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

(1,1,2)

,

(1,2,2)

, and

(-1,0,1)

. Let

\mathbf {N}

denote an arbitrary vector that is orthogonal to

P

, and

\mathbf {r} _{0}

denote the position vector of an arbitrary point contained by

P

. A point at position vector

\mathbf {r}

is contained by

P

if and only if the displacement from

\mathbf {r} _{0}

is orthogonal to

\mathbf {N}

. This yields the equation

\mathbf {N} \cdot (\mathbf {r} -\mathbf {r} _{0})=0

\iff \mathbf {N} \cdot \mathbf {r} -\mathbf {N} \cdot \mathbf {r} _{0}=0

.

The displacement from $(1,1,2)$ to $(1,2,2)$ , which is $\langle 0,1,0\rangle$ , and the displacement from $(1,1,2)$ to $(-1,0,1)$ , which is $\langle -2,-1,-1\rangle$ , are both contained by $P$ so the cross product of these two displacements forms a candidate $\mathbf {N}$ : $\mathbf {N} =\langle 0,1,0\rangle \times \langle -2,-1,-1\rangle$ $=\langle (1)(-1)-(0)(-1),(0)(-2)-(0)(-1),(0)(-1)-(1)(-2)\rangle$ $=\langle -1,0,2\rangle$

Any of $(1,1,2)$ , $(1,2,2)$ , and $(-1,0,1)$ is a candidate $\mathbf {r} _{0}$ . Let $\mathbf {r} _{0}=\langle 1,1,2\rangle$

The equation becomes

\mathbf {N} \cdot \mathbf {r} -\mathbf {N} \cdot \mathbf {r} _{0}=0

\iff \langle -1,0,2\rangle \cdot \langle x,y,z\rangle -\langle -1,0,2\rangle \cdot \langle 1,1,2\rangle

\iff -x+2z-3=0

\iff x-2z+3=0

Let

P

denote a plane that contains points

(1,1,2)

,

(1,2,2)

, and

(-1,0,1)

. Let

\mathbf {N}

denote an arbitrary vector that is orthogonal to

P

, and

\mathbf {r} _{0}

denote the position vector of an arbitrary point contained by

P

. A point at position vector

\mathbf {r}

is contained by

P

if and only if the displacement from

\mathbf {r} _{0}

is orthogonal to

\mathbf {N}

. This yields the equation

\mathbf {N} \cdot (\mathbf {r} -\mathbf {r} _{0})=0

\iff \mathbf {N} \cdot \mathbf {r} -\mathbf {N} \cdot \mathbf {r} _{0}=0

.

The displacement from $(1,1,2)$ to $(1,2,2)$ , which is $\langle 0,1,0\rangle$ , and the displacement from $(1,1,2)$ to $(-1,0,1)$ , which is $\langle -2,-1,-1\rangle$ , are both contained by $P$ so the cross product of these two displacements forms a candidate $\mathbf {N}$ : $\mathbf {N} =\langle 0,1,0\rangle \times \langle -2,-1,-1\rangle$ $=\langle (1)(-1)-(0)(-1),(0)(-2)-(0)(-1),(0)(-1)-(1)(-2)\rangle$ $=\langle -1,0,2\rangle$

Any of $(1,1,2)$ , $(1,2,2)$ , and $(-1,0,1)$ is a candidate $\mathbf {r} _{0}$ . Let $\mathbf {r} _{0}=\langle 1,1,2\rangle$

The equation becomes

\mathbf {N} \cdot \mathbf {r} -\mathbf {N} \cdot \mathbf {r} _{0}=0

\iff \langle -1,0,2\rangle \cdot \langle x,y,z\rangle -\langle -1,0,2\rangle \cdot \langle 1,1,2\rangle

\iff -x+2z-3=0

\iff x-2z+3=0

161. Find an equation of a plane parallel to the plane 2x−y+z=1 passing through the point (0,2,-2)

Let

P

denote a plane that is parallel to the plane

2x-y+z=1

and contains the point

(0,2,-2)

. Let

\mathbf {N}

denote an arbitrary vector that is orthogonal to

P

, and

\mathbf {r} _{0}

denote the position vector of an arbitrary point contained by

P

. A point at position vector

\mathbf {r}

is contained by

P

if and only if the displacement from

\mathbf {r} _{0}

is orthogonal to

\mathbf {N}

. This yields the equation

\mathbf {N} \cdot (\mathbf {r} -\mathbf {r} _{0})=0

\iff \mathbf {N} \cdot \mathbf {r} -\mathbf {N} \cdot \mathbf {r} _{0}=0

.

Any vector $\mathbf {N}$ that is orthogonal to $P$ is also orthogonal to $2x-y+z=1$ and vice versa. Since $2x-y+z=1$ $\iff \langle 2,-1,1\rangle \cdot \langle x,y,z\rangle =1$ , the coefficient vector $\langle 2,-1,1\rangle$ is orthogonal to $2x-y+z=1$ , so a candidate $\mathbf {N}$ is $\mathbf {N} =\langle 2,-1,1\rangle$ .

Since point $(0,2,-2)$ is contained by $P$ , let $\mathbf {r} _{0}=\langle 0,2,-2\rangle$ .

The equation becomes

\mathbf {N} \cdot \mathbf {r} -\mathbf {N} \cdot \mathbf {r} _{0}=0

\iff \langle 2,-1,1\rangle \cdot \langle x,y,z\rangle -\langle 2,-1,1\rangle \cdot \langle 0,2,-2\rangle =0

\iff 2x-y+z+4=0

Let

P

denote a plane that is parallel to the plane

2x-y+z=1

and contains the point

(0,2,-2)

. Let

\mathbf {N}

denote an arbitrary vector that is orthogonal to

P

, and

\mathbf {r} _{0}

denote the position vector of an arbitrary point contained by

P

. A point at position vector

\mathbf {r}

is contained by

P

if and only if the displacement from

\mathbf {r} _{0}

is orthogonal to

\mathbf {N}

. This yields the equation

\mathbf {N} \cdot (\mathbf {r} -\mathbf {r} _{0})=0

\iff \mathbf {N} \cdot \mathbf {r} -\mathbf {N} \cdot \mathbf {r} _{0}=0

.

Any vector $\mathbf {N}$ that is orthogonal to $P$ is also orthogonal to $2x-y+z=1$ and vice versa. Since $2x-y+z=1$ $\iff \langle 2,-1,1\rangle \cdot \langle x,y,z\rangle =1$ , the coefficient vector $\langle 2,-1,1\rangle$ is orthogonal to $2x-y+z=1$ , so a candidate $\mathbf {N}$ is $\mathbf {N} =\langle 2,-1,1\rangle$ .

Since point $(0,2,-2)$ is contained by $P$ , let $\mathbf {r} _{0}=\langle 0,2,-2\rangle$ .

The equation becomes

\mathbf {N} \cdot \mathbf {r} -\mathbf {N} \cdot \mathbf {r} _{0}=0

\iff \langle 2,-1,1\rangle \cdot \langle x,y,z\rangle -\langle 2,-1,1\rangle \cdot \langle 0,2,-2\rangle =0

\iff 2x-y+z+4=0

162. Find an equation of the line perpendicular to the plane x+y+2z=4 passing through the point (5,5,5).

Let

P

denote an arbitrary plane. Let

\mathbf {N}

denote an arbitrary vector that is orthogonal to

P

, and

\mathbf {r} _{0}

denote the position vector of an arbitrary point contained by

P

. A point at position vector

\mathbf {r}

is contained by

P

if and only if the displacement from

\mathbf {r} _{0}

is orthogonal to

\mathbf {N}

. This yields the equation

\mathbf {N} \cdot (\mathbf {r} -\mathbf {r} _{0})=0

\iff \mathbf {N} \cdot \mathbf {r} -\mathbf {N} \cdot \mathbf {r} _{0}=0

. Therefore the equation that defines

P

is

\mathbf {N} \cdot \mathbf {r} -\mathbf {N} \cdot \mathbf {r} _{0}=0

.

The equation $x+y+2z=4$ is equivalent to $\langle 1,1,2\rangle \cdot \langle x,y,z\rangle -4=0$ . This implies that the coefficient vector $\mathbf {N} =\langle 1,1,2\rangle$ is orthogonal to the plane defined by $x+y+2z=4$ . A line that passes through point $(5,5,5)$ and is parallel to $\mathbf {N} =\langle 1,1,2\rangle$ is parameterized by:

\mathbf {r} (t)=\langle 5,5,5\rangle +t\langle 1,1,2\rangle =\langle 5+t,5+t,5+2t\rangle

Let

P

denote an arbitrary plane. Let

\mathbf {N}

denote an arbitrary vector that is orthogonal to

P

, and

\mathbf {r} _{0}

denote the position vector of an arbitrary point contained by

P

. A point at position vector

\mathbf {r}

is contained by

P

if and only if the displacement from

\mathbf {r} _{0}

is orthogonal to

\mathbf {N}

. This yields the equation

\mathbf {N} \cdot (\mathbf {r} -\mathbf {r} _{0})=0

\iff \mathbf {N} \cdot \mathbf {r} -\mathbf {N} \cdot \mathbf {r} _{0}=0

. Therefore the equation that defines

P

is

\mathbf {N} \cdot \mathbf {r} -\mathbf {N} \cdot \mathbf {r} _{0}=0

.

The equation $x+y+2z=4$ is equivalent to $\langle 1,1,2\rangle \cdot \langle x,y,z\rangle -4=0$ . This implies that the coefficient vector $\mathbf {N} =\langle 1,1,2\rangle$ is orthogonal to the plane defined by $x+y+2z=4$ . A line that passes through point $(5,5,5)$ and is parallel to $\mathbf {N} =\langle 1,1,2\rangle$ is parameterized by:

\mathbf {r} (t)=\langle 5,5,5\rangle +t\langle 1,1,2\rangle =\langle 5+t,5+t,5+2t\rangle

163. Find an equation of the line where planes x+2y−z=1 and x+y+z=1 intersect.

To describe the line that forms the intersection between

x+2y-z=1

and

x+y+z=1

, both

x

and

y

will be expressed as functions of

z

(it can also be the case that

y

and

z

are functions of

x

, etc.). Let

E_{1}

denote equation

x+2y-z=1

and

E_{2}

denote equation

x+y+z=1

.

To find $x$ as a function of $z$ subtract 2 times $E_{2}$ from $E_{1}$ to get $E_{1}-2E_{2}:(x+2y-z)-2(x+y+z)=1-2(1)$ $\iff -x-3z=-1$ $\iff x=1-3z$

To find $y$ as a function of $z$ subtract $E_{2}$ from $E_{1}$ to get $E_{1}-E_{2}:(x+2y-z)-(x+y+z)=1-1$ $\iff y-2z=0$ $\iff y=2z$

Parameterizing

z

with

t

gives the parameterization

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

To describe the line that forms the intersection between

x+2y-z=1

and

x+y+z=1

, both

x

and

y

will be expressed as functions of

z

(it can also be the case that

y

and

z

are functions of

x

, etc.). Let

E_{1}

denote equation

x+2y-z=1

and

E_{2}

denote equation

x+y+z=1

.

To find $x$ as a function of $z$ subtract 2 times $E_{2}$ from $E_{1}$ to get $E_{1}-2E_{2}:(x+2y-z)-2(x+y+z)=1-2(1)$ $\iff -x-3z=-1$ $\iff x=1-3z$

To find $y$ as a function of $z$ subtract $E_{2}$ from $E_{1}$ to get $E_{1}-E_{2}:(x+2y-z)-(x+y+z)=1-1$ $\iff y-2z=0$ $\iff y=2z$

Parameterizing

z

with

t

gives the parameterization

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

164. Find the angle between the planes x+2y−z=1 and x+y+z=1.

Let

P

denote an arbitrary plane. Let

\mathbf {N}

denote an arbitrary vector that is orthogonal to

P

, and

\mathbf {r} _{0}

denote the position vector of an arbitrary point contained by

P

. A point at position vector

\mathbf {r}

is contained by

P

if and only if the displacement from

\mathbf {r} _{0}

is orthogonal to

\mathbf {N}

. This yields the equation

\mathbf {N} \cdot (\mathbf {r} -\mathbf {r} _{0})=0

\iff \mathbf {N} \cdot \mathbf {r} -\mathbf {N} \cdot \mathbf {r} _{0}=0

. Therefore the equation that defines

P

is

\mathbf {N} \cdot \mathbf {r} -\mathbf {N} \cdot \mathbf {r} _{0}=0

.

Let $P_{1}$ be the plane described by $x+2y-z=1$ and $P_{2}$ be the plane described by $x+y+z=1$

Since $x+2y-z=1\iff \langle 1,2,-1\rangle \cdot \langle x,y,z\rangle -1=0$ , the coefficient vector $\mathbf {N} _{1}=\langle 1,2,-1\rangle$ is orthogonal to $P_{1}$ .

Since $x+y+z=1\iff \langle 1,1,1\rangle \cdot \langle x,y,z\rangle -1=0$ , the coefficient vector $\mathbf {N} _{2}=\langle 1,1,1\rangle$ is orthogonal to $P_{2}$ .

The angle $\theta$ between $P_{1}$ and $P_{2}$ is equivalent to the angle between $\mathbf {N} _{1}$ and $\mathbf {N} _{2}$ : $\theta =\arccos \left({\frac {\mathbf {N} _{1}\cdot \mathbf {N} _{2}}{|\mathbf {N} _{1}||\mathbf {N} _{2}|}}\right)$ $=\arccos \left({\frac {\langle 1,2,-1\rangle \cdot \langle 1,1,1\rangle }{|\langle 1,2,-1\rangle ||\langle 1,1,1\rangle |}}\right)$ $=\arccos \left({\frac {1+2-1}{{\sqrt {1+4+1}}{\sqrt {1+1+1}}}}\right)$

=\arccos {\frac {2}{\sqrt {18}}}

Let

P

denote an arbitrary plane. Let

\mathbf {N}

denote an arbitrary vector that is orthogonal to

P

, and

\mathbf {r} _{0}

denote the position vector of an arbitrary point contained by

P

. A point at position vector

\mathbf {r}

is contained by

P

if and only if the displacement from

\mathbf {r} _{0}

is orthogonal to

\mathbf {N}

. This yields the equation

\mathbf {N} \cdot (\mathbf {r} -\mathbf {r} _{0})=0

\iff \mathbf {N} \cdot \mathbf {r} -\mathbf {N} \cdot \mathbf {r} _{0}=0

. Therefore the equation that defines

P

is

\mathbf {N} \cdot \mathbf {r} -\mathbf {N} \cdot \mathbf {r} _{0}=0

.

Let $P_{1}$ be the plane described by $x+2y-z=1$ and $P_{2}$ be the plane described by $x+y+z=1$

Since $x+2y-z=1\iff \langle 1,2,-1\rangle \cdot \langle x,y,z\rangle -1=0$ , the coefficient vector $\mathbf {N} _{1}=\langle 1,2,-1\rangle$ is orthogonal to $P_{1}$ .

Since $x+y+z=1\iff \langle 1,1,1\rangle \cdot \langle x,y,z\rangle -1=0$ , the coefficient vector $\mathbf {N} _{2}=\langle 1,1,1\rangle$ is orthogonal to $P_{2}$ .

The angle $\theta$ between $P_{1}$ and $P_{2}$ is equivalent to the angle between $\mathbf {N} _{1}$ and $\mathbf {N} _{2}$ : $\theta =\arccos \left({\frac {\mathbf {N} _{1}\cdot \mathbf {N} _{2}}{|\mathbf {N} _{1}||\mathbf {N} _{2}|}}\right)$ $=\arccos \left({\frac {\langle 1,2,-1\rangle \cdot \langle 1,1,1\rangle }{|\langle 1,2,-1\rangle ||\langle 1,1,1\rangle |}}\right)$ $=\arccos \left({\frac {1+2-1}{{\sqrt {1+4+1}}{\sqrt {1+1+1}}}}\right)$

=\arccos {\frac {2}{\sqrt {18}}}

165. Find the distance from the point (3,4,5) to the plane x+y+z=1.

Given a unit length vector

{\hat {\mathbf {n} }}

, consider an axis

w

oriented in the direction of

{\hat {\mathbf {n} }}

. The "

w

coordinate" is determined by orthogonally projecting points onto the

w

axis. Given a position vector

\mathbf {r}

, the expression

{\hat {\mathbf {n} }}\cdot \mathbf {r}

computes the

w

coordinate.

The equation $x+y+z=1$ is equivalent to $\langle 1,1,1\rangle \cdot \langle x,y,z\rangle =1$ $\iff {\frac {\langle 1,1,1\rangle }{|\langle 1,1,1\rangle |}}\cdot \langle x,y,z\rangle ={\frac {1}{|\langle 1,1,1\rangle |}}$ $\iff \langle 1/{\sqrt {3}},1/{\sqrt {3}},1/{\sqrt {3}}\rangle \cdot \langle x,y,z\rangle =1/{\sqrt {3}}$

Letting ${\hat {\mathbf {n} }}=\langle 1/{\sqrt {3}},1/{\sqrt {3}},1/{\sqrt {3}}\rangle$ , the plane $x+y+z=1$ consists of all points whose $w$ coordinate is $1/{\sqrt {3}}$ . The $w$ coordinate of $(3,4,5)$ is $\langle 1/{\sqrt {3}},1/{\sqrt {3}},1/{\sqrt {3}}\rangle \cdot \langle 3,4,5\rangle ={\frac {12}{\sqrt {3}}}=4{\sqrt {3}}$ .

