# Calculus/Multivariable and differential calculus:Exercises
## Parametric Equations

$P(x_{1},y_{1})$
$Q(x_{2},y_{2})$
$x=x_{1}+(x_{2}-x_{1})t{\mbox{ and }}y=y_{1}+(y_{2}-y_{1})t,{\mbox{ where }}0\leq t\leq 1$
$x=x_{1}+(x_{2}-x_{1})t{\mbox{ and }}y=y_{1}+(y_{2}-y_{1})t,{\mbox{ where }}0\leq t\leq 1$
$x=3\cos(-t),\ y=1.5\sin(-t)$
$x=3\cos(-t),\ y=1.5\sin(-t)$

## Polar Coordinates

$r=\sin(\theta )\sec ^{2}(\theta ).$
$\sin(\theta )={\frac {y}{r}}$
$\sec(\theta )={\frac {r}{x}}$
$r={\frac {y}{r}}\left({\frac {r}{x}}\right)^{2}$
$\iff r={\frac {ry}{x^{2}}}$
$\iff y=x^{2}$
$\sin(\theta )={\frac {y}{r}}$
$\sec(\theta )={\frac {r}{x}}$
$r={\frac {y}{r}}\left({\frac {r}{x}}\right)^{2}$
$\iff r={\frac {ry}{x^{2}}}$
$\iff y=x^{2}$
$x=r\cos(\theta )$
$y=r\sin(\theta )$
$r\sin(\theta )=mr\cos(\theta )+b$
$\iff (\sin(\theta )-m\cos(\theta ))r=b$
$\iff r={\frac {b}{\sin(\theta )-m\cos(\theta )}}$
$x=r\cos(\theta )$
$y=r\sin(\theta )$
$r\sin(\theta )=mr\cos(\theta )+b$
$\iff (\sin(\theta )-m\cos(\theta ))r=b$
$\iff r={\frac {b}{\sin(\theta )-m\cos(\theta )}}$
Sketch the following polar curves without using a computer.
$r=2-2\sin(\theta )$
$r^{2}=4\cos(\theta )$
$r=2\sin(5\theta )$
Sketch the following sets of points.
$\{(r,\theta ):\theta =2\pi /3\}$
$\{(r,\theta ):|\theta |\leq \pi /3{\mbox{ and }}|r|<3\}$

## Calculus in Polar Coordinates

Find points where the following curves have vertical or horizontal tangents.
$r=4\cos(\theta )$
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 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 $(4,0)$.
$r=2+2\sin(\theta )$
${\frac {dy}{d\theta }}=0$
${\frac {d}{d\theta }}(r\sin(\theta ))=0$
$\iff {\frac {dr}{d\theta }}\sin(\theta )+r\cos(\theta )=0$
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 $(-3{\sqrt {3}}/2,3/2)$.
${\frac {dy}{d\theta }}=0$
${\frac {d}{d\theta }}(r\sin(\theta ))=0$
$\iff {\frac {dr}{d\theta }}\sin(\theta )+r\cos(\theta )=0$
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 $(-3{\sqrt {3}}/2,3/2)$.
Sketch the region and find its area.
$2+\cos(\theta )$
$d\theta$
$r$
${\frac {1}{2}}r^{2}d\theta$
$A=\int _{\theta =0}^{2\pi }{\frac {1}{2}}r^{2}d\theta$
$=\int _{\theta =0}^{2\pi }{\frac {1}{2}}(2+\cos(\theta ))^{2}d\theta$
$=\int _{\theta =0}^{2\pi }(2+2\cos(\theta )+{\frac {1}{2}}\cos ^{2}(\theta ))d\theta$
$=\int _{\theta =0}^{2\pi }(2+2\cos(\theta )+{\frac {1}{4}}(\cos(2\theta )+1))d\theta$
$=\int _{\theta =0}^{2\pi }({\frac {9}{4}}+2\cos(\theta )+{\frac {1}{4}}\cos(2\theta ))d\theta$
$=({\frac {9}{4}}\theta +2\sin(\theta )+{\frac {1}{8}}\sin(2\theta )){\bigg |}_{\theta =0}^{2\pi }$
$={\frac {9}{2}}\pi$
$d\theta$
$r$
${\frac {1}{2}}r^{2}d\theta$
$A=\int _{\theta =0}^{2\pi }{\frac {1}{2}}r^{2}d\theta$
$=\int _{\theta =0}^{2\pi }{\frac {1}{2}}(2+\cos(\theta ))^{2}d\theta$
$=\int _{\theta =0}^{2\pi }(2+2\cos(\theta )+{\frac {1}{2}}\cos ^{2}(\theta ))d\theta$
$=\int _{\theta =0}^{2\pi }(2+2\cos(\theta )+{\frac {1}{4}}(\cos(2\theta )+1))d\theta$
$=\int _{\theta =0}^{2\pi }({\frac {9}{4}}+2\cos(\theta )+{\frac {1}{4}}\cos(2\theta ))d\theta$
$=({\frac {9}{4}}\theta +2\sin(\theta )+{\frac {1}{8}}\sin(2\theta )){\bigg |}_{\theta =0}^{2\pi }$
$={\frac {9}{2}}\pi$
$4\cos(2\theta )$
$r=2$
$4$
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 $A=4\int _{\theta =-\pi /6}^{\pi /6}{\frac {1}{2}}((4\cos(2\theta ))^{2}-2^{2})d\theta$ $=4\int _{\theta =-\pi /6}^{\pi /6}(8\cos ^{2}(2\theta )-2)d\theta$ $=4\int _{\theta =-\pi /6}^{\pi /6}(4(\cos(4\theta )+1)-2)d\theta$ $=4\int _{\theta =-\pi /6}^{\pi /6}(2+4\cos(4\theta ))d\theta$ $=4(2\theta +\sin(4\theta )){\bigg |}_{\theta =-\pi /6}^{\pi /6}$ $=4({\frac {2\pi }{3}}+{\sqrt {3}})$ $={\frac {8\pi }{3}}+4{\sqrt {3}}$.
$8\pi /3+4{\sqrt {3}}$
$4$
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

