FxF ()() ′ x sin Fx () cos Fx ()
= 2 FxF ()() ′ x sin Fx () cos Fx ()
49. D F t − 2 = − 2 F t − t 3 [ () () ] () () F ′ () t + F ′ () x sin 2 Fx ()
2 ⎥ = − 2 ( F () z ) F ′ () z ⎣
58. D sec x 3 ⎡ Fx () ⎤= 3sec 2 ⎣ ⎡ Fx
() ⎤ ⎦ D x ⎡ ⎣ sec Fx () ⎤ ⎦
dz
⎢ ⎣ ( F () z ) ⎥ ⎦
= 3sec 2 ⎣ ⎡ Fx () ⎦ ⎤ sec Fx () tan FxD () x [] x
d = 3 F ′ x sec 3 () Fx () tan Fx ()
51. ⎢ ( 1 + F () 2 z ) ⎥ = 21 ( + F () 2 z ) ( 1 + F
() 2 z )
dz
dz
= 21 ( + F () 2 z ) ( 2 F ′ () 2 z ) = 41 ( + F () 2 z ) F ′ () 2 z
59. gx ' () =− sin fxDfx () x () =− f ′ () x sin fx ()
g ′ () 0 =− f ′ () 0 sin f () 0 =− 2sin1 ≈− 1.683
Fy 2 d d dy ⎢
52. ⎢ y 2 + 1 ⎥
2 dy ⎢ ( () ) Fy ⎥ ⎥ ⎣ ⎦
( + 1 sec F () 2 x ) x − x ( + 1 sec F () 2 x )
60. Gx ′ =
() dx
dx
( + 1 sec F () 2 x () )
= 2 y − F ′ () y 2 y 2 = 2 y −
d 2 yF ′ y 2
2 2 ( + 1 sec F () 2 x ) − 2 xF ′ () 2 x sec F () 2 x tan F () 2 ( x Fy
dy
1 sec F () 2 x )
F ′ () y ⎟
+ 1 sec F () 0 − 0 + 1 sec F
( + 1 sec F () 0 ) ( + 1 sec F () () 0 )
Fy 2 ⎟
+ 1 sec F () 0 + 1 sec 2
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61. F ′ () x =− fxgx ()() ′ sin gx () + f ′ () x cos gx () c. DL =
2 16cos 2 2 2 F t ′ () 1 =− f ()() 1 g ′ 1 sin g () 1 + f ′ () 1 cos g t () t 1 + 49sin 2
=− () 2 1 sin 0 +− 1cos 0 =− 1 =
32 cos 2 tD t (cos 2 ) 98sin 2 t + tD t (sin 2 ) t
2 16 cos 2 2 t + 49sin 2 2 t
62. y =+ 1 x sin 3 ; x y ′ = x 3 cos 3 x + sin 3 x –64 cos 2 sin 2 t t + 196sin 2 cos 2 t = t
y ′ ( π /3 ) = 3 cos 3 + sin =−+=− π 0 π
2 16 cos 2 t + 49sin 2 t 3 3 3 − 16sin 4 t + 49sin 4 t
y −=− 1 π x − π
16 cos 2 2 t + 49sin 2 2 t
y =− π x − π /31 +
33sin 4 t
The line crosses the x-axis at x =
. 16 cos 2 2 t + 49sin 2 2 3 t
63. y = sin 2 x ; y ′ = 2sin cos x x = sin 2 x = 1 At t = : rate =
≈ 5.8 ft/sec.
x = π /4 + k π , k = ±± 0, 1, 2,...
2 3 4 3 2 2 2 69. a. (10 cos8 ,10sin 8 ) π t π 64. t y ′= ( x + 12 )( x + 1 ) x + 3 ( x + 1 )( 4 xx + 1 )
3 4 2 3 4 2 2 2 b. D t (10sin 8 ) 10 cos(8 ) π= t π tD t (8 ) π t = 2 x ( x + 1 )( x + 1 ) + 3 xx ( + 1 )( x + 1 )
=π 80 cos(8 ) π t
3 2 y 2 ′ () 1 = 222 ( )( ) + 312 ( )( ) ( ) 2 = + 32 48 80 =
At t = 1: rate = 80 π ≈ 251 cm/s
= 32 80 x − 1, y = 80 x + 31 P is rising at the rate of 251 cm/s.
2 − 3 − 3 70. a. (cos 65. 2t, sin 2t) y ′=− 2 ( x + 1 )
() 2 x =− 4 xx 2 ( + 1 )
b. (0 – cos 2 ) t 2 + ( – sin 2 ) y t 2 = 5, − 2 3 so
y ′ () 1 =− 4111 ( )( + ) =− 1/ 2
66. y ′= 32 ( x + 1 )() 2 = 62 ( x + 1 )
y ′ () 0 = 61 () =
The line crosses the x-axis at x =− 1/ 6 .
67. y ′=− 2 ( 2 x + 1 ) 2 () 2 x =− 4 xx ( + 1 )
71. 60 revolutions per minute is 120 π radians per ′
y () 1 =− 42 () =− 1/ 2
minute or 2 π radians per second.
y −=− x + , y =− x +
a. (cos 2 ,sin 2 ) π t π t
Set y = 0 and solve for x. The line crosses the
b. (0 – cos 2 ) π t 2 + ( – sin 2 ) y π t 2 = 5, 2 so x -axis at x = 3/2 .
