X1 l1 cos 1 x2 l2 cos1 2 so x x1 x2 l1 cos

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Unformatted text preview: = (x2 + 1)2 √ − 3 25 (−0.04) = 0.0048 √ 4x + 8x + 1 dx, x = 3, dx = 0.05; ∆y ≈ dy = (37/5)(0.05) = 0.37 8x + 1 A = x2 where x is the length of a side; dA = 2x dx = 2(10)(±0.1) = ±2 ft2 . ±0.1 dx = = ±0.01 so percentage error in x is ≈ ±1%; relative error in relative error in x is ≈ x 10 2x dx dx dA = = 2(±0.01) = ±0.02 so percentage error in A is ≈ ±2% =2 A is ≈ A x2 x 47. (a) V = x3 where x is the length of a side; dV = 3x2 dx = 3(25)2 (±1) = ±1875 cm3 . (b) 46. relative error in x is ≈ (a) x = 10 sin θ, y = 10 cos θ (see figure), √ π π dx = 10 cos θdθ = 10 cos π ± 180 = 10 23 ± 180 6 ≈ ±0.151 in, π π dy = −10(sin θ)dθ = −10 sin π ± 180 = −10 1 ± 180 6 2 ≈ ±0.087 in (b) 48. (a) (b) ±1 dx = = ±0.04 so percentage error in x is ≈ ±4%; relative error in V x 25 2 dV 3x dx dx is ≈ = = 3(±0.04) = ±0.12 so percentage error in V is ≈ ±12% =3 3 V x x 10 ″ x θ y √ π π dx π = (cot θ)dθ = cot ± = 3± ≈ ±0.030 x 6 180 180 so percentage error in x is ≈ ±3.0%; dy π π 1 π relative error in y is ≈ = − tan θdθ = − tan ± = −√ ± ≈ ±0.010 y 6 180 180 3 so percentage error in y is ≈ ±1.0% relative error in x is ≈ x = 25 cot θ, y = 25 csc θ (see figure); π π dx = −25 csc2 θdθ = −25 csc2 ± 3 360 π 4 ± ≈ ±0.291 cm, = −25 3 360 π π π cot ± dy = −25 csc θ cot θdθ = −25 csc 3 3 360 2 1 π √ ≈ ±0.145 cm = −25 √ ± 360 3 3 25 cm y θ x dx csc2 θ 4/3 π =− dθ = − √ ± ≈ ±0.020 so percentage x cot θ 360 1/ 3 1 dy π = − cot θdθ = − √ ± ≈ ±0.005 error in x is ≈ ±2.0%; relative error in y is ≈ y 360 3 so percentage error in y is ≈ ±0.5% relative error in x is ≈ 49. (−2k/r3 )dr dr dr dR dR = = −2 , but ≈ ±0.05 so ≈ −2(±0.05) = ±0.10; percentage error in R is R (k/r2 ) r r R ≈ ±10% 50. h = 12 sin θ thus dh = 12 cos θdθ so, with θ = 60◦ = π/3 radians and dθ = −1◦ = −π/180 radians, dh = 12 cos(π/3)(−π/180) = −π/30 ≈ −0.105 ft Supplementary Exercises 3 51. 52. 94 12 (4) sin 2θ = 4 sin 2θ thus dA = 8 cos 2θdθ so, with θ = 30◦ = π/6 radians and 4 dθ = ±15 = ±1/4◦ = ±π/720 radians, dA = 8 cos(π/3)(±π/720) = ±π/180 ≈ ±0.017 cm2 A= A = x2 where x is the length of a side; dA dx 2x dx dx = 2 , but = ≈ ±0.01 so A x2 x x dA ≈ 2(±0.01) = ±0.02; percentage error in A is ≈ ±2% A 53. V = x3 where x is the length of a side; so 54. 