$(x-x_{0})^{2}+(y-y_{0})^{2}+(z-z_{0})^{2}=r^{2}$
$(x_{0},y_{0},z_{0})$
$r$
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 $r={\sqrt {(3-1)^{2}+(4-2)^{2}+(5-0)^{2}}}={\sqrt {4+4+25}}={\sqrt {33}}$.
$(x-1)^{2}+(y-2)^{2}+z^{2}=33$
$(x-x_{0})^{2}+(y-y_{0})^{2}+(z-z_{0})^{2}=r^{2}$
$(x_{0},y_{0},z_{0})$
$r$
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 $r={\sqrt {(3-1)^{2}+(4-2)^{2}+(5-0)^{2}}}={\sqrt {4+4+25}}={\sqrt {33}}$.
$(x-1)^{2}+(y-2)^{2}+z^{2}=33$
$|\mathbf {u} +3\mathbf {v} |$
$\mathbf {u} =\langle 1,3,0\rangle$
$\mathbf {v} =\langle 3,0,2\rangle$
$\mathbf {u} +3\mathbf {v} =\langle 1,3,0\rangle +3\langle 3,0,2\rangle$ $=\langle 1,3,0\rangle +\langle 9,0,6\rangle$ $=\langle 10,3,6\rangle$.
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: $|\mathbf {u} +3\mathbf {v} |={\sqrt {10^{2}+3^{2}+6^{2}}}={\sqrt {100+9+36}}={\sqrt {145}}$.
${\sqrt {145}}$
$\langle 1,2,3\rangle$
$\langle 1,2,3\rangle$
${\sqrt {1^{2}+2^{2}+3^{2}}}={\sqrt {1+4+9}}={\sqrt {14}}$
${\frac {1}{\sqrt {14}}}\langle 1,2,3\rangle$
$\langle 1,2,3\rangle$
$-{\frac {1}{\sqrt {14}}}\langle 1,2,3\rangle$
$\langle 1,2,3\rangle$
$\pm {\frac {1}{\sqrt {14}}}\langle 1,2,3\rangle$
$\langle 1,2,3\rangle$
$\langle 1,2,3\rangle$
${\sqrt {1^{2}+2^{2}+3^{2}}}={\sqrt {1+4+9}}={\sqrt {14}}$
${\frac {1}{\sqrt {14}}}\langle 1,2,3\rangle$
$\langle 1,2,3\rangle$
$-{\frac {1}{\sqrt {14}}}\langle 1,2,3\rangle$
$\langle 1,2,3\rangle$
$\pm {\frac {1}{\sqrt {14}}}\langle 1,2,3\rangle$
$\langle 1,2,3\rangle$
$\mathbb {R} ^{3}$
$c(\mathbf {u} +\mathbf {v} )=c\mathbf {u} +c\mathbf {v}$
$c(\mathbf {u} +\mathbf {v} )=c(\langle u_{1},u_{2},u_{3}\rangle +\langle v_{1},v_{2},v_{3}\rangle )$ $=c\langle u_{1}+v_{1},u_{2}+v_{2},u_{3}+v_{3}\rangle$ $=\langle c(u_{1}+v_{1}),c(u_{2}+v_{2}),c(u_{3}+v_{3})\rangle$ $=\langle cu_{1}+cv_{1},cu_{2}+cv_{2},cu_{3}+cv_{3}\rangle$ $=\langle cu_{1},cu_{2},cu_{3}\rangle +\langle cv_{1},cv_{2},cv_{3}\rangle$ $=c\langle u_{1},u_{2},u_{3}\rangle +c\langle v_{1},v_{2},v_{3}\rangle$
$=c\mathbf {u} +c\mathbf {v}$
$c(\mathbf {u} +\mathbf {v} )=c(\langle u_{1},u_{2},u_{3}\rangle +\langle v_{1},v_{2},v_{3}\rangle )$ $=c\langle u_{1}+v_{1},u_{2}+v_{2},u_{3}+v_{3}\rangle$ $=\langle c(u_{1}+v_{1}),c(u_{2}+v_{2}),c(u_{3}+v_{3})\rangle$ $=\langle cu_{1}+cv_{1},cu_{2}+cv_{2},cu_{3}+cv_{3}\rangle$ $=\langle cu_{1},cu_{2},cu_{3}\rangle +\langle cv_{1},cv_{2},cv_{3}\rangle$ $=c\langle u_{1},u_{2},u_{3}\rangle +c\langle v_{1},v_{2},v_{3}\rangle$
$=c\mathbf {u} +c\mathbf {v}$
$3\mathbf {i} +4\mathbf {j}$
$\mathbb {R} ^{2}$
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 $\mathbf {u}$.
$\mathbf {u} ^{\perp }$
${\sqrt {(-4)^{2}+3^{2}}}={\sqrt {16+9}}={\sqrt {25}}=5$
$\mathbf {u}$
$\pm \left\langle {\frac {-4}{5}},{\frac {3}{5}}\right\rangle$
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 $\mathbf {u}$.
$\mathbf {u} ^{\perp }$
${\sqrt {(-4)^{2}+3^{2}}}={\sqrt {16+9}}={\sqrt {25}}=5$
$\mathbf {u}$
$\pm \left\langle {\frac {-4}{5}},{\frac {3}{5}}\right\rangle$
$3\mathbf {i} +4\mathbf {j}$
$\mathbb {R} ^{3}$
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 $\langle 3,4,0\rangle$.
$c=\cos(\theta )$
$\left\langle {\frac {4}{5}}c,-{\frac {3}{5}}c,\pm {\sqrt {1-c^{2}}}\right\rangle ,\ c\in [-1,1]$
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 $\langle 3,4,0\rangle$.
$c=\cos(\theta )$
$\left\langle {\frac {4}{5}}c,-{\frac {3}{5}}c,\pm {\sqrt {1-c^{2}}}\right\rangle ,\ c\in [-1,1]$
$\pi /3$
$\langle 1,2\rangle$
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 $\mathbf {v} \times \mathbf {u}$
$\mathbf {u} =\langle -4,1,1\rangle$
$\mathbf {v} =\langle 0,1,-1\rangle$
$\mathbf {u} \times \mathbf {v} =\langle (1)(-1)-(1)(1),(1)(0)-(-4)(-1),(-4)(1)-(1)(0)\rangle =\langle -2,-4,-4\rangle$
$\mathbf {v} \times \mathbf {u} =-(\mathbf {u} \times \mathbf {v} )=\langle 2,4,4\rangle$
$\mathbf {u} \times \mathbf {v} =\langle (1)(-1)-(1)(1),(1)(0)-(-4)(-1),(-4)(1)-(1)(0)\rangle =\langle -2,-4,-4\rangle$
$\mathbf {v} \times \mathbf {u} =-(\mathbf {u} \times \mathbf {v} )=\langle 2,4,4\rangle$
$\mathbf {u} =\langle 1,2,-1\rangle$
$\mathbf {v} =\langle 3,-4,6\rangle$
$\mathbf {u} \times \mathbf {v} =\langle (2)(6)-(-1)(-4),(-1)(3)-(1)(6),(1)(-4)-(2)(3)\rangle =\langle 8,-9,-10\rangle$
$\mathbf {v} \times \mathbf {u} =-(\mathbf {u} \times \mathbf {v} )=\langle -8,9,10\rangle$