16 cos 2 2 t + 49sin 2 2 t =π 2 cos 2 π t ⎜ 1 +
25 – cos 2 2 π t ⎝ ⎟ ⎠
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π 22 sin 11 π ds t so at t minutes after the hour, it makes an angle
74. From Problem 73,
radians with the vertical. By the Law of
Using a computer algebra system or graphing Cosines, the length of the elastic string is
ds
utility to view
for 0 ≤≤ t 60 , is largest
30 when t ≈ 7.5. Thus, the distance between the tips
of the hands is increasing most rapidly at about =
sin x = 2sin x cos x
x (sin x) = cos x, D x (sin 2x) = 2cos 2x, so at x 0 , At 12:15, the string is stretching at the rate of the tangent lines to y = sin x and y = sin 2x have
30 D
π sin π 2 π
≈ 0.74 cm/min
slopes of m 1 =
⎝ 2 ⎠ respectively. From Problem 40 of Section 0.7,
tan θ= 2 so at t minutes after noon it makes an angle of 1 where θ is the angle between
73. The minute hand makes 1 revolution every hour,
1 + mm 12
radians with the vertical. Similarly, at t
= –3, minutes after noon the hour hand makes an angle
the tangent lines. tan θ =
of
with the vertical. Thus, by the Law of
so θ ≈ –1.25. The curves intersect at an angle of
1.25 radians.
Cosines, the distance between the tips of the hands is
76. AB = OA sin
2 2 ⎛ π t π s t = 6 + ⋅⋅ ⎞ 8 – 2 6 8cos
D = OA cos ⋅ AB = OA cos sin 2 2 2 2
= E = D + area (semi-circle)
0.38 in./min
D cos( / 2) t
lim
= lim
t → π − E t → π − cos( / 2) t + π sin( / 2) t
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77. y = u and u = x 2 81. [ ( ( ( (0))))] ffff ′
D x y =D u y ⋅D x u = f ′ ( ( ( (0)))) fff ⋅ f ′ ( ( (0))) ff ⋅ f ′ ( (0)) f ⋅ f ′ (0)
D x (sin ) x dx sin x
c. Conjecture:
2 2 2 1 2 d f [] n () x = f '( f [ n − 1]
d [ n − 1]
80. a. DLx x ()()() = Lx ' D x x = ⋅ 2 x = ( )) x ⋅ f () x
dx
dx
b. DL x (cos 4 x ) = sec 4 xD (cos 4 x x ) = sec 4 x (4 cos 3 xD ) x (cos ) x = 4sec 4 x cos 3 x ( sin ) − x
=⋅ 4 4 ⋅ cos x ⋅− ( sin x )
cos x = –4sec sin x x =− 4 tan x
⎛ fx ⎞
⎟ = D 1 1 x 1 ⎜ fx () ⋅ ⎟ = D x ( fx ( ) ( ( )) ⋅ gx ) = fxD () x ( ( ( )) gx ) + ( ( )) gx Dfx x ()
() =− fxgx ( )( ( )) Dgx x ( ) ( ( )) + gx Dfx x ()
− fxDgx () x () Dfx x () − fxDgx () x () gx () Dfx x () − fxDgx () x () gxDfx () ()
gxDfx = () x () − fxDgx () x ()
g 2 () x
84. gx ′ () = f ′ ( f ( f ( fx () ) ) ) f ′ ( f ( fx () ) ) f ′ ( fx () ) f ′ () x gx ′ () 1 = f ′ ( f ( f ( fx () 1 ) ) ) f ′ ( f ( fx () 1 ) ) f ′ ( fx () 1 ) f ′ () x 1
= f ′ ( f ( fx () 2 ) ) f ′ ( fx () 2 ) f ′ ()() x 2 f ′ x 1 = f ′ ( fx () 1 ) f ′ ()()() x 1 f ′ x 2 f ′ x 1
2 =⎡ 2 ⎣ f ′ ()
x 1 ⎦⎣ ⎤⎡ f ′ () x 2 ⎤ ⎦
gx ′ () 2 = f ′ ( f ( f ( fx () 2 ) ) ) f ′ ( f ( fx () 2 ) ) f ′ ( fx () 2 ) f ′ () x 2
= f ′ ( f ( fx () 1 ) ) f ′ ( fx () 1 ) f ′ ()() x 1 f ′ x 2 = f ′ ( fx () 2 ) f ′ ()()() x 2 f ′ x 1 f ′ x 2
2 = 2 ⎡ ⎣ f ′ () x
1 ⎤⎡ ⎦⎣ f ′ () x 2 ⎤ ⎦ = gx ′ () 1
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2.6 Concepts Review
2 = 18(3 x + 5)(3) 162 = x + 270 dx
3. f ′ t > 0 () dy 4. = 5(3 – 5 ) (–5) x 4 = –25(3 – 5 ) x 4
