Openstax Calculus Volume 3 Solutions

Checkpoint

3.1

$r (0) = j, r (1) = −ii i + 5 j, r (−four) = 28 i - 15 j$

The domain of $r (t) = (t^{2} - 3 t) i + (4 t + 1) j$ is all real numbers.

3.3

$lim_{t \to −ii} r (t) = 3 i - 5 j - k$

3.4

$r^{'} (t) = 4 t i + 5 j$

3.5

$r^{'} (t) = (1 + ln t) i + five {due east}^{t} j - (sin t + cos t) k$

3.6

$\frac{d}{d t} [r (t) \cdot r^{'} (t)] = eight e^{iv t}$

$\begin{array}{l} \frac{d}{d t} [u (t) \times r (t)] \\ = - ({east}^{2 t} (cos t + 2 sin t) + cos ii t) i + (e^{two t} (2 t + ane) - sin 2 t) j + (t cos t + sin t - cos 2 t) k \end{array}$

3.vii

$T (t) = \frac{2 t}{\sqrt{4 t^{two} + 5}} i + \frac{ii}{\sqrt{4 t^{ii} + 5}} j + \frac{1}{\sqrt{iv t^{2} + five}} one thousand$

iii.viii

$\int_{ane}^{3} [(2 t + 4) i + (3 t^{2} - 4 t) j] d t = sixteen i + ten j$

3.ix

$r^{'} (t) = 〈 iv t, 4 t, 3 t^{2} 〉,$ so $due south = \frac{1}{27} (113^{3 / 2} - 32^{iii / 2}) \approx 37.785$

3.10

$southward = 5 t,$ or $t = s / 5 .$ Substituting this into $r (t) = 〈 3 cos t, 3 sin t, iv t 〉$ gives

$r (southward) = 〈 3 cos (\frac{south}{v}), 3 sin (\frac{due south}{v}), \frac{4 s}{5} 〉, s \geq 0.$

3.11

$κ = \frac{six}{101^{3 / 2}} \approx 0.0059$

3.12

$N (ii) = \frac{\sqrt{2}}{2} (i - j)$

3.thirteen

$κ = \frac{four}{{[1 + {(410 - 4)}^{ii}]}^{3 / 2}}$

At the bespeak $10 = 1,$ the curvature is equal to four. Therefore, the radius of the osculating circle is $\frac{one}{4} .$

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 betoken $(one, three) .$ Furthermore, the middle of the osculating circle is direct in a higher place the vertex. Therefore, the coordinates of the center are $(1, \frac{thirteen}{4}) .$ The equation of the osculating circumvolve is

${(x - 1)}^{2} + {(y - \frac{xiii}{4})}^{2} = \frac{1}{xvi} .$

3.14

$\begin{matrix} 5 (t) = r^{'} (t) = (ii t - three) i + 2 j + grand \\ a (t) = v^{'} (t) = ii i \\ v (t) = ‖ r^{'} (t) ‖ = \sqrt{{(2 t - iii)}^{2} + {two}^{2} + {one}^{2}} = \sqrt{iv t^{2} - 12 t + fourteen} \end{matrix}$

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

3.15

$\begin{matrix} five (t) & = & r^{'} (t) = 4 i + ii t j \\ a (t) & = & v^{'} (t) = 2 j \\ a_{T} & = & \frac{two t}{\sqrt{t^{2} + 4}}, a_{N} = \frac{4}{\sqrt{4 + t^{two}}} \end{matrix}$
$a_{T} (−3) = - \frac{6 \sqrt{thirteen}}{13}, a_{N} (−iii) = \frac{2 \sqrt{thirteen}}{13}$

3.17

$a = 1.224 \times 10^{ix} one thousand \approx 1,224,000 km$

Section 3.ane Exercises

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

a. $〈 \frac{\sqrt{ii}}{2}, \frac{\sqrt{2}}{2} 〉,$ b. $〈 \frac{i}{2}, \frac{\sqrt{3}}{ii} 〉,$ c. Yes, the limit as t approaches $π / 3$ is equal to $r (π / three),$ 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.

a. $〈 e^{π / 4}, \frac{\sqrt{two}}{2}, ln (\frac{π}{4}) 〉;$ b. $〈 e^{π / iv}, \frac{\sqrt{2}}{2}, ln (\frac{π}{4}) 〉;$ c. Aye

$〈 e^{π / two}, 1, ln (\frac{π}{two}) 〉$

11.

