Calculus/Multivariable and differential calculus:Exercises

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	Multivariable and differential calculus:Exercises

Parametric Equations

1. Find parametric equations describing the line segment from P(0,0) to Q(7,17).

x=7t and y=17t, where 0 ≤ t ≤ 1

2. Find parametric equations describing the line segment from

P(x_{1},y_{1})

to

Q(x_{2},y_{2})

.

x=x_{1}+(x_{2}-x_{1})t{\mbox{ and }}y=y_{1}+(y_{2}-y_{1})t,{\mbox{ where }}0\leq t\leq 1

x=x_{1}+(x_{2}-x_{1})t{\mbox{ and }}y=y_{1}+(y_{2}-y_{1})t,{\mbox{ where }}0\leq t\leq 1

3. Find parametric equations describing the ellipse centered at the origin with major axis of length 6 along the x-axis and the minor axis of length 3 along the y-axis, generated clockwise.

x=3\cos(-t),\ y=1.5\sin(-t)

x=3\cos(-t),\ y=1.5\sin(-t)

Polar Coordinates

20. Convert the equation into Cartesian coordinates:

r=\sin(\theta )\sec ^{2}(\theta ).

Making the substitutions

\sin(\theta )={\frac {y}{r}}

and

\sec(\theta )={\frac {r}{x}}

gives

r={\frac {y}{r}}\left({\frac {r}{x}}\right)^{2}

\iff r={\frac {ry}{x^{2}}}

\iff y=x^{2}

Making the substitutions

\sin(\theta )={\frac {y}{r}}

and

\sec(\theta )={\frac {r}{x}}

gives

r={\frac {y}{r}}\left({\frac {r}{x}}\right)^{2}

\iff r={\frac {ry}{x^{2}}}

\iff y=x^{2}

21. Find an equation of the line y=mx+b in polar coordinates.

Making the substitutions

x=r\cos(\theta )

and

y=r\sin(\theta )

gives

r\sin(\theta )=mr\cos(\theta )+b

\iff (\sin(\theta )-m\cos(\theta ))r=b

\iff r={\frac {b}{\sin(\theta )-m\cos(\theta )}}

Making the substitutions

x=r\cos(\theta )

and

y=r\sin(\theta )

gives

r\sin(\theta )=mr\cos(\theta )+b

\iff (\sin(\theta )-m\cos(\theta ))r=b

\iff r={\frac {b}{\sin(\theta )-m\cos(\theta )}}

Sketch the following polar curves without using a computer.

22.

r=2-2\sin(\theta )

23.

r^{2}=4\cos(\theta )

24.

r=2\sin(5\theta )

Sketch the following sets of points.

25.

\{(r,\theta ):\theta =2\pi /3\}

26.

\{(r,\theta ):|\theta |\leq \pi /3{\mbox{ and }}|r|<3\}

Calculus in Polar Coordinates

Find points where the following curves have vertical or horizontal tangents.

40.

r=4\cos(\theta )

Horizontal tangents occur at points where ${\frac {dy}{d\theta }}=0$ . This condition is equivalent to ${\frac {d}{d\theta }}(r\sin(\theta ))=0$ $\iff {\frac {dr}{d\theta }}\sin(\theta )+r\cos(\theta )=0$ .

Vertical tangents occur at points where ${\frac {dx}{d\theta }}=0$ . This condition is equivalent to ${\frac {d}{d\theta }}(r\cos(\theta ))=0$ $\iff {\frac {dr}{d\theta }}\cos(\theta )-r\sin(\theta )=0$ .

${\frac {dr}{d\theta }}=-4\sin(\theta )$

The condition for a horizontal tangent gives: $(-4\sin(\theta ))\sin(\theta )+(4\cos(\theta ))\cos(\theta )=0$ $\iff \cos ^{2}(\theta )-\sin ^{2}(\theta )=0$ $\iff \cos(2\theta )=0$ $\iff 2\theta =\pm \pi /2$ $\iff \theta =\pm \pi /4$

Horizontal tangents occur at $(r,\theta )=(2{\sqrt {2}},\pi /4),(2{\sqrt {2}},-\pi /4)$ which correspond to the Cartesian points $(2,2)$ and $(2,-2)$ .

The condition for a vertical tangent gives: $(-4\sin(\theta ))\cos(\theta )-(4\cos(\theta ))\sin(\theta )=0$ $\iff 2\sin(\theta )\cos(\theta )=0$ $\iff \sin(2\theta )=0$ $\iff 2\theta =0,\pi$ $\iff \theta =0,\pi /2$

Vertical tangents occur at $(r,\theta )=(0,\pi /2),(4,0)$ which correspond to the Cartesian points $(0,0)$ and $(4,0)$ .

Horizontal tangents at (2,2) and (2,-2); vertical tangents at (0,0) and (4,0)

Horizontal tangents occur at points where ${\frac {dy}{d\theta }}=0$ . This condition is equivalent to ${\frac {d}{d\theta }}(r\sin(\theta ))=0$ $\iff {\frac {dr}{d\theta }}\sin(\theta )+r\cos(\theta )=0$ .

Vertical tangents occur at points where ${\frac {dx}{d\theta }}=0$ . This condition is equivalent to ${\frac {d}{d\theta }}(r\cos(\theta ))=0$ $\iff {\frac {dr}{d\theta }}\cos(\theta )-r\sin(\theta )=0$ .

${\frac {dr}{d\theta }}=-4\sin(\theta )$

The condition for a horizontal tangent gives: $(-4\sin(\theta ))\sin(\theta )+(4\cos(\theta ))\cos(\theta )=0$ $\iff \cos ^{2}(\theta )-\sin ^{2}(\theta )=0$ $\iff \cos(2\theta )=0$ $\iff 2\theta =\pm \pi /2$ $\iff \theta =\pm \pi /4$

Horizontal tangents occur at $(r,\theta )=(2{\sqrt {2}},\pi /4),(2{\sqrt {2}},-\pi /4)$ which correspond to the Cartesian points $(2,2)$ and $(2,-2)$ .

The condition for a vertical tangent gives: $(-4\sin(\theta ))\cos(\theta )-(4\cos(\theta ))\sin(\theta )=0$ $\iff 2\sin(\theta )\cos(\theta )=0$ $\iff \sin(2\theta )=0$ $\iff 2\theta =0,\pi$ $\iff \theta =0,\pi /2$

Vertical tangents occur at $(r,\theta )=(0,\pi /2),(4,0)$ which correspond to the Cartesian points $(0,0)$ and $(4,0)$ .

Horizontal tangents at (2,2) and (2,-2); vertical tangents at (0,0) and (4,0)

41.

r=2+2\sin(\theta )

Horizontal tangents occur at points where

{\frac {dy}{d\theta }}=0

. This condition is equivalent to

{\frac {d}{d\theta }}(r\sin(\theta ))=0

\iff {\frac {dr}{d\theta }}\sin(\theta )+r\cos(\theta )=0

.

Vertical tangents occur at points where ${\frac {dx}{d\theta }}=0$ . This condition is equivalent to ${\frac {d}{d\theta }}(r\cos(\theta ))=0$ $\iff {\frac {dr}{d\theta }}\cos(\theta )-r\sin(\theta )=0$ .

${\frac {dr}{d\theta }}=2\cos(\theta )$

The condition for a horizontal tangent gives: $(2\cos(\theta ))\sin(\theta )+(2+2\sin(\theta ))\cos(\theta )=0$ $\iff 4\sin(\theta )\cos(\theta )+2\cos(\theta )=0$ $\iff \cos(\theta )(2\sin(\theta )+1)=0$ $\iff \cos(\theta )=0\;{\text{or}}\;\sin(\theta )=-{\frac {1}{2}}$ $\iff \theta =-5\pi /6,-\pi /2,-\pi /6,\pi /2$

Horizontal tangents occur at $(r,\theta )=(1,-5\pi /6),(0,-\pi /2),(1,-\pi /6),(4,\pi /2)$ which correspond to the Cartesian points $(-{\sqrt {3}}/2,-1/2)$ , $(0,0)$ , $({\sqrt {3}}/2,-1/2)$ , and $(0,4)$ . Point $(0,0)$ corresponds to a vertical cusp however and should be excluded leaving $(-{\sqrt {3}}/2,-1/2)$ , $({\sqrt {3}}/2,-1/2)$ , and $(0,4)$ .