= –100(3 – 5 ) (–5) x 3 = 500(3 – 5 ) x 3
dx 2
Problem Set 2.6 3 dy
3 = 1500(3 – 5 ) (–5) x 2 = –7500(3 – 5 ) x 2 dx dy
5. = 7 cos(7 ) x
2 = –7 sin(7 ) 2 x dy 3 dx
dx 3
dy 3 3
3 = –7 cos(7 ) x = –343cos(7 ) x dx dy
2. = 5 x 4 + 4 x 3
dx
= 20x +12x
6. = 3 x 2 cos( x 3 )
dx dy 2
= x 2 x 2 sin( x 2 3 3 [–3 )] 6 cos( + x x 3 ) = –9 x 4 sin( x 3 ) 6 cos( + x x 3 )
dx
dy 3 4
3 = –9 x cos( x 3 )(3 x 2 ) sin( + x 3 )(–36 x 3 ) 6 [– sin( + x x 3 )(3 x 2 )] 6 cos( + x 3 ) dx = –27 x 6 cos( x 3 ) – 36 x 3 sin( x 3 ) – 18 x 3 sin( x 3 ) 6 cos( + x 3 ) = (6 – 27 x 6 ) cos( x 3 ) – 54 x 3 sin( x 3 )
dy ( – 1)(0) – (1)(1) x
= dx
1 dy (1 – )(3) – (3 )(–1) x x
2 dy 2 ( – 1) (0) – 3[2( – 1)] x 2 x 6 dx 2
dy 3 ( x − 3 1) (0) 6(3)( − x − 1) 2 dx 3
dy 3 ( x − 3 1) (0) 2[3( − x − 2 1) ]
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9. fx ′ () = 2; () xf ′′ x = 2; (2) f ′′ = 2 (5 – )(4 ) – (2 u u u 2 )(–1)
(5 – ) (20 – 4 ) – (20 – 2 u 2 u u u 2 )2(5 – )(–1) u
() –3 θ = –2(cos θ π ) (– sin –3 θ ππ ) = 2π(cos θ π) (sin θ π)
f ′′ () θ =π 2 [(cos θ π ) ( )(cos –3 π θ π+ ) (sin θ π )(–3)(cos θ π ) (– sin –4 θ ππ )( )] =π 2 2 [(cos θ π ) − 2 + 3sin 2 θ π (cos θ π )] − 4
15. fs ′ () = s (3)(1 – s 22 ) (–2 ) (1 – s + s 23 ) = –6 (1 – s 2 s 22 ) + (1 – s 23 ) = –7 s 6 + 15 s 4 –9 s 2 + 1
f ′′ () s = –42 s 5 + 60 s 3 –18 s
(2) f ′′ = –900
( –1)2( x x + 1) – ( x + 1) 2 x 2 –2–3 x
( –1) (2 – 2) – ( x 2 x x 2 – 2 – 3)2( –1) x x ( –1)(2 – 2) – ( x x x 2 – 2 – 3)(2) x
D 3 ( x n ) = nn ( –1)( – 2) n x n –3 x n
D x 4 ( x n ) = nn ( – 1)( – 2)( – 3) n n x n –4 19. a. D x 4 (3 x 3 + x 2 –19) = 0
D n − x 1 ( x n ) = nn ( –1)( – 2)( – 3)...(2) n n x b. D 12 x (100 x 11 − 79 x 10 ) = 0
D n ( x n x ) = nn ( –1)( – 2)( – 3)...2(1) n n x 0 = n!
c. D 11 2 x 5 ( x – 3) = 0
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3t(t – 4) > 0; ( −∞ , 0) ∪ (4, ) ∞
D 4 1 x 4(3)(2) ⎜⎟ ⎛⎞=
3(x + 5)(x – 3) = 0 x = –5, x = 3
(–5) f ′′ = –24 ds 2
22. gt ′ () = 2 at + b b. 3 t 2 –18 t + 24 > 0 gt ′′ () = 2 a 3(t – 2)(t – 4) > 0
g ′ (1) = 2 a += b 3 c. 3t –18t + 24 < 0 (2, 4)
g () 1 =++= a b c 5 6t < 18
()() −+ 2 7 += c 5
t < 3; ( −∞ ,3 )
6(t + 1)(t – 1) > 0
d. a(t) = –4 < 0 for all t
The acceleration is negative for negative t.
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at () = 2 =+ 2 v (4) = –16, v(1) = 11
(12 t 2 – 84 t + 120) = 0
v (2) = 10.4, v(5) = 5
2t 3 + 32 < 0; The acceleration is not t 3
ds
31. vt 1 () = 1 = 4–6 t
negative for any positive t.
dt 2 t 3 b. 4–6 t = 2–2; t 4 – 6t = –2t + 2
4 t 2 –6 t = 0 2t(2t – 3) = 0
3 < The acceleration is not negative for 0;
any positive t.
2 t 2 –7 t += 5 0 (2t – 5)(t – 1) = 0
t = 1,
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33. a. v(t) = –32t + 48
37. vt () = 3 t 2 – 6 – 24 t
initial velocity = v 0 = 48 ft/sec
3 t d 2 2 – 6 – 24 t
b. –32t + 48 = 0
c. s = –16(1.5) 2 + 48(1.5) 256 + = 292 ft
38. Point slowing down when
The object hits the ground at t = 5.77 sec.