$2 e^{ii} i + \frac{2}{{east}^{4}} j + ii g$

xiii.

The limit does non exist considering the limit of $ln (t - 1)$ equally t approaches infinity does not exist.

xv.

$t > 0, t \neq (2 k + 1) \frac{π}{2},$ where k is an integer

17.

$t > 3, t \neq due north π,$ where n is an integer

21.

All t such that $t \in (one, \infty)$

23.

$y = ii \sqrt[3]{x},$ 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.

$10^{2} + y^{2} = nine,$ a circle centered at $(0, 0)$ with radius three, 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.

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

Notice a vector-valued role that traces out the given curve in the indicated management.

31.

For left to right, $y = x^{2},$ where t increases

33.

$(fifty, 0, 0)$

39.

One possibility is $r (t) = cos t i + sin t j + sin (iv t) k .$ 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 iii.2 Exercises

41.

$〈 iii t^{2}, 6 t, \frac{1}{2} t^{2} 〉$

43.

$〈 − e^{− t}, 3 cos (3 t), \frac{v}{\sqrt{t}} 〉$

45.

$〈 0, 0, 0 〉$

47.

$〈 \frac{−1}{{(t + ane)}^{two}}, \frac{1}{i + t^{2}}, \frac{three}{t} 〉$

49.

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

51.

$\frac{1}{\sqrt{2}} 〈 1, −one, 0 〉$

53.

$\frac{i}{\sqrt{1060.5625}} 〈 six, - \frac{3}{4}, 32 〉$

55.

$\frac{1}{\sqrt{nine {sin}^{2} (3 t) + 144 {cos}^{2} (iv t)}} 〈 0, −iii sin (3 t), 12 cos (4 t) 〉$

57.

$T (t) = \frac{−12}{13} sin (iv t) i + \frac{12}{13} cos (4 t) j + \frac{5}{thirteen} k$

59.

$〈 2 t, four t^{3}, −eight t^{7} 〉$

61.

$sin (t) + 2 t e^{t} - 4 t^{3} cos (t) + t cos (t) + t^{2} {due east}^{t} + t^{4} sin (t)$

63.

$900 t^{vii} + xvi t$

65.

Undefined or infinite

67.

$r' (t) = - b ω sin (ω t) i + b ω cos (ω t) j .$ To show orthogonality, notation that $r' (t) \cdot r (t) = 0 .$

69.

$0 i + 2 j + 4 t j$

71.

$\frac{1}{3} (10^{3 / 2} - 1)$

73.

$\begin{matrix} ‖ v (t) ‖ & = & g \\ v (t) \cdot v (t) & = & g \\ \frac{d}{d t} (5 (t) \cdot v (t)) & = & \frac{d}{d t} k = 0 \\ v (t) \cdot five' (t) + v' (t) \cdot 5 (t) & = & 0 \\ 2 v (t) \cdot 5' (t) & = & 0 \\ five (t) \cdot v' (t) & = & 0 . \end{matrix}$
The last statement implies that the velocity and acceleration are perpendicular or orthogonal.

75.

$v (t) = 〈 ane - sin t, ane - cos t 〉,$ $speed = ‖ v (t) ‖ = \sqrt{3 - 2 (sin t + cos t)}$

77.

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

79.

$r (t) = 〈 eighteen, 9 〉$ at $t = 3$

83.

$five (t) = 〈 − sin t, cos t, 1 〉$

85.

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

87.

$v (t) = 〈 − sin t, 2 cos t, 0 〉$

89.

$a (t) = 〈 - \frac{\sqrt{2}}{2}, - \sqrt{2}, 0 〉$

91.

$‖ 5 (t) ‖ = \sqrt{{sec}^{4} t + {sec}^{2} t {tan}^{2} t} = \sqrt{{sec}^{two} t ({sec}^{two} t + {tan}^{2} t)}$

95.

$〈 0, 2 sin t (t - \frac{i}{t}) - 2 cos t (1 + \frac{one}{t^{2}}), 2 sin t (i + \frac{ane}{t^{2}}) + 2 cos t (t - \frac{2}{t}) 〉$

97.