The condition for a vertical tangent gives: $(2\cos(\theta ))\cos(\theta )-(2+2\sin(\theta ))\sin(\theta )=0$ $\iff 2\cos ^{2}(\theta )-2\sin ^{2}(\theta )-2\sin(\theta )=0$ $\iff 2(1-\sin ^{2}(\theta ))-2\sin ^{2}(\theta )-2\sin(\theta )=0$ $\iff -4\sin ^{2}(\theta )-2\sin(\theta )+2=0$ $\iff 2\sin ^{2}(\theta )+\sin(\theta )-1=0$ $\iff (2\sin(\theta )-1)(\sin(\theta )+1)=0$ $\iff \sin(\theta )=-1,1/2$ $\iff \theta =-\pi /2,\pi /6,5\pi /6$

Vertical tangents occur at $(r,\theta )=(0,-\pi /2),(3,\pi /6),(3,5\pi /6)$ which correspond to the Cartesian points $(0,0)$ , $(3{\sqrt {3}}/2,3/2)$ , and $(-3{\sqrt {3}}/2,3/2)$ .

Horizontal tangents at (r,θ) = (4,π/2), (1,7π/6) and (1,-π/6); vertical tangents at (r,θ) = (3,π/6), (3,5π/6), and (0,3π/2)

Horizontal tangents occur at points where

{\frac {dy}{d\theta }}=0

. This condition is equivalent to

{\frac {d}{d\theta }}(r\sin(\theta ))=0

\iff {\frac {dr}{d\theta }}\sin(\theta )+r\cos(\theta )=0

.

Vertical tangents occur at points where ${\frac {dx}{d\theta }}=0$ . This condition is equivalent to ${\frac {d}{d\theta }}(r\cos(\theta ))=0$ $\iff {\frac {dr}{d\theta }}\cos(\theta )-r\sin(\theta )=0$ .

${\frac {dr}{d\theta }}=2\cos(\theta )$

The condition for a horizontal tangent gives: $(2\cos(\theta ))\sin(\theta )+(2+2\sin(\theta ))\cos(\theta )=0$ $\iff 4\sin(\theta )\cos(\theta )+2\cos(\theta )=0$ $\iff \cos(\theta )(2\sin(\theta )+1)=0$ $\iff \cos(\theta )=0\;{\text{or}}\;\sin(\theta )=-{\frac {1}{2}}$ $\iff \theta =-5\pi /6,-\pi /2,-\pi /6,\pi /2$

Horizontal tangents occur at $(r,\theta )=(1,-5\pi /6),(0,-\pi /2),(1,-\pi /6),(4,\pi /2)$ which correspond to the Cartesian points $(-{\sqrt {3}}/2,-1/2)$ , $(0,0)$ , $({\sqrt {3}}/2,-1/2)$ , and $(0,4)$ . Point $(0,0)$ corresponds to a vertical cusp however and should be excluded leaving $(-{\sqrt {3}}/2,-1/2)$ , $({\sqrt {3}}/2,-1/2)$ , and $(0,4)$ .

The condition for a vertical tangent gives: $(2\cos(\theta ))\cos(\theta )-(2+2\sin(\theta ))\sin(\theta )=0$ $\iff 2\cos ^{2}(\theta )-2\sin ^{2}(\theta )-2\sin(\theta )=0$ $\iff 2(1-\sin ^{2}(\theta ))-2\sin ^{2}(\theta )-2\sin(\theta )=0$ $\iff -4\sin ^{2}(\theta )-2\sin(\theta )+2=0$ $\iff 2\sin ^{2}(\theta )+\sin(\theta )-1=0$ $\iff (2\sin(\theta )-1)(\sin(\theta )+1)=0$ $\iff \sin(\theta )=-1,1/2$ $\iff \theta =-\pi /2,\pi /6,5\pi /6$

Vertical tangents occur at $(r,\theta )=(0,-\pi /2),(3,\pi /6),(3,5\pi /6)$ which correspond to the Cartesian points $(0,0)$ , $(3{\sqrt {3}}/2,3/2)$ , and $(-3{\sqrt {3}}/2,3/2)$ .

Horizontal tangents at (r,θ) = (4,π/2), (1,7π/6) and (1,-π/6); vertical tangents at (r,θ) = (3,π/6), (3,5π/6), and (0,3π/2)

Sketch the region and find its area.

42. The region inside the limaçon

2+\cos(\theta )

Given an infinitesimal wedge with angle

d\theta

and radius

r

, the area is

{\frac {1}{2}}r^{2}d\theta

. The total area is therefore

A=\int _{\theta =0}^{2\pi }{\frac {1}{2}}r^{2}d\theta

=\int _{\theta =0}^{2\pi }{\frac {1}{2}}(2+\cos(\theta ))^{2}d\theta

=\int _{\theta =0}^{2\pi }(2+2\cos(\theta )+{\frac {1}{2}}\cos ^{2}(\theta ))d\theta

=\int _{\theta =0}^{2\pi }(2+2\cos(\theta )+{\frac {1}{4}}(\cos(2\theta )+1))d\theta

=\int _{\theta =0}^{2\pi }({\frac {9}{4}}+2\cos(\theta )+{\frac {1}{4}}\cos(2\theta ))d\theta

=({\frac {9}{4}}\theta +2\sin(\theta )+{\frac {1}{8}}\sin(2\theta )){\bigg |}_{\theta =0}^{2\pi }

={\frac {9}{2}}\pi

. 9π/2

Given an infinitesimal wedge with angle

d\theta

and radius

r

, the area is

{\frac {1}{2}}r^{2}d\theta

. The total area is therefore

A=\int _{\theta =0}^{2\pi }{\frac {1}{2}}r^{2}d\theta

=\int _{\theta =0}^{2\pi }{\frac {1}{2}}(2+\cos(\theta ))^{2}d\theta

=\int _{\theta =0}^{2\pi }(2+2\cos(\theta )+{\frac {1}{2}}\cos ^{2}(\theta ))d\theta

=\int _{\theta =0}^{2\pi }(2+2\cos(\theta )+{\frac {1}{4}}(\cos(2\theta )+1))d\theta

=\int _{\theta =0}^{2\pi }({\frac {9}{4}}+2\cos(\theta )+{\frac {1}{4}}\cos(2\theta ))d\theta

=({\frac {9}{4}}\theta +2\sin(\theta )+{\frac {1}{8}}\sin(2\theta )){\bigg |}_{\theta =0}^{2\pi }

={\frac {9}{2}}\pi

. 9π/2

43. The region inside the petals of the rose

4\cos(2\theta )

and outside the circle

r=2

There are 4 petals, as seen in the image below. The area of just one of the petals needs to be computed and the multiplied by

4

.

It is first necessary to compute the angular limits of one of the petals. The petals start and end at points where $4\cos(2\theta )=2$ $\iff \cos(2\theta )={\frac {1}{2}}$ $\Leftarrow 2\theta =\pm {\frac {\pi }{3}}$ $\iff \theta =\pm {\frac {\pi }{6}}$ . The bounds on one of the petals are $\left[-{\frac {\pi }{6}},+{\frac {\pi }{6}}\right]$ .

Given an annular wedge with angle $d\theta$ , inner radius $r_{1}$ , and an outer radius of $r_{2}$ , the area is ${\frac {1}{2}}(r_{2}^{2}-r_{1}^{2})d\theta$ . The total area of all 4 petals is therefore $A=4\int _{\theta =-\pi /6}^{\pi /6}{\frac {1}{2}}((4\cos(2\theta ))^{2}-2^{2})d\theta$ $=4\int _{\theta =-\pi /6}^{\pi /6}(8\cos ^{2}(2\theta )-2)d\theta$ $=4\int _{\theta =-\pi /6}^{\pi /6}(4(\cos(4\theta )+1)-2)d\theta$ $=4\int _{\theta =-\pi /6}^{\pi /6}(2+4\cos(4\theta ))d\theta$ $=4(2\theta +\sin(4\theta )){\bigg |}_{\theta =-\pi /6}^{\pi /6}$ $=4({\frac {2\pi }{3}}+{\sqrt {3}})$ $={\frac {8\pi }{3}}+4{\sqrt {3}}$ .