$\mathbf {u} \times \mathbf {v} =\langle (2)(6)-(-1)(-4),(-1)(3)-(1)(6),(1)(-4)-(2)(3)\rangle =\langle 8,-9,-10\rangle$
$\mathbf {v} \times \mathbf {u} =-(\mathbf {u} \times \mathbf {v} )=\langle -8,9,10\rangle$
Find the area of the parallelogram with sides $\mathbf {u}$ and $\mathbf {v}$.
$\mathbf {u} =\langle -3,0,2\rangle$
$\mathbf {v} =\langle 1,1,1\rangle$
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 $|\mathbf {u} \times \mathbf {v} |$.
$\mathbf {u} \times \mathbf {v} =\langle (0)(1)-(2)(1),(2)(1)-(-3)(1),(-3)(1)-(0)(1)\rangle =\langle -2,5,-3\rangle$
$A=|\mathbf {u} \times \mathbf {v} |={\sqrt {(-2)^{2}+5^{2}+(-3)^{2}}}={\sqrt {4+25+9}}={\sqrt {38}}$
$\mathbf {u} =\langle 8,2,-3\rangle$
$\mathbf {v} =\langle 2,4,-4\rangle$
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 $|\mathbf {u} \times \mathbf {v} |$.
$\mathbf {u} \times \mathbf {v} =\langle (2)(-4)-(-3)(4),(-3)(2)-(8)(-4),(8)(4)-(2)(2)\rangle =\langle 4,26,28\rangle$
$A=|\mathbf {u} \times \mathbf {v} |={\sqrt {4^{2}+26^{2}+28^{2}}}={\sqrt {2^{2}(2^{2}+13^{2}+14^{2})}}=2{\sqrt {369}}=6{\sqrt {41}}$
$\langle 1,1,1\rangle \times \mathbf {u} =\langle 0,1,1\rangle$
$\langle 0,1,1\rangle$
$\langle 1,1,1\rangle$
$\mathbf {u}$
$\langle 1,1,1\rangle \cdot \langle 0,1,1\rangle =2\neq 0$
$\langle 1,1,1\rangle$
$\langle 0,1,1\rangle$
$\langle 1,1,1\rangle \times \mathbf {u} =\langle 0,1,1\rangle$
$\mathbf {u}$
$\emptyset =$
$\langle 0,1,1\rangle$
$\langle 1,1,1\rangle$
$\mathbf {u}$
$\langle 1,1,1\rangle \cdot \langle 0,1,1\rangle =2\neq 0$
$\langle 1,1,1\rangle$
$\langle 0,1,1\rangle$
$\langle 1,1,1\rangle \times \mathbf {u} =\langle 0,1,1\rangle$
$\mathbf {u}$
$\emptyset =$
$\langle 5,0,0\rangle$
$\langle 1,4,0\rangle$
$\langle 2,2,7\rangle$
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: $|(\mathbf {u} \times \mathbf {v} )\cdot \mathbf {w} |$.
$|(\langle 5,0,0\rangle \times \langle 1,4,0\rangle )\cdot \langle 2,2,7\rangle |$ $=|\langle (0)(0)-(0)(4),(0)(1)-(5)(0),(5)(4)-(0)(1)\rangle \cdot \langle 2,2,7\rangle |$ $=|\langle 0,0,20\rangle \cdot \langle 2,2,7\rangle |$ $=140$
$\mathbf {F} =\langle 1,2,3\rangle$
$\mathbf {r} =\langle n,0,0\rangle$
$\tau =\mathbf {r} \times \mathbf {F} =\langle n,0,0\rangle \times \langle 1,2,3\rangle =\langle (0)(3)-(0)(2),(0)(1)-(n)(3),(n)(2)-(0)(1)\rangle =\langle 0,-3n,2n\rangle$
$|\tau |={\sqrt {0^{2}+(-3n)^{2}+(2n)^{2}}}=n{\sqrt {13}}$
${\hat {\tau }}={\frac {\tau }{|\tau |}}=\langle 0,-3/{\sqrt {13}},2/{\sqrt {13}}\rangle$
$\mathbf {r} =\langle n,0,0\rangle$
$\tau =\mathbf {r} \times \mathbf {F} =\langle n,0,0\rangle \times \langle 1,2,3\rangle =\langle (0)(3)-(0)(2),(0)(1)-(n)(3),(n)(2)-(0)(1)\rangle =\langle 0,-3n,2n\rangle$
$|\tau |={\sqrt {0^{2}+(-3n)^{2}+(2n)^{2}}}=n{\sqrt {13}}$
${\hat {\tau }}={\frac {\tau }{|\tau |}}=\langle 0,-3/{\sqrt {13}},2/{\sqrt {13}}\rangle$
Prove the following identities or show them false by giving a counterexample.
$\mathbf {u} \times (\mathbf {u} \times \mathbf {v} )=\mathbf {0}$
$\mathbf {i} \times (\mathbf {i} \times \mathbf {j} )=-\mathbf {j}$
$\mathbf {i} \times (\mathbf {i} \times \mathbf {j} )=-\mathbf {j}$
$\mathbf {u} \cdot (\mathbf {v} \times \mathbf {w} )=\mathbf {w} \cdot (\mathbf {u} \times \mathbf {v} )$
$u_{1}v_{2}w_{3}-u_{1}v_{3}w_{2}+u_{2}v_{3}w_{1}-u_{2}v_{1}w_{3}+u_{3}v_{1}w_{2}-u_{3}v_{2}w_{1}$
$u_{1}v_{2}w_{3}-u_{1}v_{3}w_{2}+u_{2}v_{3}w_{1}-u_{2}v_{1}w_{3}+u_{3}v_{1}w_{2}-u_{3}v_{2}w_{1}$
$(\mathbf {u} -\mathbf {v} )\times (\mathbf {u} +\mathbf {v} )=2(\mathbf {u} \times \mathbf {v} )$
$(\mathbf {u} -\mathbf {v} )\times (\mathbf {u} +\mathbf {v} )=(\mathbf {u} -\mathbf {v} )\times \mathbf {u} +(\mathbf {u} -\mathbf {v} )\times \mathbf {v}$
$=\mathbf {u} \times \mathbf {u} -\mathbf {v} \times \mathbf {u} +\mathbf {u} \times \mathbf {v} -\mathbf {v} \times \mathbf {v}$
$=\mathbf {u} \times \mathbf {v} -\mathbf {v} \times \mathbf {u}$
$=\mathbf {u} \times \mathbf {v} +\mathbf {u} \times \mathbf {v}$
$=2(\mathbf {u} \times \mathbf {v} )$
$(\mathbf {u} -\mathbf {v} )\times (\mathbf {u} +\mathbf {v} )=(\mathbf {u} -\mathbf {v} )\times \mathbf {u} +(\mathbf {u} -\mathbf {v} )\times \mathbf {v}$
$=\mathbf {u} \times \mathbf {u} -\mathbf {v} \times \mathbf {u} +\mathbf {u} \times \mathbf {v} -\mathbf {v} \times \mathbf {v}$
$=\mathbf {u} \times \mathbf {v} -\mathbf {v} \times \mathbf {u}$
$=\mathbf {u} \times \mathbf {v} +\mathbf {u} \times \mathbf {v}$
$=2(\mathbf {u} \times \mathbf {v} )$