$T (t) = 〈 \frac{t^{ii}}{\sqrt{t^{4} + 1}}, \frac{−1}{\sqrt{t^{4} + 1}} 〉$

99.

$T (t) = \frac{ane}{3} 〈 i, 2, 2 〉$

101.

$\frac{3}{4} i + ln (ii) j + (1 - \frac{1}{e}) j$

Section 3.iii Exercises

105.

$\frac{one}{54} (37^{iii / ii} - one)$

113.

$T (0) = j,$ $N (0) = - i$

115.

$T (t) = 〈 \frac{two}{\sqrt{6}}, \frac{cos t - sin t}{\sqrt{6}}, \frac{cos t + sin t}{\sqrt{6}} 〉$

117.

$N (0) = ⟨ \frac{\sqrt{2}}{2}, 0, \frac{\sqrt{2}}{2} ⟩$

119.

$T (t) = \frac{ane}{\sqrt{4 t^{2} + 2}} < one, two t, ane >$

121.

$T (t) = \frac{1}{\sqrt{100 t^{two} + xiii}} (iii i + 10 t j + 2 k)$

123.

$T (t) = \frac{1}{\sqrt{9 t^{iv} + 76 t^{2} + 16}} ([3 t^{ii} - four] i + 10 t j)$

125.

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

127.

Arc-length function: $due south (t) = v t;$ r as a parameter of s: $r (s) = (3 - \frac{3 s}{v}) i + \frac{four s}{five} j$

129.

$r (s) = (ane + \frac{southward}{\sqrt{2}}) sin (ln (one + \frac{s}{\sqrt{ii}})) i + (1 + \frac{s}{\sqrt{two}}) cos [ln (1 + \frac{s}{\sqrt{2}})] j$

131.

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

135.

$κ \approx \frac{49.477}{{(17 + 144 t^{2})}^{three / two}}$

139.

The curvature approaches cipher.

141.

$y = 6 ten + π$ and $x + 6 y = 6 π$

143.

$x + 2 z = \frac{π}{2}$

145.

$\frac{a^{iv} b^{iv}}{{(b^{four} x^{2} + a^{4} y^{ii})}^{3 / 2}}$

149.

$\frac{one}{4} \ln (\sqrt{17} + 4) + \sqrt{17} \approx 4.647$

151.

The curvature is decreasing over this interval.

153.

$κ = \frac{6}{x^{2 / 5} (25 + 4 x^{6 / 5})}$

Section 3.4 Exercises

155.

$five (t) = (6 t) i + (ii - cos (t)) j$

157.

$v (t) = 〈 −iii sin t, 3 cos t, two t 〉,$ $a (t) = 〈 −3 cos t, −three sin t, two 〉,$ $speed = \sqrt{nine + 4 t^{2}}$

159.

$v (t) = −2 \sin t j + iii \cos t k,$ $a (t) = −2 cos t j - iii sin t k,$ $speed = \sqrt{4 {sin}^{2} t + 9 {cos}^{2} t}$

161.

$v (t) = {east}^{t} i - e^{− t} j,$ $a (t) = e^{t} i + {due east}^{− t} j,$ speed = $‖ v (t) ‖ = \sqrt{e^{ii t} + {east}^{−two t}}$

165.

$v (t) = (ω - ω cos (ω t)) i + (ω sin (ω t)) j,$
$a (t) = (ω^{ii} sin (ω t)) i + (ω^{two} cos (ω t)) j,$
$speed = \sqrt{ω^{2} - 2 ω^{ii} cos (ω t) + ω^{2} {cos}^{two} (ω t) + ω^{two} {sin}^{2} (ω t)} = \sqrt{2 ω^{2} (ane - cos (ω t))}$

167.

$‖ v (t) ‖ = \sqrt{ix + 4 t^{ii}}$

169.

$v (t) = 〈 e^{−five t} (cos t - 5 sin t), − e^{−5 t} (sin t + 5 cos t), −xx {east}^{−5 t} 〉$

171.

$a (t) = {〈 e}^{−five t} (− sin t - 5 cos t) - five e^{−5 t} (cos t - v sin t),$ $− {east}^{−5 t} (cos t - 5 sin t) + five e^{−5 t} (sin t + 5 cos t), 100 {eastward}^{−v t} 〉$

179.

