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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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 Spring '14
 The Land

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