2 (x) for all x ∈ I. 46.27 a) The function y0 (x) = 0 (for all x ∈ R) satisfies the initial conditions y(0) = y 0 (0) = 0 and also solves the homogeneous equation (46.7). Since solutions of initial value problems are unique, it follows that y(x) = y0 (x) = 0 for all x ∈ R and, by implication, y(1) = y 0 (2) = 0. b) Consider the nonhomogeneous differential equation y 00 = 2. A solution of this equation that satisfies the initial conditions y(0) = y 0 (0) = 0 is y(x) = x2 . Since in this case y(1) 6= 0 6= y 0 (2), we may infer that the conclusion in a) is not permissible under the altered assumptions in b). 23 √ 46.29 Q(t) √ 10t) and, by implication, √ = 3 cos( I(t) = −3 10 sin( 10t). 46.31 Since the exponential function does not have any roots, the result in Exercise 46.30 allows us to conclude that Wy1 ,y2 (x) 6= 0 for all x ∈ I if only Wy1 ,y2 differs from zero at a single point x0 ∈ I, which is precisely the statement of Theorem 46.9a. Chapter 47 47.1 To verify that yn = 5/4 − (−3)n /4 is a solution of the initial value problem, we observe that yn+2 + 2yn+1 − 3yn = 5 1 5 1 15 3 − (−3)n+2 + − (−3)n+1 − + (−3)n 4 4 2 2 4 4 9 3 3 = − (−3)n + (−3)n + (−3)n = 0 4 2 4 and 5 1 − = 1, 4 4 5 3 y1 = + = 2. 4 4 y0 = 47.3 a) yn = 5n /6 + 11(−1)n /6, n n b) yn = −2/2 √ n + 2n/2 , √ n c) yn = 5 cos(αn) + 3 5 sin(αn), where √ α = arccos(−2/ 5). 47.5 The answer is c). 47.7 a) yn+2 − 5yn+1 + 6yn = 0, b) yn+2 − 4 cos(3)yn+1 + 4yn = 0. 47.9 yn = 3(−2)n − 5n(−2)n /2 Chapter 48 48.13 a) L[f ](s) = 2/((s + 1)2 + 4) for s > −1, b) L[f ](s) = 24e−2s /s5 for s > 0, c) L[f ](s) = a/(s2 − a2 ) for s > |a|. 48.17 In replacing t with t + T in the defining equation f (t + T ) = f (t), we obtain f (t + 2T ) = f (t + T ) = f (t). Replacing again t with t + T yields f (t + 3T ) = f (t + T ) = f (t), and continuing in this manner we finally arrive at the conclusion that f (t + kT ) = f (t) for all nonnegative integers k. 48.29 If f ∈ Ea and g ∈ Eb for some a, b ∈ R, then both f and g are piecewise continuous on every interval [0, c] and, by implication, so is 24 Solutions f g. Furthermore, there are constants M, N ∈ R such that |f (t)e−s1 t | ≤ M for all s1 > a and |g(t)e−s2 t | ≤ N for all s2 > b. Since for all s > a + b we can find s1 > a and s2 > b such that s = s1 + s2 , it follows that |f (t)g(t)e−st | = |f (t)es1 t ||g(t)es2 t | ≤ M N for all s > a + b and all t ∈ [0, ∞). Hence f g ∈ Ea+b . 48.31 2e−2s − 2e−3s + 1 e−s − 4e−2s + L[f ](s) = −3s s(1 − e ) s2 (1 − e−3s ) 48.33 a) −d3 /ds3 2/(s2 + 4), b) d2 /ds2 (s + 3)/((s + 3)2 + 4), c) (d2 /ds2 (s + 3)/((s + 3)2 + 4))/s, d) Γ(3/2)/(s − 2)3/2 , e) e−2(s−1) (s − 1)/((s − 1)2 + 1), f ) e−2s /s 48.35 a) te−t , b) e5t /6 − e−t /6 c) The inverse transform does not exist because 6 0. lims→∞ F (s) = ∞ = Rt d) et sin(3t)/3, e) 0 sin(τ ) dτ f ) The inverse transform does not exist, because lims→∞ F (s) = 1 6= 0. g) The inverse transform does not exist, because F (s) does not converge to zero as s tends to ∞. h) The inverse transform does not exist, be6 0. cause lims→∞ ( F (s) = ∞ = 0 for 0 ≤ t < 2, i) f (t) = R t−2 τ e cos(τ ) dτ for t ≥ 2. 0 48.37 Z t L eat e−aτ f (τ ) dτ (s) 0 Z t −aτ =L e f (τ ) dτ (s − a) 0 L[e−at f (t)](s − a) = s−a 48.39 For the first equation form is ( 0 f (t) = et−3 cos(2(t − 3)) and for the second it is ( 0 f (t) = et cos(2(t − 3)) L[f ](s) . s−a the inverse trans- = for 0 ≤ t < 3, for t ≥ 3 for 0 ≤ t < 3, for t ≥ 3. 48.41 If L[f ](s) were equal to sin(s)/s, it would follow that L[f 0 ](s) = sL[f ](s) − f (0) = sin(s) − f (0), but this is impossible because sin(s) − f (0) does not converge to zero as s tends to ∞. Chapter 49 1 5 5 − + and x−1 x−2 x−3 −x/2 + 5/2 4x − 1 1/2 . + 2 g(x) = 2 + (x − 4x + 5)2 x − 4x + 5 x − 1 7/4 3i/2 3/4 49.9 f (z) = − − 2 z − i (z − i) z+i 49.13 a) L[y](s) = s/((s2 +4)(s2 −1))+1/(s−1) and y(t) = − cos(2t)/5 + e−t /10 + 11et /10. b) L[y](s) = 1/((s − 1)(s − 2)2 ) and y(t) = te2t − e2t + et . c) L[y](s) = 1/(s2 ((s − 2)2 + 1)) +(s − 5)/((s − 2)2 + 1) and y(t) = 21e2t cos(t)/25−72e2t sin(t)+t/5+4/25. 49.17 a) L[y](s) = 1/((s − 1)(s + 4)) +2/((s + 2)(s + 4)) − 3/(s + 4) and y(t) = −21e−4t /5 + et /5 + e−2t . b) L[y](s) = 1/((s + 1)(s2 + 1)) + 1/(s + 1) and y(t) = 3e−t /2 − cos(t)/2 + sin(t)/2. c) L[y](s) = 1/(2(s − 2)2 ) + 1/(2(s2 − 4)) and y(t) = te2t /2 − e−2t /8 + e2t /8. 1/200 7/2000 3/16 + − 49.19 f (x) = x + 1 (x − 1)2 x−1 37x/100 + 23/20 23x/125 + 11/20 − − (x2 + 4x + 5)2 x2 + 4x + 5 49.21 a) L[y](s) = (s + 1)/((s + 1)2 + 1)2 + 3(s + 1)/((s + 1)2 + 1) and y(t) = te−t sin(t)/2 + 3e−t cos(t). b) L[y](s) = 1/((s + 1)2 + 1)2 + 2/((s + 1)2 + 1) and y(t) = 5e−t sin(t)/2 − te−t cos(t)/2. c) L[y](s) = 2e−2s /(s(s + 3)(s − 1)) − 2e−4s /(s(s + 3)(s − 1)), and setting g(t) := et /2 + e−3t /6 − 2/3, it follows that ( 0 for 0 ≤ t < 2, y(t) = g(t − 2) for t ≥ 2 ( 0 for 0 ≤ t < 4, − g(t − 4) for t ≥ 4. 49.7 f (x) = d) L[y](s) = −3e−3s /(s2 (s+1))+3e−2s /(s2 (s+ 1)2 ), and setting g(t) := 3te−t + 6e−t + 3t − 6 and h(t) := 3e−t + 3t − 3, we may infer that ( 0 for 0 ≤ t < 2, y(t) = g(t − 2) for t ≥ 2 ( 0 for 0 ≤ t < 3, − h(t − 3) for t ≥ 3. 49.23 L[I](s) = −1/(50(s + 1)((s + 4)2 + 4)) and Solutions I(t) = e−4t cos(2t)/650 + 3e−4t sin(2t)/1300 − e−t /650. 49.25 a) Using the results and the notation of Example 49.15, it is not difficult to show that 2 1 + 2/(αs) 2 500 1 + − + αs α2 s2 1 − e−αs L[I](s) = 2 2 (s + b) + c Note: in this representation of Is (t) all terms are either constant or periodic with period α and, in particular, the apparently divergent term 200t has been replaced with the periodic term 200(t− α[t/α]). 0.05 Is(t) 0.04 0.03 and 9 −bt 3 3 e sin(ct) + I(t) = − e−bt cos(ct) + 25 100 25 1 + 200t − (Φ(t) − Φ(t − α([t/α] + 1))) 4 Z t 1 − (Φ(τ ) − Φ(τ − α([τ /α] + 1))) dτ. 2α 0 Rt b) In evaluating the integral 1/(2α) 0 Φ(τ ) dτ we obtain several terms that converge to zero (as t tends to ∞) minus the constant 0.0168115 (approximately). Consequently, the steady state solution is 3 + 200t 25 1 + Φ(t − α([t/α] + 1)) − 0.0168115 4 Z t 1 + Φ(τ − α([τ /α] + 1)) dτ. 2α 0 Is (t) = Since Φ(t − α([t/α] + 1) is periodic with period α, it follows that Z t Φ(τ − α([τ /α] + 1)) dτ 0 25 = Z + Z α[t/α] Φ(τ − α([τ /α] + 1)) dτ 0 t Φ(τ − α([τ /α] + 1)) dτ α[t/α] = [t/α] Z 0 Φ(τ ) dτ + −α = −0.0004[t/α] + Z Z t−α([t/α]+1) Φ(τ ) dτ −α t−α([t/α]+1) Φ(τ ) dτ. −α Since 0.0004/(2α) = 200α, we may conclude that 3 + 200(t − α[t/α]) 25 1 + Φ(t − α([t/α] + 1)) − 0.0168115 4 Z t−α([t/α]+1) 1 Φ(τ ) dτ. + 2α −α Is (t) = 0.02 0.01 0.00 t 0.001 0.002 -0.01 -0.02 -0.03 Chapter 50 50.5 program exercise; var k: integer; y,dx: real; begin y:=0; dx:=0.1; writeln(y); for k:=0 to 11 do begin y:=y+sqrt(1-sqr(y))*dx; writeln(y) end end. 50.7 program exercise; var k: integer; y,dx: real; begin y:=0; dx:=0.1; writeln(y); for k:=0 to 11 do begin y:=y+sqrt(1-sqr(y))*dx -y*sqr(dx)/2; writeln(y) end end. 50.9 program exercise; var k: integer; y,ys,w,dx: real; 0.003 26 Solutions begin y:=0; w:=1; dx:=pi/30; writeln(y,’ ’,w); for k:=0 to 14 do begin ys:=y; y:=y+w*dx; w:=w-ys*dx; writeln(y,’ ’,w) end end. 50.11 program fallingbodies; const gM=3.98E14; R=4E7/(2*pi); d=1.2E7; var k: integer; t,y,ys,w,dt: real; begin t:=0; y:=d+R; w:=0; dt:=100; while (y>=R) do begin writeln(t,’ ’,y); t:=t+dt; ys:=y; y:=y+w*dt -gM*sqr(dt)/(2*sqr(y)); w:=w-gM*dt/sqr(ys) +gM*w*sqr(dt)/(sqr(ys)*ys); end end. 50.13 a) y(x) = 97e−7x /392+7ex /8−x/7−6/49 and y(2) ≈ 6.057. b) Using the equations yk+1 := yk + wk ∆x, wk+1 := wk + (xk − 6wk + 7yk )∆x, we find the approximation y(2) ≈ 5.478, and the error is |6.057 − 5.478| = 0.579. c) Using the equations 1 yk+1 :=yk + wk ∆x + (xk − 6wk + 7yk )∆x2 , 2 wk+1 :=wk + (xk − 6wk + 7yk )∆x 1 + (1 − 6xk + 43wk − 42yk )∆x2 , 2 we find the approximation y(2) ≈ 5.735, and the error is |6.057 − 5.735| = 0.322. 