The distance between the plane $x+y+z=1$ and the point $(3,4,5)$ along the $w$ axis is $\left|4{\sqrt {3}}-{\frac {1}{\sqrt {3}}}\right|=\left|{\frac {12{\sqrt {3}}-{\sqrt {3}}}{3}}\right|$ $={\frac {11}{3}}{\sqrt {3}}$

The distance

{\frac {11}{3}}{\sqrt {3}}

is the distance between the point and plane along a direction that is orthogonal to the plane, and is hence the shortest distance.

Given a unit length vector

{\hat {\mathbf {n} }}

, consider an axis

w

oriented in the direction of

{\hat {\mathbf {n} }}

. The "

w

coordinate" is determined by orthogonally projecting points onto the

w

axis. Given a position vector

\mathbf {r}

, the expression

{\hat {\mathbf {n} }}\cdot \mathbf {r}

computes the

w

coordinate.

The equation $x+y+z=1$ is equivalent to $\langle 1,1,1\rangle \cdot \langle x,y,z\rangle =1$ $\iff {\frac {\langle 1,1,1\rangle }{|\langle 1,1,1\rangle |}}\cdot \langle x,y,z\rangle ={\frac {1}{|\langle 1,1,1\rangle |}}$ $\iff \langle 1/{\sqrt {3}},1/{\sqrt {3}},1/{\sqrt {3}}\rangle \cdot \langle x,y,z\rangle =1/{\sqrt {3}}$

Letting ${\hat {\mathbf {n} }}=\langle 1/{\sqrt {3}},1/{\sqrt {3}},1/{\sqrt {3}}\rangle$ , the plane $x+y+z=1$ consists of all points whose $w$ coordinate is $1/{\sqrt {3}}$ . The $w$ coordinate of $(3,4,5)$ is $\langle 1/{\sqrt {3}},1/{\sqrt {3}},1/{\sqrt {3}}\rangle \cdot \langle 3,4,5\rangle ={\frac {12}{\sqrt {3}}}=4{\sqrt {3}}$ .

The distance between the plane $x+y+z=1$ and the point $(3,4,5)$ along the $w$ axis is $\left|4{\sqrt {3}}-{\frac {1}{\sqrt {3}}}\right|=\left|{\frac {12{\sqrt {3}}-{\sqrt {3}}}{3}}\right|$ $={\frac {11}{3}}{\sqrt {3}}$

The distance

{\frac {11}{3}}{\sqrt {3}}

is the distance between the point and plane along a direction that is orthogonal to the plane, and is hence the shortest distance.

Limits And Continuity

Evaluate the following limits.

180.

\displaystyle \lim _{(x,y)\rightarrow (1,-2)}{\frac {y^{2}+2xy}{y+2x}}

\lim _{(x,y)\rightarrow (1,-2)}{\frac {y^{2}+2xy}{y+2x}}=\lim _{(x,y)\rightarrow (1,-2)}{\frac {y(y+2x)}{y+2x}}

$=\lim _{(x,y)\rightarrow (1,-2)}y$

=-2

\lim _{(x,y)\rightarrow (1,-2)}{\frac {y^{2}+2xy}{y+2x}}=\lim _{(x,y)\rightarrow (1,-2)}{\frac {y(y+2x)}{y+2x}}

$=\lim _{(x,y)\rightarrow (1,-2)}y$

=-2

181.

\displaystyle \lim _{(x,y)\rightarrow (4,5)}{\frac {{\sqrt {x+y}}-3}{x+y-9}}

\lim _{(x,y)\rightarrow (4,5)}{\frac {{\sqrt {x+y}}-3}{x+y-9}}=\lim _{(x,y)\rightarrow (4,5)}{\frac {{\sqrt {x+y}}-3}{({\sqrt {x+y}}-3)({\sqrt {x+y}}+3)}}

$=\lim _{(x,y)\rightarrow (4,5)}{\frac {1}{{\sqrt {x+y}}+3}}$

={\frac {1}{{\sqrt {4+5}}+3}}=1/6

\lim _{(x,y)\rightarrow (4,5)}{\frac {{\sqrt {x+y}}-3}{x+y-9}}=\lim _{(x,y)\rightarrow (4,5)}{\frac {{\sqrt {x+y}}-3}{({\sqrt {x+y}}-3)({\sqrt {x+y}}+3)}}

$=\lim _{(x,y)\rightarrow (4,5)}{\frac {1}{{\sqrt {x+y}}+3}}$

={\frac {1}{{\sqrt {4+5}}+3}}=1/6

At what points is the function f continuous?

182.

f(x,y)=\ln |x-y|

\{(x,y)\mid x\neq y\}

\{(x,y)\mid x\neq y\}

183.

f(x,y)=\displaystyle {\frac {\ln(x^{2}+y^{2})}{x-y+1}}

All points (x,y) except for (0,0) and the line y=x+1

Use the two-path test to show that the following limits do not exist. (A path does not have to be a straight line.)

184.

\displaystyle \lim _{(x,y)\rightarrow (0,0)}{\frac {4xy}{3x^{2}+y^{2}}}

The limit is 1 along the line y=x, and −1 along the line y=−x

185.

\displaystyle \lim _{(x,y)\rightarrow (0,0)}{\frac {y}{\sqrt {x^{2}-y^{2}}}}

The limit is 0 along the line y=0, and

1/{\sqrt {3}}

along the line x=2y

The limit is 0 along the line y=0, and

1/{\sqrt {3}}

along the line x=2y

186.

\displaystyle \lim _{(x,y)\rightarrow (0,0)}{\frac {x^{3}-y^{2}}{x^{3}+y^{2}}}

The limit is 1 along the line y=0, and −1 along the line x=0

187.

\displaystyle \lim _{(x,y)\rightarrow (0,0)}{\frac {x^{2}y^{2}+y^{6}}{x^{3}}}

The limit is 0 along any line of the form y=mx, and 2 along the parabola

x=y^{2}

The limit is 0 along any line of the form y=mx, and 2 along the parabola

x=y^{2}

Partial Derivatives

200. Find

\partial z/\partial x

if

\displaystyle z(x,y)={\frac {1}{\ln(xy)}}

By repeatedly applying the chain rule:

${\frac {\partial z}{\partial x}}={\frac {\partial }{\partial x}}\left({\frac {1}{\ln(xy)}}\right)$ $={\frac {-1}{(\ln(xy))^{2}}}{\frac {\partial }{\partial x}}(\ln(xy))$ $={\frac {-1}{(\ln(xy))^{2}}}{\frac {1}{xy}}{\frac {\partial }{\partial x}}(xy)$ $={\frac {-1}{(\ln(xy))^{2}}}{\frac {1}{xy}}y$

={\frac {-1}{x(\ln(xy))^{2}}}

By repeatedly applying the chain rule:

${\frac {\partial z}{\partial x}}={\frac {\partial }{\partial x}}\left({\frac {1}{\ln(xy)}}\right)$ $={\frac {-1}{(\ln(xy))^{2}}}{\frac {\partial }{\partial x}}(\ln(xy))$ $={\frac {-1}{(\ln(xy))^{2}}}{\frac {1}{xy}}{\frac {\partial }{\partial x}}(xy)$ $={\frac {-1}{(\ln(xy))^{2}}}{\frac {1}{xy}}y$

={\frac {-1}{x(\ln(xy))^{2}}}

201. Find all three partial derivatives of the function

\displaystyle f(x,y,z)=xe^{y^{2}+z}

\displaystyle f_{x}=e^{y^{2}+z},\ f_{y}=2xye^{y^{2}+z},\ f_{z}=f.

\displaystyle f_{x}=e^{y^{2}+z},\ f_{y}=2xye^{y^{2}+z},\ f_{z}=f.

Find the four second partial derivatives of the following functions.

202.

f(x,y)=\cos(xy)

The first derivatives are:

f_{x}=-y\sin(xy)

and

f_{y}=-x\sin(xy)

The second derivatives are:

f_{xx}=-y^{2}\cos(xy),\ f_{yy}=-x^{2}\cos(xy),\ f_{xy}=f_{yx}=-\sin(xy)-xy\cos(xy)

The first derivatives are:

f_{x}=-y\sin(xy)

and

f_{y}=-x\sin(xy)

The second derivatives are:

f_{xx}=-y^{2}\cos(xy),\ f_{yy}=-x^{2}\cos(xy),\ f_{xy}=f_{yx}=-\sin(xy)-xy\cos(xy)

203.

f(x,y)=xe^{y}

The first derivatives are:

f_{x}=e^{y}

and

f_{y}=xe^{y}

The second derivatives are:

f_{xx}=0,\ f_{yy}=xe^{y},\ f_{xy}=f_{yx}=e^{y}

The first derivatives are:

f_{x}=e^{y}

and

f_{y}=xe^{y}

The second derivatives are:

f_{xx}=0,\ f_{yy}=xe^{y},\ f_{xy}=f_{yx}=e^{y}

Chain Rule

Find $df/dt.$

220.

f(x,y)=x^{2}y-xy^{3},\ x(t)=t^{2},\ y(t)=t^{-2}

The single derivatives are:

${\frac {\partial f}{\partial x}}=2xy-y^{3}$ ; ${\frac {\partial f}{\partial y}}=x^{2}-3xy^{2}$ ; ${\frac {dx}{dt}}=2t$ ; and ${\frac {dy}{dt}}=-2t^{-3}$

The chain rule gives: ${\frac {df}{dt}}={\frac {\partial f}{\partial x}}{\frac {dx}{dt}}+{\frac {\partial f}{\partial y}}{\frac {dy}{dt}}$ $=(2xy-y^{3})(2t)+(x^{2}-3xy^{2})(-2t^{-3})$ $=(2-t^{-6})(2t)+(t^{4}-3t^{-2})(-2t^{-3})$ $=(4t-2t^{-5})+(-2t+6t^{-5})$

=2t+4t^{-5}

The single derivatives are:

${\frac {\partial f}{\partial x}}=2xy-y^{3}$ ; ${\frac {\partial f}{\partial y}}=x^{2}-3xy^{2}$ ; ${\frac {dx}{dt}}=2t$ ; and ${\frac {dy}{dt}}=-2t^{-3}$

The chain rule gives: ${\frac {df}{dt}}={\frac {\partial f}{\partial x}}{\frac {dx}{dt}}+{\frac {\partial f}{\partial y}}{\frac {dy}{dt}}$ $=(2xy-y^{3})(2t)+(x^{2}-3xy^{2})(-2t^{-3})$ $=(2-t^{-6})(2t)+(t^{4}-3t^{-2})(-2t^{-3})$ $=(4t-2t^{-5})+(-2t+6t^{-5})$

=2t+4t^{-5}

221.

f(x,y)={\sqrt {x^{2}+y^{2}}},\ x(t)=\cos(2t),\ y(t)=\sin(2t)

The single derivatives are:

${\frac {\partial f}{\partial x}}={\frac {x}{\sqrt {x^{2}+y^{2}}}}$ ; ${\frac {\partial f}{\partial y}}={\frac {y}{\sqrt {x^{2}+y^{2}}}}$ ; ${\frac {dx}{dt}}=-2\sin(2t)$ ; and ${\frac {dy}{dt}}=2\cos(2t)$

The chain rule gives: ${\frac {df}{dt}}={\frac {\partial f}{\partial x}}{\frac {dx}{dt}}+{\frac {\partial f}{\partial y}}{\frac {dy}{dt}}$ $={\frac {x}{\sqrt {x^{2}+y^{2}}}}(-2\sin(2t))+{\frac {y}{\sqrt {x^{2}+y^{2}}}}(2\cos(2t))$ $={\frac {\cos(2t)}{\sqrt {\cos ^{2}(2t)+\sin ^{2}(2t)}}}(-2\sin(2t))+{\frac {\sin(2t)}{\sqrt {\cos ^{2}(2t)+\sin ^{2}(2t)}}}(2\cos(2t))$ $=-2\cos(2t)\sin(2t)+2\sin(2t)\cos(2t)$

=0

The single derivatives are:

${\frac {\partial f}{\partial x}}={\frac {x}{\sqrt {x^{2}+y^{2}}}}$ ; ${\frac {\partial f}{\partial y}}={\frac {y}{\sqrt {x^{2}+y^{2}}}}$ ; ${\frac {dx}{dt}}=-2\sin(2t)$ ; and ${\frac {dy}{dt}}=2\cos(2t)$

The chain rule gives: ${\frac {df}{dt}}={\frac {\partial f}{\partial x}}{\frac {dx}{dt}}+{\frac {\partial f}{\partial y}}{\frac {dy}{dt}}$ $={\frac {x}{\sqrt {x^{2}+y^{2}}}}(-2\sin(2t))+{\frac {y}{\sqrt {x^{2}+y^{2}}}}(2\cos(2t))$ $={\frac {\cos(2t)}{\sqrt {\cos ^{2}(2t)+\sin ^{2}(2t)}}}(-2\sin(2t))+{\frac {\sin(2t)}{\sqrt {\cos ^{2}(2t)+\sin ^{2}(2t)}}}(2\cos(2t))$ $=-2\cos(2t)\sin(2t)+2\sin(2t)\cos(2t)$

=0

222.

\displaystyle f(x,y,z)={\frac {x-y}{y+z}},\ x(t)=t,\ \displaystyle y(t)=2t,\ z(t)=3t

The single derivatives are:

${\frac {\partial f}{\partial x}}={\frac {1}{y+z}}$ ; ${\frac {\partial f}{\partial y}}={\frac {(-1)(y+z)-(x-y)(1)}{(y+z)^{2}}}={\frac {-(x+z)}{(y+z)^{2}}}$ ; ${\frac {\partial f}{\partial z}}={\frac {y-x}{(y+z)^{2}}}$ ; ${\frac {dx}{dt}}=1$ ; ${\frac {dy}{dt}}=2$ ; and ${\frac {dz}{dt}}=3$

The chain rule gives: ${\frac {df}{dt}}={\frac {\partial f}{\partial x}}{\frac {dx}{dt}}+{\frac {\partial f}{\partial y}}{\frac {dy}{dt}}+{\frac {\partial f}{\partial z}}{\frac {dz}{dt}}$ $={\frac {1}{y+z}}(1)+{\frac {-(x+z)}{(y+z)^{2}}}(2)+{\frac {y-x}{(y+z)^{2}}}(3)$ $={\frac {(y+z)-2(x+z)+3(y-x)}{(y+z)^{2}}}$ $={\frac {-5x+4y-z}{(y+z)^{2}}}$ $={\frac {-5t+8t-3t}{(5t)^{2}}}$

=0

The single derivatives are:

${\frac {\partial f}{\partial x}}={\frac {1}{y+z}}$ ; ${\frac {\partial f}{\partial y}}={\frac {(-1)(y+z)-(x-y)(1)}{(y+z)^{2}}}={\frac {-(x+z)}{(y+z)^{2}}}$ ; ${\frac {\partial f}{\partial z}}={\frac {y-x}{(y+z)^{2}}}$ ; ${\frac {dx}{dt}}=1$ ; ${\frac {dy}{dt}}=2$ ; and ${\frac {dz}{dt}}=3$

The chain rule gives: ${\frac {df}{dt}}={\frac {\partial f}{\partial x}}{\frac {dx}{dt}}+{\frac {\partial f}{\partial y}}{\frac {dy}{dt}}+{\frac {\partial f}{\partial z}}{\frac {dz}{dt}}$ $={\frac {1}{y+z}}(1)+{\frac {-(x+z)}{(y+z)^{2}}}(2)+{\frac {y-x}{(y+z)^{2}}}(3)$ $={\frac {(y+z)-2(x+z)+3(y-x)}{(y+z)^{2}}}$ $={\frac {-5x+4y-z}{(y+z)^{2}}}$ $={\frac {-5t+8t-3t}{(5t)^{2}}}$

=0

Find $f_{s},\ f_{t}.$

223.

f(x,y)=\sin(x)\cos(2y),\ x=s+t,\ y=s-t

The single derivatives are:

${\frac {\partial f}{\partial x}}=\cos(x)\cos(2y)$ ; ${\frac {\partial f}{\partial y}}=-2\sin(x)\sin(2y)$ ; ${\frac {\partial x}{\partial s}}=1$ ; ${\frac {\partial x}{\partial t}}=1$ ; ${\frac {\partial y}{\partial s}}=1$ ; and ${\frac {\partial y}{\partial t}}=-1$

The chain rule gives: ${\frac {\partial f}{\partial s}}={\frac {\partial f}{\partial x}}{\frac {\partial x}{\partial s}}+{\frac {\partial f}{\partial y}}{\frac {\partial y}{\partial s}}$ $=(\cos(x)\cos(2y))(1)+(-2\sin(x)\sin(2y))(1)$ $=\cos(s+t)\cos(2(s-t))-2\sin(s+t)\sin(2(s-t))$

and ${\frac {\partial f}{\partial t}}={\frac {\partial f}{\partial x}}{\frac {\partial x}{\partial t}}+{\frac {\partial f}{\partial y}}{\frac {\partial y}{\partial t}}$ $=(\cos(x)\cos(2y))(1)+(-2\sin(x)\sin(2y))(-1)$ $=\cos(s+t)\cos(2(s-t))+2\sin(s+t)\sin(2(s-t))$

Therefore:

f_{s}=\cos(s+t)\cos(2s-2t)-2\sin(s+t)\sin(2s-2t)

and

f_{t}=\cos(s+t)\cos(2s-2t)+2\sin(s+t)\sin(2s-2t)

The single derivatives are:

${\frac {\partial f}{\partial x}}=\cos(x)\cos(2y)$ ; ${\frac {\partial f}{\partial y}}=-2\sin(x)\sin(2y)$ ; ${\frac {\partial x}{\partial s}}=1$ ; ${\frac {\partial x}{\partial t}}=1$ ; ${\frac {\partial y}{\partial s}}=1$ ; and ${\frac {\partial y}{\partial t}}=-1$

The chain rule gives: ${\frac {\partial f}{\partial s}}={\frac {\partial f}{\partial x}}{\frac {\partial x}{\partial s}}+{\frac {\partial f}{\partial y}}{\frac {\partial y}{\partial s}}$ $=(\cos(x)\cos(2y))(1)+(-2\sin(x)\sin(2y))(1)$ $=\cos(s+t)\cos(2(s-t))-2\sin(s+t)\sin(2(s-t))$

and ${\frac {\partial f}{\partial t}}={\frac {\partial f}{\partial x}}{\frac {\partial x}{\partial t}}+{\frac {\partial f}{\partial y}}{\frac {\partial y}{\partial t}}$ $=(\cos(x)\cos(2y))(1)+(-2\sin(x)\sin(2y))(-1)$ $=\cos(s+t)\cos(2(s-t))+2\sin(s+t)\sin(2(s-t))$

Therefore:

f_{s}=\cos(s+t)\cos(2s-2t)-2\sin(s+t)\sin(2s-2t)

and

f_{t}=\cos(s+t)\cos(2s-2t)+2\sin(s+t)\sin(2s-2t)

224.