vt () < 0
dt
e. v(5.77) ≈ –137 ft/sec;
speed = 137 − = 137 ft/sec. vt () =
d vtat ()()
dt
vt ()
34. v(t) = 48 –32t vtat ()() < when a(t) and v(t) have opposite
s = 48(1.5) –16(1.5) 2 = 36 ft
39. x () D uv = uv ′ + uv ′
D 2 x () uv = uv ′′ + uv ′′ + uv ′′ + uv ′′
b. v(1) = 16 ft/sec upward
–16t(–3 + t) = 0
∑ − ⎜⎟ D
x () uD k x () v
35. vt () = v 0 – 32 t k = 0 ⎝⎠ k v 0 – 32 t = 0 ⎛ n where ⎞
⎜ ⎟ is the binomial coefficient
0 =⎜⎟ 0 D x ( x ) D x (sin ) x – = 5280 ⎝⎠ 0
D x ( x ) D x (sin ) x +
⎜⎟ D x ( x ) D x (sin ) x = 5280 ⎝⎠ 1 ⎝⎠ 2
v 0 = 337,920 ≈ 581 ft/sec
+ ⎜⎟ D x ( x ) D x (sin ) x + ⎜⎟ D x ( x ) D x (sin ) x ⎝⎠ 3 ⎝⎠ 4
36. vt () = v 0 + 32 t = 24sin x + x 96 cos x − 72 x sin x
s = 44(3) 16(3) + 2 = 276 ft
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b. f ′′′ (2.13) ≈ 0.0271 6 y 2
– 14 xy
8. xDy 2 x + 2 xy = y 2 + x (2 ) yDy x
2.7 Concepts Review
Problem Set 2.7
1. 2 yDy –2 x = 0 Dy x x = 5 x
25 xy
+ 2–2–3 y xy 2
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11. xDy x ++ y cos( )( xy x D y x + y ) = 0 2
17. x –1/ 3 2 – y –1/ 3 y ′ –2 y ′= 0
xDy x + x cos( ) xy D y x = – – cos( ) y y xy 3 3
⎛ 2 ⎞ Dy x =
– – cos( ) y y xy
= y ′ ⎜ y –1/ 3 + 2 ⎟
x + x cos( ) xy
2 x –1/ 3
12. – sin( xy 2
)(2 xy D y x + y 2 ) = 2 yD y x + 1 y ′= 2
y –1/ 3
–2 xy sin( xy 2 2 ) 3 Dy x –2 yDy x =+ 1 y sin( xy 2 ) 2
1 + y 2 sin( xy 2
At (1, –1), y′ ==
Dy x =
2 4 –2 2 xy sin( xy )–2 y 3
Tangent line: y += 1 ( –1) x
Tangent line: –3 – ( – 1)
Tangent line: y –1 – = ( – 4) x
Tangent line: y – 1 –2( – 2) = x dy 1 1
dx
15. cos( xy xy )( ′ + y ) = y ′ 20. = x –2 / 3 –7 x 5/2 =
–7 x 5/2 dx 3 32 yx ′ [ cos( ) – 1] xy = – cos( y xy ) 3 x
y ′=
– cos( ) y xy
y cos( ) xy
21. = x
dy 1 –2 / 3 1 –4 / 3
x cos( ) – 1 xy
1 – cos( ) x xy
– dx 3 3 3 3 3 x 2 3 x 4
At ⎜ ,1, ⎞′= ⎟ y 0
⎛ π ⎞ Tangent line: y –10 x –
16. y ′ + [– sin( xy 2
)][2 xyy ′ + y 2 ]6 + x = 0
[1 – 2 xy sin( xy 2 )] = y 2 sin( xy 2 )–6 x 4
4 (3 x 2 –4) x 3 4 2 (3 x 2 –4) x 3
y 2 sin( xy 2 )–6 x
y ′=
1–2 xy sin( xy 2 )
Tangent line: y – 0 = –6(x – 1)
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26. = – (3 – 9) x –8 / 3 (3) = –5(3 – 9) x –8 / 3 34. = 1 cos( x 2 )(2 ) dx x + 6 x 2 dx
x 2 cos( x 2 )6 + 2
x 2 cos x + x 2 sin x The tangent line at (, ) x = y –
0 0 has equation 3( 3 x 2 sin ) x 4
(– x x ) which simplifies to
dy 1
30. = (1 sin 5 ) + x –3 / 4 (cos 5 )(5) x 2 x 0 – yy 0 –2– x xx 0 + y 0 + x 0 = 0. Since dx 4
x ( x 0 , y ) is on the circle, x 0 + y 0 = –3 – 4 , x
5cos 5
so the equation of the tangent line is
4 (1 sin 5 ) + x 3 – yy 0 –2 x 0 –2– x xx 0 = 3.
dy [1 cos( + x 2 + x 2 )] –3 / 4 [– sin( x 2 + x 2 )(2 x + 2)]
If (0, 0) is on the tangent line, then x 0 = –.
31. = 2 dx
4 Solve for y 0 in the equation of the circle to get ( x +
1) sin( x 2 + 2) x
+ Put these values into the equation of
x 2 )] 3 2
the tangent line to get that the tangent lines are
dy (tan 2 x + sin 2 x ) –1/ 2 (2 tan sec x 2 x + 2 sin cos ) x x
36. 16( x 2 + y 2 )(2 x + 2 yy ′ ) 100(2 – 2 = x yy ′ )
The slope of the normal line = –
At (3, 1), slope =
Normal line: y –1 = ( – 3) x
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3 + 2 2[ yy ′′ + ()]0 y ′ 2 2 6 =
′′ 9 6–8 x x yy – Slope of the normal line is 1.
48 xy − 9 x
Normal line: –
2 = 8 yy ′′ 8 y
This line includes the point (0, 0).
The slopes of the tangents are negative reciprocals, so the hyperbolas intersect at right
2( xy 2 ′′ + 2 xy ′ + 2 xy ′ + y 2 ) – 12[ yy 2 ′′ + 2()]0 yy ′ 2 = angles.