2 50.15 a) yk+1 := yk + eyk /20 ∆x 2 +yk eyk /10 ∆x2 /20 yields y(2) ≈ 3.6567. b) yk+1 := yk + (sin(yk2 ) + xk )∆x +(2yk (sin(yk2 ) + xk ) cos(yk2 ) + 1)∆x2 /2 yields y(2) ≈ 2.1912. c) Using the equations yk+1 := yk + wk ∆x 1 + (xk − wk sin(yk ) − yk )∆x2 , 2 wk+1 := wk + (xk − wk sin(yk ) − yk )∆x 1 + (1 − wk − wk2 cos(yk ))∆x2 2 1 − sin(yk )(xk − wk sin(yk ) − yk )∆x2 , 2 we find that y(2) ≈ 0.8531. Chapter 51 51.5 c2 = 0, c5 = −c2 /(4 · 5) = 0, c8 = −c5 /(7 · 8) = 0, and in general c3k+2 = 0 for all k ≥ 0. P∞ 51.7 a) Setting y(x) = k=0 ck xk and assuming the general initial conditions y(0) = z1 , y 0 (0) = z2 , we find that c0 = z1 , c1 = z2 , c2 = z1 + z2 /2, c3 = z2 /3, and in general ck+2 = (k + 1)ck+1 + 2ck − 2ck−1 + 4ck−2 (k + 2)(k + 1) for all k ≥ 2. Using this general equation to determine c4 , c5 , c6 , c7 , and c8 , we notice that there are coefficients ak and bk such that c2k = z1 /k! + ak z2 and c2k+1 = bk z2 . Given this observation, it is convenient to generate one particular solution y1 by setting z1 := 1 and z2 := 0. This yields c2k = 1/k!, c2k+1 = 0, and y1 (x) = ∞ X x2k k! k=0 2 = ex . (The reader may easily verify that this function does indeed satisfy the given differential equation.) Using the formula on p.363, it follows that a second solution is Z x 2 2 et−2t dx, y2 (x) = ex 0 and the general solution is y(x) = C1 y1 (x) + C2 y2 (x). Solutions b) Following exactly the same approach as in a), it is not difficult to see that c0 = z1 , c1 = z2 , c2 = 5z1 /2, c3 = z2 , and in general ck+2 = (k + 5)ck + 2ck−2 (k + 1)(k + 2) 27 = for all k ≥ 2. Consequently, the general solution is y(x) = z1 + z2 (x − 1) ∞ X (−1)k (z2 − z1 ) (x − 1)k + k−1 for all k ≥ 2. As we use this general equation to determine ck for a few more even and odd indices, we notice that c2k+1 = z2 /k! and c2k = ak z1 for some coefficients ak ∈ R. Setting z1 := 0 and z2 := 1, we find the solution y1 (x) = ∞ X x2k+1 k=0 k! =x ∞ X x2k k=0 k! (−1)k (z2 − z1 ) (−1)k (z2 − z1 ) − k−1 k − k=2 ∞ X k=2 (−1)k (z2 − z1 ) (x − 1)k k = z1 + z2 (x − 1) + (z2 − z1 )(x − 1) ln(x) 2 = xex . + (z2 − z1 )(ln(x) − (x − 1)) = z1 x + (z2 − z1 )x ln(x). The corresponding second solution is 2 y2 (x) = xex Chapter 52 Z x 0 −3t2 /2 e t2 dt, and the general solution is y(x) = C1 y1 (x) + C2 y2 (x). c) Since the general solution is to be represented as a power series about x0 = 1, we rewrite the given equation as follows: 0 = x2 y 00 (x) − xy 0 (x) + y(x) 2 = (x − 1)2 y 00 (x) + 2(x − 1)y 00 (x) + y 00 (x) − (x − 1)y 0 (x) − y 0 (x) + y(x). Assuming the general initial conditions y(0) = P∞ z1 , y 0 (0) = z2 and substituting k=0 ck xk for y(x), we find that c0 = z1 , c1 = z2 , and in general ck (k − 1)2 + ck+1 (2k 2 + k − 1) (k + 2)(k + 1) for all k ≥ 0. As we use this recursive equation to determine ck for the first few values k ≥ 2, we quickly notice that ck = p00 = 6/7 p01 = 1/7 p10 = 2/7 p11 = 5/7 52.7 a) 26/50, b) (25/51) · (26/50) + (26/51) · (25/50)+(25/51)·(24/50), c) (25/51)·(26/50)+ (26/51) · (25/50), d) 12/50 52.9 pAA = p pBA = p pAB = 1 − p pBB = 1 − p Chapter 53 00 = ((x − 1) + 1) y (x) − ((x − 1) + 1)y 0 (x) + y(x) ck+2 = − 52.1 (−1)k (z2 − z1 ) k(k − 1) 53.1 a12 = 2, a31 = −0.1, and a24 = 8. −2 20 6 53.9 BA = −5 8 −6 −4 22 3 53.11 If we assume A to be an m×n matrix and B to be a k × l matrix, then the size conditions can be stated as follows: a) n = k, b) l = m, c) n = k and l = m, d) m = n = k = l. 53.15 If we assume that A is an m × n matrix, B a k × l matrix, and C an i × j matrix, then the size conditions can be stated as follows: a) n = k and l = i, b) n = k = i and l = j, c) m = k and n = l = i. n 0 0 a 53.21 Dn = 0 bn 0 0 0 cn 3 −9 6 53.23 AB = 4 2 15 53.25 A+B and AB−BA are not well defined. 28 Solutions 53.27 Answers may vary, but one possible answer is 2 0 0 A = 0 1 0 . 0 1 1 53.29 From the assumption |g(x) − g(y)| < λ|x − y| we may infer that |g k+1 (x) − g k (x)| < λk |g(x) − x| for all nonnegative integers k and allPx ∈ R. Since 0 < λ < 1, the ∞ k series k=0 λ |g(x) − x| is convergent, and according to the comparison test, the series P∞ k+1 k (g (x) − g (x)) must be convergent as k=0 well. In other words, the limit of g n (x) = Pn−1 x + k=0 (g k+1 (x) − g k (x)) as n tends to ∞ exists and is finite. Denoting this limit by z (i.e., z = limn→∞ g n (x)) and observing that the condition |g(x) − g(y)| < λ|x − y| implies that g is continuous (this is a trivial consequence of Theorem 7.8), we may conclude that z is a fixed point of g because g(z) = g lim g n (x) = lim g n+1 (x) = z. n→∞ 2 1/3 If b = 1, then x = 1/3. 1 2/3 2 5/3 If b = 2, then x = −1/3. 3 1/3 54.19 In denoting the entries of Qij (λ) by qkl and those of Qij (λ)A by bkl , it follows that ( m X if k 6= j, akl bkl = pkq aql = + a if k = j λa il jl q=1 as desired. 1 0 1 54.23 and −2 1 −1 tinct reduction matrices. Chapter 54 54.3 a) x1 = 8/5, x2 = 2/5, b) x1 = 1, x2 = 1, x3 = 2 54.7 The matrix is not in standard reduced form because the rule, which says that in each leading column all but one entry must be zero, is violated. To obtain a standard reduced matrix we change the entries a12 and a33 to zero: 1 0 1 0 0 1 3 0 . 0 0 0 4 54.11 a) no solutions, b) x1 = 1 − 3t1 − 10t2 , x2 = 2 + t1 + 3t2 , x3 = t1 , x4 = 1, x5 = −t2 , x6 = t2 , c) x1 = 4, x2 = 7, x3 = 2 54.15 good: 27 out of 58, average: 19 out of 58, poor: 12 out of 58. 1 1/3 54.17 If b = 1, then x = −2/3. 0 2/3 are two dis- 54.25 In denoting the entries of C, Idm , and Idm C by cij , dij and bij respectively, it follows that m X bij = dik ckj = dii cij = cij k=1 n→∞ To prove that z is unique, we assume that g(y) = y for some y ∈ R. Since x in the preceding argument was an arbitrary real number, it follows that z = limn→∞ g n (y) = limn→∞ y = y as desired. 53.31 Answers may vary. 0 1/2 as desired. The remaining equations can be verified in a similar fashion. 54.29 A reduction matrix is −1/2 1/2 1/2 B = 3/2 −1/3 −1/2 , 1 0 −1 and the product BA is easily seen to be equal to Id. Thus Ais indeed invertible. 1/6 2/3 −1/6 54.31 A−1 = 1/6 −1/3 5/6 1/6 −1/3 −1/6 −1 −1 1 1 0 54.33 B = 1 2 1 −1 54.35 In reducing the augmented matrix below, we find the same soultions as in Exercise 54.34. 1 0 1 1 2 1 −1 1 −1 2 0 3 . 1 1 0 −1 1 0 54.37 B =Q21 (−2)Q31 (−1)Q32 (−1)P2 (1/3) · P3 (−1)Q13 (−1)Q12 (1) 4/3 −2/3 −1 1 = −2/3 1/3 1 0 −1 Solutions 1 1 1 0 −1 −1 0 1 0 0 −3 −3 −2 1 3 54.39 A−1 = . −1 −2 −4 1 3 −1 −1 0 0 1 54.41 a) Since A is the reduction matrix of A−1 , we need to find a product of elementary matrices that reduces 0 −1 1 1 −1 A−1 = 1 −1 0 1 to the identity matrix. This yields A =Q21 (−1)Q31 (1)Q32 (1)P2 (−1)Q23 (1) · Q13 (1)R12 b) Following the same approach as in a), we find that A =Q31 (7)Q21 (−2)Q32 (3)P2 (−1)P1 (17) · Q13 (−2)Q12 (−7)R13 . 54.43 If Ax1 = Ax2 = b, then A(x1 − x2 ) = Ax1 − Ax2 = b − b = 0 as desired. 1/2 0 0 0 0 0 1 0 0 0 0 0 1 0 0 54.45 a) B = 0 0 0 1 0 0 0 0 0 1 1 0 0 0 0 −1 1 0 0 0 b) C = 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 1 0 0 0 0 0 1 0 0 0 c) D = 0 0 2 0 0 0 0 0 1 0 0 −2 0 0 1 54.47 a) Yes, because the invertibility of A guarantees by definition the existence of exactly one solution. b) Yes, because the assumption b = 0 implies that x = 0 is a solution. c) Yes, for the same reason as in b). d) Yes, because the given assumption concerning the reduced matrix of A implies that A is invertible, and the argument in a) therefore applies here as well. e) Yes, for the same reason as in b). 54.49 The condition in c) guarantees the existence of infinitely many solutions, because if 29 A is not invertible, then the equation Ax = b either has no solution or infinitely many solutions. Since the assumption b = 0 implies that there is at least one solution, namely x = 0, we may therefore infer that the number of solutions is infinite. The same argument applies also to e) because the stated assumption concerning the reduced matrix of A implies that A is not invertible. Chapter 55 55.11 To apply the standard method we assign clockwise currents to each of the four loops in the network as follows: loop ABC: I1 ; loop BCD: I2 ; loop CDE: I3 ; and loop ACE: I4 . The corresponding system of linear equations is represented by the matrix equation 40 −10 0 −10 0 I1 −10 40 −15 I2 0 0 = , 0 −15 65 −20 I3 0 I4 −10 0 −20 30 5 and the solutions are I1 = 107/1406 A, I2 = 1/19 A, I3 = 63/703 A and I4 = 177/703 A. Using the probabilistic approach, we find the following values for the long term expectations of the permissible states: qAB = 703/8376, qAC = 703/4188, qBA = 163/2792, qBC = 163/1396, qBD = 163/2094, qCA = 38/349, qCB = 38/349, qCD = 76/1047, qCE = 19/349, qDB = 21/349, qDC = 21/349, qDE = 21/698, qEC := 0, and qED := 0. To verify that these results are consistent with those obtained via the standard method we can now compute the ratio of the currents for any pair of connections in the circuit. Taking, for example, the connections from A to C and B to D, we find that I4 − I1 247 qAC − qCA = = qBD − qDB 74 I2 as desired. Chapter 56 1 1 56.3 4 3 3 2 1 0 3 7 4 6 56.5 a) (A + B)t = [aij + bij ]t = [aji + bji ] = [aji ] + [bji ] = At + Bt , 30 b) (λA)t = [λaij ]t = [λaji ] = λ[aji ] = λAt , c) (At )t = ([aij ]t )t = [aji ]t = [aij ] = A. 56.13 4 − 6 = −2 56.15 The results are easily verified by observing that I(1, 2, 3) = 0, I(1, 3, 2) = 1, I(2, 1, 3) = 1, I(2, 3, 1) = 2, I(3, 1, 2) = 2, and I(3, 2, 1) = 3. 