8\pi /3+4{\sqrt {3}}

There are 4 petals, as seen in the image below. The area of just one of the petals needs to be computed and the multiplied by

4

.

It is first necessary to compute the angular limits of one of the petals. The petals start and end at points where $4\cos(2\theta )=2$ $\iff \cos(2\theta )={\frac {1}{2}}$ $\Leftarrow 2\theta =\pm {\frac {\pi }{3}}$ $\iff \theta =\pm {\frac {\pi }{6}}$ . The bounds on one of the petals are $\left[-{\frac {\pi }{6}},+{\frac {\pi }{6}}\right]$ .

Given an annular wedge with angle $d\theta$ , inner radius $r_{1}$ , and an outer radius of $r_{2}$ , the area is ${\frac {1}{2}}(r_{2}^{2}-r_{1}^{2})d\theta$ . The total area of all 4 petals is therefore $A=4\int _{\theta =-\pi /6}^{\pi /6}{\frac {1}{2}}((4\cos(2\theta ))^{2}-2^{2})d\theta$ $=4\int _{\theta =-\pi /6}^{\pi /6}(8\cos ^{2}(2\theta )-2)d\theta$ $=4\int _{\theta =-\pi /6}^{\pi /6}(4(\cos(4\theta )+1)-2)d\theta$ $=4\int _{\theta =-\pi /6}^{\pi /6}(2+4\cos(4\theta ))d\theta$ $=4(2\theta +\sin(4\theta )){\bigg |}_{\theta =-\pi /6}^{\pi /6}$ $=4({\frac {2\pi }{3}}+{\sqrt {3}})$ $={\frac {8\pi }{3}}+4{\sqrt {3}}$ .

8\pi /3+4{\sqrt {3}}

Vectors and Dot Product

60. Find an equation of the sphere with center (1,2,0) passing through the point (3,4,5)

The general equation for a sphere is

(x-x_{0})^{2}+(y-y_{0})^{2}+(z-z_{0})^{2}=r^{2}

where

(x_{0},y_{0},z_{0})

is the location of the sphere's center and

r

is the sphere's radius.

It is already known that the sphere's center is $(x_{0},y_{0},z_{0})=(1,2,0)$ . The sphere's radius is the distance between (1,2,0) and (3,4,5) which is $r={\sqrt {(3-1)^{2}+(4-2)^{2}+(5-0)^{2}}}={\sqrt {4+4+25}}={\sqrt {33}}$ .

Therefore the sphere's equation is:

(x-1)^{2}+(y-2)^{2}+z^{2}=33

.

The general equation for a sphere is

(x-x_{0})^{2}+(y-y_{0})^{2}+(z-z_{0})^{2}=r^{2}

where

(x_{0},y_{0},z_{0})

is the location of the sphere's center and

r

is the sphere's radius.

It is already known that the sphere's center is $(x_{0},y_{0},z_{0})=(1,2,0)$ . The sphere's radius is the distance between (1,2,0) and (3,4,5) which is $r={\sqrt {(3-1)^{2}+(4-2)^{2}+(5-0)^{2}}}={\sqrt {4+4+25}}={\sqrt {33}}$ .

Therefore the sphere's equation is:

(x-1)^{2}+(y-2)^{2}+z^{2}=33

.

61. Sketch the plane passing through the points (2,0,0), (0,3,0), and (0,0,4)

62. Find the value of

|\mathbf {u} +3\mathbf {v} |

if

\mathbf {u} =\langle 1,3,0\rangle

and

\mathbf {v} =\langle 3,0,2\rangle

$\mathbf {u} +3\mathbf {v} =\langle 1,3,0\rangle +3\langle 3,0,2\rangle$ $=\langle 1,3,0\rangle +\langle 9,0,6\rangle$ $=\langle 10,3,6\rangle$ .

Therefore: $|\mathbf {u} +3\mathbf {v} |={\sqrt {10^{2}+3^{2}+6^{2}}}={\sqrt {100+9+36}}={\sqrt {145}}$ .

{\sqrt {145}}

$\mathbf {u} +3\mathbf {v} =\langle 1,3,0\rangle +3\langle 3,0,2\rangle$ $=\langle 1,3,0\rangle +\langle 9,0,6\rangle$ $=\langle 10,3,6\rangle$ .

Therefore: $|\mathbf {u} +3\mathbf {v} |={\sqrt {10^{2}+3^{2}+6^{2}}}={\sqrt {100+9+36}}={\sqrt {145}}$ .

{\sqrt {145}}

63. Find all unit vectors parallel to

\langle 1,2,3\rangle

The length of

\langle 1,2,3\rangle

is

{\sqrt {1^{2}+2^{2}+3^{2}}}={\sqrt {1+4+9}}={\sqrt {14}}

. Therefore

{\frac {1}{\sqrt {14}}}\langle 1,2,3\rangle

is a unit vector that points in the same direction as

\langle 1,2,3\rangle

, and

-{\frac {1}{\sqrt {14}}}\langle 1,2,3\rangle

is a unit vector that points in the opposite direction as

\langle 1,2,3\rangle

.

\pm {\frac {1}{\sqrt {14}}}\langle 1,2,3\rangle

are the unit vectors that are parallel to

\langle 1,2,3\rangle

.

The length of

\langle 1,2,3\rangle

is

{\sqrt {1^{2}+2^{2}+3^{2}}}={\sqrt {1+4+9}}={\sqrt {14}}

. Therefore

{\frac {1}{\sqrt {14}}}\langle 1,2,3\rangle

is a unit vector that points in the same direction as

\langle 1,2,3\rangle

, and

-{\frac {1}{\sqrt {14}}}\langle 1,2,3\rangle

is a unit vector that points in the opposite direction as

\langle 1,2,3\rangle

.

\pm {\frac {1}{\sqrt {14}}}\langle 1,2,3\rangle

are the unit vectors that are parallel to

\langle 1,2,3\rangle

.

64. Prove one of the distributive properties for vectors in

\mathbb {R} ^{3}

:

c(\mathbf {u} +\mathbf {v} )=c\mathbf {u} +c\mathbf {v}

$c(\mathbf {u} +\mathbf {v} )=c(\langle u_{1},u_{2},u_{3}\rangle +\langle v_{1},v_{2},v_{3}\rangle )$ $=c\langle u_{1}+v_{1},u_{2}+v_{2},u_{3}+v_{3}\rangle$ $=\langle c(u_{1}+v_{1}),c(u_{2}+v_{2}),c(u_{3}+v_{3})\rangle$ $=\langle cu_{1}+cv_{1},cu_{2}+cv_{2},cu_{3}+cv_{3}\rangle$ $=\langle cu_{1},cu_{2},cu_{3}\rangle +\langle cv_{1},cv_{2},cv_{3}\rangle$ $=c\langle u_{1},u_{2},u_{3}\rangle +c\langle v_{1},v_{2},v_{3}\rangle$

=c\mathbf {u} +c\mathbf {v}

.

$c(\mathbf {u} +\mathbf {v} )=c(\langle u_{1},u_{2},u_{3}\rangle +\langle v_{1},v_{2},v_{3}\rangle )$ $=c\langle u_{1}+v_{1},u_{2}+v_{2},u_{3}+v_{3}\rangle$ $=\langle c(u_{1}+v_{1}),c(u_{2}+v_{2}),c(u_{3}+v_{3})\rangle$ $=\langle cu_{1}+cv_{1},cu_{2}+cv_{2},cu_{3}+cv_{3}\rangle$ $=\langle cu_{1},cu_{2},cu_{3}\rangle +\langle cv_{1},cv_{2},cv_{3}\rangle$ $=c\langle u_{1},u_{2},u_{3}\rangle +c\langle v_{1},v_{2},v_{3}\rangle$

=c\mathbf {u} +c\mathbf {v}

.