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43. Implicitly differentiate the first equation.
Implicitly differentiate the second equation. Intersection point in first quadrant: (2, 4)
Solve for the points of intersection.
(x + 3)(x – 1) = 0 x = –3, x = 1
= –2; tan (–2) θ =π+ –1 ≈ 2.034 x = –3 is extraneous, and y = –2, 2 when x = 1.
The graphs intersect at (1, –2) and (1, 2). At (1, –2): m 1 = m 1, –1 2 = 46. The equation is mv 2 – mv 2 0 = kx 2 0 – kx 2 .
At (1, 2): m 1 = –1, 1 m 2 = Differentiate implicitly with respect to t to get
2 mv = –2 kx . Since v =
44. Find the intersection points:
this simplifies
( x − 1 ) + y 2 = 1 dt
dt
47. x 2 – xy + y 2 = 16 , when y = 0,
The ellipse intersects the x-axis at (–4, 0) and Points of intersection: ⎜ ⎜ , ⎟ and ⎜ , – ⎟
x xy Implicitly differentiate the first equation. ′ 2– – y + 2 yy ′ = 0
Implicitly differentiate the second equation.
At (–4, 0), y′ = 2 2( –1) 2 x + yy′ = 0 At (4, 0), y′ = 2
1–x Tangent lines: y = 2(x + 4) and y = 2(x – 4) y ′= y
At ⎜ ⎜ ,:– ⎟ ⎟ m 1 =
tan θ =
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Problem Set 2.8
When x = 12,
2 = if x(2y – x) = 0, which occurs 2 0 xy – y
2. V =π r 3 ;3 =
4 dV
when x = 0 or y = . There are no points on
= where x = 0. If 2 y = , then
x 3 x 3 x 3 When r = 3, 3 =π 4 (3) 2 .
0.027 in./s
y == 1.
2 The tangent line is vertical at (2, 1).
The tangent line at ( x , y ) has slope – 0 0 0 , dy x dx
mi/hr
0 dt y dt
hence the equation of the tangent line is
5 x 0 When x = y 5, 26, = = (400)
dy
y – y 0 = – (– x x 0 ) which simplifies to
dt
y 0 ≈ 392 mi/h. yy + xx –( x 2 + y 0 2 0 0 0 ) = or 0 yy 0 + xx 0 = 1
2 2 1 2 r 3 3 since h ( x
0 , y 0 ) is on x + y = . If (1.25, 0) is 1 4. V =π rh ;; = r = 3 h 10 10
on the tangent line through ( x 0 , y 0 ) , x 0 = 0.8. 2
3 π h 3 dV
Put this into x + y = to get 1 y 0 = 0.6, since
y 0 > The line is 6y + 8x = 10. When x = –2, 0. dV 9 π 2 h dh
y = , so the light bulb must be
units high.
When h = 5, 3 =
100 dt
dh = 4 ≈
0.42 cm/s dt 3 π
2.8 Concepts Review
du
; 300, 400, = = = dt
5. s 2 = ( x + 300) 2 + y 2 dx
2 dy s = 2( x + 300) + 2 y
ds
dx
2. 400 mi/hr
s = ( x + 300) + y
4. negative; positive When x = 300, y = 400, s = 200 13 , so 200 13 ds = (300 300)(300) 400(400) + +
dt ds ≈ 471 mi/h dt
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40 + x =
6. y 2 = x 2 (10) ; 2 2 dy
hx
11. V = (20); , = x = 8 h
When y = 25, x ≈ 22.9, so
(2) ≈ 2.18 ft/s
dt
dt
dt x dt 22.9
When h = 3, 40 160(3) dh =
ft/min
When x = 5, y = 375 = 5 15 , so
(1) ≈ –0.258 ft/s
≈ 6.7 units/s 0.258 ft/s.
dy
The top of the ladder is moving down at
When x = 3,
8. = –4 ft /h; V =π hr ; –0.0005 = ft/h
13. A =π r ; 0.02 =
When h = 0.001 ft, V =π (0.001)(250) 2 = 62.5 π
When r = 8.1,
= –4000 + 31,250 π ≈ 94,175 ft 2 /h.
dx
dy
(The height is decreasing due to the spreading of
14. s 2 = x 2 + ( y + 2 48) ; = 30, = 24
the oil rather than the bacteria.) dt
At 2:00 p.m., x = 3(30) = 90, y = 3(24) = 72, dV 4 2 dh
3 3 dt
so s = 150.
dt When h = 4, 16 =π 4 (4) 2 dh ds 5580
dt
= 37.2 knots/h
dh
1 dt 150
= ≈ 0.0796 ft/s dt 4 π
10. y 2 = x 2 + (90) ; 5 2 dx = dt
dt When y = 150, x = 120, so dy x dx 120 = =
(5) = ft/s 4
dt y dt 150
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15. Let x be the distance from the beam to the point
opposite the lighthouse and θ be the angle 1
between the beam and the line from the
⎜⎟ = – . ⎝⎠ 2 24 lighthouse to the point opposite. Chris must lift his head at the rate of
dt
xd θ
tan θ= ; 2(2 ) = π = π rad/min, 4 1
1 dt
rad/s.
18. Let θ be the measure of the vertex angle, a be the
measure of the equal sides, and b be the measure At
, tan θ = –1 1 and sec 2 5 θ =.
