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  1. Preface
  2. 1 Parametric Equations and Polar Coordinates
    1. Introduction
    2. 1.1 Parametric Equations
    3. 1.2 Calculus of Parametric Curves
    4. 1.3 Polar Coordinates
    5. 1.4 Area and Arc Length in Polar Coordinates
    6. 1.5 Conic Sections
    7. Key Terms
    8. Key Equations
    9. Key Concepts
    10. Chapter Review Exercises
  3. 2 Vectors in Space
    1. Introduction
    2. 2.1 Vectors in the Plane
    3. 2.2 Vectors in Three Dimensions
    4. 2.3 The Dot Product
    5. 2.4 The Cross Product
    6. 2.5 Equations of Lines and Planes in Space
    7. 2.6 Quadric Surfaces
    8. 2.7 Cylindrical and Spherical Coordinates
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Chapter Review Exercises
  4. 3 Vector-Valued Functions
    1. Introduction
    2. 3.1 Vector-Valued Functions and Space Curves
    3. 3.2 Calculus of Vector-Valued Functions
    4. 3.3 Arc Length and Curvature
    5. 3.4 Motion in Space
    6. Key Terms
    7. Key Equations
    8. Key Concepts
    9. Chapter Review Exercises
  5. 4 Differentiation of Functions of Several Variables
    1. Introduction
    2. 4.1 Functions of Several Variables
    3. 4.2 Limits and Continuity
    4. 4.3 Partial Derivatives
    5. 4.4 Tangent Planes and Linear Approximations
    6. 4.5 The Chain Rule
    7. 4.6 Directional Derivatives and the Gradient
    8. 4.7 Maxima/Minima Problems
    9. 4.8 Lagrange Multipliers
    10. Key Terms
    11. Key Equations
    12. Key Concepts
    13. Chapter Review Exercises
  6. 5 Multiple Integration
    1. Introduction
    2. 5.1 Double Integrals over Rectangular Regions
    3. 5.2 Double Integrals over General Regions
    4. 5.3 Double Integrals in Polar Coordinates
    5. 5.4 Triple Integrals
    6. 5.5 Triple Integrals in Cylindrical and Spherical Coordinates
    7. 5.6 Calculating Centers of Mass and Moments of Inertia
    8. 5.7 Change of Variables in Multiple Integrals
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Chapter Review Exercises
  7. 6 Vector Calculus
    1. Introduction
    2. 6.1 Vector Fields
    3. 6.2 Line Integrals
    4. 6.3 Conservative Vector Fields
    5. 6.4 Green’s Theorem
    6. 6.5 Divergence and Curl
    7. 6.6 Surface Integrals
    8. 6.7 Stokes’ Theorem
    9. 6.8 The Divergence Theorem
    10. Key Terms
    11. Key Equations
    12. Key Concepts
    13. Chapter Review Exercises
  8. 7 Second-Order Differential Equations
    1. Introduction
    2. 7.1 Second-Order Linear Equations
    3. 7.2 Nonhomogeneous Linear Equations
    4. 7.3 Applications
    5. 7.4 Series Solutions of Differential Equations
    6. Key Terms
    7. Key Equations
    8. Key Concepts
    9. Chapter Review Exercises
  9. A | Table of Integrals
  10. B | Table of Derivatives
  11. C | Review of Pre-Calculus
  12. Answer Key
    1. Chapter 1
    2. Chapter 2
    3. Chapter 3
    4. Chapter 4
    5. Chapter 5
    6. Chapter 6
    7. Chapter 7
  13. Index

Checkpoint

3.1

r(0)=j,r(1)=−2i+5j,r(−4)=28i15jr(0)=j,r(1)=−2i+5j,r(−4)=28i15j

The domain of r(t)=(t23t)i+(4t+1)jr(t)=(t23t)i+(4t+1)j is all real numbers.