56.19 −2 56.29 x1 = −22/5, x2 = −29/5, and x3 = −3. √ √ 56.35 λ1 = 2 + 3 and λ2 = 2 − 3. 56.37 det(2A) = 2n · 3 Qn 56.39 det(A) = (−1)[n/2] k=1 an−k+1,k (where [x] denotes, as usual, the greatest iunteger less than or equal to x). 56.41 a) If A is invertible, and if v is a characteristic vector corresponding to a characteristic value λ of A, then Av = λv, or equivalently, v = λA−1 v. Since v 6= 0 (by the definition of a characteristic vector), it follows that λ cannot be zero. b) Given the result of a), we may infer that the equation Av = λv is equivalent to the equation (1/λ)v = A−1 v. In other words, 1/λ is a characteristic value of A−1 . 56.43 In applying Theorem 56.22 to the first column, we may infer the existence of constants a0 , . . . , an−1 (represented by cofactors) such that P (x) = a0 + a1 x + · · · + an−1 xn−1 . In other words, P (x) is a polynomial of degree less than or equal to n − 1. Furthermore, P (xi ) = 0 for all i ∈ {2, . . . , n} because P (xi ) is the determinant of a matrix in which the first column equals the ith column. Since the values x1 , . . . , xn are assumed to be distinct, it follows that x2 , . . . , xn are n − 1 distinct roots of P , and since the degree of P is less than or equal to n − 1, we may conclude that P (x1 ) 6= 0 (because a polynomial of degree less than or equal to n − 1 cannot have more than n − 1 distinct roots). a b 56.45 If A = is invertible, then c d 1 d −b −1 . A = ad − bc −c a 56.47 a) If Av = λv and Aw = λw, then A(v + w) = Av + Aw = λv + λw = λ(v + w), and the assumption v + w 6= 0 therefore implies that v + w is a characteristic vector of A corresponding to the characteristic value λ. Solutions b) If v is a characteristic vector and µ 6= 0, then µv 6= 0 and A(µv) = µAv = µλv = λ(µv). Thus, µv is a characteristic vector of A corresponding to the characteristic value λ. 56.49 a) Yes, because A is invertible whenever det(A) 6= 0. 2 2 b) Yes, because √ if det(A) = det(A ) = 2, then det(A) = ± 2 6= 0. c) Yes, because if det(A) det(B) = det(AB) = 2, then det(A) must be different from zero. d) No, because the zero matrix 0n×n is not invertible despite the fact that det(0n×n + Idn ) = 1 6= 0. Chapter 57 57.5 0 = λ0 − λ0 = λ(0 + 0) − λ0 = λ0 + (λ0 − λ0) = λ0 + 0 = λ0. 57.9 d) If Ax = 0 and Ay = 0, then A(x + y) = Ax + Ay = 0 + 0 = 0 and A(λx) = λAx = λ0 = 0 as desired. f ) If y1 and y2 are in U , then there are vectors x1 , x2 ∈ Fn such that Ax1 = y1 and Ax2 = y2 . Consequently, y1 + y2 is in U as well because A(x1 + x2 ) = Ax1 + Ax2 = y1 + y2 . Furthermore, λy1 is in U for all λ ∈ F, because A(λx1 ) = λAx1 = λy1 . j) If f and g are in U , then (f + g) 0 + a(f + g) = (f 0 + af ) + (g 0 + ag) = 0 + 0 = 0. Furthermore, (λf ) 0 + a(λf ) = λ(f 0 + af ) = 0. Consequently, f + g and λf are in U as well, and U therefore is a subspace. 57.27 If v = λ1 v1 + · · · + λn vn and w = µ1 v1 + · · · + µn vn , then v + w = (λ1 + µ1 )v1 + · · · + (λn + µn )vn and λ1 + µ 1 λ1 µ1 . . . .. Sβ (v + w) = = .. + .. λn + µ n λn µn = Sβ (v) + Sβ (w). Furthermore, for λ ∈ F we have λv = λλ1 v1 + · · · + λλn vn , and therefore λλ1 λ1 .. .. Sβ (λv) = . = λ . = λSβ (v). λλn λn 57.29 a) The vectors are linearly independent because 1 1 2 2 0 1 = 7 6= 0. 3 2 1 Solutions b) The vectors are linearly dependent, because R2 is two-dimensional. c) Since the first three components of the given vectors are identical with those of the vectors in a), we may conclude that the vectors are linearly independent. 57.31 If p(x) = b0 + b1 x + b2 x2 and q(x) = c0 + c1 x + c2 x2 are polynomials in P2 , then 31 in the complex vector space Cn , and therefore a1 + ib1 = · · · = an + ibn = 0. Hence a1 = · · · = an = b1 = · · · = bn = 0 as desired. Chapter 58 58.1 T (p + q)(x) m1 x100 (t) = − k1 x1 (t) + k2 (x2 (t) − x1 (t)) m2 x200 (t) = − k2 (x2 (t) − x1 (t)) = T ((b0 + c0 ) + (b1 + c1 )x + (b2 + c2 )x2 ) = ((b0 + c0 ) + 2(b2 + c2 )) + (3(b1 + c1 ) − (b2 + c2 ))x + 3(b0 + c0 )x2 m3 x300 (t) = (b0 + 2b2 ) + (3b1 − b2 )x + 3b0 x2 + (c0 + 2c2 ) + (3c1 − c2 )x + 3c0 x2 = T (p)(x) + T (q)(x), 58.7 Setting 0 0 A(t) := −1 0 0 and for a given scalar λ ∈ R, we find that T (λp)(x) = T (λb0 + λb1 x + λb2 x2 ) = (λb0 + 2λb2 ) + (3λb1 − λb2 )x + 3λb0 x2 = λ((b0 + 2b2 ) + (3b1 − b2 )x + 3b0 x2 ) = λT (p)(x). This proves that T is a linear transformation from P2 to P2 . Furthermore, setting γ := β := {1, x, x2 }, the matrix representing T is readily seen to be 1 0 2 Aβγ = 0 3 −1 . 3 0 0 57.33 Since the real numbers are a subset of the complex numbers, it follows that the scalar multiplication of a vector in Cn with a real number is well defined. Given this observation, the validity of the vector space axioms is easily verified. Furthermore, since every complex number can be written in the form a · 1 + b · i, it follows that Cn (as a real vector space) is spanned by the vectors e1 , . . . , en , ie1 , . . . , ien . Consequently, if we can show these vectors to be linearly independent, we will be able to conclude that {e1 , . . . , en , ie1 , . . . , ien } is a basis and that the dimension of Cn (as a real vector space) is 2n. So let us assume that a1 , . . . , an and b1 , . . . , bn are real numbers such that a1 e1 + · · · + an en + b1 ie1 + · · · + bn ien = 0. Then (a1 + ib1 )e1 + · · · + (an + ibn )en = 0 + k3 (x3 (t) − x2 (t)) = − k3 (x3 (t) − x2 (t)) 1 0 0 0 2 0 1 0 0 0 0 0 0 0 0 0 0 −t2 , 1 0 it follows that 0 u1 (t) u1 (t) 0 u20 (t) u2 (t) 0 0 u3 (t) = A(t) u3 (t) + et , 0 u4 (t) u4 (t) 0 u50 (t) u5 (t) 3 where u1 = x1 , u2 = x10 , u3 = x100 , u4 = x2 and u5 = x20 . 58.9 The equivalent system is c). 58.11 In denoting by A(x) the matrix 0 0 .. . 1 0 .. . 0 1 .. . ... ... .. . 0 0 .. . 0 −a0 (x) 0 −a1 (x) 0 −a2 (x) ... ... 1 −an−1 (x) , a first order system equivalent to the given differential equation is u1 (x) u10 (x) u2 (x) u20 (x) .. = A(x) .. + . . un0 (x) un (x) 0 0 .. . f (x) where u1 = y, u2 = y 0 , ..., un = y (n−1) . , 32 Solutions Chapter 59 59.3 If x(t) = (x1 (t), x2 (t)) is a solution of the system (59.6), then x100 (t) = x20 (t) = −b(t)x1 (t) − a(t)x2 (t) = −b(t)x1 (t) − a(t)x10 (t). Thus, y(t) := x1 (t) is a solution of the differential equation (59.4). Conversely, if y is a solution of (59.4), then x(t) := (y(t), y 0 (t)) is a solution of (59.6). 59.5 If x is a solution of (59.7), then Cx10 (t) Cx10 (t) (Cx) 0 (t) = ... = ... = Cx 0 (t) Cxn0 (t) b) No, because if X(0) = −Id, then X(0)2 = Id, but X(0) 6= Id. c) Yes, because if X(0)−1 = Id, then X(0) = (X(0)−1 )−1 = Id−1 = Id. 59.17 a) The definition of the determinant implies that d det(X(t)) dt X d = δ(σ)x1j1 · · · xnjn dt σ=(j1 ,...,jn )∈Sn X 0 = δ(σ)x1j x · · · xnjn + . . . 1 2j2 σ=(j1 ,...,jn )∈Sn Cxn0 (t) = CA(t)x(t) = A(t)(Cx(t)). X + 0 δ(σ)x1j1 · · · xn−1,jn−1 xnj = n σ=(j1 ,...,jn )∈Sn Thus, Cx is a solution of (59.7) as well. 59.7 Each column in the matrix f (t) g(t) f 0 (t) g 0 (t) is a solution of (59.6) because f and g are solutions of (59.4). Furthermore, since f (t0 ) = 1, f 0 (t0 ) = 0, g(t0 ) = 0, and g 0 (t0 ) = 1 (see the proof of Theorem 46.9), we may infer that f (t0 ) g(t0 ) = Id. f 0 (t0 ) g 0 (t0 ) Consequently, the matrix above is indeed the standard fundamental matrix Xt0 (t) of the system (59.6). 59.11 X(t) is a fundamental matrix because each of its column vectors is a solution of the system and det(X(0)) = −1 6= 0. Furthermore, the solution of the initial value problem is 2 x(t) = X(t)X(0)−1 3 2 cos(2t) − 3 sin(2t) = . 3 cos(2t) + 2 sin(2t) 59.13 x2 (t) 59.15 As in Exercise 59.14, the first part of each statement simply guarantees each column in X(t) to be a solution of the system. Consequently, we only need to examine in each case whether the second condition implies that X(0) = Id. a) Yes, because the assumption X(0)ei = ei implies that X(0) = X(0)Id = X(0)(e1 , . . . , en ) = (e1 , . . . , en ) = Id. 0 x11 x21 .. . xn1 ... ... .. . ... 0 x11 x1n .. x2n .. + · · · + . xn−1,1 . x0 xnn n1 ... .. . ... ... . xn−1,n 0 xnn x1n .. . 0 0 = Since X (t) = A(t)X(t), it follows that xki P n a x and, by implication, j=1 kj ji 0 0 x11 . . . x1n x21 . . . x2n .. .. .. . . . xn1 . . . xnn Pn Pn j=1 a1j xj1 . . . j=1 a1j xjn x21 ... x2n = . . .. .. .. . ... xnn xn1 xj1 . . . xjn n x21 . . . x2n X a1j . = .. = a11 det(X(t)). .. .. . . j=1 xn1 . . . xnn Similarly, we find that x11 . . . x1n 0 0 x21 . . . x2n .. .. =a22 det(X(t)) .. . . . xn1 . . . xnn .. . x11 ... x1n .. .. . .. . . =ann det(X(t)), xn−1,1 . . . xn−1,n 0 x0 ... xnn n1 Solutions Since the inverse of the matrix (v1 , v2 , v3 ) is easily seen to be and therefore, d det(X(t)) = dt n X aii (t) det(X(t)), λk1 Bk = ... 