Calculus Volume 3

# Chapter 3

### Checkpoint

3.1

$r ( 0 ) = j , r ( 1 ) = −2 i + 5 j , r ( −4 ) = 28 i − 15 j r ( 0 ) = j , r ( 1 ) = −2 i + 5 j , r ( −4 ) = 28 i − 15 j$

The domain of $r(t)=(t2−3t)i+(4t+1)jr(t)=(t2−3t)i+(4t+1)j$ is all real numbers.

3.3

$lim t → −2 r ( t ) = 3 i − 5 j − k lim t → −2 r ( t ) = 3 i − 5 j − k$

3.4

$r ′ ( t ) = 4 t i + 5 j r ′ ( t ) = 4 t i + 5 j$

3.5

$r ′ ( t ) = ( 1 + ln t ) i + 5 e t j − ( sin t + cos t ) k r ′ ( t ) = ( 1 + ln t ) i + 5 e t j − ( sin t + cos t ) k$

3.6

$d d t [ r ( t ) · r ′ ( t ) ] = 8 e 4 t d d t [ r ( t ) · r ′ ( t ) ] = 8 e 4 t$

$d d t [ u ( t ) × r ( t ) ] = − ( e 2 t ( cos t + 2 sin t ) + cos 2 t ) i + ( e 2 t ( 2 t + 1 ) − sin 2 t ) j + ( t cos t + sin t − cos 2 t ) k d d t [ u ( t ) × r ( t ) ] = − ( e 2 t ( cos t + 2 sin t ) + cos 2 t ) i + ( e 2 t ( 2 t + 1 ) − sin 2 t ) j + ( t cos t + sin t − cos 2 t ) k$

3.7

$T ( t ) = 2 t 4 t 2 + 5 i + 2 4 t 2 + 5 j + 1 4 t 2 + 5 k T ( t ) = 2 t 4 t 2 + 5 i + 2 4 t 2 + 5 j + 1 4 t 2 + 5 k$

3.8

$∫ 1 3 [ ( 2 t + 4 ) i + ( 3 t 2 − 4 t ) j ] d t = 16 i + 10 j ∫ 1 3 [ ( 2 t + 4 ) i + ( 3 t 2 − 4 t ) j ] d t = 16 i + 10 j$

3.9

$r′(t)=〈4t,4t,3t2〉,r′(t)=〈4t,4t,3t2〉,$ so $s=127(1133/2−323/2)≈37.785s=127(1133/2−323/2)≈37.785$

3.10

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

$r ( s ) = 〈 3 cos ( s 5 ) , 3 sin ( s 5 ) , 4 s 5 〉 , s ≥ 0. r ( s ) = 〈 3 cos ( s 5 ) , 3 sin ( s 5 ) , 4 s 5 〉 , s ≥ 0.$

3.11

$κ = 6 101 3 / 2 ≈ 0.0059 κ = 6 101 3 / 2 ≈ 0.0059$

3.12

$N ( 2 ) = 2 2 ( i − j ) N ( 2 ) = 2 2 ( i − j )$

3.13

$κ = 4 [ 1 + ( 4 x − 4 ) 2 ] 3 / 2 κ = 4 [ 1 + ( 4 x − 4 ) 2 ] 3 / 2$

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:

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

$( x − 1 ) 2 + ( y − 13 4 ) 2 = 1 16 . ( x − 1 ) 2 + ( y − 13 4 ) 2 = 1 16 .$

3.14

$v ( t ) = r ′ ( t ) = ( 2 t − 3 ) i + 2 j + k a ( t ) = v ′ ( t ) = 2 i v ( t ) = ‖ r ′ ( t ) ‖ = ( 2 t − 3 ) 2 + 2 2 + 1 2 = 4 t 2 − 12 t + 14 v ( t ) = r ′ ( t ) = ( 2 t − 3 ) i + 2 j + k a ( t ) = v ′ ( t ) = 2 i v ( t ) = ‖ r ′ ( t ) ‖ = ( 2 t − 3 ) 2 + 2 2 + 1 2 = 4 t 2 − 12 t + 14$

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

$a = 1.224 × 10 9 m ≈ 1,224,000 km a = 1.224 × 10 9 m ≈ 1,224,000 km$

### Section 3.1 Exercises

1 .

$f ( t ) = 3 sec t , g ( t ) = 2 tan t f ( t ) = 3 sec t , g ( t ) = 2 tan t$

3 .

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.

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 .

$〈 e π / 2 , 1 , ln ( π 2 ) 〉 〈 e π / 2 , 1 , ln ( π 2 ) 〉$

11 .

$2 e 2 i + 2 e 4 j + 2 k 2 e 2 i + 2 e 4 j + 2 k$

13 .