\displaystyle f(x,y,z)={\frac {x-z}{y+z}},\ x(t)=s+t,\ y(t)=st,\ z(t)=s-t

The single derivatives are:

${\frac {\partial f}{\partial x}}={\frac {1}{y+z}}$ ; ${\frac {\partial f}{\partial y}}={\frac {z-x}{(y+z)^{2}}}$ ; ${\frac {\partial f}{\partial z}}={\frac {(-1)(y+z)-(x-z)(1)}{(y+z)^{2}}}={\frac {-(x+y)}{(y+z)^{2}}}$ ; ${\frac {\partial x}{\partial s}}=1$ ; ${\frac {\partial x}{\partial t}}=1$ ; ${\frac {\partial y}{\partial s}}=t$ ; ${\frac {\partial y}{\partial t}}=s$ ; ${\frac {\partial z}{\partial s}}=1$ ; and ${\frac {\partial z}{\partial t}}=-1$

The chain rule gives: ${\frac {\partial f}{\partial s}}={\frac {\partial f}{\partial x}}{\frac {\partial x}{\partial s}}+{\frac {\partial f}{\partial y}}{\frac {\partial y}{\partial s}}+{\frac {\partial f}{\partial z}}{\frac {\partial z}{\partial s}}$ $={\frac {1}{y+z}}(1)+{\frac {z-x}{(y+z)^{2}}}(t)+{\frac {-(x+y)}{(y+z)^{2}}}(1)$ $={\frac {(y+z)+t(z-x)-(x+y)}{(y+z)^{2}}}$ $={\frac {(st+s-t)+t(-2t)-(s+t+st)}{(st+s-t)^{2}}}$ $={\frac {-2t-2t^{2}}{(st+s-t)^{2}}}$ $={\frac {-2t(t+1)}{(st+s-t)^{2}}}$

and ${\frac {\partial f}{\partial t}}={\frac {\partial f}{\partial x}}{\frac {\partial x}{\partial t}}+{\frac {\partial f}{\partial y}}{\frac {\partial y}{\partial t}}+{\frac {\partial f}{\partial z}}{\frac {\partial z}{\partial t}}$ $={\frac {1}{y+z}}(1)+{\frac {z-x}{(y+z)^{2}}}(s)+{\frac {-(x+y)}{(y+z)^{2}}}(-1)$ $={\frac {(y+z)+s(z-x)+(x+y)}{(y+z)^{2}}}$ $={\frac {(st+s-t)+s(-2t)+(s+t+st)}{(st+s-t)^{2}}}$ $={\frac {2s}{(st+s-t)^{2}}}$

Therefore:

f_{s}={\frac {-2t(t+1)}{(st+s-t)^{2}}}

and

f_{t}={\frac {2s}{(st+s-t)^{2}}}

The single derivatives are:

${\frac {\partial f}{\partial x}}={\frac {1}{y+z}}$ ; ${\frac {\partial f}{\partial y}}={\frac {z-x}{(y+z)^{2}}}$ ; ${\frac {\partial f}{\partial z}}={\frac {(-1)(y+z)-(x-z)(1)}{(y+z)^{2}}}={\frac {-(x+y)}{(y+z)^{2}}}$ ; ${\frac {\partial x}{\partial s}}=1$ ; ${\frac {\partial x}{\partial t}}=1$ ; ${\frac {\partial y}{\partial s}}=t$ ; ${\frac {\partial y}{\partial t}}=s$ ; ${\frac {\partial z}{\partial s}}=1$ ; and ${\frac {\partial z}{\partial t}}=-1$

The chain rule gives: ${\frac {\partial f}{\partial s}}={\frac {\partial f}{\partial x}}{\frac {\partial x}{\partial s}}+{\frac {\partial f}{\partial y}}{\frac {\partial y}{\partial s}}+{\frac {\partial f}{\partial z}}{\frac {\partial z}{\partial s}}$ $={\frac {1}{y+z}}(1)+{\frac {z-x}{(y+z)^{2}}}(t)+{\frac {-(x+y)}{(y+z)^{2}}}(1)$ $={\frac {(y+z)+t(z-x)-(x+y)}{(y+z)^{2}}}$ $={\frac {(st+s-t)+t(-2t)-(s+t+st)}{(st+s-t)^{2}}}$ $={\frac {-2t-2t^{2}}{(st+s-t)^{2}}}$ $={\frac {-2t(t+1)}{(st+s-t)^{2}}}$

and ${\frac {\partial f}{\partial t}}={\frac {\partial f}{\partial x}}{\frac {\partial x}{\partial t}}+{\frac {\partial f}{\partial y}}{\frac {\partial y}{\partial t}}+{\frac {\partial f}{\partial z}}{\frac {\partial z}{\partial t}}$ $={\frac {1}{y+z}}(1)+{\frac {z-x}{(y+z)^{2}}}(s)+{\frac {-(x+y)}{(y+z)^{2}}}(-1)$ $={\frac {(y+z)+s(z-x)+(x+y)}{(y+z)^{2}}}$ $={\frac {(st+s-t)+s(-2t)+(s+t+st)}{(st+s-t)^{2}}}$ $={\frac {2s}{(st+s-t)^{2}}}$

Therefore:

f_{s}={\frac {-2t(t+1)}{(st+s-t)^{2}}}

and

f_{t}={\frac {2s}{(st+s-t)^{2}}}

225. The volume of a pyramid with a square base is

V={\frac {1}{3}}x^{2}h

, where x is the side of the square base and h is the height of the pyramid. Suppose that

\displaystyle x(t)={\frac {t}{t+1}}

and

\displaystyle h(t)={\frac {1}{t+1}}

for

t\geq 0.

Find

V'(t).

The single derivatives are:

${\frac {\partial V}{\partial x}}={\frac {2}{3}}xh$ ; ${\frac {\partial V}{\partial h}}={\frac {1}{3}}x^{2}$ ; $x'(t)={\frac {(1)(t+1)-t(1)}{(t+1)^{2}}}={\frac {1}{(t+1)^{2}}}$ ; and $h'(t)={\frac {-1}{(t+1)^{2}}}$

The chain rule gives: $V'(t)={\frac {\partial V}{\partial x}}x'(t)+{\frac {\partial V}{\partial h}}h'(t)$ $={\frac {2}{3}}xh{\frac {1}{(t+1)^{2}}}+{\frac {1}{3}}x^{2}{\frac {-1}{(t+1)^{2}}}$ $={\frac {2t}{3(t+1)^{4}}}-{\frac {t^{2}}{3(t+1)^{4}}}$

={\frac {2t-t^{2}}{3(t+1)^{4}}}

The single derivatives are:

${\frac {\partial V}{\partial x}}={\frac {2}{3}}xh$ ; ${\frac {\partial V}{\partial h}}={\frac {1}{3}}x^{2}$ ; $x'(t)={\frac {(1)(t+1)-t(1)}{(t+1)^{2}}}={\frac {1}{(t+1)^{2}}}$ ; and $h'(t)={\frac {-1}{(t+1)^{2}}}$

The chain rule gives: $V'(t)={\frac {\partial V}{\partial x}}x'(t)+{\frac {\partial V}{\partial h}}h'(t)$ $={\frac {2}{3}}xh{\frac {1}{(t+1)^{2}}}+{\frac {1}{3}}x^{2}{\frac {-1}{(t+1)^{2}}}$ $={\frac {2t}{3(t+1)^{4}}}-{\frac {t^{2}}{3(t+1)^{4}}}$

={\frac {2t-t^{2}}{3(t+1)^{4}}}

Tangent Planes

Find an equation of a plane tangent to the given surface at the given point(s).

240.

xy\sin(z)=1,\ (1,2,\pi /6),\ (-1,-2,5\pi /6).

Start with a point

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

that is on the surface. Perturbing the

x

,

y

, and

z

coordinates by infinitesimal amounts

dx

,

dy

, and

dz

respectively, changes the value of

xy\sin(z)

by the infinitesimal amount

d(xy\sin(z))=(dx)y_{0}\sin(z_{0})+x_{0}(dy)\sin(z_{0})+x_{0}y_{0}d(\sin(z))=(y_{0}\sin(z_{0}))dx+(x_{0}\sin(z_{0}))dy+(x_{0}y_{0}\cos(z_{0}))dz

, and the value of

1

by

d(1)=0

. To remain in the surface it must be the case that

d(xy\sin(z))=d(1)

\iff (y_{0}\sin(z_{0}))dx+(x_{0}\sin(z_{0}))dy+(x_{0}y_{0}\cos(z_{0}))dz=0

.

To linearly extrapolate the condition $(y_{0}\sin(z_{0}))dx+(x_{0}\sin(z_{0}))dy+(x_{0}y_{0}\cos(z_{0}))dz=0$ to a tangent plane at $(x_{0},y_{0},z_{0})$ , replace the infinitesimal perturbations $dx$ , $dy$ , and $dz$ with large perturbations $\Delta x$ , $\Delta y$ , and $\Delta z$ to get $(y_{0}\sin(z_{0}))\Delta x+(x_{0}\sin(z_{0}))\Delta y+(x_{0}y_{0}\cos(z_{0}))\Delta z=0$ . Any point in the tangent plane at $(x_{0},y_{0},z_{0})$ can be reached by an appropriate choice of $\Delta x$ , $\Delta y$ , and $\Delta z$ where $(y_{0}\sin(z_{0}))\Delta x+(x_{0}\sin(z_{0}))\Delta y+(x_{0}y_{0}\cos(z_{0}))\Delta z=0$ . Any point $(x,y,z)$ in the tangent plane at $(x_{0},y_{0},z_{0})$ must satisfy $(y_{0}\sin(z_{0}))(x-x_{0})+(x_{0}\sin(z_{0}))(y-y_{0})+(x_{0}y_{0}\cos(z_{0}))(z-z_{0})=0$ .

The point $(x_{0},y_{0},z_{0})=(1,2,\pi /6)$ lies in the surface, and the tangent plane is $(x-1)+{\frac {1}{2}}(y-2)+{\sqrt {3}}(z-\pi /6)=0$ .

The point $(x_{0},y_{0},z_{0})=(-1,-2,5\pi /6)$ lies in the surface, and the tangent plane is $-(x+1)-{\frac {1}{2}}(y+2)-{\sqrt {3}}(z-5\pi /6)=0$ .

The tangent planes are therefore:

(x-1)+{\frac {1}{2}}(y-2)+{\sqrt {3}}(z-\pi /6)=0,\ (x+1)+{\frac {1}{2}}(y+2)+{\sqrt {3}}(z-5\pi /6)=0

Start with a point

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

that is on the surface. Perturbing the

x

,

y

, and

z

coordinates by infinitesimal amounts

dx

,

dy

, and

dz

respectively, changes the value of

xy\sin(z)

by the infinitesimal amount

d(xy\sin(z))=(dx)y_{0}\sin(z_{0})+x_{0}(dy)\sin(z_{0})+x_{0}y_{0}d(\sin(z))=(y_{0}\sin(z_{0}))dx+(x_{0}\sin(z_{0}))dy+(x_{0}y_{0}\cos(z_{0}))dz

, and the value of

1

by

d(1)=0

. To remain in the surface it must be the case that

d(xy\sin(z))=d(1)

\iff (y_{0}\sin(z_{0}))dx+(x_{0}\sin(z_{0}))dy+(x_{0}y_{0}\cos(z_{0}))dz=0

.

To linearly extrapolate the condition $(y_{0}\sin(z_{0}))dx+(x_{0}\sin(z_{0}))dy+(x_{0}y_{0}\cos(z_{0}))dz=0$ to a tangent plane at $(x_{0},y_{0},z_{0})$ , replace the infinitesimal perturbations $dx$ , $dy$ , and $dz$ with large perturbations $\Delta x$ , $\Delta y$ , and $\Delta z$ to get $(y_{0}\sin(z_{0}))\Delta x+(x_{0}\sin(z_{0}))\Delta y+(x_{0}y_{0}\cos(z_{0}))\Delta z=0$ . Any point in the tangent plane at $(x_{0},y_{0},z_{0})$ can be reached by an appropriate choice of $\Delta x$ , $\Delta y$ , and $\Delta z$ where $(y_{0}\sin(z_{0}))\Delta x+(x_{0}\sin(z_{0}))\Delta y+(x_{0}y_{0}\cos(z_{0}))\Delta z=0$ . Any point $(x,y,z)$ in the tangent plane at $(x_{0},y_{0},z_{0})$ must satisfy $(y_{0}\sin(z_{0}))(x-x_{0})+(x_{0}\sin(z_{0}))(y-y_{0})+(x_{0}y_{0}\cos(z_{0}))(z-z_{0})=0$ .

The point $(x_{0},y_{0},z_{0})=(1,2,\pi /6)$ lies in the surface, and the tangent plane is $(x-1)+{\frac {1}{2}}(y-2)+{\sqrt {3}}(z-\pi /6)=0$ .

The point $(x_{0},y_{0},z_{0})=(-1,-2,5\pi /6)$ lies in the surface, and the tangent plane is $-(x+1)-{\frac {1}{2}}(y+2)-{\sqrt {3}}(z-5\pi /6)=0$ .

The tangent planes are therefore:

(x-1)+{\frac {1}{2}}(y-2)+{\sqrt {3}}(z-\pi /6)=0,\ (x+1)+{\frac {1}{2}}(y+2)+{\sqrt {3}}(z-5\pi /6)=0

241.

z=x^{2}e^{x-y},\ (2,2,4),\ (-1,-1,1).

Start with a point

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

that is on the surface. Perturbing the

x

,

y

, and

z

coordinates by infinitesimal amounts

dx

,

dy

, and

dz

respectively, changes the value of

z

by the infinitesimal amount

dz

, and the value of

x^{2}e^{x-y}

by

d(x^{2}e^{x-y})=d(x^{2})e^{x_{0}-y_{0}}+x_{0}^{2}d(e^{x-y})

=(2x_{0}dx)e^{x_{0}-y_{0}}+x_{0}^{2}(e^{x_{0}-y_{0}}d(x-y))

=(2x_{0}e^{x_{0}-y_{0}})dx+x_{0}^{2}(e^{x_{0}-y_{0}}(dx-dy))

=(2x_{0}+x_{0}^{2})e^{x_{0}-y_{0}}dx-x_{0}^{2}e^{x_{0}-y_{0}}dy

. To remain in the surface it must be the case that

dz=d(x^{2}e^{x-y})

\iff dz=(2x_{0}+x_{0}^{2})e^{x_{0}-y_{0}}dx-x_{0}^{2}e^{x_{0}-y_{0}}dy

.