3.2
This figure is a graph of the function r(t) = (t^2-1)i + (2t-3)j, for the values of t from 0 to 3. The curve begins in the 3rd quadrant at the ordered pair (-1,-3) and increases up through the 1st quadrant. It is increasing and has arrows on the curve representing orientation to the right.
3.3

limt−2r(t)=3i5jklimt−2r(t)=3i5jk

3.4

r(t)=4ti+5jr(t)=4ti+5j

3.5

r(t)=(1+lnt)i+5etj(sint+cost)kr(t)=(1+lnt)i+5etj(sint+cost)k

3.6

ddt[r(t)·r(t)]=8e4tddt[r(t)·r(t)]=8e4t

ddt[u(t)×r(t)]=(e2t(cost+2sint)+cos2t)i+(e2t(2t+1)sin2t)j+(tcost+sintcos2t)kddt[u(t)×r(t)]=(e2t(cost+2sint)+cos2t)i+(e2t(2t+1)sin2t)j+(tcost+sintcos2t)k

3.7

T(t)=2t4t2+5i+24t2+5j+14t2+5kT(t)=2t4t2+5i+24t2+5j+14t2+5k

3.8

13[(2t+4)i+(3t24t)j]dt=16i+10j13[(2t+4)i+(3t24t)j]dt=16i+10j

3.9

r(t)=4t,4t,3t2,r(t)=4t,4t,3t2, so s=127(1133/2323/2)37.785s=127(1133/2323/2)37.785

3.10

s=5t,s=5t, or t=s/5.t=s/5. Substituting this into r(t)=3cost,3sint,4tr(t)=3cost,3sint,4t gives

r(s)=3cos(s5),3sin(s5),4s5,s0.r(s)=3cos(s5),3sin(s5),4s5,s0.

3.11

κ=61013/20.0059κ=61013/20.0059

3.12

N(2)=22(ij)N(2)=22(ij)

3.13

κ=4[1+(4x4)2]3/2κ=4[1+(4x4)2]3/2

At the point x=1,x=1, the curvature is equal to 4. Therefore, the radius of the osculating circle is 14.14.

A graph of this function appears next:

This figure is the graph of the function y = 2x^2-4x+5. The curve is a parabola opening up with vertex at (1, 3).

The vertex of this parabola is located at the point (1,3).(1,3). Furthermore, the center of the osculating circle is directly above the vertex. Therefore, the coordinates of the center are (1,134).(1,134). The equation of the osculating circle is

(x1)2+(y134)2=116.(x1)2+(y134)2=116.

3.14

v(t)=r(t)=(2t3)i+2j+ka(t)=v(t)=2iv(t)=r(t)=(2t3)2+22+12=4t212t+14v(t)=r(t)=(2t3)i+2j+ka(t)=v(t)=2iv(t)=r(t)=(2t3)2+22+12=4t212t+14

The units for velocity and speed are feet per second, and the units for acceleration are feet per second squared.

3.15
  1. v(t)=r(t)=4i+2tja(t)=v(t)=2jaT=2tt2+4,aN=2t2+4v(t)=r(t)=4i+2tja(t)=v(t)=2jaT=2tt2+4,aN=2t2+4
  2. aT(−3)=61313,aN(−3)=21313aT(−3)=61313,aN(−3)=21313
3.16

967.15 m

3.17

a=1.224×109m1,224,000kma=1.224×109m1,224,000km

Section 3.1 Exercises

1.

f(t)=3sect,g(t)=2tantf(t)=3sect,g(t)=2tant

3.


This figure is the graph of the function r(t) = 3sect i + 2tant j. The graph has two slant asymptotes. They are diagonal and pass through the origin. The curve has two parts, one to the left of the y-axis with a hyperbolic bend. Also, there is a second part of the curve to the right of the y-axis with a hyperbolic bend. The orientation is represented by arrows on the curve. Both curves have orientation that is rising.
5.

a. 22,22,22,22, b. 12,32,12,32, c. Yes, the limit as t approaches π/3π/3 is equal to r(π/3),r(π/3), d.

This figure is a graph of a circle centered at the origin. The circle has radius of 1 and has counter-clockwise orientation with arrows representing the orientation.
7.

a. eπ/4,22,ln(π4);eπ/4,22,ln(π4); b. eπ/4,22,ln(π4);eπ/4,22,ln(π4); c. Yes

9.

eπ/2,1,ln(π2)eπ/2,1,ln(π2)

11.

2e2i+2e4j+2k2e2i+2e4j+2k

13.

The limit does not exist because the limit of ln(t1)ln(t1) as t approaches infinity does not exist.

15.

t>0,t(2k+1)π2,t>0,t(2k+1)π2, where k is an integer

17.

t>3,tnπ,t>3,tnπ, where n is an integer

19.


This figure has two graphs. The first graph is labeled “cross section” and is a circle centered at the origin with radius of 1. It has counter-clockwise orientation. The section graph is labeled “side view” and is a 3 dimensional helix. The helix has counterclockwise orientation.
21.