The limit does not exist because the limit of $ln(t−1)ln(t−1)$ 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 .

$t>3,t≠nπ,t>3,t≠nπ,$ where n is an integer

19 .

21 .

All t such that $t∈(1,∞)t∈(1,∞)$

23 .

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

25 .

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

27 .

29 .

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 .

37 .

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.

### Section 3.2 Exercises

41 .

$〈 3 t 2 , 6 t , 1 2 t 2 〉 〈 3 t 2 , 6 t , 1 2 t 2 〉$

43 .

$〈 − e − t , 3 cos ( 3 t ) , 5 t 〉 〈 − e − t , 3 cos ( 3 t ) , 5 t 〉$

45 .

$〈 0 , 0 , 0 〉 〈 0 , 0 , 0 〉$

47 .

$〈 −1 ( t + 1 ) 2 , 1 1 + t 2 , 3 t 〉 〈 −1 ( t + 1 ) 2 , 1 1 + t 2 , 3 t 〉$

49 .

$〈 0 , 12 cos ( 3 t ) , cos t − t sin t 〉 〈 0 , 12 cos ( 3 t ) , cos t − t sin t 〉$

51 .

$1 2 〈 1 , −1 , 0 〉 1 2 〈 1 , −1 , 0 〉$

53 .

$1 1060.5625 〈 6 , − 3 4 , 32 〉 1 1060.5625 〈 6 , − 3 4 , 32 〉$

55 .

$1 9 sin 2 ( 3 t ) + 144 cos 2 ( 4 t ) 〈 0 , −3 sin ( 3 t ) , 12 cos ( 4 t ) 〉 1 9 sin 2 ( 3 t ) + 144 cos 2 ( 4 t ) 〈 0 , −3 sin ( 3 t ) , 12 cos ( 4 t ) 〉$

57 .

$T ( t ) = −12 13 sin ( 4 t ) i + 12 13 cos ( 4 t ) j + 5 13 k T ( t ) = −12 13 sin ( 4 t ) i + 12 13 cos ( 4 t ) j + 5 13 k$

59 .

$〈 2 t , 4 t 3 , −8 t 7 〉 〈 2 t , 4 t 3 , −8 t 7 〉$

61 .

$sin ( t ) + 2 t e t − 4 t 3 cos ( t ) + t cos ( t ) + t 2 e t + t 4 sin ( t ) sin ( t ) + 2 t e t − 4 t 3 cos ( t ) + t cos ( t ) + t 2 e t + t 4 sin ( t )$

63 .

$900 t 7 + 16 t 900 t 7 + 16 t$

65 .

1. 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 .

$0 i + 2 j + 4 t j 0 i + 2 j + 4 t j$

71 .

$1 3 ( 10 3 / 2 − 1 ) 1 3 ( 10 3 / 2 − 1 )$

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 .

$v(t)=〈1−sint,1−cost〉,v(t)=〈1−sint,1−cost〉,$ $speed=‖v(t)‖=3−2(sint+cost)speed=‖v(t)‖=3−2(sint+cost)$

77 .

$x − 1 = t , y − 1 = − t , z − 0 = 0 x − 1 = t , y − 1 = − t , z − 0 = 0$

79 .

$r(t)=〈18,9〉r(t)=〈18,9〉$ at $t=3t=3$

81 .

$161 161$

83 .

$v ( t ) = 〈 − sin t , cos t , 1 〉 v ( t ) = 〈 − sin t , cos t , 1 〉$

85 .

$a ( t ) = − cos t i − sin t j + 0 j a ( t ) = − cos t i − sin t j + 0 j$

87 .

$v ( t ) = 〈 − sin t , 2 cos t , 0 〉 v ( t ) = 〈 − sin t , 2 cos t , 0 〉$

89 .

$a ( t ) = 〈 − 2 2 , − 2 , 0 〉 a ( t ) = 〈 − 2 2 , − 2 , 0 〉$

91 .

$‖ v ( t ) ‖ = sec 4 t + sec 2 t tan 2 t = sec 2 t ( sec 2 t + tan 2 t ) ‖ v ( t ) ‖ = sec 4 t + sec 2 t tan 2 t = sec 2 t ( sec 2 t + tan 2 t )$

93 .

2

95 .

$〈 0 , 2 sin t ( t − 1 t ) − 2 cos t ( 1 + 1 t 2 ) , 2 sin t ( 1 + 1 t 2 ) + 2 cos t ( t − 2 t ) 〉 〈 0 , 2 sin t ( t − 1 t ) − 2 cos t ( 1 + 1 t 2 ) , 2 sin t ( 1 + 1 t 2 ) + 2 cos t ( t − 2 t ) 〉$

97 .