To linearly extrapolate the condition $dz=(2x_{0}+x_{0}^{2})e^{x_{0}-y_{0}}dx-x_{0}^{2}e^{x_{0}-y_{0}}dy$ to a tangent plane at $(x_{0},y_{0},z_{0})$ , replace the infinitesimal perturbations $dx$ , $dy$ , and $dz$ with large perturbations $\Delta x$ , $\Delta y$ , and $\Delta z$ to get $\Delta z=(2x_{0}+x_{0}^{2})e^{x_{0}-y_{0}}\Delta x-x_{0}^{2}e^{x_{0}-y_{0}}\Delta y$ . Any point in the tangent plane at $(x_{0},y_{0},z_{0})$ can be reached by an appropriate choice of $\Delta x$ , $\Delta y$ , and $\Delta z$ where $\Delta z=(2x_{0}+x_{0}^{2})e^{x_{0}-y_{0}}\Delta x-x_{0}^{2}e^{x_{0}-y_{0}}\Delta y$ . Any point $(x,y,z)$ in the tangent plane at $(x_{0},y_{0},z_{0})$ must satisfy $(z-z_{0})=(2x_{0}+x_{0}^{2})e^{x_{0}-y_{0}}(x-x_{0})-x_{0}^{2}e^{x_{0}-y_{0}}(y-y_{0})$ .

The point $(x_{0},y_{0},z_{0})=(2,2,4)$ lies in the surface, and the tangent plane is $z-4=8(x-2)-4(y-2)$ .

The point $(x_{0},y_{0},z_{0})=(-1,-1,1)$ lies in the surface, and the tangent plane is $z-1=-(x+1)-(y+1)$ .

The tangent planes are therefore:

-8(x-2)+4(y-2)+z-4=0,\ x+y+z+1=0

Start with a point

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

that is on the surface. Perturbing the

x

,

y

, and

z

coordinates by infinitesimal amounts

dx

,

dy

, and

dz

respectively, changes the value of

z

by the infinitesimal amount

dz

, and the value of

x^{2}e^{x-y}

by

d(x^{2}e^{x-y})=d(x^{2})e^{x_{0}-y_{0}}+x_{0}^{2}d(e^{x-y})

=(2x_{0}dx)e^{x_{0}-y_{0}}+x_{0}^{2}(e^{x_{0}-y_{0}}d(x-y))

=(2x_{0}e^{x_{0}-y_{0}})dx+x_{0}^{2}(e^{x_{0}-y_{0}}(dx-dy))

=(2x_{0}+x_{0}^{2})e^{x_{0}-y_{0}}dx-x_{0}^{2}e^{x_{0}-y_{0}}dy

. To remain in the surface it must be the case that

dz=d(x^{2}e^{x-y})

\iff dz=(2x_{0}+x_{0}^{2})e^{x_{0}-y_{0}}dx-x_{0}^{2}e^{x_{0}-y_{0}}dy

.

To linearly extrapolate the condition $dz=(2x_{0}+x_{0}^{2})e^{x_{0}-y_{0}}dx-x_{0}^{2}e^{x_{0}-y_{0}}dy$ to a tangent plane at $(x_{0},y_{0},z_{0})$ , replace the infinitesimal perturbations $dx$ , $dy$ , and $dz$ with large perturbations $\Delta x$ , $\Delta y$ , and $\Delta z$ to get $\Delta z=(2x_{0}+x_{0}^{2})e^{x_{0}-y_{0}}\Delta x-x_{0}^{2}e^{x_{0}-y_{0}}\Delta y$ . Any point in the tangent plane at $(x_{0},y_{0},z_{0})$ can be reached by an appropriate choice of $\Delta x$ , $\Delta y$ , and $\Delta z$ where $\Delta z=(2x_{0}+x_{0}^{2})e^{x_{0}-y_{0}}\Delta x-x_{0}^{2}e^{x_{0}-y_{0}}\Delta y$ . Any point $(x,y,z)$ in the tangent plane at $(x_{0},y_{0},z_{0})$ must satisfy $(z-z_{0})=(2x_{0}+x_{0}^{2})e^{x_{0}-y_{0}}(x-x_{0})-x_{0}^{2}e^{x_{0}-y_{0}}(y-y_{0})$ .

The point $(x_{0},y_{0},z_{0})=(2,2,4)$ lies in the surface, and the tangent plane is $z-4=8(x-2)-4(y-2)$ .

The point $(x_{0},y_{0},z_{0})=(-1,-1,1)$ lies in the surface, and the tangent plane is $z-1=-(x+1)-(y+1)$ .

The tangent planes are therefore:

-8(x-2)+4(y-2)+z-4=0,\ x+y+z+1=0

242.

z=\tan ^{-1}(x+y),\ (0,0,0).

Start with a point

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

that is on the surface. Perturbing the

x

,

y

, and

z

coordinates by infinitesimal amounts

dx

,

dy

, and

dz

respectively, changes the value of

z

by the infinitesimal amount

dz

, and the value of

\tan ^{-1}(x+y)

by

d(\tan ^{-1}(x+y))={\frac {1}{1+(x_{0}+y_{0})^{2}}}d(x+y)={\frac {dx+dy}{1+(x_{0}+y_{0})^{2}}}

. To remain in the surface it must be the case that

dz=d(\tan ^{-1}(x+y))

\iff dz={\frac {dx+dy}{1+(x_{0}+y_{0})^{2}}}

.

To linearly extrapolate the condition $dz={\frac {dx+dy}{1+(x_{0}+y_{0})^{2}}}$ to a tangent plane at $(x_{0},y_{0},z_{0})$ , replace the infinitesimal perturbations $dx$ , $dy$ , and $dz$ with large perturbations $\Delta x$ , $\Delta y$ , and $\Delta z$ to get $\Delta z={\frac {\Delta x+\Delta y}{1+(x_{0}+y_{0})^{2}}}$ . Any point in the tangent plane at $(x_{0},y_{0},z_{0})$ can be reached by an appropriate choice of $\Delta x$ , $\Delta y$ , and $\Delta z$ where $\Delta z={\frac {\Delta x+\Delta y}{1+(x_{0}+y_{0})^{2}}}$ . Any point $(x,y,z)$ in the tangent plane at $(x_{0},y_{0},z_{0})$ must satisfy $(z-z_{0})={\frac {(x-x_{0})+(y-y_{0})}{1+(x_{0}+y_{0})^{2}}}$ .

The point

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

lies in the surface, and the tangent plane is

z=x+y

\iff x+y-z=0

Start with a point

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

that is on the surface. Perturbing the

x

,

y

, and

z

coordinates by infinitesimal amounts

dx

,

dy

, and

dz

respectively, changes the value of

z

by the infinitesimal amount

dz

, and the value of

\tan ^{-1}(x+y)

by

d(\tan ^{-1}(x+y))={\frac {1}{1+(x_{0}+y_{0})^{2}}}d(x+y)={\frac {dx+dy}{1+(x_{0}+y_{0})^{2}}}

. To remain in the surface it must be the case that

dz=d(\tan ^{-1}(x+y))

\iff dz={\frac {dx+dy}{1+(x_{0}+y_{0})^{2}}}

.

To linearly extrapolate the condition $dz={\frac {dx+dy}{1+(x_{0}+y_{0})^{2}}}$ to a tangent plane at $(x_{0},y_{0},z_{0})$ , replace the infinitesimal perturbations $dx$ , $dy$ , and $dz$ with large perturbations $\Delta x$ , $\Delta y$ , and $\Delta z$ to get $\Delta z={\frac {\Delta x+\Delta y}{1+(x_{0}+y_{0})^{2}}}$ . Any point in the tangent plane at $(x_{0},y_{0},z_{0})$ can be reached by an appropriate choice of $\Delta x$ , $\Delta y$ , and $\Delta z$ where $\Delta z={\frac {\Delta x+\Delta y}{1+(x_{0}+y_{0})^{2}}}$ . Any point $(x,y,z)$ in the tangent plane at $(x_{0},y_{0},z_{0})$ must satisfy $(z-z_{0})={\frac {(x-x_{0})+(y-y_{0})}{1+(x_{0}+y_{0})^{2}}}$ .

The point

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

lies in the surface, and the tangent plane is

z=x+y

\iff x+y-z=0

243.

\sin(xyz)=1/2,\ (\pi ,1,1/6).

Start with a point

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

that is on the surface. Perturbing the

x

,

y

, and

z

coordinates by infinitesimal amounts

dx

,

dy

, and

dz

respectively, changes the value of

\sin(xyz)

by the infinitesimal amount

d(\sin(xyz))=\cos(x_{0}y_{0}z_{0})d(xyz)=\cos(x_{0}y_{0}z_{0})((dx)y_{0}z_{0}+x_{0}(dy)z_{0}+x_{0}y_{0}(dz))=(y_{0}z_{0}\cos(x_{0}y_{0}z_{0}))dx+(x_{0}z_{0}\cos(x_{0}y_{0}z_{0}))dy+(x_{0}y_{0}\cos(x_{0}y_{0}z_{0}))dz

, and the value of

1/2

by

d(1/2)=0

. To remain in the surface it must be the case that

d(\sin(xyz))=d(1/2)

\iff (y_{0}z_{0}\cos(x_{0}y_{0}z_{0}))dx+(x_{0}z_{0}\cos(x_{0}y_{0}z_{0}))dy+(x_{0}y_{0}\cos(x_{0}y_{0}z_{0}))dz=0

.

To linearly extrapolate the condition $(y_{0}z_{0}\cos(x_{0}y_{0}z_{0}))dx+(x_{0}z_{0}\cos(x_{0}y_{0}z_{0}))dy+(x_{0}y_{0}\cos(x_{0}y_{0}z_{0}))dz=0$ to a tangent plane at $(x_{0},y_{0},z_{0})$ , replace the infinitesimal perturbations $dx$ , $dy$ , and $dz$ with large perturbations $\Delta x$ , $\Delta y$ , and $\Delta z$ to get $(y_{0}z_{0}\cos(x_{0}y_{0}z_{0}))\Delta x+(x_{0}z_{0}\cos(x_{0}y_{0}z_{0}))\Delta y+(x_{0}y_{0}\cos(x_{0}y_{0}z_{0}))\Delta z=0$ . Any point in the tangent plane at $(x_{0},y_{0},z_{0})$ can be reached by an appropriate choice of $\Delta x$ , $\Delta y$ , and $\Delta z$ where $(y_{0}z_{0}\cos(x_{0}y_{0}z_{0}))\Delta x+(x_{0}z_{0}\cos(x_{0}y_{0}z_{0}))\Delta y+(x_{0}y_{0}\cos(x_{0}y_{0}z_{0}))\Delta z=0$ . Any point $(x,y,z)$ in the tangent plane at $(x_{0},y_{0},z_{0})$ must satisfy $(y_{0}z_{0}\cos(x_{0}y_{0}z_{0}))(x-x_{0})+(x_{0}z_{0}\cos(x_{0}y_{0}z_{0}))(y-y_{0})+(x_{0}y_{0}\cos(x_{0}y_{0}z_{0}))(z-z_{0})=0$ .

The point

(x_{0},y_{0},z_{0})=(\pi ,1,1/6)

lies in the surface, and the tangent plane is

{\frac {\sqrt {3}}{12}}(x-\pi )+{\frac {\pi {\sqrt {3}}}{12}}(y-1)+{\frac {\pi {\sqrt {3}}}{2}}(z-1/6)=0

\iff (x-\pi )+\pi (y-1)+6\pi (z-1/6)=0

\iff x+\pi y+6\pi z=3\pi

Start with a point

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

that is on the surface. Perturbing the

x

,

y

, and

z

coordinates by infinitesimal amounts

dx

,

dy

, and

dz

respectively, changes the value of

\sin(xyz)

by the infinitesimal amount

d(\sin(xyz))=\cos(x_{0}y_{0}z_{0})d(xyz)=\cos(x_{0}y_{0}z_{0})((dx)y_{0}z_{0}+x_{0}(dy)z_{0}+x_{0}y_{0}(dz))=(y_{0}z_{0}\cos(x_{0}y_{0}z_{0}))dx+(x_{0}z_{0}\cos(x_{0}y_{0}z_{0}))dy+(x_{0}y_{0}\cos(x_{0}y_{0}z_{0}))dz

, and the value of

1/2

by

d(1/2)=0

. To remain in the surface it must be the case that

d(\sin(xyz))=d(1/2)

\iff (y_{0}z_{0}\cos(x_{0}y_{0}z_{0}))dx+(x_{0}z_{0}\cos(x_{0}y_{0}z_{0}))dy+(x_{0}y_{0}\cos(x_{0}y_{0}z_{0}))dz=0

.

To linearly extrapolate the condition $(y_{0}z_{0}\cos(x_{0}y_{0}z_{0}))dx+(x_{0}z_{0}\cos(x_{0}y_{0}z_{0}))dy+(x_{0}y_{0}\cos(x_{0}y_{0}z_{0}))dz=0$ to a tangent plane at $(x_{0},y_{0},z_{0})$ , replace the infinitesimal perturbations $dx$ , $dy$ , and $dz$ with large perturbations $\Delta x$ , $\Delta y$ , and $\Delta z$ to get $(y_{0}z_{0}\cos(x_{0}y_{0}z_{0}))\Delta x+(x_{0}z_{0}\cos(x_{0}y_{0}z_{0}))\Delta y+(x_{0}y_{0}\cos(x_{0}y_{0}z_{0}))\Delta z=0$ . Any point in the tangent plane at $(x_{0},y_{0},z_{0})$ can be reached by an appropriate choice of $\Delta x$ , $\Delta y$ , and $\Delta z$ where $(y_{0}z_{0}\cos(x_{0}y_{0}z_{0}))\Delta x+(x_{0}z_{0}\cos(x_{0}y_{0}z_{0}))\Delta y+(x_{0}y_{0}\cos(x_{0}y_{0}z_{0}))\Delta z=0$ . Any point $(x,y,z)$ in the tangent plane at $(x_{0},y_{0},z_{0})$ must satisfy $(y_{0}z_{0}\cos(x_{0}y_{0}z_{0}))(x-x_{0})+(x_{0}z_{0}\cos(x_{0}y_{0}z_{0}))(y-y_{0})+(x_{0}y_{0}\cos(x_{0}y_{0}z_{0}))(z-z_{0})=0$ .

The point

(x_{0},y_{0},z_{0})=(\pi ,1,1/6)

lies in the surface, and the tangent plane is

{\frac {\sqrt {3}}{12}}(x-\pi )+{\frac {\pi {\sqrt {3}}}{12}}(y-1)+{\frac {\pi {\sqrt {3}}}{2}}(z-1/6)=0

\iff (x-\pi )+\pi (y-1)+6\pi (z-1/6)=0

\iff x+\pi y+6\pi z=3\pi

Maximum And Minimum Problems

Find critical points of the function f. When possible, determine whether each critical point corresponds to a local maximum, a local minimum, or a saddle point.

How to find local minimums, local maximums, and saddle points for a function with an unconstrained domain.

Start with a candidate point $(x,y)=(x_{0},y_{0})$ , and envision that the $x$ and $y$ coordinates are changing at rates of $m$ and $n$ respectively: ${\frac {dx}{dt}}=m$ and ${\frac {dy}{dt}}=n$ . The rate of change in $f$ is ${\frac {df}{dt}}={\frac {\partial f}{\partial x}}{\frac {dx}{dt}}+{\frac {\partial f}{\partial y}}{\frac {dy}{dt}}$ $={\frac {\partial f}{\partial x}}m+{\frac {\partial f}{\partial y}}n$ .

$(x,y)=(x_{0},y_{0})$ is a local minimum only if ${\frac {df}{dt}}\geq 0$ for all $(m,n)$ . This occurs iff ${\frac {\partial f}{\partial x}}={\frac {\partial f}{\partial y}}=0$ .
$(x,y)=(x_{0},y_{0})$ is a local maximum only if ${\frac {df}{dt}}\leq 0$ for all $(m,n)$ . This occurs iff ${\frac {\partial f}{\partial x}}={\frac {\partial f}{\partial y}}=0$ .

Points where ${\frac {\partial f}{\partial x}}={\frac {\partial f}{\partial y}}=0$ are "critical points" and may contain a local minimum, a local maximum, or a saddle point. It is then needed to classify the critical point.

The second derivative is ${\frac {d^{2}f}{dt^{2}}}={\frac {d}{dt}}\left({\frac {\partial f}{\partial x}}\right)m+{\frac {d}{dt}}\left({\frac {\partial f}{\partial y}}\right)n$ $=\left({\frac {\partial ^{2}f}{\partial x^{2}}}m+{\frac {\partial ^{2}f}{\partial y\partial x}}n\right)m+\left({\frac {\partial ^{2}f}{\partial x\partial y}}m+{\frac {\partial ^{2}f}{\partial ^{2}y}}n\right)n$ $={\frac {\partial ^{2}f}{\partial x^{2}}}m^{2}+2{\frac {\partial ^{2}f}{\partial y\partial x}}mn+{\frac {\partial ^{2}f}{\partial y^{2}}}n^{2}$ .

A critical point is a local minimum iff ${\frac {d^{2}f}{dt^{2}}}\geq 0$ for all $(m,n)$ .
A critical point is a local maximum iff ${\frac {d^{2}f}{dt^{2}}}\leq 0$ for all $(m,n)$ .
A critical point that is neither a local minimum nor a local maximum is a saddle point.