0 ... .. . ... 0 .. . λkn for all nonnegative integers k, it follows that P∞ k 0 k=0 λ1 /k! . . . .. .. .. eB = . . P∞ . k 0 ... k=0 λn /k! λ 1 ... 0 e .. .. . . .. = . . b) eA 1 0 = 0 0 it follows that 0 3/2 −9/2 6 eAt = et 0 −2 0 −1 3 0 −4 8 1 5/2 −7/2 0 . + e2t 0 3 −6 + e3t 0 0 0 1 −2 0 0 0 −21et /2 + 16e2t − 9e3t /2 14et − 12e2t b) x(t) = t 2t 7e − 4e 60.23 a) eA 0 −1/2 3/2 1 −2 , [γij ] = 0 1 5/2 −7/2 Chapter 60 0 i=1 as desired. Pn b) Setting a(t) := i=1 aii (t), the result in a) implies that det(X(t)) is a solution of the differential equation y 0 = a(t)y. Consequently, we can find a constant C ∈ R such that det(X(t)) = CeA(t) . Since the exponential function is never zero, we may infer that det(X(t)) is different from zero for all t unless C = 0, in which case it is equal to zero for all t. In particular, if det(X(t0 )) 6= 0 for a single point t0 , then det(X(t)) 6= 0 for all t. 60.5 Since 33 ... 1 = 0 0 1 4 1 2 0 1 0 0 eλn 1 4 1 2 0 1 25/6 5 2 1 60.25 eA(t−2) = 1 t − 2 t2 − 3t + 2 2t3 /3 − 5t2 /2 + 14/3 0 1 2t − 4 2t2 − 7t + 6 0 0 1 2t − 4 0 0 0 1 60.27 a) The characteristic values of A are λ1 = 1, λ2 = 2, λ3 = 3, and corresponding characteristic vectors are −3 −4 1 v1 = 4 , v2 = 3 , v3 = 0 . 2 1 0 60.29 Let v be a characteristic vector of A corresponding to the characteristic value λi . Then every column vector of Pi (A) is a scalar multiple of v (see Theorems 60.20 and 60.19), and the familiar properties of the determinant therefore imply that det(Pi (A)) = 0 (see Theorem 56.17b, c, e). 60.31 To give an example of how the entries cij of eAt can be computed, we will examine the entry c25 (assuming, of course, that n ≥ 5). In taking a few powers of A, we readily observe that the entry in the second row and fifth column of Ak assumes the value 0 for k < 3 and the value k3 λk−3 for k ≥ 3. Thus, c25 = ∞ k−3 k X k λ t k=3 3 k! = ∞ t3 X (λt)k t3 eλt = . 3! k! 3! k=0 In general, we find that ( 0 cij = j−i λt t e /(j − i)! if i > j, if i ≤ j. 60.33 Since e2A v = Idv + 2Av + 22 A2 v/2 + 23 A3 v/6 + 24 A4 v/24 = v − 2v + 2v − 4v/3 + 2v/3 = v/3, we may infer that the components of e2A v are x1 = 1/3, x2 = 1/3 and x3 = 2/3. 60.35 x(t) = 2(t−1) + 4e3(t−1) −e 1 2(t−1) −e + e3(t−1) + 2(t − 1)e3(t−1) 3 2(t−1) e − e3(t−1) + 2(t − 1)e3(t−1) 34 Chapter 61 61.3 x1 (t) = 1/4−8et +35e2t /4−2tet −5te2t /2, x2 (t) = −1/2+3tet −5e2t /2+tet , x3 (t) = 1/4− 10et + 35e2t /4 − 3tet − 5te2t /2. 61.5 x2 (t) ≈ −0.06eλ1 t + 1.35eλ2 t + 0.49eλ3 t − et − e2t + 0.22, x3 (t) ≈ 0.02eλ1 t − 2.13eλ2 t + 0.22eλ3 t + et + e2t − 0.11. 61.7 y(x) = −9 cos(2x)/40 − 11 sin(2x)/20 + 8ex /5 − x2 /4 − x/2 − 3/8 61.9 x1 (t) = −e2t /3 + 46e3t /27 − t/9 − 10/27, x2 (t) = −e2t /3+35e3t /81−13t/27+23te3t /27− 8/81, x3 (t) = e2t /3 − 34e3t /81 + 2t/27 + 23te3t /27 + 7/81. Chapter 62 62.3 L[x1 ](s) = −(3s + 13)/(2(s2 + 5s + 4)) + 5/(2(s + 1)), L[x2 ](s) = (2s − 1)/(s2 + 5s + 4), L[x3 ](s) = (s + 14)/(s2 + 5s + 4) and x1 (t) = 5e−t /2 − 3e−2t cos(t)/2 − 7e−2t sin(t)/2, x2 (t) = 2e−2t cos(t) − 5e−2t sin(t)/2, x3 (t) = e−2t cos(t) − 12e−2t sin(t). 62.7 a) L[x1 ](s) = 2/(s − 2) − 1/(s + 1), L[x2 ](s) = −1/(s + 1), L[x3 ](s) = 0 and x1 (t) = −e−t + 2e2t , x2 (t) = −e−t , x3 (t) = 0. b) L[x1 ](s) = 1/(s − 3) + 1/(s + 1) + 2/(s + 1)2 , L[x2 ](s) = 1/(s − 3) + 3/(s + 1)2 , L[x3 ](s) = 1/(s − 3) − 1/(s + 1) + 1/(s + 1)2 and x1 (t) = e−t + e3t + 2te−t , x2 (t) = e3t + 3te−t , x3 (t) = −e−t + e3t + te−t . 62.9 L[x1 ](s) = (2s + 1)/(6(s2 + 1)2 ) −(s − 1)/(9(s2 + 1)) − (8s + 1)/(9(s2 + 4)), L[x2 ](s) = (2s + 1)/(6(s2 + 1)2 ) +(s − 1)/(9(s2 + 1)) + (8s + 1)/(9(s2 + 4)) and x1 (t) = − cos(t)/9 − t cos(t)/12 − 8 cos(2t)/9 + 7 sin(t)/36 + t sin(t)/6 − sin(2t)/18, x2 (t) = cos(t)/9 − t cos(t)/6 + 8 cos(2t)/9 + sin(t)/18 + t sin(t)/3 + sin(2t)/18. Chapter 63 63.9 With the substitutions u1 = I1 , u2 = I10 , u3 = I2 , u4 = I20 , u5 = I3 , and u6 = I30 the network is described by the equations Solutions u1 (C1 + C2 )u3 R2 u4 R2 u6 − − + , L2 C1 L2 C1 C2 L2 L2 u50 = u6 , u40 = u60 = R2 u4 u5 (R1 + R2 )u6 − − . L3 C3 L3 L3 Setting λ1 := −433856.14, λ2 := −89369.25, λ3 := −5389.48, λ4 := −1784.48, a := −18133.65, b := 23771.07, and ω := 1000 and using the formula in (61.24), we find the following solutions: I1 (t) ≈ 7.47 · 10−12 eλ1 t + 0.00014eλ2 t − 0.0015eλ3 t − 0.00044eλ4 t + 0.0018 cos(ωt) − 0.000056eat cos(bt) + 0.00051 sin(ωt) − 1.48 · 10−6 cos2 (bt) sin(ωt)+0.000098eat sin(bt)−1.48· 10−6 sin(ωt) sin2 (bt) A, I2 (t) ≈ 1.09 · 10−9 eλ1 t + 6.98 · 10−6 eλ2 t − 0.00071eλ3 t − 0.00036eλ4 t + 0.00086 cos(ωt) + 0.00021eat cos(bt) + 0.00034 sin(ωt) − 3.72 · 10−6 cos2 (bt) sin(ωt)−0.000018eat sin(bt)−3.72· 10−6 sin(ωt) sin2 (bt) A, and I3 (t) ≈ −6.03 · 10−9 eλ1 t + 4.66 · 10−6 eλ2 t − 0.00068eλ3 t + 0.00045eλ4 t + 0.00012 cos(ωt) + 0.00011eat cos(bt) − 0.00012 sin(ωt) − 2.03 · 10−6 cos2 (bt) sin(ωt)−0.000011eat sin(bt)−2.03· 10−6 sin(ωt) sin2 (bt) A. Chapter 64 64.7 The functions in the first row of U(x) are y1 (x) = cos(x), y2 (x) = x cos(x), y3 (x) = x2 cos(x), y4 (x) = sin(x), y5 (x) = x sin(x), y6 (x) = x2 sin(x), and the solution is y(x) = 2 cos(x)+x cos(x)/2−x2 cos(x)/2−3 sin(x)/2+ 2x sin(x). 64.15 y(x) = 9e−x /25 + 11ex /25 + cos(2x)/5 + x cos(2x)/20 − 13 sin(2x)/200 64.17 y(x) = C1 e2x + C2 xe2x + C3 e2x cos(x) + C4 e2x sin(x) 64.19 From the transform L[y](s) = 10/(s − 2) −33/(s − 2)2 + (99 − 19s)/(21((s − 2)2 + 1)) + 1/(2(s − 1)) we find the same solution as in Exercise 64.18. Chapter 65 u10 = u2 , 65.1 See Figure S.9 u1 R1 u2 u3 E0 − + + , L1 C1 L1 L1 C1 L1 u30 = u4 , 65.15 π/4 √ 65.21 37/ 89 u20 = − 65.23 4x + y − 3z = −3 Solutions z = v1 w1 + v2 w2 = v · w. c) According to b),√ we may infer that √ kAvk = Av · Av = v · v = kvk. d) Since v/kvk is a vector of length 1, there must exist an angle φ ∈ [0, 2π) such that v/kvk = cos(φ)e1 + sin(φ)e2 . Using the definition of A, we thus obtain: cos(θ) cos(φ) − sin(θ) sin(φ) Av = kvk sin(θ) cos(φ) + cos(θ) sin(φ) cos(θ + φ) = kvk . sin(θ + φ) 3 2 1 x y 0 -1 -2 -1 0 1 1 2 0 -1 -2 Figure S.9: graph for Exercise 65.1. 65.27 The vector v × w points downward in the direction that a right screw would move in if turned in the direction from v to w. z x w 0 v 35 y v w -1 Consequently, Av is a counterclockwise rotation of v by θ. Note: the fact that Av is a rotation of v by θ also follows from the result in c) in conjunction with the observation that the angle between Av and v is cos(θ)kvk2 Av · v = arccos arccos kAvkkvk kvk2 = θ. 0 1 65.29 √ 2 2 1 -2 0 3 e1 e2 65.31 v × w = −1 3 2 1 e3 −6 0 = −2 −2 −7 65.33 cube: b), rectangular box: a) 65.35 L intersects only the lines given in c) and d). √ 65.37 38 65.39 If θ is the angle between v and w, then Pw (v) = kvk cos(θ)w/kwk = (v · w)w/kwk2 . 65.41 (a + b)/2 + (c − (a + b)/2)/3 = (a + c)/2 + (b − (a + c)/2)/3 = (b + c)/2 + (a − (b + c)/2)/3 = (a + b + c)/3 65.43 a sphere of radius 2 centered at (−1, 3, 2) √ 65.45 arccos(−1/ 84) √ 65.47 7/ 18 65.49 a) cos(θ) sin(θ) cos(θ) − sin(θ) t AA= sin(θ) cos(θ) − sin(θ) cos(θ) 1 0 = , 0 1 w1 v = b) A 1 · A v w 2 2 cos(θ)v1 − sin(θ)v2 cos(θ)w1 − sin(θ)w2 · sin(θ)v1 cos(θ)v2 sin(θ)w1 cos(θ)w2 However, this alternative argument does not allow us to draw any conclusions concerning the orientation of the rotation. Chapter 66 66.3 Using the given values for α, β, and m, it follows that 3mg − cos(β) −19.21 ≈ , Fa = sin(β) 58.86 2 sin(β) 3mg cos(β) 19.21 Fb = ≈ , sin(β) 58.86 2 sin(β) mg − cos(α) −18.80 Fp = ≈ , sin(α) 19.62 2 sin(α) mg cos(α) 18.80 ≈ . Fr = sin(α) 19.62 2 sin(α) Furthermore, according to Newton’s third fundamental law of mechanics, the magnitude of the forces at A and B is k − Fa k = k − Fb k = 61.92 kg · m/s2 . p 66.5 d = 100 − g 2 /4 ≈ 8.71 m 66.7 The force increases, and the relevant explanation is given at the end of Example 66.1. Chapter 67 67.29 Since t is the first component of r(t), as defined in Example 67.11, we may infer that 36 Solutions y t1 = t2 whenever r(t1 ) = r(t2 ). Thus r is simple. 5 4 67.31 Using a computer to solve for λ the equation 2λ 15 = L = sinh δg δgd 2λ 15λ ≈ sinh 9.81 7 · 9.81 15λ 3 a) 2 1 -4 -3 -2 -1 x 1 2 3 -1 b) -2 c) yields λ ≈ 1.993. Hence 4 -3 -4 f (x) ≈ 1.524 cosh(0.656x) − 7.653, s 2 kg · m d ≈ 10.01 . kFk = λ 1 + f 0 2 s2 Figure S.10: graphs for Exercise 67.39. y x 2π 67.35 If the length of the rope is given to be L, then λ is the solution of the equation 2λ sinh 1= δgL δgd 2λ . y x In setting x := δgL/2λ and y := δgd/2λ, this equation can be rewritten in the form x = sinh(y). As L approaches d, or equivalently, as x approaches y, the equation x = sinh(y) is transformed into the equation x = sinh(x). Since the only solution of the latter equation is x = 0, it follows that x = δgL/2λ must approach zero as L approaches d, or equivalently, λ must tend p ∞. By implication, the magnitude kFk = λ 1 + f 0 (d/2)2 = λ cosh(δgd/2λ) must increase to ∞ as well. This observation is intuitively not surprising, because one would expect the force needed to stretch a rope of positive density into a perfectly straight line to be infinite. 