θ of the base. Observe that b = a 2 sin and the dx = 5
dt 4 θ
(4 ) 15.71 π≈ km/min
height of the triangle is cos . a
2 A = (100) sin 2 θ = 5000sin ; θ =
When θ= , = and x = ≈ 7322. = 5000 cos θ
π ⎞⎛ 1 ⎞ When θ = , 5000 cos =
≈ 433 cm 2 min . ≈ –1740 ft/s or –1186 mi/h
The plane’s ground speed is 1186 mi/h.
19. Let p be the point on the bridge directly above
17. a. Let x
be the distance along the ground from
the railroad tracks. If a is the distance between p
da
the light pole to Chris, and let s be the
and the automobile, then
= 66 ft/s. If l is the
distance from Chris to the tip of his shadow.
dt
By similar triangles, =
, so s =
6 30 x
distance between the train and the point directly
4 below p, then dl = 88 ft/s. The distance from the
. = ft/s, hence 2
dt 4 dt dt
train to p is 100 2 + l 2 , while the distance from
ds 1 = ft/s no matter how far from the light
p to the automobile is a. The distance between
dt 2
the train and automobile is
pole Chris is.
2 2 ⎜ 2 100 + l ⎟ = a ++ l 100 .
b. Let l = x + s, then
dl dx ds
= + = ft/s. 2 =
a da + l dl c. The angular rate at which Chris must lift his
dt
dt
. After 10 seconds, a = 660
head to follow his shadow is the same as the
a 2 ++ l 2 100 2
rate at which the angle that the light makes
and l = 880, so
with the ground is decreasing. Let θ be the dD 660(66) 880(88) angle that the light makes with the ground at +
≈ 110 ft/s.
the tip of Chris' shadow.
2 . dt = ft/s s dt dt 2
π When s = 6, θ = , so
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48 10 + s
20. V =π⋅ ha ( + + ab b ); 20, a = b =+ 20, By similar triangles,
+ 10 h + 400 ⎟ ⎟ (for t > 1), so s =
=π ⎜ ⎜ 1200 h + 15 h + ⎟ ⎟ =
t 20 (1 – t ) – (10 t – 40)(–2 ) t
dh The ball hits the ground when t = 2,
The shadow is moving
≈ 13.33 ft/s.
When h = 30 and dV = 2000, 9 dt
675 ⎞ dh 3025 π dh 2 ⎛
≈ 0.84 cm/min.
π h 3 π h 3 At 7:00 a.m., h = 15,
(40 (15) π −π (15) )( 3) 2 −≈− 1125 π≈− 3534. dV dh dh =π dt 16
Webster City residents used water at the rate of
dh 3 2400 + 3534 = 5934 ft /h. When h = 3, –2 = [16 (3) – (3) ] π π 2
dt
25. Assuming that the tank is now in the shape of an
dh = –2 ≈ –0.016 ft/hr
upper hemisphere with radius r, we again let t be dt 39 π
the number of hours past midnight and h be the height of the water at time t. The volume, V, of
22. s 2 = a 2 + b 2 − 2 ab cos ; θ water in the tank at that time is given by
rad/h
dt
41 – 40 cos θ and so V = 16000 π − (20 − h ) 2 ( 40 + h )
2 s = 40sin θ
ds
from which
(20 − h ) 40 ( + h )
3 Thus Webster City residents were using water at ds
dt
the rate of 2400 1649 + = 4049 cubic feet per ≈ 18 in./hr
dt
hour at 7:00 A.M.
23. Let P be the point on the ground where the ball
26. The amount of water used by Webster City can hits. Then the distance from P to the bottom of
be found by:
the light pole is 10 ft. Let s be the distance usage = beginning amount added amount + between P and the shadow of the ball. The height
remaining amount
of the ball t seconds after it is dropped is
64 –16 . t 2 Thus the usage is ≈ π (20) (9) 2400(12) 2 + − π (20) (10.5) 2 ≈ 26,915 ft 3
over the 12 hour period.
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27. a. Let dx
be the distance from the bottom of the wall to the end of the ladder on the ground, so = 2 ft/s. Let y dt
be the height of the opposite end of the ladder. By similar triangles,
(144 + x 23/2 ) dt
When the ladder makes an angle of 60° with the ground, x = 43 and
dy
⋅= 2 –1.125 ft/s.
dt 2 dt ⎜ (144 + x 23/2 ) dt ⎟ dt ⎜ ⎝ ⎠ ⎝ (144 + x 23/2 ) ⎟ dt (144 + x 23/2 ⎠ ) dt 2 dx
+ x ⎢ 23/2 –216(144 ) dx dt + 216 x 3 () 2 144 + x 2 (2 ) x dx ⎤ dt ⎥ dx
–216(144 + x 2 ) 648 + x 2 ⎛ dx 2 ⎞
432 x 2 – 31,104 ⎛ dx 2 ⎞
(144 + x 25/2 )
⎝ dt ⎠
(144 + x 25/2 ) ⎝ dt ⎠
When the ladder makes an angle of 60° with the ground,
dy 2 432 48 – 31,104 ⋅
2 5/2 (2) ≈ –0.08 ft/s 2 dt (144 48) +
dV
28. a. If the ball has radius 6 in., the volume of the
29. = k (4 π r 2 )
water in the tank is
This is the same as in Problem 21, so
again –0.016 ft/hr.
b. If the ball has radius 2 ft, and the height of
b. If the original volume was V 0 , the volume
the water in the tank is h feet with 2 ≤≤, h 3 8
after 1 hour is
the part of the ball in the water has volume V 0 . The original radius
⎣ V ⎢ 3 ⎦ ⎥ 3 was 0 0 while the radius after 1
The volume of water in the tank is
8 3 2 dr
π h 3 (6 – ) hh 2 π
hour is r 1 = 3 V 0 ⋅
= r 0 . Since is
= – r 0 unit/hr. The snowball
will take 3 hours to melt completely.