## Calculus of Vector-Valued Functions

$\mathbf {r} (t)=\langle te^{-t},t\ln t,t\cos(t)\rangle$
$\langle e^{-t}-te^{-t},\ln(t)+1,\cos(t)-t\sin(t)\rangle$
$\langle e^{-t}-te^{-t},\ln(t)+1,\cos(t)-t\sin(t)\rangle$
$\mathbf {r} (t)=\langle 2t^{4},6t^{3/2},10/t\rangle$
$t=1$
${\dot {\mathbf {r} }}(t)=\langle 8t^{3},9t^{1/2},-10/t^{2}\rangle$
$t=1$
${\dot {\mathbf {r} }}(1)=\langle 8,9,-10\rangle$
${\dot {\mathbf {r} }}(t)=\langle 8t^{3},9t^{1/2},-10/t^{2}\rangle$
$t=1$
${\dot {\mathbf {r} }}(1)=\langle 8,9,-10\rangle$
$\mathbf {r} (t)=\langle t,2,2/t\rangle ,\ t\neq 0$
${\dot {\mathbf {r} }}(t)=\langle 1,0,-2/t^{2}\rangle$
$|{\dot {\mathbf {r} }}(t)|={\sqrt {1^{2}+0^{2}+(-2/t^{2})^{2}}}={\sqrt {1+4/t^{4}}}={\frac {\sqrt {t^{4}+4}}{t^{2}}}$
$\mathbf {T} (t)={\frac {{\dot {\mathbf {r} }}(t)}{|{\dot {\mathbf {r} }}(t)|}}={\frac {\langle t^{2},0,-2\rangle }{\sqrt {t^{4}+4}}}$
${\dot {\mathbf {r} }}(t)=\langle 1,0,-2/t^{2}\rangle$
$|{\dot {\mathbf {r} }}(t)|={\sqrt {1^{2}+0^{2}+(-2/t^{2})^{2}}}={\sqrt {1+4/t^{4}}}={\frac {\sqrt {t^{4}+4}}{t^{2}}}$
$\mathbf {T} (t)={\frac {{\dot {\mathbf {r} }}(t)}{|{\dot {\mathbf {r} }}(t)|}}={\frac {\langle t^{2},0,-2\rangle }{\sqrt {t^{4}+4}}}$
$\mathbf {r} (t)=\langle \sin(t),\cos(t),e^{-t}\rangle ,\ t\in [0,\pi ]$
$t=0$
${\dot {\mathbf {r} }}(t)=\langle \cos(t),-\sin(t),-e^{-t}\rangle$
$|{\dot {\mathbf {r} }}(t)|={\sqrt {(\cos(t))^{2}+(-\sin(t))^{2}+(-e^{-t})^{2}}}={\sqrt {1+e^{-2t}}}$ so the unit tangent vector is $\mathbf {T} (t)={\frac {{\dot {\mathbf {r} }}(t)}{|{\dot {\mathbf {r} }}(t)|}}={\frac {\langle \cos(t),-\sin(t),-e^{-t}\rangle }{\sqrt {1+e^{-2t}}}}$
$t=0$
$\mathbf {T} (0)={\frac {\langle 1,0,-1\rangle }{\sqrt {2}}}$
${\dot {\mathbf {r} }}(t)=\langle \cos(t),-\sin(t),-e^{-t}\rangle$
$|{\dot {\mathbf {r} }}(t)|={\sqrt {(\cos(t))^{2}+(-\sin(t))^{2}+(-e^{-t})^{2}}}={\sqrt {1+e^{-2t}}}$ so the unit tangent vector is $\mathbf {T} (t)={\frac {{\dot {\mathbf {r} }}(t)}{|{\dot {\mathbf {r} }}(t)|}}={\frac {\langle \cos(t),-\sin(t),-e^{-t}\rangle }{\sqrt {1+e^{-2t}}}}$
$t=0$
$\mathbf {T} (0)={\frac {\langle 1,0,-1\rangle }{\sqrt {2}}}$
$\mathbf {r}$
$\mathbf {r} '(t)=\langle {\sqrt {t}},\cos(\pi t),4/t\rangle$
$\mathbf {r} (1)=\langle 2,3,4\rangle$
$t>0$
$\mathbf {r} (t)$
$\mathbf {r} (t)=\mathbf {r} (1)+\int _{u=1}^{t}\mathbf {r} '(u)du$
$\mathbf {r} (t)=\mathbf {r} (1)+\int _{u=1}^{t}\mathbf {r} '(u)du$ $=\langle 2,3,4\rangle +\int _{u=1}^{t}\langle {\sqrt {u}},\cos(\pi u),4/u\rangle du$ $=\langle 2,3,4\rangle +\left\langle {\frac {2}{3}}u^{3/2},{\frac {\sin(\pi u)}{\pi }},4\ln(u)\right\rangle {\Bigg |}_{u=1}^{t}$ $=\langle 2,3,4\rangle +\left\langle {\frac {2}{3}}(t^{3/2}-1),{\frac {\sin(\pi t)}{\pi }},4\ln(t)\right\rangle$
$=\left\langle {\frac {2t^{3/2}+4}{3}},{\frac {\sin(\pi t)}{\pi }}+3,4\ln(t)+4\right\rangle$
$t>0$
$\mathbf {r} (t)$
$\mathbf {r} (t)=\mathbf {r} (1)+\int _{u=1}^{t}\mathbf {r} '(u)du$
$\mathbf {r} (t)=\mathbf {r} (1)+\int _{u=1}^{t}\mathbf {r} '(u)du$ $=\langle 2,3,4\rangle +\int _{u=1}^{t}\langle {\sqrt {u}},\cos(\pi u),4/u\rangle du$ $=\langle 2,3,4\rangle +\left\langle {\frac {2}{3}}u^{3/2},{\frac {\sin(\pi u)}{\pi }},4\ln(u)\right\rangle {\Bigg |}_{u=1}^{t}$ $=\langle 2,3,4\rangle +\left\langle {\frac {2}{3}}(t^{3/2}-1),{\frac {\sin(\pi t)}{\pi }},4\ln(t)\right\rangle$
$=\left\langle {\frac {2t^{3/2}+4}{3}},{\frac {\sin(\pi t)}{\pi }}+3,4\ln(t)+4\right\rangle$
$\displaystyle \int _{0}^{\ln 2}(e^{-t}\mathbf {i} +2e^{2t}\mathbf {j} -4e^{t}\mathbf {k} )dt$
$\int _{t=0}^{\ln 2}\langle e^{-t},2e^{2t},-4e^{t}\rangle dt$
$=\langle -e^{-t},e^{2t},-4e^{t}\rangle {\bigg |}_{t=0}^{\ln 2}$ $=\langle -1/2-(-1),4-1,-8-(-4)\rangle$
$=\langle 1/2,3,-4\rangle$
$\int _{t=0}^{\ln 2}\langle e^{-t},2e^{2t},-4e^{t}\rangle dt$
$=\langle -e^{-t},e^{2t},-4e^{t}\rangle {\bigg |}_{t=0}^{\ln 2}$ $=\langle -1/2-(-1),4-1,-8-(-4)\rangle$
$=\langle 1/2,3,-4\rangle$

## Motion in Space

$\mathbf {r} (t)=\langle 3\sin(t),5\cos(t),4\sin(t)\rangle$
$\mathbf {v} =\langle 3\cos(t),-5\sin(t),4\cos(t)\rangle$
$|\mathbf {v} |=5$
$\mathbf {a} =\langle -3\sin(t),-5\cos(t),-4\sin(t)\rangle$
$\mathbf {v} =\langle 3\cos(t),-5\sin(t),4\cos(t)\rangle$
$|\mathbf {v} |=5$
$\mathbf {a} =\langle -3\sin(t),-5\cos(t),-4\sin(t)\rangle$
$t\geq 0$
$\mathbf {a} (t)=\langle e^{-t},1\rangle ,\ \mathbf {v} (0)=\langle 1,0\rangle ,\ \mathbf {r} (0)=\langle 0,0\rangle$
$\mathbf {v} (t)=\mathbf {v} (0)+\int _{u=0}^{t}\mathbf {a} (u)du$
$=\langle 1,0\rangle +\int _{u=0}^{t}\langle e^{-u},1\rangle du$ $=\langle 1,0\rangle +\langle -e^{-u},u\rangle {\bigg |}_{u=0}^{t}$ $=\langle 1,0\rangle +\langle -e^{-t}-(-1),t-0\rangle$ $=\langle 2-e^{-t},t\rangle$
$\mathbf {r} (t)=\mathbf {r} (0)+\int _{u=0}^{t}\mathbf {v} (u)du$ $=\langle 0,0\rangle +\int _{u=0}^{t}\langle 2-e^{-u},u\rangle du$ $=\langle 2u+e^{-u},{\frac {1}{2}}u^{2}\rangle {\bigg |}_{u=0}^{t}$ $=\langle (2t+e^{-t})-1,{\frac {1}{2}}t^{2}-0\rangle$
$=\langle e^{-t}+2t-1,t^{2}/2\rangle$
$\mathbf {v} (t)=\mathbf {v} (0)+\int _{u=0}^{t}\mathbf {a} (u)du$
$=\langle 1,0\rangle +\int _{u=0}^{t}\langle e^{-u},1\rangle du$ $=\langle 1,0\rangle +\langle -e^{-u},u\rangle {\bigg |}_{u=0}^{t}$ $=\langle 1,0\rangle +\langle -e^{-t}-(-1),t-0\rangle$ $=\langle 2-e^{-t},t\rangle$
$\mathbf {r} (t)=\mathbf {r} (0)+\int _{u=0}^{t}\mathbf {v} (u)du$ $=\langle 0,0\rangle +\int _{u=0}^{t}\langle 2-e^{-u},u\rangle du$ $=\langle 2u+e^{-u},{\frac {1}{2}}u^{2}\rangle {\bigg |}_{u=0}^{t}$ $=\langle (2t+e^{-t})-1,{\frac {1}{2}}t^{2}-0\rangle$
$=\langle e^{-t}+2t-1,t^{2}/2\rangle$