While it will not be shown here, ${\frac {d^{2}f}{dt^{2}}}$ can attain both positive and negative values iff ${\frac {\partial ^{2}f}{\partial x^{2}}}{\frac {\partial ^{2}f}{\partial y^{2}}}-\left({\frac {\partial ^{2}f}{\partial y\partial x}}\right)^{2}<0$ .

A critical point is a local minimum if ${\frac {\partial ^{2}f}{\partial x^{2}}}{\frac {\partial ^{2}f}{\partial y^{2}}}-\left({\frac {\partial ^{2}f}{\partial y\partial x}}\right)^{2}\geq 0$ and ${\frac {\partial ^{2}f}{\partial x^{2}}},{\frac {\partial ^{2}f}{\partial y^{2}}}\geq 0$
A critical point is a local maximum if ${\frac {\partial ^{2}f}{\partial x^{2}}}{\frac {\partial ^{2}f}{\partial y^{2}}}-\left({\frac {\partial ^{2}f}{\partial y\partial x}}\right)^{2}\geq 0$ and ${\frac {\partial ^{2}f}{\partial x^{2}}},{\frac {\partial ^{2}f}{\partial y^{2}}}\leq 0$
A critical point is a saddle point if ${\frac {\partial ^{2}f}{\partial x^{2}}}{\frac {\partial ^{2}f}{\partial y^{2}}}-\left({\frac {\partial ^{2}f}{\partial y\partial x}}\right)^{2}<0$

The quantity ${\frac {\partial ^{2}f}{\partial x^{2}}}{\frac {\partial ^{2}f}{\partial y^{2}}}-\left({\frac {\partial ^{2}f}{\partial y\partial x}}\right)^{2}$ will be called the "discriminant".

260.

f(x,y)=x^{4}+2y^{2}-4xy

The first order derivatives are:

{\frac {\partial f}{\partial x}}=4x^{3}-4y

and

{\frac {\partial f}{\partial y}}=4y-4x

. Finding the critical points is done by solving the equations

\left\{{\begin{array}{c}4x^{3}-4y=0\\4y-4x=0\end{array}}\right.

$4y-4x=0\iff y=x$ . Substituting $x$ in place of $y$ in the first equation gives $4x^{3}-4x=0\iff x(x^{2}-1)=0\iff x=-1,0,+1$ . This gives the critical points $(x,y)=(-1,-1),(0,0),(1,1)$ .

The second order derivatives are ${\frac {\partial ^{2}f}{\partial x^{2}}}=12x^{2}$ ; ${\frac {\partial ^{2}f}{\partial y^{2}}}=4$ ; and ${\frac {\partial ^{2}f}{\partial y\partial x}}=-4$ . The discriminant is: ${\frac {\partial ^{2}f}{\partial x^{2}}}{\frac {\partial ^{2}f}{\partial y^{2}}}-\left({\frac {\partial ^{2}f}{\partial y\partial x}}\right)^{2}=(12x^{2})(4)-(-4)^{2}=48x^{2}-16$ .

For critical point $(-1,-1)$ , the discriminant is $32$ and ${\frac {\partial ^{2}f}{\partial x^{2}}}=12$ so $(-1,-1)$ is a local minimum.

For critical point $(0,0)$ , the discriminant is $-16$ so $(-1,-1)$ is a saddle point.

For critical point $(1,1)$ , the discriminant is $32$ and ${\frac {\partial ^{2}f}{\partial x^{2}}}=12$ so $(1,1)$ is a local minimum.

Local minima at (1,1) and (−1,−1), saddle at (0,0)

The first order derivatives are:

{\frac {\partial f}{\partial x}}=4x^{3}-4y

and

{\frac {\partial f}{\partial y}}=4y-4x

. Finding the critical points is done by solving the equations

\left\{{\begin{array}{c}4x^{3}-4y=0\\4y-4x=0\end{array}}\right.

$4y-4x=0\iff y=x$ . Substituting $x$ in place of $y$ in the first equation gives $4x^{3}-4x=0\iff x(x^{2}-1)=0\iff x=-1,0,+1$ . This gives the critical points $(x,y)=(-1,-1),(0,0),(1,1)$ .

The second order derivatives are ${\frac {\partial ^{2}f}{\partial x^{2}}}=12x^{2}$ ; ${\frac {\partial ^{2}f}{\partial y^{2}}}=4$ ; and ${\frac {\partial ^{2}f}{\partial y\partial x}}=-4$ . The discriminant is: ${\frac {\partial ^{2}f}{\partial x^{2}}}{\frac {\partial ^{2}f}{\partial y^{2}}}-\left({\frac {\partial ^{2}f}{\partial y\partial x}}\right)^{2}=(12x^{2})(4)-(-4)^{2}=48x^{2}-16$ .

For critical point $(-1,-1)$ , the discriminant is $32$ and ${\frac {\partial ^{2}f}{\partial x^{2}}}=12$ so $(-1,-1)$ is a local minimum.

For critical point $(0,0)$ , the discriminant is $-16$ so $(-1,-1)$ is a saddle point.

For critical point $(1,1)$ , the discriminant is $32$ and ${\frac {\partial ^{2}f}{\partial x^{2}}}=12$ so $(1,1)$ is a local minimum.

Local minima at (1,1) and (−1,−1), saddle at (0,0)

261.

f(x,y)=\tan ^{-1}(xy)

The first order derivatives are:

{\frac {\partial f}{\partial x}}={\frac {1}{1+(xy)^{2}}}y={\frac {y}{1+(xy)^{2}}}

and

{\frac {\partial f}{\partial y}}={\frac {1}{1+(xy)^{2}}}x={\frac {x}{1+(xy)^{2}}}

. Finding the critical points is done by solving the equations

\left\{{\begin{array}{c}y/(1+(xy)^{2})=0\\x/(1+(xy)^{2})=0\end{array}}\right.

${\frac {y}{1+(xy)^{2}}}=0\iff y=0$ and ${\frac {x}{1+(xy)^{2}}}=0\iff x=0$ . So the only critical point is $(x,y)=(0,0)$ .

The second order derivatives are ${\frac {\partial ^{2}f}{\partial x^{2}}}=-{\frac {y}{(1+(xy)^{2})^{2}}}(2xy^{2})={\frac {-2xy^{3}}{(1+(xy)^{2})^{2}}}$ ; ${\frac {\partial ^{2}f}{\partial y^{2}}}=-{\frac {x}{(1+(xy)^{2})^{2}}}(2x^{2}y)={\frac {-2x^{3}y}{(1+(xy)^{2})^{2}}}$ ; and ${\frac {\partial ^{2}f}{\partial y\partial x}}={\frac {(1)(1+(xy)^{2})-(y)(2x^{2}y)}{(1+(xy)^{2})^{2}}}={\frac {1-(xy)^{2}}{(1+(xy)^{2})^{2}}}$ . The discriminant is ${\frac {\partial ^{2}f}{\partial x^{2}}}{\frac {\partial ^{2}f}{\partial y^{2}}}-\left({\frac {\partial ^{2}f}{\partial y\partial x}}\right)^{2}=\left({\frac {-2xy^{3}}{(1+(xy)^{2})^{2}}}\right)\left({\frac {-2x^{3}y}{(1+(xy)^{2})^{2}}}\right)-\left({\frac {1-(xy)^{2}}{(1+(xy)^{2})^{2}}}\right)^{2}$ $={\frac {4(xy)^{4}-(1-(xy)^{2})^{2}}{(1+(xy)^{2})^{4}}}$ $={\frac {5(xy)^{4}+2(xy)^{2}-1}{(1+(xy)^{2})^{4}}}$

For the critical point $(0,0)$ , the discriminant is $-1$ so $(0,0)$ is a saddle point.

Saddle at (0,0)

The first order derivatives are:

{\frac {\partial f}{\partial x}}={\frac {1}{1+(xy)^{2}}}y={\frac {y}{1+(xy)^{2}}}

and

{\frac {\partial f}{\partial y}}={\frac {1}{1+(xy)^{2}}}x={\frac {x}{1+(xy)^{2}}}

. Finding the critical points is done by solving the equations

\left\{{\begin{array}{c}y/(1+(xy)^{2})=0\\x/(1+(xy)^{2})=0\end{array}}\right.

${\frac {y}{1+(xy)^{2}}}=0\iff y=0$ and ${\frac {x}{1+(xy)^{2}}}=0\iff x=0$ . So the only critical point is $(x,y)=(0,0)$ .

The second order derivatives are ${\frac {\partial ^{2}f}{\partial x^{2}}}=-{\frac {y}{(1+(xy)^{2})^{2}}}(2xy^{2})={\frac {-2xy^{3}}{(1+(xy)^{2})^{2}}}$ ; ${\frac {\partial ^{2}f}{\partial y^{2}}}=-{\frac {x}{(1+(xy)^{2})^{2}}}(2x^{2}y)={\frac {-2x^{3}y}{(1+(xy)^{2})^{2}}}$ ; and ${\frac {\partial ^{2}f}{\partial y\partial x}}={\frac {(1)(1+(xy)^{2})-(y)(2x^{2}y)}{(1+(xy)^{2})^{2}}}={\frac {1-(xy)^{2}}{(1+(xy)^{2})^{2}}}$ . The discriminant is ${\frac {\partial ^{2}f}{\partial x^{2}}}{\frac {\partial ^{2}f}{\partial y^{2}}}-\left({\frac {\partial ^{2}f}{\partial y\partial x}}\right)^{2}=\left({\frac {-2xy^{3}}{(1+(xy)^{2})^{2}}}\right)\left({\frac {-2x^{3}y}{(1+(xy)^{2})^{2}}}\right)-\left({\frac {1-(xy)^{2}}{(1+(xy)^{2})^{2}}}\right)^{2}$ $={\frac {4(xy)^{4}-(1-(xy)^{2})^{2}}{(1+(xy)^{2})^{4}}}$ $={\frac {5(xy)^{4}+2(xy)^{2}-1}{(1+(xy)^{2})^{4}}}$

For the critical point $(0,0)$ , the discriminant is $-1$ so $(0,0)$ is a saddle point.

Saddle at (0,0)

262.

f(x,y)=2xye^{-x^{2}-y^{2}}

The first order derivatives are:

{\frac {\partial f}{\partial x}}=2ye^{-x^{2}-y^{2}}+(2xy)e^{-x^{2}-y^{2}}(-2x)=(2y-4x^{2}y)e^{-x^{2}-y^{2}}

and

{\frac {\partial f}{\partial y}}=2xe^{-x^{2}-y^{2}}+(2xy)e^{-x^{2}-y^{2}}(-2y)=(2x-4xy^{2})e^{-x^{2}-y^{2}}

. Finding the critical points is done by solving the equations

\left\{{\begin{array}{c}(2y-4x^{2}y)e^{-x^{2}-y^{2}}=0\\(2x-4xy^{2})e^{-x^{2}-y^{2}}=0\end{array}}\right.

$\left\{{\begin{array}{c}(2y-4x^{2}y)e^{-x^{2}-y^{2}}=0\\(2x-4xy^{2})e^{-x^{2}-y^{2}}=0\end{array}}\right.\iff \left\{{\begin{array}{c}2y(1-2x^{2})=0\\2x(1-2y^{2})=0\end{array}}\right.$ $\iff \left\{{\begin{array}{c}y=0\;{\text{or}}\;x=-1/{\sqrt {2}}\;{\text{or}}\;x=1/{\sqrt {2}}\\x=0\;{\text{or}}\;y=-1/{\sqrt {2}}\;{\text{or}}\;y=1/{\sqrt {2}}\end{array}}\right.$ . The critical points are $(x,y)=(0,0),(-1/{\sqrt {2}},-1/{\sqrt {2}}),(-1/{\sqrt {2}},1/{\sqrt {2}}),(1/{\sqrt {2}},-1/{\sqrt {2}}),(1/{\sqrt {2}},1/{\sqrt {2}})$ .

The second order derivatives are: ${\frac {\partial ^{2}f}{\partial x^{2}}}=(-8xy)e^{-x^{2}-y^{2}}+(2y-4x^{2}y)e^{-x^{2}-y^{2}}(-2x)=(8x^{3}y-12xy)e^{-x^{2}-y^{2}}$ ; ${\frac {\partial ^{2}f}{\partial y^{2}}}=(-8xy)e^{-x^{2}-y^{2}}+(2x-4xy^{2})e^{-x^{2}-y^{2}}(-2y)=(8xy^{3}-12xy)e^{-x^{2}-y^{2}}$ ; and ${\frac {\partial ^{2}f}{\partial y\partial x}}=(2-4x^{2})e^{-x^{2}-y^{2}}+(2y-4x^{2}y)e^{-x^{2}-y^{2}}(-2y)=(8x^{2}y^{2}-4x^{2}-4y^{2}+2)e^{-x^{2}-y^{2}}=2(2x^{2}-1)(2y^{2}-1)e^{-x^{2}-y^{2}}$ . The discriminant is ${\frac {\partial ^{2}f}{\partial x^{2}}}{\frac {\partial ^{2}f}{\partial y^{2}}}-\left({\frac {\partial ^{2}f}{\partial y\partial x}}\right)^{2}=((8x^{3}y-12xy)e^{-x^{2}-y^{2}})((8xy^{3}-12xy)e^{-x^{2}-y^{2}})-(2(2x^{2}-1)(2y^{2}-1)e^{-x^{2}-y^{2}})^{2}$ $=(16x^{2}y^{2}(2x^{2}-3)(2y^{2}-3)-4(2x^{2}-1)^{2}(2y^{2}-1)^{2})e^{-2x^{2}-2y^{2}}$ .

For critical point $(0,0)$ , the discriminant is $-4(-1)(-1)=-4$ so $(0,0)$ is a saddle point.

For critical points $(-1/{\sqrt {2}},-1/{\sqrt {2}})$ and $(1/{\sqrt {2}},1/{\sqrt {2}})$ , the discriminant is $4(-2)(-2)e^{-2}=16e^{-2}$ and ${\frac {\partial ^{2}f}{\partial x^{2}}}=(2-6)e^{-1}=-4e^{-1}$ so $(-1/{\sqrt {2}},-1/{\sqrt {2}})$ and $(1/{\sqrt {2}},1/{\sqrt {2}})$ are local maximums.

For critical points $(-1/{\sqrt {2}},1/{\sqrt {2}})$ and $(1/{\sqrt {2}},-1/{\sqrt {2}})$ , the discriminant is $4(-2)(-2)e^{-2}=16e^{-2}$ and ${\frac {\partial ^{2}f}{\partial x^{2}}}=(-2+6)e^{-1}=4e^{-1}$ so $(-1/{\sqrt {2}},1/{\sqrt {2}})$ and $(1/{\sqrt {2}},-1/{\sqrt {2}})$ are local minimums.

Saddle at (0,0), local maxima at

(\pm 1/{\sqrt {2}},\pm 1/{\sqrt {2}}),

local minima at

(\pm 1/{\sqrt {2}},\mp 1/{\sqrt {2}})

The first order derivatives are:

{\frac {\partial f}{\partial x}}=2ye^{-x^{2}-y^{2}}+(2xy)e^{-x^{2}-y^{2}}(-2x)=(2y-4x^{2}y)e^{-x^{2}-y^{2}}

and

{\frac {\partial f}{\partial y}}=2xe^{-x^{2}-y^{2}}+(2xy)e^{-x^{2}-y^{2}}(-2y)=(2x-4xy^{2})e^{-x^{2}-y^{2}}

. Finding the critical points is done by solving the equations

\left\{{\begin{array}{c}(2y-4x^{2}y)e^{-x^{2}-y^{2}}=0\\(2x-4xy^{2})e^{-x^{2}-y^{2}}=0\end{array}}\right.

$\left\{{\begin{array}{c}(2y-4x^{2}y)e^{-x^{2}-y^{2}}=0\\(2x-4xy^{2})e^{-x^{2}-y^{2}}=0\end{array}}\right.\iff \left\{{\begin{array}{c}2y(1-2x^{2})=0\\2x(1-2y^{2})=0\end{array}}\right.$ $\iff \left\{{\begin{array}{c}y=0\;{\text{or}}\;x=-1/{\sqrt {2}}\;{\text{or}}\;x=1/{\sqrt {2}}\\x=0\;{\text{or}}\;y=-1/{\sqrt {2}}\;{\text{or}}\;y=1/{\sqrt {2}}\end{array}}\right.$ . The critical points are $(x,y)=(0,0),(-1/{\sqrt {2}},-1/{\sqrt {2}}),(-1/{\sqrt {2}},1/{\sqrt {2}}),(1/{\sqrt {2}},-1/{\sqrt {2}}),(1/{\sqrt {2}},1/{\sqrt {2}})$ .