All t such that t(1,)t(1,)

23.

y=2x3,y=2x3, a variation of the cube-root function

This figure is the graph of y = 2 times the cube root of x. It is an increasing function passing through the origin. The curve becomes more vertical near the origin. It has orientation to the right represented with arrows on the curve.
25.

x2+y2=9,x2+y2=9, a circle centered at (0,0)(0,0) with radius 3, and a counterclockwise orientation

This figure is the graph of x^2 + y^2 = 9. It is a circle centered at the origin with radius 3. It has orientation counter-clockwise represented with arrows on the curve.
27.


This figure is the graph of r(t) = 2cost^2 i + (2 – the square root of t) j. The curve spirals in the first quadrant, touching the y-axis. As the curve gets closer to the x-axis, the spirals become tighter. It has the look of a spring being compressed. The arrows on the curve represent orientation going downward.
29.


This figure has two graphs. The first is 3 dimensional and is a curve making a figure eight on its side inside of a box. The box represents the first octant. The second graph is 2 dimensional. It represents the same curve from a “view in the yt plane”. The horizontal axis is labeled “t”. The curve is connected and crosses over itself in the first quadrant resembling a figure eight.


Find a vector-valued function that traces out the given curve in the indicated direction.

31.

For left to right, y=x2,y=x2, where t increases

33.

(50,0,0)(50,0,0)

35.


This figure is a graph in the 3 dimensional coordinate system. It is a curve starting at the middle of the box and curving towards the upper left corner The box represents an octant of the coordinate system.
37.


This figure has two graphs. The first is 3 dimensional and is a connected curve with counter-clockwise orientation inside of a box. The second graph is 3 dimensional. It represents the same curve from different view of the box. From the side of the box the curve is connected and has depth to it.
39.

One possibility is r(t)=costi+sintj+sin(4t)k.r(t)=costi+sintj+sin(4t)k. By increasing the coefficient of t in the third component, the number of turning points will increase.

This figure is a 3 dimensional graph. It is a connected curve inside of a box. The curve has orientation. As the orientation travels around the curve, it does go up and down in depth.

Section 3.2 Exercises

41.

3t2,6t,12t23t2,6t,12t2

43.

et,3cos(3t),5tet,3cos(3t),5t

45.

0,0,00,0,0

47.

−1(t+1)2,11+t2,3t−1(t+1)2,11+t2,3t

49.

0,12cos(3t),costtsint0,12cos(3t),costtsint

51.

121,−1,0121,−1,0

53.

11060.56256,34,3211060.56256,34,32

55.

19sin2(3t)+144cos2(4t)0,−3sin(3t),12cos(4t)19sin2(3t)+144cos2(4t)0,−3sin(3t),12cos(4t)

57.

T(t)=−1213sin(4t)i+1213cos(4t)j+513kT(t)=−1213sin(4t)i+1213cos(4t)j+513k

59.

2t,4t3,−8t72t,4t3,−8t7

61.

sin(t)+2tet4t3cos(t)+tcos(t)+t2et+t4sin(t)sin(t)+2tet4t3cos(t)+tcos(t)+t2et+t4sin(t)

63.

900t7+16t900t7+16t

65.

  1. This figure is a graph of a curve in 3 dimensions. The curve has asymptotes and from the above view, the curve resembles the secant function.
  2. Undefined or infinite
67.

r'(t)=bωsin(ωt)i+bωcos(ωt)j.r'(t)=bωsin(ωt)i+bωcos(ωt)j. To show orthogonality, note that r'(t)·r(t)=0.r'(t)·r(t)=0.

69.

0i+2j+4tj0i+2j+4tj

71.

13(103/21)13(103/21)

73.


v(t)=kv(t)·v(t)=kddt(v(t)·v(t))=ddtk=0v(t)·v'(t)+v'(t)·v(t)=02v(t)·v'(t)=0v(t)·v'(t)=0.v(t)=kv(t)·v(t)=kddt(v(t)·v(t))=ddtk=0v(t)·v'(t)+v'(t)·v(t)=02v(t)·v'(t)=0v(t)·v'(t)=0.
The last statement implies that the velocity and acceleration are perpendicular or orthogonal.