## Length of Curves

Find the length of the following curves.
$\mathbf {r} (t)=\langle 4\cos(3t),4\sin(3t)\rangle ,\ t\in [0,2\pi /3].$
$dt$
$ds=\left|{\frac {d\mathbf {r} }{dt}}\right|dt=|\langle -12\sin(3t),12\cos(3t)\rangle |dt=12dt$.
The total length is therefore:
$L=\int _{t=0}^{2\pi /3}ds=\int _{t=0}^{2\pi /3}12dt=12t{\Bigg |}_{t=0}^{2\pi /3}=8\pi$
$dt$
$ds=\left|{\frac {d\mathbf {r} }{dt}}\right|dt=|\langle -12\sin(3t),12\cos(3t)\rangle |dt=12dt$.
The total length is therefore:
$L=\int _{t=0}^{2\pi /3}ds=\int _{t=0}^{2\pi /3}12dt=12t{\Bigg |}_{t=0}^{2\pi /3}=8\pi$
$\mathbf {r} (t)=\langle 2+3t,1-4t,3t-4\rangle ,\ t\in [1,6].$
$dt$
$ds=\left|{\frac {d\mathbf {r} }{dt}}\right|dt=|\langle 3,-4,3\rangle |dt={\sqrt {9+16+9}}\cdot dt={\sqrt {34}}\cdot dt$.
The total length is therefore:
$L=\int _{t=1}^{6}ds=\int _{t=1}^{6}{\sqrt {34}}\cdot dt={\sqrt {34}}\cdot t{\Bigg |}_{t=1}^{6}=5{\sqrt {34}}$
$dt$
$ds=\left|{\frac {d\mathbf {r} }{dt}}\right|dt=|\langle 3,-4,3\rangle |dt={\sqrt {9+16+9}}\cdot dt={\sqrt {34}}\cdot dt$.
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

$\mathbf {r} (t)=\langle t^{2},2t^{2},4t^{2}\rangle \ t\in [1,4].$
$dt$
$ds=\left|{\frac {d\mathbf {r} }{dt}}\right|dt=|\langle 2t,4t,8t\rangle |dt=|t|{\sqrt {4+16+64}}\cdot dt={\sqrt {84}}\cdot tdt=2{\sqrt {21}}\cdot tdt$
Given an upper bound of $u\in [1,4]$, the arc length swept out from $t=1$ to $t=u$ is: $s=\int _{t=1}^{u}ds=\int _{t=1}^{u}2{\sqrt {21}}\cdot tdt={\sqrt {21}}\cdot t^{2}{\bigg |}_{t=1}^{u}={\sqrt {21}}(u^{2}-1)$
$0$
$15{\sqrt {21}}$
$s\in [0,15{\sqrt {21}}]$
$t$
$s$
$u={\sqrt {{\frac {s}{\sqrt {21}}}+1}}$
$\displaystyle \mathbf {r} (s)=\left({\frac {s}{\sqrt {21}}}+1\right)\langle 1,2,4\rangle$
$dt$
$ds=\left|{\frac {d\mathbf {r} }{dt}}\right|dt=|\langle 2t,4t,8t\rangle |dt=|t|{\sqrt {4+16+64}}\cdot dt={\sqrt {84}}\cdot tdt=2{\sqrt {21}}\cdot tdt$
Given an upper bound of $u\in [1,4]$, the arc length swept out from $t=1$ to $t=u$ is: $s=\int _{t=1}^{u}ds=\int _{t=1}^{u}2{\sqrt {21}}\cdot tdt={\sqrt {21}}\cdot t^{2}{\bigg |}_{t=1}^{u}={\sqrt {21}}(u^{2}-1)$
$0$
$15{\sqrt {21}}$
$s\in [0,15{\sqrt {21}}]$
$t$
$s$
$u={\sqrt {{\frac {s}{\sqrt {21}}}+1}}$
$\displaystyle \mathbf {r} (s)=\left({\frac {s}{\sqrt {21}}}+1\right)\langle 1,2,4\rangle$
$\mathbf {r} (t)=\langle t^{2},t\rangle .$
${\frac {d\mathbf {r} }{dt}}=\langle 2t,1\rangle$
$\mathbf {T} (t)={\frac {d\mathbf {r} }{ds}}={\frac {d\mathbf {r} /dt}{|d\mathbf {r} /dt|}}=\displaystyle {\frac {\langle 2t,1\rangle }{\sqrt {4t^{2}+1}}}$
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: $\mathbf {N} (t)$ $={\frac {d\mathbf {T} /dt}{|d\mathbf {T} /dt|}}$ $={\frac {{\frac {2}{(4t^{2}+1)^{3/2}}}\langle 1,-2t\rangle }{{\frac {2}{(4t^{2}+1)^{3/2}}}{\sqrt {4t^{2}+1}}}}$
$=\displaystyle {\frac {\langle 1,-2t\rangle }{\sqrt {4t^{2}+1}}}$
${\frac {d\mathbf {r} }{dt}}=\langle 2t,1\rangle$
$\mathbf {T} (t)={\frac {d\mathbf {r} }{ds}}={\frac {d\mathbf {r} /dt}{|d\mathbf {r} /dt|}}=\displaystyle {\frac {\langle 2t,1\rangle }{\sqrt {4t^{2}+1}}}$
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