The second order derivatives are: ${\frac {\partial ^{2}f}{\partial x^{2}}}=(-8xy)e^{-x^{2}-y^{2}}+(2y-4x^{2}y)e^{-x^{2}-y^{2}}(-2x)=(8x^{3}y-12xy)e^{-x^{2}-y^{2}}$ ; ${\frac {\partial ^{2}f}{\partial y^{2}}}=(-8xy)e^{-x^{2}-y^{2}}+(2x-4xy^{2})e^{-x^{2}-y^{2}}(-2y)=(8xy^{3}-12xy)e^{-x^{2}-y^{2}}$ ; and ${\frac {\partial ^{2}f}{\partial y\partial x}}=(2-4x^{2})e^{-x^{2}-y^{2}}+(2y-4x^{2}y)e^{-x^{2}-y^{2}}(-2y)=(8x^{2}y^{2}-4x^{2}-4y^{2}+2)e^{-x^{2}-y^{2}}=2(2x^{2}-1)(2y^{2}-1)e^{-x^{2}-y^{2}}$ . The discriminant is ${\frac {\partial ^{2}f}{\partial x^{2}}}{\frac {\partial ^{2}f}{\partial y^{2}}}-\left({\frac {\partial ^{2}f}{\partial y\partial x}}\right)^{2}=((8x^{3}y-12xy)e^{-x^{2}-y^{2}})((8xy^{3}-12xy)e^{-x^{2}-y^{2}})-(2(2x^{2}-1)(2y^{2}-1)e^{-x^{2}-y^{2}})^{2}$ $=(16x^{2}y^{2}(2x^{2}-3)(2y^{2}-3)-4(2x^{2}-1)^{2}(2y^{2}-1)^{2})e^{-2x^{2}-2y^{2}}$ .

For critical point $(0,0)$ , the discriminant is $-4(-1)(-1)=-4$ so $(0,0)$ is a saddle point.

For critical points $(-1/{\sqrt {2}},-1/{\sqrt {2}})$ and $(1/{\sqrt {2}},1/{\sqrt {2}})$ , the discriminant is $4(-2)(-2)e^{-2}=16e^{-2}$ and ${\frac {\partial ^{2}f}{\partial x^{2}}}=(2-6)e^{-1}=-4e^{-1}$ so $(-1/{\sqrt {2}},-1/{\sqrt {2}})$ and $(1/{\sqrt {2}},1/{\sqrt {2}})$ are local maximums.

For critical points $(-1/{\sqrt {2}},1/{\sqrt {2}})$ and $(1/{\sqrt {2}},-1/{\sqrt {2}})$ , the discriminant is $4(-2)(-2)e^{-2}=16e^{-2}$ and ${\frac {\partial ^{2}f}{\partial x^{2}}}=(-2+6)e^{-1}=4e^{-1}$ so $(-1/{\sqrt {2}},1/{\sqrt {2}})$ and $(1/{\sqrt {2}},-1/{\sqrt {2}})$ are local minimums.

Saddle at (0,0), local maxima at

(\pm 1/{\sqrt {2}},\pm 1/{\sqrt {2}}),

local minima at

(\pm 1/{\sqrt {2}},\mp 1/{\sqrt {2}})

Find absolute maximum and minimum values of the function f on the set R.

How to find candidate points for the absolute minimum and maximum of a function with a constrained domain.

Let $f:\mathbb {R} ^{2}\to \mathbb {R}$ denote the function for which the absolute minimum and maximum is sought. Let the domain be constrained to all points $(x,y)$ where $g(x,y)=0$ where $g(x,y)$ is an appropriate function over $\mathbb {R} ^{2}$ .

Start with a candidate point $(x,y)=(x_{0},y_{0})$ where $g(x_{0},y_{0})=0$ , and envision that the $x$ and $y$ coordinates are changing at rates of $m$ and $n$ respectively: ${\frac {dx}{dt}}=m$ and ${\frac {dy}{dt}}=n$ . The rate of change in $f$ is ${\frac {df}{dt}}={\frac {\partial f}{\partial x}}{\frac {dx}{dt}}+{\frac {\partial f}{\partial y}}{\frac {dy}{dt}}$ $={\frac {\partial f}{\partial x}}m+{\frac {\partial f}{\partial y}}n$ . Since it is required that $g(x,y)=0$ , it must be the case that ${\frac {dg}{dt}}=0\iff {\frac {\partial g}{\partial x}}{\frac {dx}{dt}}+{\frac {\partial g}{\partial y}}{\frac {dy}{dt}}=0$ $\iff {\frac {\partial g}{\partial x}}m+{\frac {\partial g}{\partial y}}n=0$ .

$(x,y)=(x_{0},y_{0})$ is a local minimum or maximum only if ${\frac {df}{dt}}={\frac {\partial f}{\partial x}}m+{\frac {\partial f}{\partial y}}n=0$ for all $(m,n)$ where ${\frac {\partial g}{\partial x}}m+{\frac {\partial g}{\partial y}}n=0$ . This occurs iff the gradient $\left\langle {\frac {\partial f}{\partial x}},{\frac {\partial f}{\partial y}}\right\rangle$ is parallel to the gradient $\left\langle {\frac {\partial g}{\partial x}},{\frac {\partial g}{\partial y}}\right\rangle$ . This condition can be quantified by $\left\langle {\frac {\partial f}{\partial x}},{\frac {\partial f}{\partial y}}\right\rangle =\lambda \left\langle {\frac {\partial g}{\partial x}},{\frac {\partial g}{\partial y}}\right\rangle$ where factor $\lambda$ is a "Lagrange multiplier".

Points $(x,y)$ where $g(x,y)=0$ and $\left\langle {\frac {\partial f}{\partial x}},{\frac {\partial f}{\partial y}}\right\rangle =\lambda \left\langle {\frac {\partial g}{\partial x}},{\frac {\partial g}{\partial y}}\right\rangle$ for some $\lambda$ are candidate points for the absolute minimum or maximum. If the domain has any corners, then these corners are also candidate points.

263.

f(x,y)=x^{2}+y^{2}-2y+1,\ R=\{(x,y)\mid x^{2}+y^{2}\leq 4\}

Candidate points will be derived from two sources: Critical points of the function

f(x,y)=x^{2}+y^{2}-2y+1

assuming an unconstrained domain, and candidate points assuming the restriction

g(x,y)=x^{2}+y^{2}-4=0

.

The first order derivatives of $f(x,y)$ are ${\frac {\partial f}{\partial x}}=2x$ and ${\frac {\partial f}{\partial y}}=2y-2$ , so the only critical point where ${\frac {\partial f}{\partial x}}={\frac {\partial f}{\partial y}}=0$ occurs at $(x,y)=(0,1)$ . This critical point lies in $R$ so it remains a valid candidate. $f(0,1)=0$ .

The first order derivatives of $g(x,y)$ are ${\frac {\partial g}{\partial x}}=2x$ and ${\frac {\partial g}{\partial y}}=2y$ . Candidate points assuming the restriction $g(x,y)=0$ must satisfy $\left\{{\begin{array}{c}x^{2}+y^{2}-4=0\\2x=\lambda (2x)\\2y-2=\lambda (2y)\end{array}}\right.$ for some $\lambda$ . These equations are equivalent to $\left\{{\begin{array}{c}x^{2}+y^{2}-4=0\\x=0\;{\text{or}}\;\lambda =1\\\lambda =(y-1)/y\end{array}}\right.$ . If $x=0$ , then the only restriction left on $y$ is $y^{2}-4=0$ and $y\neq 0$ . This gives two candidate points $(0,-2),(0,2)$ . If $\lambda =1$ , then $1=(y-1)/y\iff y=y-1\iff 0=-1$ which is never true. Hence the only valid candidate points derived by restricting the domain to $g(x,y)=x^{2}+y^{2}-4=0$ are $(0,-2),(0,2)$ . $f(0,-2)=9$ and $f(0,2)=1$ .

In total, the candidates are $f(0,1)=0$ , $f(0,-2)=9$ , and $f(0,2)=1$ .

The absolute minimum of $0$ occurs at $(0,1)$ , and the absolute maximum of $9$ occurs at $(0,-2)$ .

Maximum of 9 at (0,−2) and minimum of 0 at (0,1)

Candidate points will be derived from two sources: Critical points of the function

f(x,y)=x^{2}+y^{2}-2y+1

assuming an unconstrained domain, and candidate points assuming the restriction

g(x,y)=x^{2}+y^{2}-4=0

.

The first order derivatives of $f(x,y)$ are ${\frac {\partial f}{\partial x}}=2x$ and ${\frac {\partial f}{\partial y}}=2y-2$ , so the only critical point where ${\frac {\partial f}{\partial x}}={\frac {\partial f}{\partial y}}=0$ occurs at $(x,y)=(0,1)$ . This critical point lies in $R$ so it remains a valid candidate. $f(0,1)=0$ .

The first order derivatives of $g(x,y)$ are ${\frac {\partial g}{\partial x}}=2x$ and ${\frac {\partial g}{\partial y}}=2y$ . Candidate points assuming the restriction $g(x,y)=0$ must satisfy $\left\{{\begin{array}{c}x^{2}+y^{2}-4=0\\2x=\lambda (2x)\\2y-2=\lambda (2y)\end{array}}\right.$ for some $\lambda$ . These equations are equivalent to $\left\{{\begin{array}{c}x^{2}+y^{2}-4=0\\x=0\;{\text{or}}\;\lambda =1\\\lambda =(y-1)/y\end{array}}\right.$ . If $x=0$ , then the only restriction left on $y$ is $y^{2}-4=0$ and $y\neq 0$ . This gives two candidate points $(0,-2),(0,2)$ . If $\lambda =1$ , then $1=(y-1)/y\iff y=y-1\iff 0=-1$ which is never true. Hence the only valid candidate points derived by restricting the domain to $g(x,y)=x^{2}+y^{2}-4=0$ are $(0,-2),(0,2)$ . $f(0,-2)=9$ and $f(0,2)=1$ .

In total, the candidates are $f(0,1)=0$ , $f(0,-2)=9$ , and $f(0,2)=1$ .

The absolute minimum of $0$ occurs at $(0,1)$ , and the absolute maximum of $9$ occurs at $(0,-2)$ .

Maximum of 9 at (0,−2) and minimum of 0 at (0,1)

264.

f(x,y)=x^{2}+y^{2}-2x-2y,

R is a closed triangle with vertices (0,0), (2,0), and (0,2).

Triangle

R

is defined by the constraints

x\geq 0

,

y\geq 0

, and

x+y\leq 2

.

Candidate points for the absolute minimum and maximum will be derived from 5 sources:

Critical points of the function $f(x,y)=x^{2}+y^{2}-2x-2y$ assuming an unconstrained domain.
Candidate points assuming the restriction $g_{1}(x,y)=x=0$ .
Candidate points assuming the restriction $g_{2}(x,y)=y=0$ .
Candidate points assuming the restriction $g_{3}(x,y)=x+y-2=0$ .
The vertex points $(0,0),(2,0),(0,2)$ .

The first order derivatives of $f(x,y)$ are ${\frac {\partial f}{\partial x}}=2x-2$ and ${\frac {\partial f}{\partial y}}=2y-2$ , so the only critical point where ${\frac {\partial f}{\partial x}}={\frac {\partial f}{\partial y}}=0$ is $(x,y)=(1,1)$ . The critical point $(1,1)$ lies in $R$ so it remains a valid candidate. $f(1,1)=-2$ .

The first order derivatives of $g_{1}(x,y)$ are ${\frac {\partial g_{1}}{\partial x}}=1$ and ${\frac {\partial g_{1}}{\partial y}}=0$ . Candidate points assuming the restriction $g_{1}(x,y)=0$ must satisfy $\left\{{\begin{array}{c}x=0\\2x-2=\lambda (1)\\2y-2=\lambda (0)\end{array}}\right.$ for some $\lambda$ . These equations are equivalent to $\left\{{\begin{array}{c}x=0\\\lambda =-2\\y=1\end{array}}\right.$ . This yields the candidate point $(0,1)$ . Point $(0,1)$ lies in $R$ so it remains a valid candidate. $f(0,1)=-1$ .

The first order derivatives of $g_{2}(x,y)$ are ${\frac {\partial g_{2}}{\partial x}}=0$ and ${\frac {\partial g_{2}}{\partial y}}=1$ . Candidate points assuming the restriction $g_{2}(x,y)=0$ must satisfy $\left\{{\begin{array}{c}y=0\\2x-2=\lambda (0)\\2y-2=\lambda (1)\end{array}}\right.$ for some $\lambda$ . These equations are equivalent to $\left\{{\begin{array}{c}y=0\\x=1\\\lambda =-2\end{array}}\right.$ . This yields the candidate point $(1,0)$ . Point $(1,0)$ lies in $R$ so it remains a valid candidate. $f(1,0)=-1$ .

The first order derivatives of $g_{3}(x,y)$ are ${\frac {\partial g_{3}}{\partial x}}=1$ and ${\frac {\partial g_{3}}{\partial y}}=1$ . Candidate points assuming the restriction $g_{3}(x,y)=0$ must satisfy $\left\{{\begin{array}{c}x+y-2=0\\2x-2=\lambda (1)\\2y-2=\lambda (1)\end{array}}\right.$ for some $\lambda$ . The second equation yields $\lambda =2x-2$ . Substituting $2x-2$ in place of $\lambda$ in the third equation gives $2y-2=2x-2\iff y=x$ . Substituting $x$ in place of $y$ in the first equation gives $x+x-2=0\iff x=1$ , which then yields $y=1$ and $\lambda =0$ . This yields the candidate point $(1,1)$ . Point $(1,1)$ lies in $R$ so it remains a valid candidate. $f(1,1)=-2$ .

Finally, we add the vertices $(0,0),(2,0),(0,2)$ to the lineup of candidate points.

Evaluating $f(x,y)=x^{2}+y^{2}-2x-2y$ at each candidate point gives $f(1,1)=-2$ ; $f(0,1)=-1$ ; $f(1,0)=-1$ ; $f(0,0)=0$ ; $f(2,0)=0$ ; and $f(0,2)=0$ . The absolute minimum of $-2$ occurs at $(1,1)$ , while the absolute maximum of $0$ occurs at all of $(0,0),(2,0),(0,2)$ .

Maximum of 0 at (2,0), (0,2), and (0,0); minimum of −2 at (1,1)

Triangle

R

is defined by the constraints

x\geq 0

,

y\geq 0

, and

x+y\leq 2

.

Candidate points for the absolute minimum and maximum will be derived from 5 sources:

Critical points of the function $f(x,y)=x^{2}+y^{2}-2x-2y$ assuming an unconstrained domain.
Candidate points assuming the restriction $g_{1}(x,y)=x=0$ .
Candidate points assuming the restriction $g_{2}(x,y)=y=0$ .
Candidate points assuming the restriction $g_{3}(x,y)=x+y-2=0$ .
The vertex points $(0,0),(2,0),(0,2)$ .

The first order derivatives of $f(x,y)$ are ${\frac {\partial f}{\partial x}}=2x-2$ and ${\frac {\partial f}{\partial y}}=2y-2$ , so the only critical point where ${\frac {\partial f}{\partial x}}={\frac {\partial f}{\partial y}}=0$ is $(x,y)=(1,1)$ . The critical point $(1,1)$ lies in $R$ so it remains a valid candidate. $f(1,1)=-2$ .

The first order derivatives of $g_{1}(x,y)$ are ${\frac {\partial g_{1}}{\partial x}}=1$ and ${\frac {\partial g_{1}}{\partial y}}=0$ . Candidate points assuming the restriction $g_{1}(x,y)=0$ must satisfy $\left\{{\begin{array}{c}x=0\\2x-2=\lambda (1)\\2y-2=\lambda (0)\end{array}}\right.$ for some $\lambda$ . These equations are equivalent to $\left\{{\begin{array}{c}x=0\\\lambda =-2\\y=1\end{array}}\right.$ . This yields the candidate point $(0,1)$ . Point $(0,1)$ lies in $R$ so it remains a valid candidate. $f(0,1)=-1$ .

The first order derivatives of $g_{2}(x,y)$ are ${\frac {\partial g_{2}}{\partial x}}=0$ and ${\frac {\partial g_{2}}{\partial y}}=1$ . Candidate points assuming the restriction $g_{2}(x,y)=0$ must satisfy $\left\{{\begin{array}{c}y=0\\2x-2=\lambda (0)\\2y-2=\lambda (1)\end{array}}\right.$ for some $\lambda$ . These equations are equivalent to $\left\{{\begin{array}{c}y=0\\x=1\\\lambda =-2\end{array}}\right.$ . This yields the candidate point $(1,0)$ . Point $(1,0)$ lies in $R$ so it remains a valid candidate. $f(1,0)=-1$ .

The first order derivatives of $g_{3}(x,y)$ are ${\frac {\partial g_{3}}{\partial x}}=1$ and ${\frac {\partial g_{3}}{\partial y}}=1$ . Candidate points assuming the restriction $g_{3}(x,y)=0$ must satisfy $\left\{{\begin{array}{c}x+y-2=0\\2x-2=\lambda (1)\\2y-2=\lambda (1)\end{array}}\right.$ for some $\lambda$ . The second equation yields $\lambda =2x-2$ . Substituting $2x-2$ in place of $\lambda$ in the third equation gives $2y-2=2x-2\iff y=x$ . Substituting $x$ in place of $y$ in the first equation gives $x+x-2=0\iff x=1$ , which then yields $y=1$ and $\lambda =0$ . This yields the candidate point $(1,1)$ . Point $(1,1)$ lies in $R$ so it remains a valid candidate. $f(1,1)=-2$ .