When h = 3,
(–2) ≈ –0.018 ft/hr.
dt 36 π
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30. PV = k
Problem Set 2.9
1. dy = (2x + 1)dx
dP
At t = 6.5, P ≈ 67,
3. dy = –4(2 x + –5 3) (2) dx = –8(2 x + 3) –5 dx dt
dV V dP = 300 –
(–30) 134 ≈ in. 3 /min
P dt
31. Let l be the distance along the ground from the
4. dy = –2(3 x 2 ++ x –3 1) (6 x + 1) dx
brother to the tip of the shadow. The shadow is
= –2(6 x + 1)(3 x 2 ++ x 1) –3 dx
controlled by both siblings when =
or
5. dy = 3(sin x + cos ) (cos – sin ) x 2 x x dx
l = 6. Again using similar triangles, this occurs 6
when y= , so y = 40. Thus, the girl controls
6. dy = 3(tan x + 2 1) (sec 2 x dx )
20 3 the tip of the shadow when y ≥
40 and the boy
= 3sec 2 x (tan x + 1) 2 dx
controls it when y < 40. Let x be the distance along the ground from the
7. dy = – (7 x 2 + x 3 –1) –5 / 2 (14 x + 3) dx
dx
light pole to the girl.
dt
20 5 4 =− (14 x + 3)(7 x 2 + 3 x − 1) − 52 dx
When y ≥ 40,
or y = x . 2
or y = ( x + 4). + sin 2 )[10 x x +
20 3 20 8. dy = 2( x 10 9 1
⋅ (cos 2 )(2)] x dx
When y < 40,
x = 30 when y = 40. Thus,
10. a. dy = 3 x dx 2 = 3(0.5) (1) 2 = 0.75
⎪⎩ 17 dt Hence, the tip of the shadow is moving at the rate
b. dy = 3 x dx 2 = 3(–1) (0.75) 2 = 2.25
4 16 of (4) = ft/s when the girl is at least 30 feet
from the light pole, and it is moving
20 80 (4) = ft/s when the girl is less than 30 ft
17 17 from the light pole.
2.9 Concepts Review
12. a. dy = – 2 = – 2 = –0.5
3. Δx is small.
4. larger ; smaller
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20. y = 3 x dy 1 ;; = x –2 / 3 1 dx = dx
3 3 32 x x = 27, dx = –0.09
21. V =π r 3 ; 5, 0.125 r = dr =
dV =π 4 r dr 2 =π 4 (5) (0.125) 2 ≈ 39.27 cm 3
14. a. Δ= y (1.5) – (0.5) 3 3 = 3.25
22. V = x 3 ; 40, 0.5 x = 3 dx =
b. Δ= y (–0.25) – (–1) 3 3 = 0.984375 dV = 3 x dx 2 = 3( 40) (0.5) 17.54 3 2 ≈ in. 3
15. a. Δ= y – = – 23. V =π r 3 ; 6 ft r = = 72in., –0.3 dr =
dV =π 4 r dr 2 =π 4 (72) (–0.3) 2 ≈ –19,543
b. Δ= y
+= –0.3 4 –1.25 2 V ≈π (72) –19,543 3
2 2 ≈ 1,543,915 in 3 ≈ 893 ft 16. a. 3 Δ= y [(2.5) – 3] – [(2) – 3] = 2.25 dy = 2xdx = 2(2)(0.5) = 2
24. V =π rhr 2 ; 6 ft = = 72in., dr =− 0.05,
b. Δ= y [(2.88) – 3] – [(3) – 3] 2 2 = –0.7056 h = 8ft = 96in.
dy = 2xdx = 2(3)(–0.12) = –0.72
dV =π 2 rhdr =π 2 (72)(96)( 0.05) − ≈− 2171in. 3
17. a. Δ= y
About 9.4 gal of paint are needed.
dy = (4 x 3 + 2) dx = [4(2) 3 + 2](1) = 34 25. C = 2 π r ; r = 4000 mi = 21,120,000 ft, dr = 2
dC = 2 π dr = π 2 (2) = 4 π ≈ 12.6 ft
b. Δ= y [(2.005) 4 + 2(2.005)] – [(2) 4 + 2(2)]
≈ 0.1706 L
26. T =π 2 ; 4, –0.03 L = dL =
dy = (4 x 3 + 2) dx = [4(2) 3 + 2](0.005) = 0.17
18. y = x dy ;; = dx x = 400, dx = 2 32
The time change in 24 hours is 402 ≈ 400 + dy = + 20 0.05 = 20.05 (0.0083)(60)(60)(24) ≈ 717 sec
19. y = x dy ;; = dx x = 36, dx = –0.1 27. V =π=π r (10) ≈ 4189
dV =π 4 r dr 2 =π 4 (10) (0.05) 2 ≈ 62.8 The dy =
volume is 4189 ± 62.8 cm 3 .