65. Find all unit vectors orthogonal to

3\mathbf {i} +4\mathbf {j}

in

\mathbb {R} ^{2}

Rotating $\mathbf {u} =\langle 3,4\rangle$ $90^{\circ }$ counterclockwise gives $\mathbf {u} ^{\perp }=\langle -4,3\rangle$ . $\mathbf {u} ^{\perp }$ is orthogonal to $\mathbf {u}$ , and the normalization of $\mathbf {u} ^{\perp }$ and its negative are the only unit vectors that are orthogonal to $\mathbf {u}$ .

The magnitude of

\mathbf {u} ^{\perp }

is

{\sqrt {(-4)^{2}+3^{2}}}={\sqrt {16+9}}={\sqrt {25}}=5

so the only unit vectors that are orthogonal to

\mathbf {u}

are

\pm \left\langle {\frac {-4}{5}},{\frac {3}{5}}\right\rangle

.

Rotating $\mathbf {u} =\langle 3,4\rangle$ $90^{\circ }$ counterclockwise gives $\mathbf {u} ^{\perp }=\langle -4,3\rangle$ . $\mathbf {u} ^{\perp }$ is orthogonal to $\mathbf {u}$ , and the normalization of $\mathbf {u} ^{\perp }$ and its negative are the only unit vectors that are orthogonal to $\mathbf {u}$ .

The magnitude of

\mathbf {u} ^{\perp }

is

{\sqrt {(-4)^{2}+3^{2}}}={\sqrt {16+9}}={\sqrt {25}}=5

so the only unit vectors that are orthogonal to

\mathbf {u}

are

\pm \left\langle {\frac {-4}{5}},{\frac {3}{5}}\right\rangle

.

66. Find all unit vectors orthogonal to

3\mathbf {i} +4\mathbf {j}

in

\mathbb {R} ^{3}

All vectors $\mathbf {u} =\langle u_{1},u_{2},u_{3}\rangle$ that are orthogonal to $\langle 3,4,0\rangle$ must satisfy $\langle 3,4,0\rangle \cdot \mathbf {u} =0$ $\iff 3u_{1}+4u_{2}=0$ $\iff u_{2}=-{\frac {3}{4}}u_{1}$ .

The set of possible values of $\mathbf {u}$ is $\left\{\left\langle x,-{\frac {3}{4}}x,z\right\rangle {\bigg |}x,z\in \mathbb {R} \right\}$ . The restriction that $|\mathbf {u} |=1$ becomes ${\sqrt {u_{1}^{2}+u_{2}^{2}+u_{3}^{2}}}=1$ $\iff x^{2}+{\frac {9}{16}}x^{2}+z^{2}=1$ $\iff {\frac {25}{16}}x^{2}+z^{2}=1$ $\iff \left({\frac {x}{4/5}}\right)^{2}+z^{2}=1$ .

The set of possible $x$ and $z$ is an ellipse with radii $4/5$ and $1$ . One possible parameterization of $x$ and $z$ is $x={\frac {4}{5}}\cos(\theta )$ and $z=\sin(\theta )$ where $\theta \in \mathbb {R}$ . This parameterization yields $\mathbf {u} =\left\langle {\frac {4}{5}}\cos(\theta ),-{\frac {3}{5}}\cos(\theta ),\sin(\theta )\right\rangle$ where $\theta \in \mathbb {R}$ as the complete set of unit vectors that are orthogonal to $\langle 3,4,0\rangle$ .

Re-parameterizing by letting

c=\cos(\theta )

gives the set

\left\langle {\frac {4}{5}}c,-{\frac {3}{5}}c,\pm {\sqrt {1-c^{2}}}\right\rangle ,\ c\in [-1,1]

All vectors $\mathbf {u} =\langle u_{1},u_{2},u_{3}\rangle$ that are orthogonal to $\langle 3,4,0\rangle$ must satisfy $\langle 3,4,0\rangle \cdot \mathbf {u} =0$ $\iff 3u_{1}+4u_{2}=0$ $\iff u_{2}=-{\frac {3}{4}}u_{1}$ .

The set of possible values of $\mathbf {u}$ is $\left\{\left\langle x,-{\frac {3}{4}}x,z\right\rangle {\bigg |}x,z\in \mathbb {R} \right\}$ . The restriction that $|\mathbf {u} |=1$ becomes ${\sqrt {u_{1}^{2}+u_{2}^{2}+u_{3}^{2}}}=1$ $\iff x^{2}+{\frac {9}{16}}x^{2}+z^{2}=1$ $\iff {\frac {25}{16}}x^{2}+z^{2}=1$ $\iff \left({\frac {x}{4/5}}\right)^{2}+z^{2}=1$ .

The set of possible $x$ and $z$ is an ellipse with radii $4/5$ and $1$ . One possible parameterization of $x$ and $z$ is $x={\frac {4}{5}}\cos(\theta )$ and $z=\sin(\theta )$ where $\theta \in \mathbb {R}$ . This parameterization yields $\mathbf {u} =\left\langle {\frac {4}{5}}\cos(\theta ),-{\frac {3}{5}}\cos(\theta ),\sin(\theta )\right\rangle$ where $\theta \in \mathbb {R}$ as the complete set of unit vectors that are orthogonal to $\langle 3,4,0\rangle$ .

Re-parameterizing by letting

c=\cos(\theta )

gives the set

\left\langle {\frac {4}{5}}c,-{\frac {3}{5}}c,\pm {\sqrt {1-c^{2}}}\right\rangle ,\ c\in [-1,1]

67. Find all unit vectors that make an angle of

\pi /3

with the vector

\langle 1,2\rangle

The angle that $\langle 1,2\rangle$ makes with the x-axis is $\arctan(2)$ counterclockwise.

Making a both a clockwise and a counterclockwise rotation of $\pi /3$ gives $\langle \cos(\arctan(2)\mp \pi /3),\sin(\arctan(2)\mp \pi /3)\rangle$ $=\langle \cos(\arctan(2))\cos(\mp \pi /3)-\sin(\arctan(2))\sin(\mp \pi /3),\sin(\arctan(2))\cos(\mp \pi /3)+\cos(\arctan(2))\sin(\mp \pi /3)\rangle$ $=\left\langle {\frac {1}{\sqrt {1+2^{2}}}}\cdot {\frac {1}{2}}-{\frac {2}{\sqrt {1+2^{2}}}}\cdot {\frac {\mp {\sqrt {3}}}{2}},{\frac {2}{\sqrt {1+2^{2}}}}\cdot {\frac {1}{2}}+{\frac {1}{\sqrt {1+2^{2}}}}\cdot {\frac {\mp {\sqrt {3}}}{2}}\right\rangle$ $=\left\langle {\frac {1}{2{\sqrt {5}}}}\pm {\frac {\sqrt {3}}{\sqrt {5}}},{\frac {1}{\sqrt {5}}}\mp {\frac {\sqrt {3}}{2{\sqrt {5}}}}\right\rangle$ $={\frac {1}{2{\sqrt {5}}}}\left\langle 1\pm 2{\sqrt {3}},2\mp {\sqrt {3}}\right\rangle$

={\frac {\sqrt {5}}{10}}\left\langle 1\pm 2{\sqrt {3}},\ 2\mp {\sqrt {3}}\right\rangle

The angle that $\langle 1,2\rangle$ makes with the x-axis is $\arctan(2)$ counterclockwise.

Making a both a clockwise and a counterclockwise rotation of $\pi /3$ gives $\langle \cos(\arctan(2)\mp \pi /3),\sin(\arctan(2)\mp \pi /3)\rangle$ $=\langle \cos(\arctan(2))\cos(\mp \pi /3)-\sin(\arctan(2))\sin(\mp \pi /3),\sin(\arctan(2))\cos(\mp \pi /3)+\cos(\arctan(2))\sin(\mp \pi /3)\rangle$ $=\left\langle {\frac {1}{\sqrt {1+2^{2}}}}\cdot {\frac {1}{2}}-{\frac {2}{\sqrt {1+2^{2}}}}\cdot {\frac {\mp {\sqrt {3}}}{2}},{\frac {2}{\sqrt {1+2^{2}}}}\cdot {\frac {1}{2}}+{\frac {1}{\sqrt {1+2^{2}}}}\cdot {\frac {\mp {\sqrt {3}}}{2}}\right\rangle$ $=\left\langle {\frac {1}{2{\sqrt {5}}}}\pm {\frac {\sqrt {3}}{\sqrt {5}}},{\frac {1}{\sqrt {5}}}\mp {\frac {\sqrt {3}}{2{\sqrt {5}}}}\right\rangle$ $={\frac {1}{2{\sqrt {5}}}}\left\langle 1\pm 2{\sqrt {3}},2\mp {\sqrt {3}}\right\rangle$

={\frac {\sqrt {5}}{10}}\left\langle 1\pm 2{\sqrt {3}},\ 2\mp {\sqrt {3}}\right\rangle

Cross Product

Find $\mathbf {u} \times \mathbf {v}$ and $\mathbf {v} \times \mathbf {u}$

80.

