Calculus Volume 3

# Chapter 3

### Checkpoint

3.1

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

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

3.3

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

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

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

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

3.7

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

3.8

$∫13[(2t+4)i+(3t2−4t)j]dt=16i+10j∫13[(2t+4)i+(3t2−4t)j]dt=16i+10j$

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

$r(s)=〈3cos(s5),3sin(s5),4s5〉,s≥0.r(s)=〈3cos(s5),3sin(s5),4s5〉,s≥0.$

3.11

$κ=61013/2≈0.0059κ=61013/2≈0.0059$

3.12

$N(2)=22(i−j)N(2)=22(i−j)$

3.13

$κ=4[1+(4x−4)2]3/2κ=4[1+(4x−4)2]3/2$

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

$(x−1)2+(y−134)2=116.(x−1)2+(y−134)2=116.$

3.14

$v(t)=r′(t)=(2t−3)i+2j+ka(t)=v′(t)=2iv(t)=‖r′(t)‖=(2t−3)2+22+12=4t2−12t+14v(t)=r′(t)=(2t−3)i+2j+ka(t)=v′(t)=2iv(t)=‖r′(t)‖=(2t−3)2+22+12=4t2−12t+14$

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

$a=1.224×109m≈1,224,000kma=1.224×109m≈1,224,000km$

### Section 3.1 Exercises

1.

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

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.

$〈eπ/2,1,ln(π2)〉〈eπ/2,1,ln(π2)〉$

11.

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

13.

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

$〈3t2,6t,12t2〉〈3t2,6t,12t2〉$

43.

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

45.

$〈0,0,0〉〈0,0,0〉$

47.

$〈−1(t+1)2,11+t2,3t〉〈−1(t+1)2,11+t2,3t〉$

49.

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

51.

$12〈1,−1,0〉12〈1,−1,0〉$

53.

$11060.5625〈6,−34,32〉11060.5625〈6,−34,32〉$

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.

$〈2t,4t3,−8t7〉〈2t,4t3,−8t7〉$

61.

$sin(t)+2tet−4t3cos(t)+tcos(t)+t2et+t4sin(t)sin(t)+2tet−4t3cos(t)+tcos(t)+t2et+t4sin(t)$

63.

$900t7+16t900t7+16t$

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.

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

71.

$13(103/2−1)13(103/2−1)$

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.

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

79.

$r(t)=〈18,9〉r(t)=〈18,9〉$ at $t=3t=3$

81.

$161161$

83.

$v(t)=〈−sint,cost,1〉v(t)=〈−sint,cost,1〉$

85.

$a(t)=−costi−sintj+0ja(t)=−costi−sintj+0j$

87.

$v(t)=〈−sint,2cost,0〉v(t)=〈−sint,2cost,0〉$

89.

$a(t)=〈−22,−2,0〉a(t)=〈−22,−2,0〉$

91.

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

93.

2

95.

$〈0,2sint(t−1t)−2cost(1+1t2),2sint(1+1t2)+2cost(t−2t)〉〈0,2sint(t−1t)−2cost(1+1t2),2sint(1+1t2)+2cost(t−2t)〉$

97.

$T(t)=〈t2t4+1,−1t4+1〉T(t)=〈t2t4+1,−1t4+1〉$

99.

$T(t)=13〈1,2,2〉T(t)=13〈1,2,2〉$

101.

$34i+ln(2)j+(1−1e)j34i+ln(2)j+(1−1e)j$

### Section 3.3 Exercises

103.

$8585$

105.

$154(373/2−1)154(373/2−1)$

107.

Length $=2π=2π$

109.

$6π6π$

111.

$e−1ee−1e$

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.

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

121.

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

123.

$T(t)=19t4+76t2+16([3t2−4]i+10tj)T(t)=19t4+76t2+16([3t2−4]i+10tj)$

125.

$N(t)=〈−sint,0,−cost〉N(t)=〈−sint,0,−cost〉$

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.

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

131.

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

133.

$1212$

135.

$κ≈49.477(17+144t2)3/2κ≈49.477(17+144t2)3/2$

137.

$122122$

139.

The curvature approaches zero.

141.

$y=6x+πy=6x+π$ and $x+6y=6πx+6y=6π$

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.

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

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.

$t=4t=4$

165.

$v(t)=(ω−ωcos(ωt))i+(ωsin(ωt))j,v(t)=(ω−ωcos(ωt))i+(ωsin(ωt))j,$
$a(t)=(ω2sin(ωt))i+(ω2cos(ωt))j,a(t)=(ω2sin(ωt))i+(ω2cos(ωt))j,$
$speed=ω2−2ω2cos(ωt)+ω2cos2(ωt)+ω2sin2(ωt)=2ω2(1−cos(ωt))speed=ω2−2ω2cos(ωt)+ω2cos2(ωt)+ω2sin2(ωt)=2ω2(1−cos(ωt))$

167.

$‖v(t)‖=9+4t2‖v(t)‖=9+4t2$

169.

$v(t)=〈e−5t(cost−5sint),−e−5t(sint+5cost),−20e−5t〉v(t)=〈e−5t(cost−5sint),−e−5t(sint+5cost),−20e−5t〉$

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.

$r(t)=0i+(16t3+4.5t−143)j+(t36−12t+13)kr(t)=0i+(16t3+4.5t−143)j+(t36−12t+13)k$

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.

$aT=0,aT=0,$ $aN=12π2aN=12π2$

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.

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

### 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.

$r(t)=〈t,2−t28,−2−t28〉r(t)=〈t,2−t28,−2−t28〉$

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.

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

217.

$372+112sinh−1(6)372+112sinh−1(6)$

219.

$r(t(s))=cos(2s65)i+8s65j−sin(2s65)kr(t(s))=cos(2s65)i+8s65j−sin(2s65)k$

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 $t=1,t=1,$ $r(1)=〈1,0,0〉r(1)=〈1,0,0〉$ 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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