3x2 dx dx dx dV = , but ≈ ±0.02 =3 V x3 x x dV ≈ 3(±0.02) = ±0.06; percentage error in V is ≈ ±6%. V dV 4πr2 dr dr dV dr dr = = 3 , but ≈ ±0.03 so 3 ≈ ±0.03, ≈ ±0.01; maximum permissible 3 /3 V 4πr r V r r percentage error in r is ≈ ±1%. (πD/2)dD dD dA dA 1 πD2 where D is the diameter of the circle; = =2 , but ≈ ±0.01 so 4 A πD2 /4 D A dD dD ≈ ±0.01, ≈ ±0.005; maximum permissible percentage error in D is ≈ ±0.5%. 2 D D 55. A= 56. V = x3 where x is the length of a side; approximate ∆V by dV if x = 1 and dx = ∆x = 0.02, dV = 3x2 dx = 3(1)2 (0.02) = 0.06 in3 . 57. V = volume of cylindrical rod = πr2 h = πr2 (15) = 15πr2 ; approximate ∆V by dV if r = 2.5 and dr = ∆r = 0.001. dV = 30πr dr = 30π (2.5)(0.001) ≈ 0.236 cm3 . 58. π dP 1 dL 1 2π 1 2π √ = so the relative error in P ≈ the relative P = √ L, dP = √ √ dL = √ √ dL, g g2 L P 2L 2 gL 1 error in L. Thus the percentage error in P is ≈ the percentage error in L. 2 60. (a) α = ∆L/(L∆T ) = 0.006/(40 × 10) = 1.5 × 10−5/◦ C (b) 59. ∆L = 2.3 × 10−5 (180)(25) ≈ 0.1 cm, so the pole is about 180.1 cm long. ∆V = 7.5 × 10−4 (4000)(−20) = −60 gallons; the truck delivers 4000 − 60 = 3940 gallons. CHAPTER 3 SUPPLEMENTARY EXERCISES 4. (a) = lim h→0 (b) 9 − 4(x + h) − h −4h dy = lim dx h→0 h( √ 9 − 4x (9 − 4(x + h) − (9 − 4x) √ h( 9 − 4(x + h) + 9 − 4x) −2 −4 =√ =√ √ 2 9 − 4x 9 − 4x 9 − 4(x + h) + 9 − 4x) = lim h→0 x x+h − dy x + h + 1 x + 1 = lim (x + h)(x + 1) − x(x + h + 1) = lim h→0 dx h→0 h h(x + h + 1)(x + 1) 1 h = = lim h→0 h(x + h + 1)(x + 1) (x + 1)2 95 5. 6. 7. Chapter 3 Set f (x) = 0: f (x) = 6(2)(2x + 7)5 (x − 2)5 + 5(2x + 7)6 (x − 2)4 = 0, so 2x + 7 = 0 or x − 2 = 0 or, factoring out (2x + 7)5 (x − 2)4 , 12(x − 2) + 5(2x + 7) = 0. This reduces to x = −7/2, x = 2, or 22x + 11 = 0, so the tangent line is horizontal at x = −7/2, 2, −1/2. 4(x2 + 2x)(x − 3)3 − (2x + 2)(x − 3)4 , and a fraction can equal zero only if its (x2 + 2x)2 numerator equals zero. So either x−3 = 0 or, after factoring out (x−3)3 , 4(x2 +2x)−(2x+2)(x−3) = 0, √ √ −6 ± 36 − 4 · 3 2 = −3 ± 6. So 2x + 12x + 6 = 0, whose roots are (by the quadratic formula) x = 2 √ the tangent line is horizontal at x = 3, −3 ± 6. Set f (x) = 0: f (x) = √ 3 Set f (x) = √ (x − 1)2 + 2 3x + 1(x − 1) = 0. If x = 1 then y = 0. If x = 1 then divide out 2 3x + 1 √ √ x − 1 and multiply through by 2 3x + 1 (at points where f is differentiable we must have 3x + 1 = 0) to obtain 3(x − 1) + 4(3x + 1) = 0, or 15x + 1 = 0. So the tangent line is horizontal at x = 1, −1/15. 2 2 2 3x + 1 3x + 1 3x + 1 d 3x + 1 x2 (3) − (3x + 1)(2x) 3x2 + 2x =3 = −3 =...
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