\mathbf {u} =\langle -4,1,1\rangle

and

\mathbf {v} =\langle 0,1,-1\rangle

\mathbf {u} \times \mathbf {v} =\langle (1)(-1)-(1)(1),(1)(0)-(-4)(-1),(-4)(1)-(1)(0)\rangle =\langle -2,-4,-4\rangle

\mathbf {v} \times \mathbf {u} =-(\mathbf {u} \times \mathbf {v} )=\langle 2,4,4\rangle

\mathbf {u} \times \mathbf {v} =\langle (1)(-1)-(1)(1),(1)(0)-(-4)(-1),(-4)(1)-(1)(0)\rangle =\langle -2,-4,-4\rangle

\mathbf {v} \times \mathbf {u} =-(\mathbf {u} \times \mathbf {v} )=\langle 2,4,4\rangle

81.

\mathbf {u} =\langle 1,2,-1\rangle

and

\mathbf {v} =\langle 3,-4,6\rangle

\mathbf {u} \times \mathbf {v} =\langle (2)(6)-(-1)(-4),(-1)(3)-(1)(6),(1)(-4)-(2)(3)\rangle =\langle 8,-9,-10\rangle

\mathbf {v} \times \mathbf {u} =-(\mathbf {u} \times \mathbf {v} )=\langle -8,9,10\rangle

\mathbf {u} \times \mathbf {v} =\langle (2)(6)-(-1)(-4),(-1)(3)-(1)(6),(1)(-4)-(2)(3)\rangle =\langle 8,-9,-10\rangle

\mathbf {v} \times \mathbf {u} =-(\mathbf {u} \times \mathbf {v} )=\langle -8,9,10\rangle

Find the area of the parallelogram with sides $\mathbf {u}$ and $\mathbf {v}$ .

82.

\mathbf {u} =\langle -3,0,2\rangle

and

\mathbf {v} =\langle 1,1,1\rangle

The cross product of vectors $\mathbf {u}$ and $\mathbf {v}$ is a vector with length $|\mathbf {u} ||\mathbf {v} |\sin(\theta )$ where $\theta$ is the angle between $\mathbf {u}$ and $\mathbf {v}$ . $|\mathbf {u} ||\mathbf {v} |\sin(\theta )$ is the area of the parallelogram with sides $\mathbf {u}$ and $\mathbf {v}$ , so this area is found by computing $|\mathbf {u} \times \mathbf {v} |$ .

$\mathbf {u} \times \mathbf {v} =\langle (0)(1)-(2)(1),(2)(1)-(-3)(1),(-3)(1)-(0)(1)\rangle =\langle -2,5,-3\rangle$

A=|\mathbf {u} \times \mathbf {v} |={\sqrt {(-2)^{2}+5^{2}+(-3)^{2}}}={\sqrt {4+25+9}}={\sqrt {38}}

The cross product of vectors $\mathbf {u}$ and $\mathbf {v}$ is a vector with length $|\mathbf {u} ||\mathbf {v} |\sin(\theta )$ where $\theta$ is the angle between $\mathbf {u}$ and $\mathbf {v}$ . $|\mathbf {u} ||\mathbf {v} |\sin(\theta )$ is the area of the parallelogram with sides $\mathbf {u}$ and $\mathbf {v}$ , so this area is found by computing $|\mathbf {u} \times \mathbf {v} |$ .

$\mathbf {u} \times \mathbf {v} =\langle (0)(1)-(2)(1),(2)(1)-(-3)(1),(-3)(1)-(0)(1)\rangle =\langle -2,5,-3\rangle$

A=|\mathbf {u} \times \mathbf {v} |={\sqrt {(-2)^{2}+5^{2}+(-3)^{2}}}={\sqrt {4+25+9}}={\sqrt {38}}

83.

\mathbf {u} =\langle 8,2,-3\rangle

and

\mathbf {v} =\langle 2,4,-4\rangle

The cross product of vectors $\mathbf {u}$ and $\mathbf {v}$ is a vector with length $|\mathbf {u} ||\mathbf {v} |\sin(\theta )$ where $\theta$ is the angle between $\mathbf {u}$ and $\mathbf {v}$ . $|\mathbf {u} ||\mathbf {v} |\sin(\theta )$ is the area of the parallelogram with sides $\mathbf {u}$ and $\mathbf {v}$ , so this area is found by computing $|\mathbf {u} \times \mathbf {v} |$ .

$\mathbf {u} \times \mathbf {v} =\langle (2)(-4)-(-3)(4),(-3)(2)-(8)(-4),(8)(4)-(2)(2)\rangle =\langle 4,26,28\rangle$

A=|\mathbf {u} \times \mathbf {v} |={\sqrt {4^{2}+26^{2}+28^{2}}}={\sqrt {2^{2}(2^{2}+13^{2}+14^{2})}}=2{\sqrt {369}}=6{\sqrt {41}}

The cross product of vectors $\mathbf {u}$ and $\mathbf {v}$ is a vector with length $|\mathbf {u} ||\mathbf {v} |\sin(\theta )$ where $\theta$ is the angle between $\mathbf {u}$ and $\mathbf {v}$ . $|\mathbf {u} ||\mathbf {v} |\sin(\theta )$ is the area of the parallelogram with sides $\mathbf {u}$ and $\mathbf {v}$ , so this area is found by computing $|\mathbf {u} \times \mathbf {v} |$ .

$\mathbf {u} \times \mathbf {v} =\langle (2)(-4)-(-3)(4),(-3)(2)-(8)(-4),(8)(4)-(2)(2)\rangle =\langle 4,26,28\rangle$

A=|\mathbf {u} \times \mathbf {v} |={\sqrt {4^{2}+26^{2}+28^{2}}}={\sqrt {2^{2}(2^{2}+13^{2}+14^{2})}}=2{\sqrt {369}}=6{\sqrt {41}}

84. Find all vectors that satisfy the equation

\langle 1,1,1\rangle \times \mathbf {u} =\langle 0,1,1\rangle

The cross product is orthogonal to both multiplicand vectors.

\langle 0,1,1\rangle

should be orthogonal to both

\langle 1,1,1\rangle

and

\mathbf {u}

. However,

\langle 1,1,1\rangle \cdot \langle 0,1,1\rangle =2\neq 0

so

\langle 1,1,1\rangle

and

\langle 0,1,1\rangle

are not orthogonal. The equation

\langle 1,1,1\rangle \times \mathbf {u} =\langle 0,1,1\rangle

is never true, and therefore the set of vectors

\mathbf {u}

that satisfy the equation is

\emptyset =

"None".

The cross product is orthogonal to both multiplicand vectors.

\langle 0,1,1\rangle

should be orthogonal to both

\langle 1,1,1\rangle

and

\mathbf {u}

. However,

\langle 1,1,1\rangle \cdot \langle 0,1,1\rangle =2\neq 0

so

\langle 1,1,1\rangle

and

\langle 0,1,1\rangle

are not orthogonal. The equation

\langle 1,1,1\rangle \times \mathbf {u} =\langle 0,1,1\rangle

is never true, and therefore the set of vectors

\mathbf {u}

that satisfy the equation is

\emptyset =

"None".

85. Find the volume of the parallelepiped with edges given by position vectors

\langle 5,0,0\rangle

,

\langle 1,4,0\rangle

, and

\langle 2,2,7\rangle

The volume of a parallelepiped with edges defined by the vectors $\mathbf {u}$ , $\mathbf {v}$ , and $\mathbf {w}$ is the absolute value of the scalar triple product: $|(\mathbf {u} \times \mathbf {v} )\cdot \mathbf {w} |$ .