$T ( t ) = 〈 t 2 t 4 + 1 , −1 t 4 + 1 〉 T ( t ) = 〈 t 2 t 4 + 1 , −1 t 4 + 1 〉$

99 .

$T ( t ) = 1 3 〈 1 , 2 , 2 〉 T ( t ) = 1 3 〈 1 , 2 , 2 〉$

101 .

$3 4 i + ln ( 2 ) j + ( 1 − 1 e ) j 3 4 i + ln ( 2 ) j + ( 1 − 1 e ) j$

### Section 3.3 Exercises

103 .

$8 5 8 5$

105 .

$1 54 ( 37 3 / 2 − 1 ) 1 54 ( 37 3 / 2 − 1 )$

107 .

Length $=2π=2π$

109 .

$6 π 6 π$

111 .

$e − 1 e e − 1 e$

113 .

$T(0)=j,T(0)=j,$ $N(0)=−iN(0)=−i$

115 .

$T ( t ) = 〈 26,cost−sint6, cost+sint6 〉 T ( t ) = 〈 26,cost−sint6, cost+sint6 〉$

117 .

119 .

$T ( t ) = 1 4 t 2 + 2 < 1 , 2 t , 1 > T ( t ) = 1 4 t 2 + 2 < 1 , 2 t , 1 >$

121 .

$T ( t ) = 1 100 t 2 + 13 ( 3 i + 10 t j + 2 k ) T ( t ) = 1 100 t 2 + 13 ( 3 i + 10 t j + 2 k )$

123 .

$T ( t ) = 1 9 t 4 + 76 t 2 + 16 ( [ 3 t 2 − 4 ] i + 10 t j ) T ( t ) = 1 9 t 4 + 76 t 2 + 16 ( [ 3 t 2 − 4 ] i + 10 t j )$

125 .

$N ( t ) = 〈 − sin t , 0 , − cos t 〉 N ( t ) = 〈 − sin t , 0 , − cos t 〉$

127 .

Arc-length function: $s(t)=5t;s(t)=5t;$ r as a parameter of s: $r(s)=(3−3s5)i+4s5jr(s)=(3−3s5)i+4s5j$

129 .

$r ( s ) = ( 1 + s 2 ) sin ( ln ( 1 + s 2 ) ) i + ( 1 + s 2 ) cos [ ln ( 1 + s 2 ) ] j r ( s ) = ( 1 + s 2 ) sin ( ln ( 1 + s 2 ) ) i + ( 1 + s 2 ) cos [ ln ( 1 + s 2 ) ] j$

131 .

The maximum value of the curvature occurs at $x=1.x=1.$

133 .

$1 2 1 2$

135 .

$κ ≈ 49.477 ( 17 + 144 t 2 ) 3 / 2 κ ≈ 49.477 ( 17 + 144 t 2 ) 3 / 2$

137 .

$1 2 2 1 2 2$

139 .

The curvature approaches zero.

141 .

$y=6x+πy=6x+π$ and $x+6y=6πx+6y=6π$

143 .

$x + 2 z = π 2 x + 2 z = π 2$

145 .

$a 4 b 4 ( b 4 x 2 + a 4 y 2 ) 3 / 2 a 4 b 4 ( b 4 x 2 + a 4 y 2 ) 3 / 2$

147 .

$10 10 3 10 10 3$

149 .

$38 3 38 3$

151 .

The curvature is decreasing over this interval.

153 .

$κ = 6 x 2 / 5 ( 25 + 4 x 6 / 5 ) κ = 6 x 2 / 5 ( 25 + 4 x 6 / 5 )$

### Section 3.4 Exercises

155 .

$v ( t ) = ( 6 t ) i + ( 2 − cos ( t ) ) j v ( t ) = ( 6 t ) i + ( 2 − cos ( t ) ) j$

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 .

$v(t)=−2sintj+3costk,v(t)=−2sintj+3costk,$ $a(t)=−2costj−3sintk,a(t)=−2costj−3sintk,$ $speed=4sin2t+9cos2tspeed=4sin2t+9cos2t$

161 .

$v(t)=eti−e−tj,v(t)=eti−e−tj,$ $a(t)=eti+e−tj,a(t)=eti+e−tj,$ $‖v(t)‖e2t+e−2t‖v(t)‖e2t+e−2t$

163 .

$t = 4 t = 4$

165 .

$v ( t ) = ( ω − ω cos ( ω t ) ) i + ( ω sin ( ω t ) ) j , v ( t ) = ( ω − ω cos ( ω t ) ) i + ( ω sin ( ω t ) ) j ,$
$a ( t ) = ( ω 2 sin ( ω t ) ) i + ( ω 2 cos ( ω t ) ) j , a ( t ) = ( ω 2 sin ( ω t ) ) i + ( ω 2 cos ( ω t ) ) j ,$
$speed = ω 2 − 2 ω 2 cos ( ω t ) + ω 2 cos 2 ( ω t ) + ω 2 sin 2 ( ω t ) = 2 ω 2 ( 1 − cos ( ω t ) ) speed = ω 2 − 2 ω 2 cos ( ω t ) + ω 2 cos 2 ( ω t ) + ω 2 sin 2 ( ω t ) = 2 ω 2 ( 1 − cos ( ω t ) )$

167 .