Finally, we add the vertices $(0,0),(2,0),(0,2)$ to the lineup of candidate points.

Evaluating $f(x,y)=x^{2}+y^{2}-2x-2y$ at each candidate point gives $f(1,1)=-2$ ; $f(0,1)=-1$ ; $f(1,0)=-1$ ; $f(0,0)=0$ ; $f(2,0)=0$ ; and $f(0,2)=0$ . The absolute minimum of $-2$ occurs at $(1,1)$ , while the absolute maximum of $0$ occurs at all of $(0,0),(2,0),(0,2)$ .

Maximum of 0 at (2,0), (0,2), and (0,0); minimum of −2 at (1,1)

Finding the locations and shortest distances between two surfaces.

Consider two surfaces $\sigma _{1}$ and $\sigma _{2}$ in 3D space defined by the equations $f_{1}(x,y,z)=0$ and $f_{2}(x,y,z)=0$ respectively. Given a point $(x_{1},y_{1},z_{1})$ from $\sigma _{1}$ and a point $(x_{2},y_{2},z_{2})$ from $\sigma _{2}$ , if $(x_{1},y_{1},z_{1})$ and $(x_{2},y_{2},z_{2})$ are the points that minimize the distance between $\sigma _{1}$ and $\sigma _{2}$ , then it must be the case that the displacement $\langle x_{2}-x_{1},y_{2}-y_{1},z_{2}-z_{1}\rangle$ is perpendicular to both surfaces. The gradient vector $\left\langle {\frac {\partial f_{1}}{\partial x}},{\frac {\partial f_{1}}{\partial y}},{\frac {\partial f_{1}}{\partial z}}\right\rangle$ is orthogonal to $\sigma _{1}$ , and the gradient vector $\left\langle {\frac {\partial f_{2}}{\partial x}},{\frac {\partial f_{2}}{\partial y}},{\frac {\partial f_{2}}{\partial z}}\right\rangle$ is orthogonal to $\sigma _{2}$ . The displacement vector must be parallel to both gradient vectors: $\langle x_{2}-x_{1},y_{2}-y_{1},z_{2}-z_{1}\rangle =\lambda _{1}\left\langle {\frac {\partial f_{1}}{\partial x}},{\frac {\partial f_{1}}{\partial y}},{\frac {\partial f_{1}}{\partial z}}\right\rangle =\lambda _{2}\left\langle {\frac {\partial f_{2}}{\partial x}},{\frac {\partial f_{2}}{\partial y}},{\frac {\partial f_{2}}{\partial z}}\right\rangle$ for some $\lambda _{1}$ and $\lambda _{2}$ .

Candidate points for the shortest distance between two surfaces must satisfy the following 8 equations: $\left\{{\begin{array}{c}f_{1}(x_{1},y_{1},z_{1})=0\\f_{2}(x_{2},y_{2},z_{2})=0\\x_{2}-x_{1}=\lambda _{1}{\frac {\partial f_{1}}{\partial x}}\;{\text{and}}\;y_{2}-y_{1}=\lambda _{1}{\frac {\partial f_{1}}{\partial y}}\;{\text{and}}\;z_{2}-z_{1}=\lambda _{1}{\frac {\partial f_{1}}{\partial z}}\\x_{2}-x_{1}=\lambda _{2}{\frac {\partial f_{2}}{\partial x}}\;{\text{and}}\;y_{2}-y_{1}=\lambda _{2}{\frac {\partial f_{2}}{\partial y}}\;{\text{and}}\;z_{2}-z_{1}=\lambda _{2}{\frac {\partial f_{2}}{\partial z}}\end{array}}\right.$ for some $\lambda _{1}$ and $\lambda _{2}$ .

265. Find the point on the plane x−y+z=2 closest to the point (1,1,1).

The plane is defined by the equation

f(x,y,z)=x-y+z-2=0

, and a normal vector to the plane is

\left\langle {\frac {\partial f}{\partial x}},{\frac {\partial f}{\partial y}},{\frac {\partial f}{\partial z}}\right\rangle =\langle 1,-1,1\rangle

. The closest point

(x,y,z)

on the plane to the point

(1,1,1)

is the point where the displacement

\langle x-1,y-1,z-1\rangle

is parallel to

\langle 1,-1,1\rangle

. The following equations must be satisfied:

\left\{{\begin{array}{c}x-y+z-2=0\\x-1=\lambda (1)\;{\text{and}}\;y-1=\lambda (-1)\;{\text{and}}\;z-1=\lambda (1)\end{array}}\right.

.

The second equation gives $\lambda =x-1$ , and replacing $\lambda$ with $x-1$ in the third and forth equations gives $y-1=1-x\iff y=2-x$ and $z-1=x-1\iff z=x$ respectively. In the first equation, replacing $y$ with $2-x$ and $z$ with $x$ gives $x-(2-x)+x-2=0\iff 3x-4=0\iff x=4/3$ . This gives in turn $y=2/3$ and $z=4/3$ .

The only candidate point for the closest distance is

(4/3,2/3,4/3)

, so therefore the point on the plane

x-y+z=2

that is closest to the point

(1,1,1)

is

(4/3,2/3,4/3)

.

The plane is defined by the equation

f(x,y,z)=x-y+z-2=0

, and a normal vector to the plane is

\left\langle {\frac {\partial f}{\partial x}},{\frac {\partial f}{\partial y}},{\frac {\partial f}{\partial z}}\right\rangle =\langle 1,-1,1\rangle

. The closest point

(x,y,z)

on the plane to the point

(1,1,1)

is the point where the displacement

\langle x-1,y-1,z-1\rangle

is parallel to

\langle 1,-1,1\rangle

. The following equations must be satisfied:

\left\{{\begin{array}{c}x-y+z-2=0\\x-1=\lambda (1)\;{\text{and}}\;y-1=\lambda (-1)\;{\text{and}}\;z-1=\lambda (1)\end{array}}\right.

.

The second equation gives $\lambda =x-1$ , and replacing $\lambda$ with $x-1$ in the third and forth equations gives $y-1=1-x\iff y=2-x$ and $z-1=x-1\iff z=x$ respectively. In the first equation, replacing $y$ with $2-x$ and $z$ with $x$ gives $x-(2-x)+x-2=0\iff 3x-4=0\iff x=4/3$ . This gives in turn $y=2/3$ and $z=4/3$ .

The only candidate point for the closest distance is

(4/3,2/3,4/3)

, so therefore the point on the plane

x-y+z=2

that is closest to the point

(1,1,1)

is

(4/3,2/3,4/3)

.

266. Find the point on the surface

z=x^{2}+y^{2}+10

closest to the plane

x+2y-z=0.

The surface is defined by the equation

f_{1}(x,y,z)=x^{2}+y^{2}-z+10=0

, and the plane is defined by the equation

f_{2}(x,y,z)=x+2y-z=0

. Given a point

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

from the surface and a point

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

from the plane, these two points are closest to each other only if (but not always if) the displacement vector

\langle x_{2}-x_{1},y_{2}-y_{1},z_{2}-z_{1}\rangle

is a parallel to the surface normal vector

\left\langle {\frac {\partial f_{1}}{\partial x}},{\frac {\partial f_{1}}{\partial y}},{\frac {\partial f_{1}}{\partial z}}\right\rangle =\langle 2x_{1},2y_{1},-1\rangle

and the plane normal vector

\left\langle {\frac {\partial f_{2}}{\partial x}},{\frac {\partial f_{2}}{\partial y}},{\frac {\partial f_{2}}{\partial z}}\right\rangle =\langle 1,2,-1\rangle

. There must exist factors

\lambda _{1}

and

\lambda _{2}

such that the following 8 equations hold:

\left\{{\begin{array}{c}x_{1}^{2}+y_{1}^{2}-z_{1}+10=0\\x_{2}+2y_{2}-z_{2}=0\\x_{2}-x_{1}=\lambda _{1}(2x_{1})\;{\text{and}}\;y_{2}-y_{1}=\lambda _{1}(2y_{1})\;{\text{and}}\;z_{2}-z_{1}=\lambda _{1}(-1)\\x_{2}-x_{1}=\lambda _{2}(1)\;{\text{and}}\;y_{2}-y_{1}=\lambda _{2}(2)\;{\text{and}}\;z_{2}-z_{1}=\lambda _{2}(-1)\end{array}}\right.

The fifth equation is equivalent to $\lambda _{1}=-(z_{2}-z_{1})$ , and the eighth equation is equivalent to $\lambda _{2}=-(z_{2}-z_{1})$ . Eliminating via substitution $\lambda _{1}$ and $\lambda _{2}$ in all of the other equations gives: $\left\{{\begin{array}{c}x_{1}^{2}+y_{1}^{2}-z_{1}+10=0\\x_{2}+2y_{2}-z_{2}=0\\x_{2}-x_{1}=-2x_{1}(z_{2}-z_{1})\;{\text{and}}\;y_{2}-y_{1}=-2y_{1}(z_{2}-z_{1})\\x_{2}-x_{1}=-(z_{2}-z_{1})\;{\text{and}}\;y_{2}-y_{1}=-2(z_{2}-z_{1})\end{array}}\right.$

If $z_{2}-z_{1}=0$ , then $x_{2}-x_{1}=0$ and $y_{2}-y_{1}=0$ , and then $(x_{1},y_{1},z_{1})$ and $(x_{2},y_{2},z_{2})$ are the same point which corresponds to an intersection between the surface and the plane. While it will not be shown here, it is relatively simple to show that the surface and plane fail to intersect. Excluding the possibility that $z_{2}-z_{1}=0$ , the equations $x_{2}-x_{1}=-2x_{1}(z_{2}-z_{1})$ and $x_{2}-x_{1}=-(z_{2}-z_{1})$ together imply that $-2x_{1}=-1\iff x_{1}=1/2$ ; and the equations $y_{2}-y_{1}=-2y_{1}(z_{2}-z_{1})$ and $y_{2}-y_{1}=-2(z_{2}-z_{1})$ together imply that $-2y_{1}=-2\iff y_{1}=1$ . The values $x_{1}=1/2$ and $y_{1}=1$ give $z_{1}=x_{1}^{2}+y_{1}^{2}+10=(1/2)^{2}+1^{2}+10=11.25$ . Hence the point on the surface $z=x^{2}+y^{2}+10$ that is closest to the plane $x+2y-z=0$ is $(x_{1},y_{1},z_{1})=(0.5,1,11.25)$ .

From $x_{2}-x_{1}=-(z_{2}-z_{1})$ and $y_{2}-y_{1}=-2(z_{2}-z_{1})$ , it follows that $x_{2}-0.5=-(z_{2}-11.25)\iff x_{2}=11.75-z_{2}$ and $y_{2}-1=-2(z_{2}-11.25)\iff y_{2}=23.5-2z_{2}$ . In the equation $x_{2}+2y_{2}-z_{2}=0$ , eliminating $x_{2}$ and $y_{2}$ via substitution gives $(11.75-z_{2})+2(23.5-2z_{2})-z_{2}=0\iff 6z_{2}=58.75\iff z_{2}=9.791{\overline {6}}$ . Hence $x_{2}=1.958{\overline {3}}$ and $y_{2}=3.91{\overline {6}}$ . The corresponding closest point on the plane is $(x_{2},y_{2},z_{2})=(1.958{\overline {3}},3.91{\overline {6}},9.791{\overline {6}})$ .

The closest point on the surface is

(0.5,1,11.25)

.

The surface is defined by the equation

f_{1}(x,y,z)=x^{2}+y^{2}-z+10=0

, and the plane is defined by the equation

f_{2}(x,y,z)=x+2y-z=0

. Given a point

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

from the surface and a point

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

from the plane, these two points are closest to each other only if (but not always if) the displacement vector

\langle x_{2}-x_{1},y_{2}-y_{1},z_{2}-z_{1}\rangle

is a parallel to the surface normal vector

\left\langle {\frac {\partial f_{1}}{\partial x}},{\frac {\partial f_{1}}{\partial y}},{\frac {\partial f_{1}}{\partial z}}\right\rangle =\langle 2x_{1},2y_{1},-1\rangle

and the plane normal vector

\left\langle {\frac {\partial f_{2}}{\partial x}},{\frac {\partial f_{2}}{\partial y}},{\frac {\partial f_{2}}{\partial z}}\right\rangle =\langle 1,2,-1\rangle

. There must exist factors

\lambda _{1}

and

\lambda _{2}

such that the following 8 equations hold:

\left\{{\begin{array}{c}x_{1}^{2}+y_{1}^{2}-z_{1}+10=0\\x_{2}+2y_{2}-z_{2}=0\\x_{2}-x_{1}=\lambda _{1}(2x_{1})\;{\text{and}}\;y_{2}-y_{1}=\lambda _{1}(2y_{1})\;{\text{and}}\;z_{2}-z_{1}=\lambda _{1}(-1)\\x_{2}-x_{1}=\lambda _{2}(1)\;{\text{and}}\;y_{2}-y_{1}=\lambda _{2}(2)\;{\text{and}}\;z_{2}-z_{1}=\lambda _{2}(-1)\end{array}}\right.

The fifth equation is equivalent to $\lambda _{1}=-(z_{2}-z_{1})$ , and the eighth equation is equivalent to $\lambda _{2}=-(z_{2}-z_{1})$ . Eliminating via substitution $\lambda _{1}$ and $\lambda _{2}$ in all of the other equations gives: $\left\{{\begin{array}{c}x_{1}^{2}+y_{1}^{2}-z_{1}+10=0\\x_{2}+2y_{2}-z_{2}=0\\x_{2}-x_{1}=-2x_{1}(z_{2}-z_{1})\;{\text{and}}\;y_{2}-y_{1}=-2y_{1}(z_{2}-z_{1})\\x_{2}-x_{1}=-(z_{2}-z_{1})\;{\text{and}}\;y_{2}-y_{1}=-2(z_{2}-z_{1})\end{array}}\right.$

If $z_{2}-z_{1}=0$ , then $x_{2}-x_{1}=0$ and $y_{2}-y_{1}=0$ , and then $(x_{1},y_{1},z_{1})$ and $(x_{2},y_{2},z_{2})$ are the same point which corresponds to an intersection between the surface and the plane. While it will not be shown here, it is relatively simple to show that the surface and plane fail to intersect. Excluding the possibility that $z_{2}-z_{1}=0$ , the equations $x_{2}-x_{1}=-2x_{1}(z_{2}-z_{1})$ and $x_{2}-x_{1}=-(z_{2}-z_{1})$ together imply that $-2x_{1}=-1\iff x_{1}=1/2$ ; and the equations $y_{2}-y_{1}=-2y_{1}(z_{2}-z_{1})$ and $y_{2}-y_{1}=-2(z_{2}-z_{1})$ together imply that $-2y_{1}=-2\iff y_{1}=1$ . The values $x_{1}=1/2$ and $y_{1}=1$ give $z_{1}=x_{1}^{2}+y_{1}^{2}+10=(1/2)^{2}+1^{2}+10=11.25$ . Hence the point on the surface $z=x^{2}+y^{2}+10$ that is closest to the plane $x+2y-z=0$ is $(x_{1},y_{1},z_{1})=(0.5,1,11.25)$ .

From $x_{2}-x_{1}=-(z_{2}-z_{1})$ and $y_{2}-y_{1}=-2(z_{2}-z_{1})$ , it follows that $x_{2}-0.5=-(z_{2}-11.25)\iff x_{2}=11.75-z_{2}$ and $y_{2}-1=-2(z_{2}-11.25)\iff y_{2}=23.5-2z_{2}$ . In the equation $x_{2}+2y_{2}-z_{2}=0$ , eliminating $x_{2}$ and $y_{2}$ via substitution gives $(11.75-z_{2})+2(23.5-2z_{2})-z_{2}=0\iff 6z_{2}=58.75\iff z_{2}=9.791{\overline {6}}$ . Hence $x_{2}=1.958{\overline {3}}$ and $y_{2}=3.91{\overline {6}}$ . The corresponding closest point on the plane is $(x_{2},y_{2},z_{2})=(1.958{\overline {3}},3.91{\overline {6}},9.791{\overline {6}})$ .

The closest point on the surface is

(0.5,1,11.25)

.

Double Integrals over Rectangular Regions

Evaluate the given integral over the region R.

280.

\displaystyle \iint _{R}(x^{2}+xy)dA,\ R=\{(x,y)\mid x\in [1,2],\ y\in [-1,1]\}

14/3

14/3

281.

\displaystyle \iint _{R}(xy\sin(x^{2}))dA,\ R=\{(x,y)\mid x\in [0,{\sqrt {\pi /2}}],\ y\in [0,1]\}

1/4

1/4

282.

\displaystyle \iint _{R}{\frac {x}{(1+xy)^{2}}}dA,\ R=\{(x,y)\mid x\in [0,4],\ y\in [1,2]\}

\ln(5/3)

\ln(5/3)

Evaluate the given iterated integrals.

283.

\displaystyle \int _{0}^{2}\int _{0}^{1}x^{5}y^{2}e^{x^{3}y^{3}}dydx

(e^{8}-9)/9

(e^{8}-9)/9

284.