The range is approximately 886.29 m.

181.

$five = 42.xvi$ m/sec

183.

$r (t) = 0 i + (\frac{1}{vi} t^{3} + 4.five t - \frac{14}{3}) j + (\frac{t^{iii}}{6} - \frac{i}{2} t + \frac{1}{three}) k$

185.

$a_{T} = 0,$ $a_{N} = a ω^{2}$

187.

$a_{T} = \sqrt{three} {eastward}^{t},$ $a_{Northward} = \sqrt{two} {east}^{t}$

189.

$a_{T} = 2 t,$ $a_{N} = 2$

191.

$a_{T} \frac{6 t + 12 t^{3}}{\sqrt{1 + t^{4} + t^{ii}}},$ $a_{N} = 6 \sqrt{\frac{1 + 4 t^{2} + t^{4}}{ane + t^{ii} + t^{4}}}$

193.

$a_{T} = 0,$ $a_{N} = 12 π^{ii}$

195.

$r (t) = (\frac{−ane}{m} cos t + c + \frac{i}{g}) i + (\frac{− sin t}{grand} + (v_{0} + \frac{one}{m}) t) j$

201.

$a_{T} = 0.43 {m/sec}^{2},$
$a_{N} = 2.46 {chiliad/sec}^{2}$

Review Exercises

203.

False, $\frac{d}{d t} [u (t) \times u (t)] = 0$

205.

Simulated, it is $| r^{'} (t) |$

211.

$r (t) = 〈 t, 2 - \frac{t^{2}}{eight}, −2 - \frac{t^{two}}{eight} 〉$

213.

$u^{'} (t) = 〈 ii t, 2, 20 t^{iv} 〉,$ $u ″ (t) = 〈 2, 0, 80 t^{3} 〉,$ $\frac{d}{d t} [u^{'} (t) \times u (t)] = 〈 −480 t^{3} - 160 t^{4}, 24 + 75 t^{2}, 12 + 4 t 〉,$ $\frac{d}{d t} [u (t) \times u^{'} (t)] = 〈 480 t^{3} + 160 t^{4}, −24 - 75 t^{2}, −12 - 4 t 〉,$ $\frac{d}{d t} [u (t) \cdot u^{'} (t)] = 720 t^{8} - 9600 t^{3} + 6 t^{two} + 4,$ unit of measurement tangent vector: $T (t) = \frac{2 t}{\sqrt{400 t^{8} + 4 t^{2} + 4}} i + \frac{2}{\sqrt{400 t^{8} + 4 t^{two} + iv}} j + \frac{20 t^{4}}{\sqrt{400 t^{8} + iv t^{2} + 4}} k$

215.

$\frac{ln {(4)}^{2}}{2} i + 2 j + \frac{2 (2 + \sqrt{two})}{π} k$

217.

$\frac{\sqrt{37}}{2} + \frac{ane}{12} {sinh}^{−1} (6)$

219.

$r (t (s)) = cos (\frac{2 s}{\sqrt{65}}) i + \frac{8 s}{\sqrt{65}} j - sin (\frac{two due south}{\sqrt{65}}) g$

221.

$\frac{{eastward}^{2 t}}{{(e^{ii t} + one)}^{2}}$

223.

$a_{T} = \frac{{eastward}^{2 t}}{\sqrt{ane + e^{2 t}}},$ $a_{N} = \frac{\sqrt{ii e^{2 t} + four e^{ii t} sin t cos t + 1}}{\sqrt{1 + {due east}^{2 t}}}$

225.

$v (t) = 〈 two t, \frac{1}{t}, π cos (π t) 〉$ m/sec, $a (t) = 〈 2, - \frac{1}{t^{ii}}, − π^{2} sin (π t) 〉 {chiliad/sec}^{ii},$ $speed = \sqrt{iv t^{two} + \frac{ane}{t^{ii}} + {π^{two} cos}^{2} (π t)}$ m/sec; at $t = 1,$ $r (i) = 〈 one, 0, 0 〉$ m, $v (i) = 〈 2, 1, –π 〉$ grand/sec, $a (one) = 〈 2, −1, 0 〉$ m/sec², and $speed = \sqrt{5 + π^{2}}$ m/sec