3 cosh(t) 3/ cos(t) 67.37 r(t) = or r(t) = 4 sinh(t) 4 tan(t) (answers may vary). 67.39 See Figure S.10. 2 1 67.41 tangent line: s(t) = 4 + t 4 0 4 R 2π q (1 + cos(θ))2 + sin2 (θ) dθ 67.43 a) L = 0 R 2π = 2 0 | cos(θ/2)| dθ = 8 (see Figure S.11). R π/2 q b) L = −π/2 cos2 (θ) + sin2 (θ) dθ R π/2 = −π/2 dθ = π (see Figure S.11). 1 Figure S.11: graphs for Exercise 67.43a, b. Rπ √ Rπ√ c) L = 0 θ4 + 4θ2 dθ = 0 θ θ2 + 4 dθ √ R π2 +4 √ 3 u/2 du = ( π 2 + 4 − 8)/3 = 4 y x 2 π 67.45 If we denote by θ the angle between the positive x-axis and the line segment from the center of the circle at (1/2, 0) to the center of the wheel, then a parameterization of the path of P is 1/2 + cos(θ) + cos(2θ)/2 r(θ) := . sin(θ) + sin(2θ)/2 Furthermore, using the double angle formulae for sine and cosine, we find that cos(θ)(1 + cos(θ)) r(θ) = . sin(θ)(1 + cos(θ)) Since this representation of r is identical with the one given in Example 67.32, we may infer Solutions 69.11 C = 0. 8 C = C -2 = 6 C= 2 C= 1 -2 = C = level curves of f 69.25 x -1 x C = -1 C 2 C because r1 − r2 is obviously parallel to F12 . √ 68.17 t = (6 + 3 4 + 2g)/g 2 = 0 + (r1 − p) × F12 + 0 + (r2 − p) × F21 = (r1 − r2 ) × F12 = 0 = = = r10 × m1 r10 + (r1 − p) × m1 r100 + r20 × m2 r20 + (r2 − p) × m2 r200 y C 1 M 0 0. 6 = C y C 68.11 Since d kr − r0 k2 = 2(r − r0 ) · r 0 = 0, dt we may conclude that kr − r0 k is constant and, by implication, the curve described by r is contained in a sphere centered at r0 . 68.13 Since d/dt n·r = n·r 0 = 0, it follows that n · r is constant. Therefore, n · r = n · r(0) or, equivalently, n·(r−r(0)) = 0. This shows that r traces out a curve that is contained in the plane described by the equation n · (r − r(0)) = 0. 68.15 Using Newton’s second and third laws, we see that m1 r100 + m2 r200 F12 + F21 = C 00 = m1 + m2 m1 + m2 F12 − F12 = = 0, m1 + m2 and similarly, x 69.13 = and, by implication, r ! r 2g π g 0 kr (10)k ≈ ≈ 0.086 m/s. sin 10 20 6 3 0. 4 = 0. 2 y C p (π/40) cos( 2g/3 t) , it 68.9 With r(t) ≈ −3/2 follows that p p −(π/20) g/6 sin(10 2g/3) r 0 (10) ≈ 0 69.1 F = −γmM (r0 − r)/kr0 − rk3 = Chapter 68 Chapter 69 C that the curve described by r is indeed the polar graph shown in Figure 67.9. √ 67.47 center: (1/6, −1/6, 1/3); radius: 210/6 67.49 a) not parallel and no intersection, b) parallel, but no intersection, c) not parallel, but intersection at the point (6/5, −13/5, 7/5) 67.51 c) and d) are simple because the ycoordinate 3t3 is invertible on R, and b) is simple as well because the given x- and ycoordinates are both invertible on (0, 3). By contrast, a) is not simple, because r(−1) = r(1). 37 level curves of g y x 1 3 69.31 With h(x, y) = x, f (x) = x, and g(y) = e y 3 , we find that Fe (x) = ln |x| and G(y) = 2 −1/(2y ). Consequently, the equation of the flow lines is ln |x| + 1/(2y 2 ) = C. 69.33 a) z C=1 C=0 C = -1 38 Solutions b) 69.39 a) = 0 y C 2 2 C = 1 1 1/2 C= C= x -2 c) c) -1 0 1 2 y 3 C=2 2 C=1 1 C=0 π/2 x C = -1 -1 d) C = −1 C=0 69.41 a) b) c) d) e) f) C=1 69.35 graph contour diagram a b b d c a d c 69.37 a) If P is an antiderivative of g/f , then P (x) − y = C is the xy-equation of the flow lines. b) Here the equation is x − Q(y) = C, where Q is an antiderivative of f /g. 69.43 Proceeding exactly as in Exercise 69.42, we obtain 4t + 1 r(t) = −e−3t 4t − 3 (answers may vary). 69.45 b) f (x, y) := x − ey , c) f (x, y) := y − ex (answers may vary). 69.47 b) and d). Solutions Chapter 71 71.5 ∂f /∂x = 2xyz 2 cos(x2 y), ∂f /∂y = x2 z 2 cos(x2 y), and ∂f /∂z = 2z sin(x2 y). 71.7 equation: −2x + 2y + z = 2, perpendicular vector: n = −2e1 + 2e2 + e3 71.23 If f and g are continuous at r0 , then f (r) and g(r) approach f (r0 ) and g(r0 ), respectively, as r approaches r0 . Consequently, the sum of f (r) and g(r) must approach the sum of f (r0 ) and g(r0 ). In other words, f + g is continuous at r0 . Analogous arguments also serve to establish the continuity of λf , f g and f /g. Furthermore, if r is a continuous parameterization, then r(t) approaches r(t0 ) as t approaches t0 . Consequently, the assumption of continuity on f allows us to infer that f ◦ r is continuous, because as t approaches t0 the values f (r(t)) must approach f (r(t0 )). This completes the proof of Theorem 71.21. Turning our attention now to Theorem 71.22, we first observe that f clearly is continuous at a point r0 ∈ D if and only if lim |f (r0 + u) − f (r0 )| = 0. u→0 Since |f (r0 + u) − f (r0 )| = |f (r0 + u) − f (r0 ) − ∇f (r0 ) · u + ∇f (r0 ) · u|, the assumption of differentiability on f implies that lim |f (r0 + u) − f (r0 )| = lim |∇f (r0 ) · u| = 0, u→0 u→0 as desired. √ 71.27 ∂u f (1, −3) = −3/ 17 71.31 V (r) = −γmM/kr − r0 k 71.33 f (x, y) = x2 y 2 − x2 /2 + C 71.35 Only G is a gradient field. 71.37 f (r) = − ln(krk) + C 71.43 a) yexy sin(x) + exy cos(x) + 2xy, b) xexy sin(x) + x2 , c) yz cos(xyz) cos(x2 y 2 z 2 ) −2xy 2 z 2 sin(xyz) sin(x2 y 2 z 2 ), d) xy cos(xyz) cos(x2 y 2 z 2 ) −2x2 y 2 z sin(xyz) sin(x2 y 2 z 2 ) 71.45 Here the condition to be satisfied is −∂f /∂x 1 −∂f /∂y k 1 , 1 2 39 or equivalently, −2∇f = e1 + e2 . Since 2 y −1 ∇f = , 2yx + 2 it follows that −2(y 2 − 1) = 1 and −2(2yx + 2) = 1, and the points whose coordinates simultaneously satisfy both of√ these equations √ are√ easily seen to be (−5 2/4, 1/ 2) and √ (5 2/4, −1/ 2). 71.47 ∂u f = 0. Intuitive explanation: since the curve described by the parameterization r(θ) = (R cos(θ), R sin(θ)) is a level curve of f and since u = r 0 (θ)/kr 0 (θ)k is tangent to this level curve, it is natural to expect the directional derivative of f in the direction of u to be zero. 71.49 Setting F(x, y) := (x − y)e1 + (x + y)e2 and denoting by θ the angle between F(x, y) and r := (x, y), it follows that cos(θ) = 1 F(x, y) · r = √ , kF(x, y)kkrk 2 and therefore, θ = π/4. Applying now the techniques developed in Chapter 60 to solve the initial value problem 0 x 1 −1 x x(0) 1 = , = , 1 1 y0 y y(0) −1 we find the parameterization t x(t) e (cos(t) + sin(t)) r(t) = = t . y(t) e (sin(t) − cos(t)) √ 71.51 ∂u f = 18/ 5 71.53 In each case the flow lines are to be drawn in such a way that they meet the given level curves at a right angle at all points. 71.55 a) ∂u f (r) = ∇f (r) · u = k∇f (r)k ≥ 0. b) ∇f is a vector in R2 , but the graph of f is a subset of R3 and, by implication, any vector tangent to the graph must be in R3 as well. c) Same mistake as in b). d) The derivative of a multivariable function is a vector (the gradient of f ), not a number. e) The stated conditions imply that |f (r + u) − f (r) − ∇f (r) · u| kuk |∇f (r) · u| = lim k∇f (r)k| cos(θ)| = lim u→0 u→0 kuk √ = lim 5| cos(θ)|, 0 = lim u→0 u→0 40 where θ is the angle between ∇f (r) and u. However, the limit of | cos(θ)| does not exist because the value of θ is entirely arbitrary (u may approach 0 from any direction). f ) ∂u f (r(t)) = ∇f (r(t))·r 0 (t) = r 0 (t)·r 0 (t) ≥ 0. g) If r parameterizes a level curve, the function f ◦ r is constant and ∂u f (r(t)) = ∇f (r(t)) · r 0 (t) = d/dt f (r(t)) = 0 6= −2. h) On an equipotential surface the potential energy remains constant and so does the sum of the potential and kintetic energies (by the law of the conservation of energy). Consequently, in contradiction to the stated assumption K(1) = 2 6= 4 = K(3), the kinetic energy must remain constant as well. 71.57 71.59 Using the well known values R ≈ 1.5 · 1011 m, T ≈ 365.25 · 24 · 3600 s, and γ ≈ 6.672 · 10−11 m3 s−2 kg−1 , it follows that M ≈ 4π 2 R3 /(γT 2 ) ≈ 2.005 · 1030 kg. Chapter 72 72.1 If either q > 0 and Q > 0 or q < 0 and Q < 0, then qQ > 0, and F(r) points in the direction from q to Q, thus modeling an attractive force. Otherwise qQ will be negative, and F(r) will represent a repellent force pointing away from Q. 72.13 With r(t) = (3t, 1, 2 − t) for t ∈ [0, 1] the value of the path integral turns out to be 24. 