$‖ v ( t ) ‖ = 9 + 4 t 2 ‖ v ( t ) ‖ = 9 + 4 t 2$

169 .

$v ( t ) = 〈 e −5 t ( cos t − 5 sin t ) , − e −5 t ( sin t + 5 cos t ) , −20 e −5 t 〉 v ( t ) = 〈 e −5 t ( cos t − 5 sin t ) , − e −5 t ( sin t + 5 cos t ) , −20 e −5 t 〉$

171 .

$a(t)=〈e−5t(−sint−5cost)−5e−5t(cost−5sint),a(t)=〈e−5t(−sint−5cost)−5e−5t(cost−5sint),$ $−e−5t(cost−5sint)+5e−5t(sint+5cost),100e−5t〉−e−5t(cost−5sint)+5e−5t(sint+5cost),100e−5t〉$

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 .

$r ( t ) = 0 i + ( 1 6 t 3 + 4.5 t − 14 3 ) j + ( t 3 6 − 1 2 t + 1 3 ) k r ( t ) = 0 i + ( 1 6 t 3 + 4.5 t − 14 3 ) j + ( t 3 6 − 1 2 t + 1 3 ) k$

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 .

$aT=0,aT=0,$ $aN=12π2aN=12π2$

195 .

$r ( t ) = ( −1 m cos t + c + 1 m ) i + ( − sin t m + ( v 0 + 1 m ) t ) j r ( t ) = ( −1 m cos t + c + 1 m ) i + ( − sin t m + ( v 0 + 1 m ) t ) j$

197 .

10.94 km/sec

201 .

$a T = 0.43 m/sec 2 , a T = 0.43 m/sec 2 ,$
$a N = 2.46 m/sec 2 a N = 2.46 m/sec 2$

### Review Exercises

203 .

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

205 .

False, it is $|r′(t)||r′(t)|$

207 .

$t<4,t<4,$ $t≠nπ2t≠nπ2$

209 .

211 .

$r ( t ) = 〈 t , 2 − t 2 8 , −2 − t 2 8 〉 r ( t ) = 〈 t , 2 − t 2 8 , −2 − t 2 8 〉$

213 .

$u′(t)=〈2t,2,20t4〉,u′(t)=〈2t,2,20t4〉,$ $u″(t)=〈2,0,80t3〉,u″(t)=〈2,0,80t3〉,$ $ddt[u′(t)×u(t)]=〈−480t3−160t4,24+75t2,12+4t〉,ddt[u′(t)×u(t)]=〈−480t3−160t4,24+75t2,12+4t〉,$ $ddt[u(t)×u′(t)]=〈480t3+160t4,−24−75t2,−12−4t〉,ddt[u(t)×u′(t)]=〈480t3+160t4,−24−75t2,−12−4t〉,$ $ddt[u(t)·u′(t)]=720t8−9600t3+6t2+4,ddt[u(t)·u′(t)]=720t8−9600t3+6t2+4,$ 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 .

$ln ( 4 ) 2 2 i + 2 j + 2 ( 2 + 2 ) π k ln ( 4 ) 2 2 i + 2 j + 2 ( 2 + 2 ) π k$

217 .

$37 2 + 1 12 sinh −1 ( 6 ) 37 2 + 1 12 sinh −1 ( 6 )$

219 .

$r ( t ( s ) ) = cos ( 2 s 65 ) i + 8 s 65 j − sin ( 2 s 65 ) k r ( t ( s ) ) = cos ( 2 s 65 ) i + 8 s 65 j − sin ( 2 s 65 ) k$

221 .

$e 2 t ( e 2 t + 1 ) 2 e 2 t ( e 2 t + 1 ) 2$

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 $t=1,t=1,$ $r(1)=〈1,0,0〉r(1)=〈1,0,0〉$ m, $v(1)=〈2,−1,1〉v(1)=〈2,−1,1〉$ m/sec, $a(1)=〈2,−1,0〉a(1)=〈2,−1,0〉$ m/sec2, and $speed=6speed=6$ m/sec

227 .

$r(t)=v0t−g2t2j,r(t)=v0t−g2t2j,$ $r(t)=〈v0(cosθ)t,v0(sinθ)t,−g2t2〉r(t)=〈v0(cosθ)t,v0(sinθ)t,−g2t2〉$

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