$(1,1,2),\ (1,2,2),\ (-1,0,1).$
$P$
$(1,1,2)$
$(1,2,2)$
$(-1,0,1)$
$\mathbf {N}$
$P$
$\mathbf {r} _{0}$
$P$
$\mathbf {r}$
$P$
$\mathbf {r} _{0}$
$\mathbf {N}$
$\mathbf {N} \cdot (\mathbf {r} -\mathbf {r} _{0})=0$
$\iff \mathbf {N} \cdot \mathbf {r} -\mathbf {N} \cdot \mathbf {r} _{0}=0$
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 $\mathbf {r} _{0}=\langle 1,1,2\rangle$
$\mathbf {N} \cdot \mathbf {r} -\mathbf {N} \cdot \mathbf {r} _{0}=0$
$\iff \langle -1,0,2\rangle \cdot \langle x,y,z\rangle -\langle -1,0,2\rangle \cdot \langle 1,1,2\rangle$
$\iff -x+2z-3=0$
$\iff x-2z+3=0$
$P$
$(1,1,2)$
$(1,2,2)$
$(-1,0,1)$
$\mathbf {N}$
$P$
$\mathbf {r} _{0}$
$P$
$\mathbf {r}$
$P$
$\mathbf {r} _{0}$
$\mathbf {N}$
$\mathbf {N} \cdot (\mathbf {r} -\mathbf {r} _{0})=0$
$\iff \mathbf {N} \cdot \mathbf {r} -\mathbf {N} \cdot \mathbf {r} _{0}=0$
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 $\mathbf {r} _{0}=\langle 1,1,2\rangle$
$\mathbf {N} \cdot \mathbf {r} -\mathbf {N} \cdot \mathbf {r} _{0}=0$
$\iff \langle -1,0,2\rangle \cdot \langle x,y,z\rangle -\langle -1,0,2\rangle \cdot \langle 1,1,2\rangle$
$\iff -x+2z-3=0$
$\iff x-2z+3=0$
$P$
$2x-y+z=1$
$(0,2,-2)$
$\mathbf {N}$
$P$
$\mathbf {r} _{0}$
$P$
$\mathbf {r}$
$P$
$\mathbf {r} _{0}$
$\mathbf {N}$
$\mathbf {N} \cdot (\mathbf {r} -\mathbf {r} _{0})=0$
$\iff \mathbf {N} \cdot \mathbf {r} -\mathbf {N} \cdot \mathbf {r} _{0}=0$
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 $\mathbf {r} _{0}=\langle 0,2,-2\rangle$.
$\mathbf {N} \cdot \mathbf {r} -\mathbf {N} \cdot \mathbf {r} _{0}=0$
$\iff \langle 2,-1,1\rangle \cdot \langle x,y,z\rangle -\langle 2,-1,1\rangle \cdot \langle 0,2,-2\rangle =0$
$\iff 2x-y+z+4=0$
$P$
$2x-y+z=1$
$(0,2,-2)$
$\mathbf {N}$
$P$
$\mathbf {r} _{0}$
$P$
$\mathbf {r}$
$P$
$\mathbf {r} _{0}$
$\mathbf {N}$
$\mathbf {N} \cdot (\mathbf {r} -\mathbf {r} _{0})=0$
$\iff \mathbf {N} \cdot \mathbf {r} -\mathbf {N} \cdot \mathbf {r} _{0}=0$
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 $\mathbf {r} _{0}=\langle 0,2,-2\rangle$.
$\mathbf {N} \cdot \mathbf {r} -\mathbf {N} \cdot \mathbf {r} _{0}=0$
$\iff \langle 2,-1,1\rangle \cdot \langle x,y,z\rangle -\langle 2,-1,1\rangle \cdot \langle 0,2,-2\rangle =0$
$\iff 2x-y+z+4=0$
$P$
$\mathbf {N}$
$P$
$\mathbf {r} _{0}$
$P$
$\mathbf {r}$
$P$
$\mathbf {r} _{0}$
$\mathbf {N}$
$\mathbf {N} \cdot (\mathbf {r} -\mathbf {r} _{0})=0$
$\iff \mathbf {N} \cdot \mathbf {r} -\mathbf {N} \cdot \mathbf {r} _{0}=0$
$P$
$\mathbf {N} \cdot \mathbf {r} -\mathbf {N} \cdot \mathbf {r} _{0}=0$
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:
$\mathbf {r} (t)=\langle 5,5,5\rangle +t\langle 1,1,2\rangle =\langle 5+t,5+t,5+2t\rangle$
$P$
$\mathbf {N}$
$P$
$\mathbf {r} _{0}$
$P$
$\mathbf {r}$
$P$
$\mathbf {r} _{0}$
$\mathbf {N}$
$\mathbf {N} \cdot (\mathbf {r} -\mathbf {r} _{0})=0$
$\iff \mathbf {N} \cdot \mathbf {r} -\mathbf {N} \cdot \mathbf {r} _{0}=0$
$P$
$\mathbf {N} \cdot \mathbf {r} -\mathbf {N} \cdot \mathbf {r} _{0}=0$
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:
$\mathbf {r} (t)=\langle 5,5,5\rangle +t\langle 1,1,2\rangle =\langle 5+t,5+t,5+2t\rangle$
$x+2y-z=1$
$x+y+z=1$
$x$
$y$
$z$
$y$
$z$
$x$
$E_{1}$
$x+2y-z=1$
$E_{2}$
$x+y+z=1$
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 $E_{1}-E_{2}:(x+2y-z)-(x+y+z)=1-1$ $\iff y-2z=0$ $\iff y=2z$
$z$
$t$
$\mathbf {r} (t)=\langle 1-3t,2t,t\rangle$
$x+2y-z=1$
$x+y+z=1$
$x$
$y$
$z$
$y$
$z$
$x$
$E_{1}$
$x+2y-z=1$
$E_{2}$
$x+y+z=1$
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 $E_{1}-E_{2}:(x+2y-z)-(x+y+z)=1-1$ $\iff y-2z=0$ $\iff y=2z$
$z$
$t$
$\mathbf {r} (t)=\langle 1-3t,2t,t\rangle$
$P$
$\mathbf {N}$
$P$
$\mathbf {r} _{0}$
$P$
$\mathbf {r}$
$P$
$\mathbf {r} _{0}$
$\mathbf {N}$
$\mathbf {N} \cdot (\mathbf {r} -\mathbf {r} _{0})=0$
$\iff \mathbf {N} \cdot \mathbf {r} -\mathbf {N} \cdot \mathbf {r} _{0}=0$
$P$
$\mathbf {N} \cdot \mathbf {r} -\mathbf {N} \cdot \mathbf {r} _{0}=0$
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 $\mathbf {N} _{2}$: $\theta =\arccos \left({\frac {\mathbf {N} _{1}\cdot \mathbf {N} _{2}}{|\mathbf {N} _{1}||\mathbf {N} _{2}|}}\right)$ $=\arccos \left({\frac {\langle 1,2,-1\rangle \cdot \langle 1,1,1\rangle }{|\langle 1,2,-1\rangle ||\langle 1,1,1\rangle |}}\right)$ $=\arccos \left({\frac {1+2-1}{{\sqrt {1+4+1}}{\sqrt {1+1+1}}}}\right)$
$=\arccos {\frac {2}{\sqrt {18}}}$
$P$
$\mathbf {N}$
$P$
$\mathbf {r} _{0}$
$P$
$\mathbf {r}$
$P$
$\mathbf {r} _{0}$
$\mathbf {N}$
$\mathbf {N} \cdot (\mathbf {r} -\mathbf {r} _{0})=0$
$\iff \mathbf {N} \cdot \mathbf {r} -\mathbf {N} \cdot \mathbf {r} _{0}=0$
$P$
$\mathbf {N} \cdot \mathbf {r} -\mathbf {N} \cdot \mathbf {r} _{0}=0$
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 $\mathbf {N} _{2}$: $\theta =\arccos \left({\frac {\mathbf {N} _{1}\cdot \mathbf {N} _{2}}{|\mathbf {N} _{1}||\mathbf {N} _{2}|}}\right)$ $=\arccos \left({\frac {\langle 1,2,-1\rangle \cdot \langle 1,1,1\rangle }{|\langle 1,2,-1\rangle ||\langle 1,1,1\rangle |}}\right)$ $=\arccos \left({\frac {1+2-1}{{\sqrt {1+4+1}}{\sqrt {1+1+1}}}}\right)$
$=\arccos {\frac {2}{\sqrt {18}}}$
${\hat {\mathbf {n} }}$
$w$
${\hat {\mathbf {n} }}$
$w$
$w$
$\mathbf {r}$
${\hat {\mathbf {n} }}\cdot \mathbf {r}$
$w$
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 $\left|4{\sqrt {3}}-{\frac {1}{\sqrt {3}}}\right|=\left|{\frac {12{\sqrt {3}}-{\sqrt {3}}}{3}}\right|$ $={\frac {11}{3}}{\sqrt {3}}$
${\frac {11}{3}}{\sqrt {3}}$
${\hat {\mathbf {n} }}$
$w$
${\hat {\mathbf {n} }}$
$w$
$w$
$\mathbf {r}$
${\hat {\mathbf {n} }}\cdot \mathbf {r}$
$w$
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 $\left|4{\sqrt {3}}-{\frac {1}{\sqrt {3}}}\right|=\left|{\frac {12{\sqrt {3}}-{\sqrt {3}}}{3}}\right|$ $={\frac {11}{3}}{\sqrt {3}}$
${\frac {11}{3}}{\sqrt {3}}$