72.15 a) zero, b) negative, c) negative, d) positive, e) negative, f ) zero 72.17 = −exy + C, R a) V (x, y) −3 b) C F · dc = e − e2 , c) −e2 + C + 10 = V (1, 2) + 10 = V (−1, 3) + 2 −3 2 mv p /2 = −e + C + 2v , and therefore, v = −3 2 (e − e + 10)/2. R2 72.19 a) 1 F(r(t)) · r 0 (t) dt = 0 Solutions b) Setting α := arctan(3/2), we define √ 13 cos(t) √ s(t) := 13 sin(t) for t ∈ [α, α + 2π] and 2(t − α − 2π + 1) s(t) := 3(t − α − 2π + 1) for t ∈ [α+2π, α+2π +1]. Then s(α) = (2, 3) = r(1), s(α + 2π + 1) = (4, 6) = r(2), and Z α+2π+1 F(s(t)) · s 0 (t) dt α = Z α+2π 13 dt + α Z α+2π+1 0 dt = 13 6= 0. α+2π Rb 72.21 a) FALSE because a F(r(t)) · r 0 (t) dt = Rb 0 kr (t)k2 dt ≥ 0. a b) FALSE for the same reason as in a). c) POSSIBLY TRUE/POSSIBLY FALSE: set√ ting r(t) := (cos(t), sin(t))/ 2π for t ∈ [0, 2π] and s(t) := (cos(2t), sin(2t)) for t ∈ [0, π], it is easy to see that r and s describe closed flow lines of the vector fields F(x, y) := (−y, x) and G(x, y) := 2(−y, x) respectively. However, the corresponding path integral is equal to 1 only for F and r, but not for G and s. d) FALSE because there are no closed flow lines in gradient fields. e) POSSIBLY TRUE/POSSIBLY FALSE: the examples in c) apply also in this case. f ) FALSE because the path integral of a gradient field over a closed curve is always equal to zero. g) FALSE: F and r (or G and s) in c) provide a counterexample. h) TRUE because the path integral of a gradient field over a closed curve is indeed always equal to zero. i) FALSE: the vector field F(x, y) = (y, 0) defined on R2 serves as a counterexample. 72.23 the parameterization in b) Chapter 73 73.3 See Figure S.12. 73.13 Since the solutions of the quadratic equa√ tion ax2 + bx + c = 0 are (−b ± b2 − 4ac)/2a, it follows that real solutions exist if and only if b2 − 4ac ≥ 0 or, equivalently, ac − b2 /4 ≤ 0. Solutions global maximum local maximum global minimum 41 From the equality of the second components, we may infer that either y = 0 and, by implication, x = ±a, or λ = b2 . In the latter case, the equality of the first components in the vector equation above implies that x = a2 x0 /(a2 − b2 ). Given the constraint g(x, y) = x2 /a2 + y 2 /b2 = 1, it is easy to see that the corresponding value of y satisfies the equation a2 x20 2 2 . y =b 1− 2 (a − b2 )2 Figure S.12: graph for Exercise 73.3. Furthermore, for both equations in (73.12) it is the case that ac = ∂ 2 f /∂x2 (r0 )∂ 2 f /∂y 2 (r0 ) and b2 /4 = (∂ 2 f /∂y∂x(r0 ))2 . Consequently, the term shown in (73.13) is indeed the common discriminant of the equations in (73.12). √ 73.15 The √ critical points are (0, 0), (0, 1/ 2) and (0, −1/ 2).√Furthermore, since √ Df (0, 0) = 8 > 0, Df (0, 1/ 2) = Df (0, −1/ 2) = −14 < 0 and ∂ 2 f /∂x2 (x, y) = −2(y 4 − y 2 + 2) < 0, we may√conclude that f has saddle points at (0, ±1/ 2) and a local maximum at (0, 0). 73.21 Setting f (x, y) := (x − x0 )2 + (y − y0 )2 and g(x, y) := y − mx − b yields 2(x − x0 ) −λm = ∇f = λ∇g = . λ 2(y − y0 ) In eliminating λ and using the constraint g(x, y) = 0, we find that (x − x0 )/m = −(y − y0 ) = y0 − mx − b, or equivalently, x = (x0 − mb + my0 )/(1 + m2 ). The corresponding value for y is (mx0 + m2 y0 + b)/(1 + m2 ). Substituting x and y back into the defining equation for f yields f (x, y) (my0 − mb − m2 x0 )2 (mx0 + b − y0 )2 + 2 2 (1 + m ) (1 + m2 )2 (mx0 + b − y0 )2 = . 1 + m2 = Finally, in taking the square root we find the same formula for the minimal distance as in Chapter 12. 73.25 In setting f (x, y) := (x − x0 )2 + y 2 and g(x, y) = x2 /a2 + y 2 /b2 , we obtain 2(x − x0 ) 2x/a2 =λ . 2y 2y/b2 Substituting the values for x and y 2 back into the defining equation for f yields x20 2 . f (x, y) = b 1 − 2 a − b2 To determine the minimal distance, we need to compare this value of f with f (a, 0) = (a − x0 )2 and f (−a, x0 ) = (a + x0 )2 . In fact, we only need to decide whether the value of f above is less than or greater than (a − x0 )2 because the assumption x0 > 0 implies that (a − x0 )2 < (a + x0 )2 . Since x20 2 2 (a − x0 ) − b 1 − 2 a − b2 2 b2 a2 a − x − , = 2 0 a − b2 a and since the term on the right-hand side of this equation is less than or greater than zero depending on whether a < b or a > b, we may infer that the minimal distance from the ellipse to the point (x0 , 0) is ( if a < b, a − x0 dmin = 2 2 2 2 b (1 − x0 /(a − b )) if a > b. 73.27 Setting f (a, b, c) := (1 − (a − b + c))2 + (−1 − (a + b + c))2 + (0 − (4a + 2b + c))2 + (6 − (25a + 5b + c))2 , it is easy to see that the equation ∇f = 0 is equivalent to the matrix equation 643 131 31 a 150 131 31 7 b = 28 . 31 7 4 c 6 42 Solutions Multiplying both sides with the inverse of the matrix on the left yields a = 5/12, b = −49/60, and c = −3/10 (see Figure S.13 for the graph). Note: the symmetric arrangement of the coefficients in the matrix above is not coincidental— think about it. y 10 8 6 4 2 x -2 2 4 ∂2 f ((2n + 1)π/2, (2m + 1)π/2) = 2 ∂x ( −1 if n and m are both even or both odd, 1 otherwise, 6 -2 Figure S.13: graph for Exercise 73.27. 73.29 Since a parameterization of the plane is s , t r(s, t) := D/C − As/C − Bt/C it follows that the minimal distance is the minimum of f (s, t) := kr(s, t) − pk, where p = x0 e1 + y0 e2 + z0 e3 . In setting the gradient of f equal to 0 we find that AD + (B 2 + C 2 )x0 − ABy0 − ACz0 , A2 + B 2 + C 2 BD + (A2 + C 2 )y0 − BAx0 − BCz0 t= , A2 + B 2 + C 2 s= and the corresponding value of f is dmin = f (s, t) = 1)π/2, (2m + 1)π/2), where n and m are understood to be arbitrary integers. Since Df (x, y) = sin2 (x) sin2 (y) − cos2 (x) cos2 (y), we may infer that f has a saddle point at each point of the form (nπ, mπ) because Df (nπ, mπ) = −1 < 0. Furthermore, f assumes a local extremum at each point of the form ((2n+1)π/2, (2m+1)π/2) because Df ((2n + 1)π/2, (2m + 1)π/2) = 1 > 0. Since |Ax0 + By0 + Cz0 − D| √ . A2 + B 2 + C 2 73.31 Since the area of the triangle is easily seen to be (bx+ay +ab)/2, we define f (x, y) := (bx+ ay + ab)/2, and the constraint is represented by the equation g(x, y) := x2 /a2 + y 2 /b2 = 1. In solving for x, y, and λ the equations ∇f (x, y) = λ∇g(x, √ y) = 1, we find that x = √ y) and g(x, Consequently, the maxia/ 2 and y = b/ √ 2. √ √ mal area is f (a/ 2, b/ 2) = ( 2 + 1)ab/2. 73.33 a) The critical points are (0, 0), (2, 0) and (3, 0). Since Df (0, 0) < 0, Df (2, 0) > 0, Df (3, 0) < 0, and ∂ 2 /∂x2 f (2, 0) > 0, we may infer that f has a local minimum at (2, 0) and saddle points at (0, 0) and (3, 0). b) The set of critical points consists of the points of the form (nπ, mπ) and ((2n + it follows that f assumes a local maximum at ((2n + 1)π/2, (2m + 1)π/2) whenever n and m are both even or both odd and otherwise a local minimum. 73.35 There are no local extrema, but f has saddle points at (0, 0, 0) and all points of the form (2, y, 1), (−2, y, 1), (2, y, −1) and (−2, y, −1) (where y is understood to be an arbitrary real number). 73.37 If S = (x0 , y0 , z0 ), E = (x1 , y1 , z1 ), and R = (x, y, 0) then the distance that the light ray travels from S to E via R is q D(x, y) = (x − x0 )2 + (y − y0 )2 + z02 q + (x − x1 )2 + (y − y1 )2 + z12 . Setting q (x − x0 )2 + (y − y0 )2 + z02 , q f1 (x, y) := (x − x1 )2 + (y − y1 )2 + z12 , f0 (x, y) := it is not difficult to show that (x − x0 )/f0 (x, y) ∇D(x, y) = (y − y0 )/f0 (x, y) (x − x1 )/f1 (x, y) . + (y − y1 )/f1 (x, y) Consequently, (x, y) is a critical point of D only if y − y0 x − x0 f0 (x, y) = =− , x − x1 f1 (x, y) y − y1 that is, only if (x − x0 )(y − y1 ) − (y − y0 )(x − x1 ) = 0. Solutions 43 Given this result, we may conclude that the third component of the vector x − x0 x − x1 n = y − y0 × y − y1 −z0 −z1 e) Since the first and second components of the particle’s velocity vector are not affected by the passage through the potential barrier, we may infer that x2 = v1+ T and y2 = v2+ T . Given this observation, it follows that is zero. In other words, n is parallel to the xyplane and, by implication, the plane containing S, E, and R is perpendicular to the xy-plane. 