$|(\langle 5,0,0\rangle \times \langle 1,4,0\rangle )\cdot \langle 2,2,7\rangle |$ $=|\langle (0)(0)-(0)(4),(0)(1)-(5)(0),(5)(4)-(0)(1)\rangle \cdot \langle 2,2,7\rangle |$ $=|\langle 0,0,20\rangle \cdot \langle 2,2,7\rangle |$ $=140$

The volume of a parallelepiped with edges defined by the vectors $\mathbf {u}$ , $\mathbf {v}$ , and $\mathbf {w}$ is the absolute value of the scalar triple product: $|(\mathbf {u} \times \mathbf {v} )\cdot \mathbf {w} |$ .

$|(\langle 5,0,0\rangle \times \langle 1,4,0\rangle )\cdot \langle 2,2,7\rangle |$ $=|\langle (0)(0)-(0)(4),(0)(1)-(5)(0),(5)(4)-(0)(1)\rangle \cdot \langle 2,2,7\rangle |$ $=|\langle 0,0,20\rangle \cdot \langle 2,2,7\rangle |$ $=140$

86. A wrench has a pivot at the origin and extends along the x-axis. Find the magnitude and the direction of the torque at the pivot when the force

\mathbf {F} =\langle 1,2,3\rangle

is applied to the wrench n units away from the origin.

The moment arm is

\mathbf {r} =\langle n,0,0\rangle

, so the torque applied is

\tau =\mathbf {r} \times \mathbf {F} =\langle n,0,0\rangle \times \langle 1,2,3\rangle =\langle (0)(3)-(0)(2),(0)(1)-(n)(3),(n)(2)-(0)(1)\rangle =\langle 0,-3n,2n\rangle

The magnitude of the torque is

|\tau |={\sqrt {0^{2}+(-3n)^{2}+(2n)^{2}}}=n{\sqrt {13}}

. The torque's direction is

{\hat {\tau }}={\frac {\tau }{|\tau |}}=\langle 0,-3/{\sqrt {13}},2/{\sqrt {13}}\rangle

.

The moment arm is

\mathbf {r} =\langle n,0,0\rangle

, so the torque applied is

\tau =\mathbf {r} \times \mathbf {F} =\langle n,0,0\rangle \times \langle 1,2,3\rangle =\langle (0)(3)-(0)(2),(0)(1)-(n)(3),(n)(2)-(0)(1)\rangle =\langle 0,-3n,2n\rangle

The magnitude of the torque is

|\tau |={\sqrt {0^{2}+(-3n)^{2}+(2n)^{2}}}=n{\sqrt {13}}

. The torque's direction is

{\hat {\tau }}={\frac {\tau }{|\tau |}}=\langle 0,-3/{\sqrt {13}},2/{\sqrt {13}}\rangle

.

Prove the following identities or show them false by giving a counterexample.

87.

\mathbf {u} \times (\mathbf {u} \times \mathbf {v} )=\mathbf {0}

False:

\mathbf {i} \times (\mathbf {i} \times \mathbf {j} )=-\mathbf {j}

False:

\mathbf {i} \times (\mathbf {i} \times \mathbf {j} )=-\mathbf {j}

88.

\mathbf {u} \cdot (\mathbf {v} \times \mathbf {w} )=\mathbf {w} \cdot (\mathbf {u} \times \mathbf {v} )

True: Once expressed in component form, both sides evaluate to

u_{1}v_{2}w_{3}-u_{1}v_{3}w_{2}+u_{2}v_{3}w_{1}-u_{2}v_{1}w_{3}+u_{3}v_{1}w_{2}-u_{3}v_{2}w_{1}

True: Once expressed in component form, both sides evaluate to

u_{1}v_{2}w_{3}-u_{1}v_{3}w_{2}+u_{2}v_{3}w_{1}-u_{2}v_{1}w_{3}+u_{3}v_{1}w_{2}-u_{3}v_{2}w_{1}

89.

(\mathbf {u} -\mathbf {v} )\times (\mathbf {u} +\mathbf {v} )=2(\mathbf {u} \times \mathbf {v} )

True:

(\mathbf {u} -\mathbf {v} )\times (\mathbf {u} +\mathbf {v} )=(\mathbf {u} -\mathbf {v} )\times \mathbf {u} +(\mathbf {u} -\mathbf {v} )\times \mathbf {v}

=\mathbf {u} \times \mathbf {u} -\mathbf {v} \times \mathbf {u} +\mathbf {u} \times \mathbf {v} -\mathbf {v} \times \mathbf {v}

=\mathbf {u} \times \mathbf {v} -\mathbf {v} \times \mathbf {u}

=\mathbf {u} \times \mathbf {v} +\mathbf {u} \times \mathbf {v}

=2(\mathbf {u} \times \mathbf {v} )

True:

(\mathbf {u} -\mathbf {v} )\times (\mathbf {u} +\mathbf {v} )=(\mathbf {u} -\mathbf {v} )\times \mathbf {u} +(\mathbf {u} -\mathbf {v} )\times \mathbf {v}

=\mathbf {u} \times \mathbf {u} -\mathbf {v} \times \mathbf {u} +\mathbf {u} \times \mathbf {v} -\mathbf {v} \times \mathbf {v}

=\mathbf {u} \times \mathbf {v} -\mathbf {v} \times \mathbf {u}

=\mathbf {u} \times \mathbf {v} +\mathbf {u} \times \mathbf {v}

=2(\mathbf {u} \times \mathbf {v} )

Calculus of Vector-Valued Functions

100. Differentiate

\mathbf {r} (t)=\langle te^{-t},t\ln t,t\cos(t)\rangle

.

\langle e^{-t}-te^{-t},\ln(t)+1,\cos(t)-t\sin(t)\rangle

\langle e^{-t}-te^{-t},\ln(t)+1,\cos(t)-t\sin(t)\rangle

101. Find a tangent vector for the curve

\mathbf {r} (t)=\langle 2t^{4},6t^{3/2},10/t\rangle

at the point

t=1

.

{\dot {\mathbf {r} }}(t)=\langle 8t^{3},9t^{1/2},-10/t^{2}\rangle

so a possible a tangent vector at

t=1

is

{\dot {\mathbf {r} }}(1)=\langle 8,9,-10\rangle

{\dot {\mathbf {r} }}(t)=\langle 8t^{3},9t^{1/2},-10/t^{2}\rangle

so a possible a tangent vector at

t=1

is

{\dot {\mathbf {r} }}(1)=\langle 8,9,-10\rangle

102. Find the unit tangent vector for the curve

\mathbf {r} (t)=\langle t,2,2/t\rangle ,\ t\neq 0

.

{\dot {\mathbf {r} }}(t)=\langle 1,0,-2/t^{2}\rangle

|{\dot {\mathbf {r} }}(t)|={\sqrt {1^{2}+0^{2}+(-2/t^{2})^{2}}}={\sqrt {1+4/t^{4}}}={\frac {\sqrt {t^{4}+4}}{t^{2}}}

so the unit tangent vector is

\mathbf {T} (t)={\frac {{\dot {\mathbf {r} }}(t)}{|{\dot {\mathbf {r} }}(t)|}}={\frac {\langle t^{2},0,-2\rangle }{\sqrt {t^{4}+4}}}

{\dot {\mathbf {r} }}(t)=\langle 1,0,-2/t^{2}\rangle

|{\dot {\mathbf {r} }}(t)|={\sqrt {1^{2}+0^{2}+(-2/t^{2})^{2}}}={\sqrt {1+4/t^{4}}}={\frac {\sqrt {t^{4}+4}}{t^{2}}}

so the unit tangent vector is

\mathbf {T} (t)={\frac {{\dot {\mathbf {r} }}(t)}{|{\dot {\mathbf {r} }}(t)|}}={\frac {\langle t^{2},0,-2\rangle }{\sqrt {t^{4}+4}}}

103. Find the unit tangent vector for the curve

\mathbf {r} (t)=\langle \sin(t),\cos(t),e^{-t}\rangle ,\ t\in [0,\pi ]

at the point

t=0

.