\displaystyle \int _{1}^{4}\int _{0}^{2}e^{y{\sqrt {x}}}dydx

e^{4}-e^{2}-2

e^{4}-e^{2}-2

Double Integrals over General Regions

Evaluate the following integrals.

300.

\displaystyle \iint _{R}xydA,

R is bounded by x=0, y=2x+1, and y=5−2x.

2

2

301.

\displaystyle \iint _{R}(x+y)dA,

R is in the first quadrant and bounded by x=0,

y=x^{2},

and

y=8-x^{2}.

152/3

152/3

Use double integrals to compute the volume of the given region.

302. The solid in the first octant bound by the coordinate planes and the surface

z=8-x^{2}-2y^{2}.

4\pi {\sqrt {2}}

4\pi {\sqrt {2}}

303. The solid beneath the cylinder

z=y^{2}

and above the region

R=\{(x,y)\mid y\in [0,1],\ x\in [y,1]\}.

1/12

1/12

304. The solid bounded by the paraboloids

z=x^{2}+y^{2}

and

z=50-x^{2}-y^{2}.

625\pi

625\pi

Double Integrals in Polar Coordinates

320. Evaluate

\displaystyle \iint _{R}2xydA

for

R=\{(r,\theta )\mid r\in [1,3],\ \theta \in [0,\pi /2]\}

20

20

321. Find the average value of the function

f(r,\theta )=1/r^{2}

over the region

\{(r,\theta )\mid r\in [2,4]\}.

\displaystyle {\frac {\ln 2}{6}}

\displaystyle {\frac {\ln 2}{6}}

322. Evaluate

\displaystyle \int _{0}^{3}\int _{0}^{\sqrt {9-x^{2}}}{\sqrt {x^{2}+y^{2}}}dydx.

9\pi /2

9\pi /2

323. Evaluate

\displaystyle \iint _{R}{\frac {x-y}{x^{2}+y^{2}+1}}dA

if R is the unit disk centered at the origin.

0

0

Triple Integrals

340. Evaluate

\displaystyle \int _{1}^{\ln 8}\int _{0}^{\ln 4}\int _{0}^{\ln 2}e^{-x-y-2z}dxdydz.

\displaystyle {\frac {3}{16}}\left(e^{-2}-8^{-2}\right)

\displaystyle {\frac {3}{16}}\left(e^{-2}-8^{-2}\right)

In the following exercises, sketching the region of integration may be helpful.

341. Find the volume of the solid in the first octant bounded by the plane 2x+3y+6z=12 and the coordinate planes.

8

8

342. Find the volume of the solid in the first octant bounded by the cylinder

z=\sin(y)

for

y\in [0,\pi ]

, and the planes y=x and x=0.

\pi

\pi

343. Evaluate

\displaystyle \int _{0}^{1}\int _{y}^{2-y}\int _{0}^{2-x-y}xydzdxdy.

2/15

2/15

344. Rewrite the integral

\displaystyle \int _{0}^{1}\int _{-2}^{2}\int _{0}^{\sqrt {4-y^{2}}}dzdydx

in the order dydzdx.

\displaystyle \int _{0}^{1}\int _{0}^{2}\int _{-{\sqrt {4-z^{2}}}}^{\sqrt {4-z^{2}}}dydzdx

\displaystyle \int _{0}^{1}\int _{0}^{2}\int _{-{\sqrt {4-z^{2}}}}^{\sqrt {4-z^{2}}}dydzdx

Cylindrical And Spherical Coordinates

360. Evaluate the integral in cylindrical coordinates:

\displaystyle \int _{0}^{3}\int _{0}^{\sqrt {9-x^{2}}}\int _{0}^{\sqrt {x^{2}+y^{2}}}{\frac {1}{\sqrt {x^{2}+y^{2}}}}dzdydx

9\pi /4

9\pi /4

361. Find the mass of the solid cylinder

D=\{(r,\theta ,z)\mid r\in [0,3],\ z\in [0,2]\}

given the density function

\delta (r,\theta ,z)=5e^{-r^{2}}

10\pi (1-e^{-9})

10\pi (1-e^{-9})

362. Use a triple integral to find the volume of the region bounded by the plane z=0 and the hyperboloid

z={\sqrt {17}}-{\sqrt {1+x^{2}+y^{2}}}

\displaystyle {\frac {2\pi (1+7{\sqrt {17}})}{3}}

\displaystyle {\frac {2\pi (1+7{\sqrt {17}})}{3}}

363. If D is a unit ball, use a triple integral in spherical coordinates to evaluate

\iiint _{D}(x^{2}+y^{2}+z^{2})^{5/2}dV

\pi /2

\pi /2

364. Find the mass of a solid cone

\{(\rho ,\phi ,\theta )\mid \phi \leq \pi /3,\ z\in [0,4]\}

if the density function is

\delta (\rho ,\phi ,\theta )=5-z

128\pi

128\pi

365. Find the volume of the region common to two cylinders:

x^{2}+z^{2}=1,\ y^{2}+z^{2}=1

16/3

16/3

Center of Mass and Centroid

380. Find the center of mass for three particles located in space at (1,2,3), (0,0,1), and (1,1,0), with masses 2, 1, and 1 respectively.

{\frac {\langle 3,5,7\rangle }{4}}

{\frac {\langle 3,5,7\rangle }{4}}

381. Find the center of mass for a piece of wire with the density

\rho (x)=1+\sin(x)

for

x\in [0,\pi ].

\pi /2

\pi /2

382. Find the center of mass for a piece of wire with the density

\rho (x)=2-x^{2}/16

for

x\in [0,4].

9/5

9/5

383. Find the centroid of the region in the first quadrant bounded by the coordinate axes and

x^{2}+y^{2}=16.

\left({\frac {16}{3\pi }},{\frac {16}{3\pi }}\right)

\left({\frac {16}{3\pi }},{\frac {16}{3\pi }}\right)

384. Find the centroid of the region in the first quadrant bounded by

y=\ln(x)

,

y=0

, and

x=e

.

((e^{2}+1)/4,e/2-1)

((e^{2}+1)/4,e/2-1)

385. Find the center of mass for the region

\{(x,y)\mid x\in [0,4],y\in [0,2]\}

, with the density

\rho (x,y)=1+x/2.

(7/3,1)

(7/3,1)

386. Find the center of mass for the triangular plate with vertices (0,0), (0,4), and (4,0), with density

\rho (x,y)=1+x+y.

(16/11,16/11)

(16/11,16/11)

Vector Fields

One can sketch two-dimensional vector fields by plotting vector values, flow curves, and/or equipotential curves.

401. Find and sketch the gradient field

\mathbf {F} =\nabla \phi

for the potential function

\phi (x,y)={\sqrt {x^{2}+y^{2}}}

.

\mathbf {F} =\left\langle {\frac {x}{\sqrt {x^{2}+y^{2}}}},{\frac {y}{\sqrt {x^{2}+y^{2}}}}\right\rangle

\mathbf {F} =\left\langle {\frac {x}{\sqrt {x^{2}+y^{2}}}},{\frac {y}{\sqrt {x^{2}+y^{2}}}}\right\rangle

402. Find and sketch the gradient field

\mathbf {F} =\nabla \phi

for the potential function

\phi (x,y)=\sin(x)\sin(y)

for

|x|\leq \pi

and

|y|\leq \pi

.

\nabla \phi (x,y)=\langle \cos(x)\sin(y),\sin(x)\cos(y)\rangle

\nabla \phi (x,y)=\langle \cos(x)\sin(y),\sin(x)\cos(y)\rangle

403. Find the gradient field

\mathbf {F} =\nabla \phi

for the potential function

\phi (x,y,z)=e^{-z}\sin(x+y)

\mathbf {F} =e^{-z}\left\langle \cos(x+y),\cos(x+y),-\sin(x+y)\right\rangle

\mathbf {F} =e^{-z}\left\langle \cos(x+y),\cos(x+y),-\sin(x+y)\right\rangle

Line Integrals

420. Evaluate

\int _{C}(x^{2}+y^{2})ds

if C is the line segment from (0,0) to (5,5)

{\frac {250{\sqrt {2}}}{3}}

{\frac {250{\sqrt {2}}}{3}}

421. Evaluate

\int _{C}(x^{2}+y^{2})ds

if C is the circle of radius 4 centered at the origin

128\pi

128\pi

422. Evaluate

\int _{C}(y-z)ds

if C is the helix

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

-2{\sqrt {10}}\pi ^{2}

-2{\sqrt {10}}\pi ^{2}

423. Evaluate

\int _{C}\mathbf {F} \cdot d\mathbf {r}

if

\mathbf {F} =\langle x,y\rangle

and C is the arc of the parabola

\mathbf {r} (t)=\langle 4t,t^{2}\rangle ,\ t\in [0,1]

17/2

17/2

424. Find the work required to move an object from (1,1,1) to (8,4,2) along a straight line in the force field

\displaystyle \mathbf {F} ={\frac {\langle x,y,z\rangle }{x^{2}+y^{2}+z^{2}}}

\ln(2{\sqrt {7}})

\ln(2{\sqrt {7}})

Conservative Vector Fields

Determine if the following vector fields are conservative on $\mathbb {R} ^{2}.$

440.

\langle -y,x+y\rangle

No

441.

\langle 2x^{3}+xy^{2},2y^{3}+x^{2}y\rangle

Yes

Determine if the following vector fields are conservative on their respective domains in $\mathbb {R} ^{3}.$ When possible, find the potential function.

442.

\langle y,x,1\rangle

\phi (x,y,z)=xy+z

\phi (x,y,z)=xy+z

443.

\langle x^{3},2y,-z^{3}\rangle

\phi (x,y,z)=(x^{4}+4y^{2}-z^{4})/4

\phi (x,y,z)=(x^{4}+4y^{2}-z^{4})/4

Green's Theorem

460. Evaluate the circulation of the field

\mathbf {F} =\langle 2xy,x^{2}-y^{2}\rangle

over the boundary of the region above y=0 and below y=x(2-x) in two different ways, and compare the answers.

0

0

461. Evaluate the circulation of the field

\mathbf {F} =\langle 0,x^{2}+y^{2}\rangle

over the unit circle centered at the origin in two different ways, and compare the answers.

0

0

462. Evaluate the flux of the field

\mathbf {F} =\langle y,-x\rangle

over the square with vertices (0,0), (1,0), (1,1), and (0,1) in two different ways, and compare the answers.

0

0

Divergence And Curl

480. Find the divergence of

\langle 2x,4y,-3z\rangle

3

3

481. Find the divergence of

\displaystyle {\frac {\langle x,y,z\rangle }{1+x^{2}+y^{2}}}

\displaystyle {\frac {x^{2}+y^{2}+3}{(1+x^{2}+y^{2})^{2}}}

\displaystyle {\frac {x^{2}+y^{2}+3}{(1+x^{2}+y^{2})^{2}}}

482. Find the curl of

\langle x^{2}-y^{2},xy,z\rangle

\langle 0,0,3y\rangle

\langle 0,0,3y\rangle

483. Find the curl of

\langle z^{2}\sin(y),xz^{2}\cos(y),2xz\sin(y)\rangle

\mathbf {0}

\mathbf {0}

484. Prove that the general rotation field

\mathbf {F} =\mathbf {a} \times \mathbf {r}

, where

\mathbf {a}

is a non-zero constant vector and

\mathbf {r} =\langle x,y,z\rangle

, has zero divergence, and the curl of

\mathbf {F}

is

2\mathbf {a}

.

If

\mathbf {a} =\langle a_{1},a_{2},a_{3}\rangle

, then

$\mathbf {F} =\mathbf {a} \times \mathbf {r} =\langle a_{2}z-a_{3}y,a_{3}x-a_{1}z,a_{1}y-a_{2}x\rangle =\langle f,g,h\rangle$ , and then

$\nabla \cdot \mathbf {F} =\mathbf {f} _{x}+\mathbf {g} _{y}+\mathbf {h} _{z}=0+0+0=0,$

\nabla \times \mathbf {F} =\langle h_{y}-g_{z},f_{z}-h_{x},g_{x}-f_{y}\rangle =\langle a_{1}+a_{1},a_{2}+a_{2},a_{3}+a_{3}\rangle =2\mathbf {a} .

If

\mathbf {a} =\langle a_{1},a_{2},a_{3}\rangle

, then

$\mathbf {F} =\mathbf {a} \times \mathbf {r} =\langle a_{2}z-a_{3}y,a_{3}x-a_{1}z,a_{1}y-a_{2}x\rangle =\langle f,g,h\rangle$ , and then

$\nabla \cdot \mathbf {F} =\mathbf {f} _{x}+\mathbf {g} _{y}+\mathbf {h} _{z}=0+0+0=0,$

\nabla \times \mathbf {F} =\langle h_{y}-g_{z},f_{z}-h_{x},g_{x}-f_{y}\rangle =\langle a_{1}+a_{1},a_{2}+a_{2},a_{3}+a_{3}\rangle =2\mathbf {a} .

Surface Integrals

500. Give a parametric description of the plane

2x-4y+3z=16.

\langle u,v,(16-2u+4v)/3\rangle ,\ u,v\in \mathbb {R}

\langle u,v,(16-2u+4v)/3\rangle ,\ u,v\in \mathbb {R}

501. Give a parametric description of the hyperboloid

z^{2}=1+x^{2}+y^{2}.

\langle {\sqrt {v^{2}-1}}\cos(u),{\sqrt {v^{2}-1}}\sin(u),v\rangle ,\ u\in [0,2\pi ],\ |v|\geq 1

\langle {\sqrt {v^{2}-1}}\cos(u),{\sqrt {v^{2}-1}}\sin(u),v\rangle ,\ u\in [0,2\pi ],\ |v|\geq 1

502. Integrate

f(x,y,z)=xy

over the portion of the plane z=2−x−y in the first octant.

2/{\sqrt {3}}

2/{\sqrt {3}}

503. Integrate

f(x,y,z)=x^{2}+y^{2}

over the paraboloid

z=x^{2}+y^{2},\ z\in [0,4].

{\frac {(391{\sqrt {17}}+1)\pi }{60}}

{\frac {(391{\sqrt {17}}+1)\pi }{60}}

504. Find the flux of the field

\mathbf {F} =\langle x,y,z\rangle

across the surface of the cone

z^{2}=x^{2}+y^{2},\ z\in [0,1],

with normal vectors pointing in the positive z direction.

0

0

505. Find the flux of the field

\mathbf {F} =\langle -y,z,1\rangle

across the surface

y=x^{2},\ z\in [0,4],\ x\in [0,1],

with normal vectors pointing in the positive y direction.

-10

-10

Stokes' Theorem

520. Use a surface integral to evaluate the circulation of the field

\mathbf {F} =\langle x^{2}-z^{2},y,2xz\rangle

on the boundary of the plane

z=4-x-y

in the first octant.

{\frac {-128}{3}}

{\frac {-128}{3}}

521. Use a surface integral to evaluate the circulation of the field

\mathbf {F} =\langle y^{2},-z^{2},x\rangle

on the circle

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

15\pi

15\pi

522. Use a line integral to find

\iint _{S}(\nabla \times F)\cdot \mathbf {n} dS

where

\mathbf {F} =\langle x,y,z\rangle

,

S

is the upper half of the ellipsoid

{\frac {x^{2}}{4}}+{\frac {y^{2}}{9}}+z^{2}=1

, and

\mathbf {n}

points in the direction of the z-axis.

0

0

523. Use a line integral to find

\iint _{S}(\nabla \times F)\cdot \mathbf {n} dS

where

\mathbf {F} =\langle 2y,-z,x-y-z\rangle

,

S

is the part of the sphere

x^{2}+y^{2}+z^{2}=25

for

3\leq z\leq 5

, and

\mathbf {n}

points in the direction of the z-axis.

-32\pi

-32\pi

Divergence Theorem

Compute the net outward flux of the given field across the given surface.

540.

\mathbf {F} =\langle x,-2y,3z\rangle

,

S

is a sphere of radius

{\sqrt {6}}

centered at the origin.

16{\sqrt {6}}\pi

16{\sqrt {6}}\pi

541.

\mathbf {F} =\langle x,2y,z\rangle

,

S

is the boundary of the tetrahedron in the first octant bounded by

x+y+z=1

2/3

2/3

542.

\mathbf {F} =\langle y+z,x+z,x+y\rangle

,

S

is the boundary of the cube

\{(x,y,z)\mid |x|\leq 1,|y|\leq 1,|z|\leq 1\}

0

0

543.

\mathbf {F} =\langle x,y,z\rangle

,

S

is the surface of the region bounded by the paraboloid

z=4-x^{2}-y^{2}

and the xy-plane.

24\pi

24\pi

544.

\mathbf {F} =\langle z-x,x-y,2y-z\rangle

,

S

is the boundary of the region between the concentric spheres of radii 2 and 4, centered at the origin.

-224\pi

-224\pi

545.

\mathbf {F} =\langle x,2y,3z\rangle

,

S

is the boundary of the region between the cylinders

x^{2}+y^{2}=1

and

x^{2}+y^{2}=4

and cut off by planes

z=0

and

z=8

144\pi

144\pi

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