73.39 a) According to (73.31) and (73.32), we may infer that m m kv+ k2 + V+ = kv+ − we3 k2 + V− , 2 2 or equivalently, m(x1 − x2 ) mx1 + = mv1+ − mv1+ = 0, t+ T − t+ 2(V+ − V− ) = w2 + 2v3+ w. m Solving for w yields r 2(V+ − V− ) + , w = −v3 ± (v3+ )2 + m and similarly, my1 m(y1 − y2 ) + = mv2+ − mv2+ = 0. t+ T − t+ In other words, the first two components of ∇f (x1 , y1 , t+ ) are equal to zero. Furthermore, since krx1 ,y1 − r0 k = kr1 − r0 k = kv+ kt+ , kr2 − rx1 ,y1 k = kr2 − r1 k = kv− kt− , we may infer that mkr2 − rx1 ,y1 k2 mkrx1 ,y1 − r0 k2 − 2 2(T − t+ ) 2t2+ m m = kv− k2 − kv+ k2 2 2 = V+ − V− (by (73.31)). but since w must be negative, it follows that r 2(V+ − V− ) + w = −v3 − (v3+ )2 + . m b) Since T = t+ + t− , we obtain Z T Z t+ m kv+ k2 − V+ dt L(t) dt = 2 0 0 Z T m + kv− k2 − V− dt 2 t+ m = kv+ k2 − V+ t+ 2 m kv+ + we3 k2 − V− t− . + 2 c) Performing a calculation analogous to the one in b), we obtain Z T mkrx,y − r0 k2 Lx,y,τ (t) dt = − τ V+ 2τ 0 mkr2 − rx,y k2 − (T − τ )V− . + 2(T − τ ) d) In denoting the first and second components of r2 by x2 and y2 respectively, we find ∇f to be equal to mx m(x − x2 ) + τ T −τ my m(y − y ) 2 . + τ T − τ 2 2 mkr2 − rx,y k mkrx,y − r0 k − + V− − V+ 2(T − τ )2 2τ 2 Consequently, the third component of ∇f (x1 , y1 , t+ ) is zero as well, and (x1 , y1 , t+ ) has thus been shown to be a critical point of f . Chapter 74 74.7 For n = 1, we have m1 X lim m1 →∞ ∆x1,i1 →0 i1 =1 f (ri1 )∆x1,i1 = lim m X m→∞ ∆xi →0 i=1 R1R2R3 f (xi )∆xi . xy 2 z 3 dz dy dx = 27 √ √ R 3 74.25 2(4 4 − x2 − 2 4 − x2 /3) dx R = R (32 cos2 (t) − 64 cos4 (t)/3) dt = (16 cos(2t)/3 + 8t − 8 cos(4t)/3) dt = 8 sin(2t)/3 + 8t − 2 sin(4t)/3 + C = 16 sin(t) cos(t)/3 + 8t −8 sin(t) cos(t)(2 cos2 (t) − 1)/3 √ +C = 8 arcsin(x/2) + (x3 + 2x) 4 − x2 /3 + C R 2 R √4−y2 74.27 −2 √ 2 (x2 + y 2 ) dx dy − 4−y R1R1√ 74.29 0 y 1 − x2 dx dy R 1 R π/2 = 0 arcsin(y) (cos(2t) + 1)/2 dt dy 74.19 0 0 0 44 Solutions R1 π/2 (sin(t) cos(t) + t)/2|arcsin(y) dy p R01 = 0 (−y 1 − y 2 − arcsin(y) + π/2)/2 dy R 1 R √1−x2 R √1−x2 −y2 74.31 −1 −√1−x2 √ 2 2 dz dy dx − 1−x −y R 1 R √1−x2 p √ = −1 − 1−x2 2 1 − x2 − y 2 dy dx R 1 R π/2 2(1 − x2 ) cos2 (t) dt dx −1 −π/2 R1 = −1 π(1 − x2 ) dx = 4π/3 = = 1/3 74.37 By symmetry, the first and second components of the force on q must be zero. Using the given values for q, d, and δ, and setting ∆x := ∆y := 0.1 and xk := yk := −2 + 0.1k, an approximation for the third component is obtained as follows: 40 40 1 1 XX F3 ≈ q 3 ∆x∆y πε0 i=1 j=1 x2i + yj2 + 4 ≈ 1.04687 . πε0 Alternatively, we can choose xk and yk to be the midpoints of the subintervals in the partition of [−2, 2] along the x- and y-axes. So in setting xk := yk := −2 + 0.1(2k − 1)/2 we find the (presumably improved) estimate R π/2 R 1 y sin(xy) dx dy 74.43 a) 0 0 R π/2 = 0 (1 − cos(x)) dx = π/2 − 1, R 1 R π/2 b) 0 0 y sin(xy) dy dx R1 = 0 (−π cos(πx/2)/(2x) + sin(πx/2)/x2 ) dx = limc→0 − sin(πx/2)/x|1c = π/2 − 1 74.45 a) 1/2, b) 1/4, c) 1/8, d) 1/2n 74.47 of f , it follows that R 1 R 1 Given the definition P∞ f (x, y) dx dy = n=1 (2n /2n −2n /2n ) = 0 0 0R R P∞ 1 1 and 0 0 f (x, y) dy dx = 1 + n=1 (−2n /2n + 2n+1 /2n+1 ) = 1. However, this surprising observation does not contradict the result stated in 74.10 because f is not bounded on [0, 1] × [0, 1]. 74.49 First row beginning from the left: R 2 R 2−x/2 f (x, y) dy dx, R02 Rx/2 R1 R1 2 f (x, y) dy dx − −1 0 f (x, y) dy dx, −2 0 R 1 R |y| f (x, y) dx dy. −1 −|y| Second row beginning from the left: R 1 R 1−|x| f (x, y) dy dx, −1 −1+|x| R 3π/4 R sin(x) f (x, y) dy dx, − cos(x) R3Rx R−π/4 1Rx f (x, y) dy dx + 1 √x2 −1 f (x, y) dy dx. 0 0 74.51 a) z 1 40 40 1 1 XX F3 ≈ q 3 ∆x∆y πε0 i=1 j=1 x2i + yj2 + 4 ≈ 1.04735 . πε0 74.39 a) y 1 1 x z b) z 4 y 3 2 2 2 x 1 c) 0 z y 2 x R b) R S f (r) dA = 0·2+1·2+2·2+3·2+4·8 = 44, c) S f (r) dA = 0 · 25/2 + 1 · 25/2 = 25/2 R 74.41 R a) S f (r) dA = 12, b) RS f (r) dA = 81 for S = [0, 3] × [0, 3] × [0, 3] andR S f (r) dA = 288 for S = [0, 4]×[0, 4]×[0, 4]. c) S f (r) dA = 3n3 (n − 1)/2 y -1 2 x 1 d) See Figure S.14. Solutions z 45 c) y 1 2 1/2 3 π/4 x y 1 x d) 1 y 1 Figure S.14: graph for Exercise 74.51d. 74.53 left to right: R 2 R 1 RFrom 1−y f (x, y, z) dz dy dx, R04 R01 R0 1−|y| f (x, y, z) dz dy dx, 0 −1 |y|−1 √ R 3 R 2 R 1+ 1−y2 √ 2 f (x, y, z) dz dy dx. 0 0 1− x 0 e) 1−y 1 z Chapter 75 75.3 R3R0 75.5 R 2π R ∞ 0 −π/2 0 0 r2 (cos(θ) − sin(θ)) dr dθ = 18 r/(1 + r2 )2 dr dθ = π y 75.7 From left to right: R π/6 R 2/ cos(θ) f (r cos(θ), r sin(θ))r dr dθ, R0π/2 R02 cos(θ) f (r cos(θ), r sin(θ))r dr dθ. 0 0 √ 75.13 r = 5, θ = arctan(2), z = 3 R 2π R 2 R 3 75.15 0 0 2 (r2 + z)r dr dz dθ = 75π 75.17 R 2 R 2π R z/2 0 0 0 x f ) The region is a torus with centerline radius 2 and inner radius 1. f (r cos(θ), r sin(θ), z)r dr dθ dz 75.27 a) y 2 1 x 0 b) 1 2 3 y 3 R1Rπ 75.29 a) 0 0 f (g(s, t)) ds dt, R1R1 b) 0 0 6f (g(s, t)) ds dt, R1R2 c) 0 0 f (g(s, t)) ds dt, R π/4 R 1 d) 0 sf (g(s, t))/ cos2 (t) ds dt, R 1 R 1 R0 2π e) 0 0 0 f (g(θ, r, z))rz 2 dθ dr dz, R 1 R 2π R 2π f ) 0 0 0 f (g(θ, φ, r))r(2r+cos(φ)) dθ dφ dr R π/2 R 1 75.31 a) −π/2 0 r2 cos(θ) dr dθ, R π R −1/ cos(θ) 3 2 b) 3π/4 0 r sin (θ) dr dθ, R π/2 R 1/ cos(θ) c) −π/2 0 r/(1 + r2 ) dr dθ, R π/2 R cos(θ) 3 d) −π/2 0 r dr dθ 2 1 x 0 1 2 R α/2 R 2π R R 2 ρ sin(φ) dρ dθ dφ 75.33 V = 0 0 0 = 2πR3 (1 − cos(α/2))/3. Furthermore, for α = 2π we have cos(α/2) = −1, and therefore, the volume of a ball of radius R is 4πR3 /3. 46 Solutions 75.35 a) z 1 y x 1 1 R 2π R 1 R 1−r2 r3 cos(θ) sin(θ) dz dr dθ 0 0 0 R π/2 R g(φ) 4 R 2π c) 0 0 ρ cos(θ) sin(θ) sin3 (φ) dρ dφ dθ 0 with q − cos(φ) + 1 + 3 sin2 (φ) . g(φ) = 2 sin2 (φ) b) R R πhR2 /3, W x dx dy dz = W y dx dy dz = R 2π R R R h R and W z dx dy dz = 0 0 hr/R zr dz dr dθ πh2 R2 /2. Consequently, the center of mass at (0, 0, 3h/4). p R z(t) √ 76.3 Since −2 −z(t) = 0 1/ −z dz p Rt 0 Rtp z (t)/ −z(t) dt = − 0 4g/3 sin(α) dt 0p − 4g/3 sin(α)t, we may infer that z(t) −g sin2 (α)t2 /3 as desired. 75.39 Setting h(θ, r, z) := f (r cos(θ), r sin(θ), z), we findR theR following integrals: 2π 2 R 4 a) 0 0 r2 h(θ, r, z)r dz dr dθ, R 2π R π/2 R cos(z) h(θ, r, z)r dz dr dθ, b) 0 −π/2 0 R 2π R 2 R √2−r2 c) 0 0 0 h(θ, r, z)r dz dr dθ, R 2π R 1 R 3 d) 0 0 3r h(θ, r, z)r dz dr dθ. 75.41 From left to right: Jg (x, ρ, σ) = 3(1 − x/2)2 (1 − ρ), Jg (y, ρ, θ) = ρf (y)2√ , Jg (θ, ρ, σ) = 5σ 2 ρ/ 2. 75.43 cylindrical coordinates: R 2π R 1 In R 2−2r f (g(θ, r, z))r dz dr dθ, 0 0 0 where g(θ, r, z) represents the transformation to cylindrical coordinates. In spherical coordinates: R 2π R π/2 R 1/(cos(φ)+2 sin(φ)) h(θ, φ, ρ) dρ dφ dθ, 0 0 0 where h(θ, φ, ρ) = f (g(θ, φ, ρ))ρ2 sin(φ). Here g(θ, φ, ρ) represents the transformation to spherical coordinates. Chapter 76 76.1 Using cylindrical coordinates, we find R 2π R R R h R that W dx dy dz = 0 0 hr/R r dz dr dθ = = = = 76.5 In performing an integration in spherical coordinates, we easily determine the moment of inertia I to have the value mR2 /5. The corresponding kinetic energy is K(t) = 3mkr 0 (t)k2 /5. Using these results, a calculation analogous to that for a rolling wheel (as discussed in Chapter 76) yields 75.37 4s cos(π/6) − 2t sin(π/6) a) g(s, t) = = 4s sin(π/6) + 2 cos(π/6)t √ 2 3s√− t for (s, t) ∈ [0, 1] × [0, 1] (answers 2s + 3t may vary), b) Jg = 8, √ √ R1R1 c) 0 0 8(2 3s − t)(2s + 3t) ds dt 0, = is 5g sin(α) cos(α)t2 , 12 5g z(t) = − sin2 (α)t2 . 12 y(t) = Chapter 77 77.3 R 2π R √2 0 Z 0 re−r 2 /2 /(2π) dr dθ = 1 − e−1 2 810 e−(x−800) /800 √ dx ≈ 0.533 20 2π 780 77.13 Using the result of Exercise 77.12 with = 0 and σ1 =R 1, we obtain: √ Rµ1µ+σ 2 1 p(x) dx = −1 e−x /2 / 2π dx ≈ 0.68, µ−σ √ R2 R µ+2σ 2 p(x) dx = −2 e−x /2 / 2π dx ≈ 0.95, µ−2σ √ R µ+3σ R3 2 p(x) dx = −3 e−x /2 / 2π dx ≈ 0.997. µ−3σ 77.11 Chapter 78 78.3 1 1−1 0−1 p(s, t) = 0 + s 2 − 0 + t 2 − 0 2 3−2 4−2 1−t = 2s + 2t (answers may vary). 2 + s + 2t x 78.9 p(x, y) = y for any point x2 + y 2 2 (x, y) ∈ R whose coordinates satisfy the inequality x2 + y 2 ≤ 1. Solutions a cos(θ) sin(φ) 78.13 p(θ, φ) = b sin(θ) sin(φ) for (θ, φ) ∈ c cos(φ) [0, 2π] × [0, π] (answers may vary). 78.15 Example 78.2: the plane is oriented in the direction of v × w = ∂p/∂s × ∂p/∂t. Example 78.5: the positive normal vector rh cos(θ)/R ∂ ∂ p× p = rh sin(θ)/R ∂θ ∂r −r induces a downward orientation toward the outside of the cone. Example 78.7: since the z-component of the positive normal vector −∂f /∂x ∂ ∂ p× p = −∂f /∂y ∂x ∂y 1 is positive, it follows that the surface represented by the graph of f is oriented upward. Example 78.10: in this case the positive normal vector R cos(θ) ∂ ∂ p× p = R sin(θ) ∂θ ∂z 0 induces an orientation toward the outside of the cylinder. 78.23 Using the parameterization 1 + cos(θ) sin(φ) p(φ, θ) = 2 + sin(θ) sin(φ) 1 + cos(φ) with cos(θ) sin(φ) ∂ ∂ p× p = sin(φ) sin(θ) sin(φ) , ∂φ ∂θ cos(φ) it follows that ∂ ∂ p× p F(p(φ, θ)) · ∂φ ∂θ 1 + cos(θ) sin(φ) cos(θ) sin(φ) · sin(θ) sin(φ) 0 = sin(φ) 0 cos(φ) = cos(θ) sin2 (φ) + cos2 (θ) sin3 (φ), and therefore, Z F · dA S(p) Z 2π = 0 Z π F(p(φ, θ)) · 0 ∂ 4π ∂ p× p dφ dθ = . ∂φ ∂θ 3 47 78.29 Example 78.5: Z 2π Z R r 2 2 r h A= + r2 dr dθ R2 0 0 p = πR h2 + R2 . Example 78.10: Z 2π Z A= 0 h R dz dθ = 2πRh. 0 78.31 Substituting the components of p into the given equation for the plane yields 2r cos(θ) + r sin(θ) + r = 2, or equivalently, r= 2 . 