{\dot {\mathbf {r} }}(t)=\langle \cos(t),-\sin(t),-e^{-t}\rangle

$|{\dot {\mathbf {r} }}(t)|={\sqrt {(\cos(t))^{2}+(-\sin(t))^{2}+(-e^{-t})^{2}}}={\sqrt {1+e^{-2t}}}$ so the unit tangent vector is $\mathbf {T} (t)={\frac {{\dot {\mathbf {r} }}(t)}{|{\dot {\mathbf {r} }}(t)|}}={\frac {\langle \cos(t),-\sin(t),-e^{-t}\rangle }{\sqrt {1+e^{-2t}}}}$

At

t=0

:

\mathbf {T} (0)={\frac {\langle 1,0,-1\rangle }{\sqrt {2}}}

{\dot {\mathbf {r} }}(t)=\langle \cos(t),-\sin(t),-e^{-t}\rangle

$|{\dot {\mathbf {r} }}(t)|={\sqrt {(\cos(t))^{2}+(-\sin(t))^{2}+(-e^{-t})^{2}}}={\sqrt {1+e^{-2t}}}$ so the unit tangent vector is $\mathbf {T} (t)={\frac {{\dot {\mathbf {r} }}(t)}{|{\dot {\mathbf {r} }}(t)|}}={\frac {\langle \cos(t),-\sin(t),-e^{-t}\rangle }{\sqrt {1+e^{-2t}}}}$

At

t=0

:

\mathbf {T} (0)={\frac {\langle 1,0,-1\rangle }{\sqrt {2}}}

104. Find

\mathbf {r}

if

\mathbf {r} '(t)=\langle {\sqrt {t}},\cos(\pi t),4/t\rangle

and

\mathbf {r} (1)=\langle 2,3,4\rangle

.

For an arbitrary

t>0

the position

\mathbf {r} (t)

can be computed by the integral

\mathbf {r} (t)=\mathbf {r} (1)+\int _{u=1}^{t}\mathbf {r} '(u)du

.

$\mathbf {r} (t)=\mathbf {r} (1)+\int _{u=1}^{t}\mathbf {r} '(u)du$ $=\langle 2,3,4\rangle +\int _{u=1}^{t}\langle {\sqrt {u}},\cos(\pi u),4/u\rangle du$ $=\langle 2,3,4\rangle +\left\langle {\frac {2}{3}}u^{3/2},{\frac {\sin(\pi u)}{\pi }},4\ln(u)\right\rangle {\Bigg |}_{u=1}^{t}$ $=\langle 2,3,4\rangle +\left\langle {\frac {2}{3}}(t^{3/2}-1),{\frac {\sin(\pi t)}{\pi }},4\ln(t)\right\rangle$

=\left\langle {\frac {2t^{3/2}+4}{3}},{\frac {\sin(\pi t)}{\pi }}+3,4\ln(t)+4\right\rangle

For an arbitrary

t>0

the position

\mathbf {r} (t)

can be computed by the integral

\mathbf {r} (t)=\mathbf {r} (1)+\int _{u=1}^{t}\mathbf {r} '(u)du

.

$\mathbf {r} (t)=\mathbf {r} (1)+\int _{u=1}^{t}\mathbf {r} '(u)du$ $=\langle 2,3,4\rangle +\int _{u=1}^{t}\langle {\sqrt {u}},\cos(\pi u),4/u\rangle du$ $=\langle 2,3,4\rangle +\left\langle {\frac {2}{3}}u^{3/2},{\frac {\sin(\pi u)}{\pi }},4\ln(u)\right\rangle {\Bigg |}_{u=1}^{t}$ $=\langle 2,3,4\rangle +\left\langle {\frac {2}{3}}(t^{3/2}-1),{\frac {\sin(\pi t)}{\pi }},4\ln(t)\right\rangle$

=\left\langle {\frac {2t^{3/2}+4}{3}},{\frac {\sin(\pi t)}{\pi }}+3,4\ln(t)+4\right\rangle

105. Evaluate

\displaystyle \int _{0}^{\ln 2}(e^{-t}\mathbf {i} +2e^{2t}\mathbf {j} -4e^{t}\mathbf {k} )dt

\int _{t=0}^{\ln 2}\langle e^{-t},2e^{2t},-4e^{t}\rangle dt

$=\langle -e^{-t},e^{2t},-4e^{t}\rangle {\bigg |}_{t=0}^{\ln 2}$ $=\langle -1/2-(-1),4-1,-8-(-4)\rangle$

=\langle 1/2,3,-4\rangle

\int _{t=0}^{\ln 2}\langle e^{-t},2e^{2t},-4e^{t}\rangle dt

$=\langle -e^{-t},e^{2t},-4e^{t}\rangle {\bigg |}_{t=0}^{\ln 2}$ $=\langle -1/2-(-1),4-1,-8-(-4)\rangle$

=\langle 1/2,3,-4\rangle

Motion in Space

120. Find velocity, speed, and acceleration of an object if the position is given by

\mathbf {r} (t)=\langle 3\sin(t),5\cos(t),4\sin(t)\rangle

.

\mathbf {v} =\langle 3\cos(t),-5\sin(t),4\cos(t)\rangle

,

|\mathbf {v} |=5

,

\mathbf {a} =\langle -3\sin(t),-5\cos(t),-4\sin(t)\rangle

\mathbf {v} =\langle 3\cos(t),-5\sin(t),4\cos(t)\rangle

,

|\mathbf {v} |=5

,

\mathbf {a} =\langle -3\sin(t),-5\cos(t),-4\sin(t)\rangle

121. Find the velocity and the position vectors for

t\geq 0

if the acceleration is given by

\mathbf {a} (t)=\langle e^{-t},1\rangle ,\ \mathbf {v} (0)=\langle 1,0\rangle ,\ \mathbf {r} (0)=\langle 0,0\rangle

.

\mathbf {v} (t)=\mathbf {v} (0)+\int _{u=0}^{t}\mathbf {a} (u)du

$=\langle 1,0\rangle +\int _{u=0}^{t}\langle e^{-u},1\rangle du$ $=\langle 1,0\rangle +\langle -e^{-u},u\rangle {\bigg |}_{u=0}^{t}$ $=\langle 1,0\rangle +\langle -e^{-t}-(-1),t-0\rangle$ $=\langle 2-e^{-t},t\rangle$

$\mathbf {r} (t)=\mathbf {r} (0)+\int _{u=0}^{t}\mathbf {v} (u)du$ $=\langle 0,0\rangle +\int _{u=0}^{t}\langle 2-e^{-u},u\rangle du$ $=\langle 2u+e^{-u},{\frac {1}{2}}u^{2}\rangle {\bigg |}_{u=0}^{t}$ $=\langle (2t+e^{-t})-1,{\frac {1}{2}}t^{2}-0\rangle$

=\langle e^{-t}+2t-1,t^{2}/2\rangle

\mathbf {v} (t)=\mathbf {v} (0)+\int _{u=0}^{t}\mathbf {a} (u)du

$=\langle 1,0\rangle +\int _{u=0}^{t}\langle e^{-u},1\rangle du$ $=\langle 1,0\rangle +\langle -e^{-u},u\rangle {\bigg |}_{u=0}^{t}$ $=\langle 1,0\rangle +\langle -e^{-t}-(-1),t-0\rangle$ $=\langle 2-e^{-t},t\rangle$

$\mathbf {r} (t)=\mathbf {r} (0)+\int _{u=0}^{t}\mathbf {v} (u)du$ $=\langle 0,0\rangle +\int _{u=0}^{t}\langle 2-e^{-u},u\rangle du$ $=\langle 2u+e^{-u},{\frac {1}{2}}u^{2}\rangle {\bigg |}_{u=0}^{t}$ $=\langle (2t+e^{-t})-1,{\frac {1}{2}}t^{2}-0\rangle$

=\langle e^{-t}+2t-1,t^{2}/2\rangle

Length of Curves

Find the length of the following curves.

140.

\mathbf {r} (t)=\langle 4\cos(3t),4\sin(3t)\rangle ,\ t\in [0,2\pi /3].

For an infinitesimal step

dt

, the length traversed is approximately

$ds=\left|{\frac {d\mathbf {r} }{dt}}\right|dt=|\langle -12\sin(3t),12\cos(3t)\rangle |dt=12dt$ .