## Limits And Continuity

Evaluate the following limits.
$\displaystyle \lim _{(x,y)\rightarrow (1,-2)}{\frac {y^{2}+2xy}{y+2x}}$
$\lim _{(x,y)\rightarrow (1,-2)}{\frac {y^{2}+2xy}{y+2x}}=\lim _{(x,y)\rightarrow (1,-2)}{\frac {y(y+2x)}{y+2x}}$
$=\lim _{(x,y)\rightarrow (1,-2)}y$
$=-2$
$\lim _{(x,y)\rightarrow (1,-2)}{\frac {y^{2}+2xy}{y+2x}}=\lim _{(x,y)\rightarrow (1,-2)}{\frac {y(y+2x)}{y+2x}}$
$=\lim _{(x,y)\rightarrow (1,-2)}y$
$=-2$
$\displaystyle \lim _{(x,y)\rightarrow (4,5)}{\frac {{\sqrt {x+y}}-3}{x+y-9}}$
$\lim _{(x,y)\rightarrow (4,5)}{\frac {{\sqrt {x+y}}-3}{x+y-9}}=\lim _{(x,y)\rightarrow (4,5)}{\frac {{\sqrt {x+y}}-3}{({\sqrt {x+y}}-3)({\sqrt {x+y}}+3)}}$
$=\lim _{(x,y)\rightarrow (4,5)}{\frac {1}{{\sqrt {x+y}}+3}}$
$={\frac {1}{{\sqrt {4+5}}+3}}=1/6$
$\lim _{(x,y)\rightarrow (4,5)}{\frac {{\sqrt {x+y}}-3}{x+y-9}}=\lim _{(x,y)\rightarrow (4,5)}{\frac {{\sqrt {x+y}}-3}{({\sqrt {x+y}}-3)({\sqrt {x+y}}+3)}}$
$=\lim _{(x,y)\rightarrow (4,5)}{\frac {1}{{\sqrt {x+y}}+3}}$
$={\frac {1}{{\sqrt {4+5}}+3}}=1/6$
At what points is the function f continuous?
$f(x,y)=\ln |x-y|$
$\{(x,y)\mid x\neq y\}$
$\{(x,y)\mid x\neq y\}$
$f(x,y)=\displaystyle {\frac {\ln(x^{2}+y^{2})}{x-y+1}}$
Use the two-path test to show that the following limits do not exist. (A path does not have to be a straight line.)
$\displaystyle \lim _{(x,y)\rightarrow (0,0)}{\frac {4xy}{3x^{2}+y^{2}}}$
$\displaystyle \lim _{(x,y)\rightarrow (0,0)}{\frac {y}{\sqrt {x^{2}-y^{2}}}}$
$1/{\sqrt {3}}$
$1/{\sqrt {3}}$
$\displaystyle \lim _{(x,y)\rightarrow (0,0)}{\frac {x^{3}-y^{2}}{x^{3}+y^{2}}}$
$\displaystyle \lim _{(x,y)\rightarrow (0,0)}{\frac {x^{2}y^{2}+y^{6}}{x^{3}}}$
$x=y^{2}$
$x=y^{2}$

## Partial Derivatives

$\partial z/\partial x$
$\displaystyle z(x,y)={\frac {1}{\ln(xy)}}$
${\frac {\partial z}{\partial x}}={\frac {\partial }{\partial x}}\left({\frac {1}{\ln(xy)}}\right)$ $={\frac {-1}{(\ln(xy))^{2}}}{\frac {\partial }{\partial x}}(\ln(xy))$ $={\frac {-1}{(\ln(xy))^{2}}}{\frac {1}{xy}}{\frac {\partial }{\partial x}}(xy)$ $={\frac {-1}{(\ln(xy))^{2}}}{\frac {1}{xy}}y$
$={\frac {-1}{x(\ln(xy))^{2}}}$
${\frac {\partial z}{\partial x}}={\frac {\partial }{\partial x}}\left({\frac {1}{\ln(xy)}}\right)$ $={\frac {-1}{(\ln(xy))^{2}}}{\frac {\partial }{\partial x}}(\ln(xy))$ $={\frac {-1}{(\ln(xy))^{2}}}{\frac {1}{xy}}{\frac {\partial }{\partial x}}(xy)$ $={\frac {-1}{(\ln(xy))^{2}}}{\frac {1}{xy}}y$
$={\frac {-1}{x(\ln(xy))^{2}}}$
$\displaystyle f(x,y,z)=xe^{y^{2}+z}$
$\displaystyle f_{x}=e^{y^{2}+z},\ f_{y}=2xye^{y^{2}+z},\ f_{z}=f.$
$\displaystyle f_{x}=e^{y^{2}+z},\ f_{y}=2xye^{y^{2}+z},\ f_{z}=f.$
Find the four second partial derivatives of the following functions.
$f(x,y)=\cos(xy)$
$f_{x}=-y\sin(xy)$
$f_{y}=-x\sin(xy)$
The second derivatives are:
$f_{xx}=-y^{2}\cos(xy),\ f_{yy}=-x^{2}\cos(xy),\ f_{xy}=f_{yx}=-\sin(xy)-xy\cos(xy)$
$f_{x}=-y\sin(xy)$
$f_{y}=-x\sin(xy)$
The second derivatives are:
$f_{xx}=-y^{2}\cos(xy),\ f_{yy}=-x^{2}\cos(xy),\ f_{xy}=f_{yx}=-\sin(xy)-xy\cos(xy)$
$f(x,y)=xe^{y}$
$f_{x}=e^{y}$
$f_{y}=xe^{y}$
The second derivatives are:
$f_{xx}=0,\ f_{yy}=xe^{y},\ f_{xy}=f_{yx}=e^{y}$
$f_{x}=e^{y}$
$f_{y}=xe^{y}$
The second derivatives are:
$f_{xx}=0,\ f_{yy}=xe^{y},\ f_{xy}=f_{yx}=e^{y}$