35.9 ≈ 36 + dy = 6 – 0.0083 5.9917 = The absolute error is ≈ 62.8 while the relative error is 62.8 / 4189 ≈ 0.015 or 1.5% .
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28. V =π rh 2 =π (3) (12) 2 ≈ 339 33. Using the approximation fx ( +Δ≈ x ) fx () + f '( ) x Δ x
dV = 24 π rdr = π 24 (3)(0.0025) ≈ 0.565
3 we let x = 3.05 and Δ=− x 0.05 The volume is 339 ± 0.565 in. . We can rewrite
the above form as
The absolute error is ≈ 0.565 while the relative
fx () ≈ fx ( +Δ− x ) fx '( ) Δ x
error is 0.565 / 339 ≈ 0.0017 or 0.17% .
which gives
f (3.05) ≈ f (3) − f '(3.05)( 0.05) −
34. From similar triangles, the radius at height h is
ds =
45, 602sin θθ d 2 1 4
2 45, 602 – 45, 602 cos θ
h . Thus, V =π rh 2 = π h 3 , so
dV = π (100)( 1) −≈− 50 cm
45, 602 – 45, 602 cos 0.53
s ≈ 79.097 ± 0.729 cm
27 cm , 3 so there is The absolute error is ≈ 0.729 while the relative
The ice cube has volume 3 3 =
room for the ice cube without the cup
error is 0.729 / 79.097 ≈ 0.0092 or 0.92% .
overflowing.
30. A = ab sin θ = (151)(151) sin 0.53 5763.33 ≈ 35. V =π rh +π r
A = sin ; θθ =
d θ 0.53, 0.005 = V = 100 π+π r 2 r 3 ; 10, 0.1 r = dr =
22,801 dA =
(cos ) θθ d dV = (200 π+π r 4 r 2 ) dr
A ≈ 5763.33 ± 49.18 cm 2 36. The percent increase in mass is
The absolute error is ≈ 49.18 while the relative
error is 49.18 / 5763.33 0.0085 or 0.85% ≈ .
31. y = 3 x 2 –2 x + 11; 2, 0.001 x = dx =
mv ⎛
dy = (6x – 2)dx = [6(2) – 2](0.001) = 0.01
2 = so with Δx = 0.001, 6,
⎟ dv = Δ y – dy ≤ (6)(0.001) = 0.000003 m c c 2 ⎟
⎟ dv
32. Using the approximation
dv
fx ( +Δ≈ x ) fx () + f '( ) x Δ x v = 0.9c, dv = 0.02c
we let x = 1.02 and Δ=− x 0.02 . We can rewrite
the above form as
c − 0.81 c 0.19
fx () ≈ fx ( +Δ− x ) fx '( ) Δ x The percent increase in mass is about 9.5.
which gives
f (1.02) ≈ f (1) − f '(1.02)( 0.02) − = + 10 12(0.02) 10.24 =
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37. fx () = x 2 ; '( ) f x = 2; xa = 2 39. hx () = sin ; '( ) xhx = cos ; xa = 0 The linear approximation is then
The linear approximation is then
Lx () = f (2) + f '(2)( x − 2) Lx () =+ 0 1( x − 0) = x =+ 4 4( x − 2) = 4 x − 4
40. Fx () = 3 x + 4; '( ) Fx = 3; a = 3
38. gx () = x 2 cos ; '( ) xgx =− x 2 sin x + x 2 cos x The linear approximation is then
3) 13 3 + x − 9 The linear approximation is then
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45. f () x = mx + b ; f ′ () x = m ()
41. fx = 1 − x 2 ;
The linear approximation is then
L () x = ma + b + m ( x − a ) = am + b + mx − ma
(2) − x
mx + b f ()() x = L x
46. Lx () − fx () = a +
The linear approximation is then
Lx () =+ 10 ( x − 0 ) = 1 x
ax + a
47. The linear approximation to fx () at a is
Lx () = fa () + fax '( )( − a )
= a 2 + 2( ax − a )
42. gx () =
= 2 ax − a 2
) −− x ( 2 x ) 1 + x 2 1 Thus,
g ' () x =
( 2 ax − − a )
2 fx () − Lx () = x 2 −
= x 2 − 2 ax + a 2
The linear approximation is then
L () x = + ⎛− ⎜ x ⎟ = x −
2 20 1 ⎞ 20 4 = ( x − a ) 2
48. f ()( x = 1 + x ) , f ′ ()( x = α 1 + x − 1 ) , a = 0
The linear approximation is then
L () x = 1 + α () x = α x + 1
43. h () x = x sec x ; h ′ () x = sec x + x sec x tan x , a = 0
The linear approximation is then
L () x = 0 + 1 ( x − 0 ) = x −5
44. G () x = x + sin 2 x ; G ′ () x = 1 + 2 cos 2 x , a = π / 2 5
The linear approximation is then π
L () x = + () − 1 ⎜ ⎛− x ⎟ = − x + π 2 ⎝
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49. a. lim ε () h = lim ( fx ( + h ) − fx () − f ′ () xh )
= fx () − fx () − f ′ () x 0 = 0 ε () h ⎡ fx ( + h ) − fx ()
b. lim
= lim ⎢
h ()
= f ′ () x − f ′ () x = 0
2.10 Chapter Review