75.

v(t)=1sint,1cost,v(t)=1sint,1cost, speed=v(t)=42(sint+cost)speed=v(t)=42(sint+cost)

77.

x1=t,y1=t,z0=0x1=t,y1=t,z0=0

79.

r(t)=18,9r(t)=18,9 at t=3t=3

81.

593593

83.

v(t)=sint,cost,1v(t)=sint,cost,1

85.

a(t)=costisintj+0ja(t)=costisintj+0j

87.

v(t)=sint,2cost,0v(t)=sint,2cost,0

89.

a(t)=22,2,0a(t)=22,2,0

91.

v(t)=sec4t+sec2ttan2t=sec2t(sec2t+tan2t)v(t)=sec4t+sec2ttan2t=sec2t(sec2t+tan2t)

93.

2

95.

0,2sint(t1t)2cost(1+1t2),2sint(1+1t2)+2cost(t2t)0,2sint(t1t)2cost(1+1t2),2sint(1+1t2)+2cost(t2t)

97.

T(t)=t2t4+1,−1t4+1T(t)=t2t4+1,−1t4+1

99.

T(t)=131,2,2T(t)=131,2,2

101.

34i+ln(2)j+(11e)j34i+ln(2)j+(11e)j

Section 3.3 Exercises

103.

8585

105.

154(373/21)154(373/21)

107.

Length =2π=2π

109.

6π6π

111.

e1ee1e

113.

T(0)=j,T(0)=j, N(0)=iN(0)=i

115.

T(t)=26,costsint6, cost+sint6T(t)=26,costsint6, cost+sint6

117.

N(0)=22,0,22N(0)=22,0,22

119.

T(t)=14t2+2<1,2t,1>T(t)=14t2+2<1,2t,1>

121.

T(t)=1100t2+13(3i+10tj+2k)T(t)=1100t2+13(3i+10tj+2k)

123.

T(t)=19t4+76t2+16([3t24]i+10tj)T(t)=19t4+76t2+16([3t24]i+10tj)

125.

N(t)=sint,0,costN(t)=sint,0,cost

127.

Arc-length function: s(t)=5t;s(t)=5t; r as a parameter of s: r(s)=(33s5)i+4s5jr(s)=(33s5)i+4s5j

129.

r(s)=(1+s2)sin(ln(1+s2))i+(1+s2)cos[ln(1+s2)]jr(s)=(1+s2)sin(ln(1+s2))i+(1+s2)cos[ln(1+s2)]j

131.

The maximum value of the curvature occurs at x=54.x=54.

133.

1212

135.

κ49.477(17+144t2)3/2κ49.477(17+144t2)3/2

137.

122122

139.

The curvature approaches zero.

141.

y=6x+πy=6x+π and x+6=6πx+6=6π

143.

x+2z=π2x+2z=π2

145.

a4b4(b4x2+a4y2)3/2a4b4(b4x2+a4y2)3/2

147.

1010310103

149.

383383

151.

The curvature is decreasing over this interval.

153.

κ=6x2/5(25+4x6/5)κ=6x2/5(25+4x6/5)

Section 3.4 Exercises

155.

v(t)=(6t)i+(2cos(t))jv(t)=(6t)i+(2cos(t))j

157.

v(t)=−3sint,3cost,2t,v(t)=−3sint,3cost,2t, a(t)=−3cost,−3sint,2,a(t)=−3cost,−3sint,2, speed=9+4t2speed=9+4t2

159.

v(t)=−2sintj+3costk,v(t)=−2sintj+3costk, a(t)=−2costj3sintk,a(t)=−2costj3sintk, speed=4sin2(t)+9cos2(t)speed=4sin2(t)+9cos2(t)

161.

v(t)=etietj,v(t)=etietj, a(t)=eti+etj,a(t)=eti+etj, v(t)e2t+e−2tv(t)e2t+e−2t

163.

t=4t=4

165.

v(t)=(ωωcos(ωt))i+(ωsin(ωt))j,v(t)=(ωωcos(ωt))i+(ωsin(ωt))j,
a(t)=(ω2sin(ωt))i+(ω2cos(ωt))j,a(t)=(ω2sin(ωt))i+(ω2cos(ωt))j,
speed=ω22ω2cos(ωt)+ω2cos2(ωt)+ω2sin2(ωt)=2ω2(1cos(ωt))speed=ω22ω2cos(ωt)+ω2cos2(ωt)+ω2sin2(ωt)=2ω2(1cos(ωt))

167.

v(t)=9+4t2v(t)=9+4t2

169.

v(t)=e−5t(cost5sint),e−5t(sint+5cost),−20e−5tv(t)=e−5t(cost5sint),e−5t(sint+5cost),−20e−5t

171.

a(t)=e−5t(sint5cost)5e−5t(cost5sint),a(t)=e−5t(sint5cost)5e−5t(cost5sint), e−5t(cost5sint)+5e−5t(sint+5cost),100e−5te−5t(cost5sint)+5e−5t(sint+5cost),100e−5t

173.