The total length is therefore:

L=\int _{t=0}^{2\pi /3}ds=\int _{t=0}^{2\pi /3}12dt=12t{\Bigg |}_{t=0}^{2\pi /3}=8\pi

For an infinitesimal step

dt

, the length traversed is approximately

$ds=\left|{\frac {d\mathbf {r} }{dt}}\right|dt=|\langle -12\sin(3t),12\cos(3t)\rangle |dt=12dt$ .

The total length is therefore:

L=\int _{t=0}^{2\pi /3}ds=\int _{t=0}^{2\pi /3}12dt=12t{\Bigg |}_{t=0}^{2\pi /3}=8\pi

141.

\mathbf {r} (t)=\langle 2+3t,1-4t,3t-4\rangle ,\ t\in [1,6].

For an infinitesimal step

dt

, the length traversed is approximately

$ds=\left|{\frac {d\mathbf {r} }{dt}}\right|dt=|\langle 3,-4,3\rangle |dt={\sqrt {9+16+9}}\cdot dt={\sqrt {34}}\cdot dt$ .

The total length is therefore:

L=\int _{t=1}^{6}ds=\int _{t=1}^{6}{\sqrt {34}}\cdot dt={\sqrt {34}}\cdot t{\Bigg |}_{t=1}^{6}=5{\sqrt {34}}

For an infinitesimal step

dt

, the length traversed is approximately

$ds=\left|{\frac {d\mathbf {r} }{dt}}\right|dt=|\langle 3,-4,3\rangle |dt={\sqrt {9+16+9}}\cdot dt={\sqrt {34}}\cdot dt$ .

The total length is therefore:

L=\int _{t=1}^{6}ds=\int _{t=1}^{6}{\sqrt {34}}\cdot dt={\sqrt {34}}\cdot t{\Bigg |}_{t=1}^{6}=5{\sqrt {34}}

Parametrization and Normal Vectors

142. Find a description of the curve that uses arc length as a parameter:

\mathbf {r} (t)=\langle t^{2},2t^{2},4t^{2}\rangle \ t\in [1,4].

For an infinitesimal step

dt

, the length traversed is approximately

ds=\left|{\frac {d\mathbf {r} }{dt}}\right|dt=|\langle 2t,4t,8t\rangle |dt=|t|{\sqrt {4+16+64}}\cdot dt={\sqrt {84}}\cdot tdt=2{\sqrt {21}}\cdot tdt

Given an upper bound of $u\in [1,4]$ , the arc length swept out from $t=1$ to $t=u$ is: $s=\int _{t=1}^{u}ds=\int _{t=1}^{u}2{\sqrt {21}}\cdot tdt={\sqrt {21}}\cdot t^{2}{\bigg |}_{t=1}^{u}={\sqrt {21}}(u^{2}-1)$

The arc length spans a range from

0

to

15{\sqrt {21}}

. For an arc length of

s\in [0,15{\sqrt {21}}]

, the upper bound on

t

that generates an arc length of

s

is

u={\sqrt {{\frac {s}{\sqrt {21}}}+1}}

, and the point at which this upper bound occurs is:

\displaystyle \mathbf {r} (s)=\left({\frac {s}{\sqrt {21}}}+1\right)\langle 1,2,4\rangle

For an infinitesimal step

dt

, the length traversed is approximately

ds=\left|{\frac {d\mathbf {r} }{dt}}\right|dt=|\langle 2t,4t,8t\rangle |dt=|t|{\sqrt {4+16+64}}\cdot dt={\sqrt {84}}\cdot tdt=2{\sqrt {21}}\cdot tdt

Given an upper bound of $u\in [1,4]$ , the arc length swept out from $t=1$ to $t=u$ is: $s=\int _{t=1}^{u}ds=\int _{t=1}^{u}2{\sqrt {21}}\cdot tdt={\sqrt {21}}\cdot t^{2}{\bigg |}_{t=1}^{u}={\sqrt {21}}(u^{2}-1)$

The arc length spans a range from

0

to

15{\sqrt {21}}

. For an arc length of

s\in [0,15{\sqrt {21}}]

, the upper bound on

t

that generates an arc length of

s

is

u={\sqrt {{\frac {s}{\sqrt {21}}}+1}}

, and the point at which this upper bound occurs is:

\displaystyle \mathbf {r} (s)=\left({\frac {s}{\sqrt {21}}}+1\right)\langle 1,2,4\rangle

143. Find the unit tangent vector T and the principal unit normal vector N for the curve

\mathbf {r} (t)=\langle t^{2},t\rangle .

Check that T⋅N=0.

A tangent vector is

{\frac {d\mathbf {r} }{dt}}=\langle 2t,1\rangle

. Normalizing this vector to get the unit tangent vector gives:

$\mathbf {T} (t)={\frac {d\mathbf {r} }{ds}}={\frac {d\mathbf {r} /dt}{|d\mathbf {r} /dt|}}=\displaystyle {\frac {\langle 2t,1\rangle }{\sqrt {4t^{2}+1}}}$

A vector that has the direction of the principal unit normal vector is ${\frac {d\mathbf {T} }{dt}}=\left\langle {\frac {2}{\sqrt {4t^{2}+1}}}-{\frac {8t^{2}}{(4t^{2}+1)^{3/2}}},{\frac {-4t}{(4t^{2}+1)^{3/2}}}\right\rangle$ $={\frac {1}{(4t^{2}+1)^{3/2}}}\langle 2(4t^{2}+1)-8t^{2},-4t\rangle$ $={\frac {1}{(4t^{2}+1)^{3/2}}}\langle 2,-4t\rangle$ $={\frac {2}{(4t^{2}+1)^{3/2}}}\langle 1,-2t\rangle$

Normalizing ${\frac {d\mathbf {T} }{dt}}$ gives the principal unit normal vector: $\mathbf {N} (t)$ $={\frac {d\mathbf {T} /dt}{|d\mathbf {T} /dt|}}$ $={\frac {{\frac {2}{(4t^{2}+1)^{3/2}}}\langle 1,-2t\rangle }{{\frac {2}{(4t^{2}+1)^{3/2}}}{\sqrt {4t^{2}+1}}}}$

=\displaystyle {\frac {\langle 1,-2t\rangle }{\sqrt {4t^{2}+1}}}

A tangent vector is

{\frac {d\mathbf {r} }{dt}}=\langle 2t,1\rangle

. Normalizing this vector to get the unit tangent vector gives:

$\mathbf {T} (t)={\frac {d\mathbf {r} }{ds}}={\frac {d\mathbf {r} /dt}{|d\mathbf {r} /dt|}}=\displaystyle {\frac {\langle 2t,1\rangle }{\sqrt {4t^{2}+1}}}$

A vector that has the direction of the principal unit normal vector is ${\frac {d\mathbf {T} }{dt}}=\left\langle {\frac {2}{\sqrt {4t^{2}+1}}}-{\frac {8t^{2}}{(4t^{2}+1)^{3/2}}},{\frac {-4t}{(4t^{2}+1)^{3/2}}}\right\rangle$ $={\frac {1}{(4t^{2}+1)^{3/2}}}\langle 2(4t^{2}+1)-8t^{2},-4t\rangle$ $={\frac {1}{(4t^{2}+1)^{3/2}}}\langle 2,-4t\rangle$ $={\frac {2}{(4t^{2}+1)^{3/2}}}\langle 1,-2t\rangle$

Normalizing ${\frac {d\mathbf {T} }{dt}}$ gives the principal unit normal vector: $\mathbf {N} (t)$ $={\frac {d\mathbf {T} /dt}{|d\mathbf {T} /dt|}}$ $={\frac {{\frac {2}{(4t^{2}+1)^{3/2}}}\langle 1,-2t\rangle }{{\frac {2}{(4t^{2}+1)^{3/2}}}{\sqrt {4t^{2}+1}}}}$

=\displaystyle {\frac {\langle 1,-2t\rangle }{\sqrt {4t^{2}+1}}}

Equations of Lines And Planes

160. Find an equation of a plane passing through points

(1,1,2),\ (1,2,2),\ (-1,0,1).

Let

P

denote a plane that contains points