2 cos(θ) + sin(θ) + 1 Consequently, a parameterization for the curve of intersection is cos(θ) 2 sin(θ) r(θ) = 2 cos(θ) + sin(θ) + 1 1 for θ ∈ [0, 2π]. Since the angle between the normal vector n = 2e1 +e2 +e3 and the positive z-axis (i.e., the axis of symmetry of the cone) is √ easily seen to be arccos(1/ 6) ≈ 65.9◦ > 45◦ , the curve described by r must be a hyperbola. 78.33 surface area √ 3 paraboloid π( 17 − 1)/6 spherical cap 2πRh √ √ solid of revolution 2π( 2 + ln(1 + 2)) 78.35 First row on the left: positive; first row on the right: zero; second row on the left: zero; second row on the right: positive. Chapter 79 79.3 div F(r) = 5krk2 79.7 The parameterization of... r cos(θ) ...the upper annulus is p1 (r, θ) = r sin(θ) 2 for (r, θ) ∈ [1, 2] × [0, 2π]. r cos(θ) ...the lower annulus is p2 (θ, r) = r sin(θ) 0 48 Solutions for (θ, r) ∈ [0, 2π] × [1, 2]. 2 cos(θ) ...the outer cylinder is p3 (θ, z) = 2 sin(θ) z for (θ, z) ∈ [0, 2π] × [0, 2]. cos(θ) ...the inner cylinder is p4 (z, θ) = sin(θ) z for (z, θ) ∈ [0, 2] × [0, 2π]. The corresponding normal vectors are positive 0 ∂ ∂ p1 × p1 = 0, ∂r ∂θ r 0 ∂ ∂ p2 × p 2 = 0 , ∂θ ∂r −r 2 cos(θ) ∂ ∂ p3 × p3 = 2 sin(θ) , ∂θ ∂z 0 − cos(θ) ∂ ∂ p4 × p4 = − sin(θ) , ∂z ∂θ 0 and therefore, Z Z 2 Z 2π r cos(θ) 0 r sin(θ) · 0 dθ dr F · dA = ∂D 1 0 0 r Z 2π Z 2 r cos(θ) 0 r sin(θ) · 0 dθ dr + 0 1 0 −r Z 2 Z 2π 2 cos(θ) 2 cos(θ) 2 sin(θ) · 2 sin(θ) dθ dz + 0 0 z 0 Z 2π Z 2 cos(θ) − cos(θ) sin(θ) · − sin(θ) dz dθ + 0 0 0 0 = 0 + 0 + 16π − 4π = 12π. Since div F(r) is easily seen to be equal to 2 for all r, we find the same value as above by integrating the divergence over D: Z 2π Z 2 Z 2 Z div F(r) dV = 2r dz dr dθ = 12π. D 0 1 0 79.9 a) The surface of V consists of the following six faces: S1 = {0} × [0, 1] × [0, 1], S2 = {1} × [0, 1] × [0, 1], S3 = [0, 1] × {0} × [0, 1], S4 = [0, 1] × {0} × [0, 1], S5 = [0, 1] × [0, 1] × {0}, S6 = [0, 1] × [0, 1] × {1}. On S1 , S2 , S3 , and S4 the positive normal vector is parallel to either the x- or the y-axis. Consequently, the dot product of F(r) with the positive normal vector is zero on each of these four faces because F(r) points in the direction of the z-axis and is thus perpendicular to the positive normal. Furthermore, on S5 we have z = 0 and, by implication, F(r) = 0 for all r ∈ S5 . In order to determine the remaining flux integral over S6 we introduce the parameterization x p(x, y) := y for (x, y) ∈ [0, 1] × [0, 1]. 1 In computing the positive normal vector, we find that ∂p/∂x × ∂p/∂y = e3 , and therefore, Z Z 1Z 1 F · dA = dx dy = 1. 0 ∂V 0 Since div F(r) = 1, it also follows that Z Z 1Z 1Z 1 div F(r) dV = dx dy dz = 1 0 V 0 0 as predicted by the divergence theorem. 2 cos(θ) sin(φ) b) In setting p(φ, θ) := 2 sin(θ) sin(φ) for 2 cos(φ) (φ, θ) ∈ [0, π] × [0, 2π], we find the positive normal vector to be ∂p/∂φ × ∂p/∂θ = 2 sin(φ)p, and therefore, Z 2π Z π Z F · dA = 2kpk2 p · p sin(φ) dφ dθ ∂V (p) = Z 0 2π 0 Z 0 π 32 sin(φ) dφ dθ = 128π. 0 Using the result of Exercise 79.3, this value is also found for the integral of the divergence: Z Z div F(r) dV = 5krk2 dV V = Z 2π 0 Z π 0 V Z 2 5ρ4 sin(φ) dρ dφ dθ = 128π. 0 c) In this case the divergence theorem does not apply, because F is not defined on the entire region V . Note: kF(x, y, z)k tends to ∞ as x approaches 2. 79.11 a) div F is a scalar, while G is a vector. b) The divergence operator cannot be applied to multivariable functions, only to vector fields. 79.13 Setting F(r) := v, we may infer that div F(r) = 0, and therefore, Exercise 79.12 implies that div(f v) = v · ∇f . Solutions 79.15 Setting f (x, y, z) := (x2 + y 2 + z 2 )/2 and g(x, y, z) := (x3 + y 3 + z 3 )/3, it follows that F = ∇f and G = ∇g. Consequently, the result of Exercise 79.14 implies that Z (F × G) · dA ∂D(p) Z = div(∇f × ∇g) dV = 0. D 79.17 Using the results of Exercises 79.10 and 79.16, it follows that div(f ∇f ) = ∇f · ∇f + f div ∇f 2 ∂2 ∂2 ∂ = k∇f k2 + f f + f + f ∂x2 ∂y 2 ∂z 2 = k∇f k2 . Chapter 80 1−x 80.3 curl F = y − 1 0 80.5 Given the parameterization r cos(θ) p(r, θ) := r sin(θ) 2−r for (r, θ) ∈ [1, 2] × [0, 2π], we find that r cos(θ) ∂ ∂ p× p = r sin(θ) , ∂r ∂θ r and therefore, Z curl F · dA S(p) = Z = Z 2π Z 0 2 1 2π Z 0 1 r cos(θ) 1 · r sin(θ) dr dθ 1 r 2 (r cos(θ) + r sin(θ) + r) dr dθ = 3π. 1 The same value we also obtain from the sum of the path integrals of F along the oriented boundary curves C1 and C2 : Z Z F · dc + F · dc C1 C2 49 0 −2 sin(t) 2 cos(t) · 2 cos(t) dt = 0 2 sin(t) 0 Z 2π 1 − sin(t) cos(t) · − cos(t) dt + 0 − sin(t) 0 Z 2π Z 2π 4 cos2 (t) dt + (− sin(t) − cos2 (t)) dt = Z 2π 0 0 = 4π − π = 3π. 80.9 a) F is not a gradient field because curl F = 2ze1 − 2xe2 − 2ye3 6= 0. b) F is a gradient field because curl F = 0. c) F is not a gradient field because curl F = 2ze1 − 2xe2 − 2ye3 6= 0. 80.11 Using the parameterization 2 cos(θ) p(θ, z) := 2 sin(θ) z for (θ, z) ∈ [0, π] × [0, 2], the positive normal vector is 2 cos(θ) ∂ ∂ p× p = 2 sin(θ) . ∂θ ∂z 0 Since curl F = 2ye1 + 2ze2 + 2xe3 , we find that Z curl F · dA S(p) Z π = Z 0 =8 2 (8 sin(θ) cos(θ) + 4z sin(θ)) dz dθ 0 Z π sin(θ) dθ = 16. 0 The boundary curve of the semicylinder described by p consists of four segments C1 , C2 , C3 , and C4 that may be described by the following parameterizations: 2 cos(t) r1 (t) := 2 sin(t) for t ∈ [0, π], 0 −2 r2 (t) := 0 for t ∈ [0, 2], t −2 cos(t) r3 (t) := 2 sin(t) for t ∈ [0, π], 2 2 r4 (t) := 0 for t ∈ [0, 2]. 2−t 50 Solutions The corresponding path integral is Z Z Z Z F · dc + F · dc + F · dc + F · dc C1 C2 C3 C4 Z π 0 −2 sin(t) 4 cos2 (t) · 2 cos(t) dt = 0 0 4 sin2 (t) 2 Z 2 t 0 4 · 0 dt + 0 0 1 Z π 4 2 sin(t) 4 cos2 (t) · 2 cos(t) dt + 0 0 4 sin2 (t) Z 2 (2 − t)2 0 4 · 0 dt + 0 0 −1 Z π 8 cos3 (t) dt + 0 = 0 Z π + (8 sin(t) + 8 cos3 (t)) dt + 0 = 16, A.2 sin(2α) = sin(α + α) = sin(α) cos(α) + cos(α) sin(α) = 2 sin(α) cos(α), cos(2α) = cos(α + α) = cos(α) cos(α) − sin(α) sin(α) = cos2 (α) − sin2 (α), and the alternative representations of the double angle formula for cosine are easily deduced from the trigonometric theorem of Pythagoras. Furthermore, tan(2α) = 2 tan(α) tan(α) + tan(α) = . 1 − tan(α) tan(α) 1 − tan2 (α) A.3 Considering the diagram in Figure S.15, it follows that c2 = h2 + (b cos(γ) − a)2 = b2 sin2 (α) + b2 cos2 (γ) − 2ab cos(γ) + a2 = a2 + b2 − 2ab cos(γ), as desired. With regard to the law of sine, the 0 α as desired. 80.13 Using Stokes’ theorem and the result of Exercise 80.12b, we may infer that Z (f ∇g + g∇f ) · dc C Z = curl(f ∇g + g∇f ) · dA S Z = (∇f × ∇g + ∇g × ∇f ) · dA ZS = 0 · dA = 0. b c h β γ a Figure S.15: obtuse triangle. same diagram also allows us to infer that h/b h/c sin(π − β) sin(β) sin(γ) = = = = , c c b b b S 80.15 Using the first of Maxwell’s equations, we may infer that curl E = − ∂ ∂ B = − (B0 + tB1 ) = −B1 ∂t ∂t is independent of t. and similarly, it can be shown that sin(α)/a = sin(β)/b. A.4 r2 = 3r2 sin(θ) cos(θ) √ A.6 (0, 2): r = 2, θ = π/2, √ (1, 4): r = 17, θ = −3): arctan(4), (−2, 2): r = 8, θ = 3π/4, (4,√ r = 5, θ = 2π−arctan(3/4), (−2, −4): r = 20, θ = π + arctan(2) Appendix A A.1 cot(α) + cot(β) 1 = cot(α + β) cot(α) cot(β) − 1 tan(α) tan(β) cot(α) + cot(β) = · tan(α) tan(β) cot(α) cot(β) − 1 tan(α) + tan(β) = 1 − tan(α) tan(β) tan(α + β) = Appendix B r s B.1 and s = q/n, then √ a a = √ a) If r = p/mmn √ mn mn rmn smn pn qm pn+qm a = a a = a = √a pn+qm mn a = ar+s . The second identity concerning the quotient of ar and as immediately follows from the first by observing that 1/as = a−s . b) 82/3 = 4 and 16−3/4 = 1/8. Solutions Appendix C 51 x2 y2 +√ =1 13 + 7 13 − 2 C.2 If we set u := x − x0 and v := y − y0 , then the given equation is equivalent to the equation u2 /a2 + v 2 /b2 = 1 which in turn describes an ellipse in standard position in a uv-coordinate system centered at (x0 , y0 ). C.1 √ = = = = C.3 y 2 = 4x/5 C.4 Given the equation in (C.2), we may infer that p p (x − c)2 + y 2 = (x + c)2 + y 2 ± 2a, and therefore, p (x − c)2 = (x + c)2 ± 4a (x + c)2 + y 2 + 4a2 . Rearranging the terms and squaring both sides yields 16a2 ((x + c)2 + y 2 ) = (4a2 + 4xc)2 , or equivalently, (a2 − c2 )x2 + a2 y 2 = a2 (a2 − c2 ). Thus, x2 y2 − = 1. a2 c2 − a2 C.5 2x2 2y 2 √ −√ =1 11 − 117 117 − 9 Appendix D √ D.1 The roots are at x = (3 ± 89)/8, and the global minimum is at (3/8, −19/2). √ ! √ ! 3 − 89 3 + 89 x− D.2 f (x) = 4 x − 8 8 Appendix E E.2 P1 For2 n = 1 the equation is valid because k=1 k = 1 = 1·2·3/6. Assuming the equation to be valid for a given n, we find that n+1 X k=1 k 2 = (n + 1)2 + n(n + 1)(2n + 1) 6 (n + 1)(6n + 6 + n(2n + 1)) 6 (n + 1)(2n2 + 7n + 6) 6 (n + 1)(n + 2)(2n + 3) 6 (n + 1)((n + 1) + 1)(2(n + 1) + 1) , 6 = (n + 1)2 + n X k=1 k2 as desired.

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