44.185 sec

175.

t=88.37t=88.37 sec

177.

88.37 sec

179.

The range is approximately 886.29 m.

181.

v=42.16v=42.16 m/sec

183.

r(t)=0i+(16t3+4.5t143)j+(t3612t+13)kr(t)=0i+(16t3+4.5t143)j+(t3612t+13)k

185.

aT=0,aT=0, aN=aω2aN=aω2

187.

aT=3et,aT=3et, aN=2etaN=2et

189.

aT=2t,aT=2t, aN=4+2t2aN=4+2t2

191.

aT6t+12t31+t4+t2,aT6t+12t31+t4+t2, aN=61+4t2+t41+t2+t4aN=61+4t2+t41+t2+t4

193.

aT=0,aT=0, aN=23πaN=23π

195.

r(t)=(−1mcost+c+1m)i+(sintm+(v0+1m)t)jr(t)=(−1mcost+c+1m)i+(sintm+(v0+1m)t)j

197.

10.94 km/sec

201.

aT=0.43m/sec2,aT=0.43m/sec2,
aN=2.46m/sec2aN=2.46m/sec2

Chapter Review Exercises

203.

False, ddt[u(t)×u(t)]=0ddt[u(t)×u(t)]=0

205.

False, it is |r(t)||r(t)|

207.

t<4,t<4, tnπ2tnπ2

209.


This figure is a curve in 3 dimensions. It is inside of a box. The box represents an octant. The curve begins in the center of the bottom of the box and spirals to the top of the box, increasing radius as it goes.
211.

r(t)=t,2t28,−2t28r(t)=t,2t28,−2t28

213.

u(t)=2t,2,20t4,u(t)=2t,2,20t4, u(t)=2,0,80t3,u(t)=2,0,80t3, ddt[u(t)×u(t)]=−480t3160t4,24+75t2,12+4t,ddt[u(t)×u(t)]=−480t3160t4,24+75t2,12+4t, ddt[u(t)×u(t)]=480t3+160t4,−2475t2,−124t,ddt[u(t)×u(t)]=480t3+160t4,−2475t2,−124t, ddt[u(t)·u(t)]=720t89600t3+6t2+4,ddt[u(t)·u(t)]=720t89600t3+6t2+4, unit tangent vector: T(t)=2t400t8+4t2+4i+2400t8+4t2+4j+20t4400t8+4t2+4kT(t)=2t400t8+4t2+4i+2400t8+4t2+4j+20t4400t8+4t2+4k

215.

ln(4)22i+2j+2(2+2)πkln(4)22i+2j+2(2+2)πk

217.

372+112sinh−1(6)372+112sinh−1(6)

219.

r(t(s))=cos(2s65)i+8s65jsin(2s65)kr(t(s))=cos(2s65)i+8s65jsin(2s65)k

221.

e2t(e2t+1)2e2t(e2t+1)2

223.

aT=e2t1+e2t,aT=e2t1+e2t, aN=2e2t+4e2tsintcost+11+e2taN=2e2t+4e2tsintcost+11+e2t

225.

v(t)=2t,1t,cos(πt)v(t)=2t,1t,cos(πt) m/sec, a(t)=2,1t2,sin(πt)m/sec2,a(t)=2,1t2,sin(πt)m/sec2, speed=4t2+1t2+cos2(πt)speed=4t2+1t2+cos2(πt) m/sec; at t=1,t=1, r(1)=1,0,0r(1)=1,0,0 m, v(1)=2,−1,1v(1)=2,−1,1 m/sec, a(1)=2,−1,0a(1)=2,−1,0 m/sec2, and speed=6speed=6 m/sec

227.

r(t)=v0tg2t2j,r(t)=v0tg2t2j, r(t)=v0(cosθ)t,v0(sinθ)t,g2t2r(t)=v0(cosθ)t,v0(sinθ)t,g2t2

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