## Chain Rule

Find $df/dt.$
$f(x,y)=x^{2}y-xy^{3},\ x(t)=t^{2},\ y(t)=t^{-2}$
${\frac {\partial f}{\partial x}}=2xy-y^{3}$; ${\frac {\partial f}{\partial y}}=x^{2}-3xy^{2}$; ${\frac {dx}{dt}}=2t$; and ${\frac {dy}{dt}}=-2t^{-3}$
The chain rule gives: ${\frac {df}{dt}}={\frac {\partial f}{\partial x}}{\frac {dx}{dt}}+{\frac {\partial f}{\partial y}}{\frac {dy}{dt}}$ $=(2xy-y^{3})(2t)+(x^{2}-3xy^{2})(-2t^{-3})$ $=(2-t^{-6})(2t)+(t^{4}-3t^{-2})(-2t^{-3})$ $=(4t-2t^{-5})+(-2t+6t^{-5})$
$=2t+4t^{-5}$
${\frac {\partial f}{\partial x}}=2xy-y^{3}$; ${\frac {\partial f}{\partial y}}=x^{2}-3xy^{2}$; ${\frac {dx}{dt}}=2t$; and ${\frac {dy}{dt}}=-2t^{-3}$
The chain rule gives: ${\frac {df}{dt}}={\frac {\partial f}{\partial x}}{\frac {dx}{dt}}+{\frac {\partial f}{\partial y}}{\frac {dy}{dt}}$ $=(2xy-y^{3})(2t)+(x^{2}-3xy^{2})(-2t^{-3})$ $=(2-t^{-6})(2t)+(t^{4}-3t^{-2})(-2t^{-3})$ $=(4t-2t^{-5})+(-2t+6t^{-5})$
$=2t+4t^{-5}$
$f(x,y)={\sqrt {x^{2}+y^{2}}},\ x(t)=\cos(2t),\ y(t)=\sin(2t)$
${\frac {\partial f}{\partial x}}={\frac {x}{\sqrt {x^{2}+y^{2}}}}$; ${\frac {\partial f}{\partial y}}={\frac {y}{\sqrt {x^{2}+y^{2}}}}$; ${\frac {dx}{dt}}=-2\sin(2t)$; and ${\frac {dy}{dt}}=2\cos(2t)$
The chain rule gives: ${\frac {df}{dt}}={\frac {\partial f}{\partial x}}{\frac {dx}{dt}}+{\frac {\partial f}{\partial y}}{\frac {dy}{dt}}$ $={\frac {x}{\sqrt {x^{2}+y^{2}}}}(-2\sin(2t))+{\frac {y}{\sqrt {x^{2}+y^{2}}}}(2\cos(2t))$ $={\frac {\cos(2t)}{\sqrt {\cos ^{2}(2t)+\sin ^{2}(2t)}}}(-2\sin(2t))+{\frac {\sin(2t)}{\sqrt {\cos ^{2}(2t)+\sin ^{2}(2t)}}}(2\cos(2t))$ $=-2\cos(2t)\sin(2t)+2\sin(2t)\cos(2t)$
$=0$
${\frac {\partial f}{\partial x}}={\frac {x}{\sqrt {x^{2}+y^{2}}}}$; ${\frac {\partial f}{\partial y}}={\frac {y}{\sqrt {x^{2}+y^{2}}}}$; ${\frac {dx}{dt}}=-2\sin(2t)$; and ${\frac {dy}{dt}}=2\cos(2t)$
The chain rule gives: ${\frac {df}{dt}}={\frac {\partial f}{\partial x}}{\frac {dx}{dt}}+{\frac {\partial f}{\partial y}}{\frac {dy}{dt}}$ $={\frac {x}{\sqrt {x^{2}+y^{2}}}}(-2\sin(2t))+{\frac {y}{\sqrt {x^{2}+y^{2}}}}(2\cos(2t))$ $={\frac {\cos(2t)}{\sqrt {\cos ^{2}(2t)+\sin ^{2}(2t)}}}(-2\sin(2t))+{\frac {\sin(2t)}{\sqrt {\cos ^{2}(2t)+\sin ^{2}(2t)}}}(2\cos(2t))$ $=-2\cos(2t)\sin(2t)+2\sin(2t)\cos(2t)$
$=0$
$\displaystyle f(x,y,z)={\frac {x-y}{y+z}},\ x(t)=t,\ \displaystyle y(t)=2t,\ z(t)=3t$
${\frac {\partial f}{\partial x}}={\frac {1}{y+z}}$; ${\frac {\partial f}{\partial y}}={\frac {(-1)(y+z)-(x-y)(1)}{(y+z)^{2}}}={\frac {-(x+z)}{(y+z)^{2}}}$; ${\frac {\partial f}{\partial z}}={\frac {y-x}{(y+z)^{2}}}$; ${\frac {dx}{dt}}=1$; ${\frac {dy}{dt}}=2$; and ${\frac {dz}{dt}}=3$
The chain rule gives: ${\frac {df}{dt}}={\frac {\partial f}{\partial x}}{\frac {dx}{dt}}+{\frac {\partial f}{\partial y}}{\frac {dy}{dt}}+{\frac {\partial f}{\partial z}}{\frac {dz}{dt}}$ $={\frac {1}{y+z}}(1)+{\frac {-(x+z)}{(y+z)^{2}}}(2)+{\frac {y-x}{(y+z)^{2}}}(3)$ $={\frac {(y+z)-2(x+z)+3(y-x)}{(y+z)^{2}}}$ $={\frac {-5x+4y-z}{(y+z)^{2}}}$ $={\frac {-5t+8t-3t}{(5t)^{2}}}$
$=0$
${\frac {\partial f}{\partial x}}={\frac {1}{y+z}}$; ${\frac {\partial f}{\partial y}}={\frac {(-1)(y+z)-(x-y)(1)}{(y+z)^{2}}}={\frac {-(x+z)}{(y+z)^{2}}}$; ${\frac {\partial f}{\partial z}}={\frac {y-x}{(y+z)^{2}}}$; ${\frac {dx}{dt}}=1$; ${\frac {dy}{dt}}=2$; 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 $f_{s},\ f_{t}.$
$f(x,y)=\sin(x)\cos(2y),\ x=s+t,\ y=s-t$
${\frac {\partial f}{\partial x}}=\cos(x)\cos(2y)$; ${\frac {\partial f}{\partial y}}=-2\sin(x)\sin(2y)$; ${\frac {\partial x}{\partial s}}=1$; ${\frac {\partial x}{\partial t}}=1$; ${\frac {\partial y}{\partial s}}=1$; 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:
$f_{s}=\cos(s+t)\cos(2s-2t)-2\sin(s+t)\sin(2s-2t)$
$f_{t}=\cos(s+t)\cos(2s-2t)+2\sin(s+t)\sin(2s-2t)$
${\frac {\partial f}{\partial x}}=\cos(x)\cos(2y)$; ${\frac {\partial f}{\partial y}}=-2\sin(x)\sin(2y)$; ${\frac {\partial x}{\partial s}}=1$; ${\frac {\partial x}{\partial t}}=1$; ${\frac {\partial y}{\partial s}}=1$; 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:
$f_{s}=\cos(s+t)\cos(2s-2t)-2\sin(s+t)\sin(2s-2t)$
$f_{t}=\cos(s+t)\cos(2s-2t)+2\sin(s+t)\sin(2s-2t)$
$\displaystyle f(x,y,z)={\frac {x-z}{y+z}},\ x(t)=s+t,\ y(t)=st,\ z(t)=s-t$
${\frac {\partial f}{\partial x}}={\frac {1}{y+z}}$; ${\frac {\partial f}{\partial y}}={\frac {z-x}{(y+z)^{2}}}$; ${\frac {\partial f}{\partial z}}={\frac {(-1)(y+z)-(x-z)(1)}{(y+z)^{2}}}={\frac {-(x+y)}{(y+z)^{2}}}$; ${\frac {\partial x}{\partial s}}=1$; ${\frac {\partial x}{\partial t}}=1$; ${\frac {\partial y}{\partial s}}=t$; ${\frac {\partial y}{\partial t}}=s$; ${\frac {\partial z}{\partial s}}=1$; 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:
$f_{s}={\frac {-2t(t+1)}{(st+s-t)^{2}}}$
$f_{t}={\frac {2s}{(st+s-t)^{2}}}$
${\frac {\partial f}{\partial x}}={\frac {1}{y+z}}$; ${\frac {\partial f}{\partial y}}={\frac {z-x}{(y+z)^{2}}}$; ${\frac {\partial f}{\partial z}}={\frac {(-1)(y+z)-(x-z)(1)}{(y+z)^{2}}}={\frac {-(x+y)}{(y+z)^{2}}}$; ${\frac {\partial x}{\partial s}}=1$; ${\frac {\partial x}{\partial t}}=1$; ${\frac {\partial y}{\partial s}}=t$; ${\frac {\partial y}{\partial t}}=s$; ${\frac {\partial z}{\partial s}}=1$; 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:
$f_{s}={\frac {-2t(t+1)}{(st+s-t)^{2}}}$
$f_{t}={\frac {2s}{(st+s-t)^{2}}}$
$V={\frac {1}{3}}x^{2}h$
$\displaystyle x(t)={\frac {t}{t+1}}$
$\displaystyle h(t)={\frac {1}{t+1}}$
$t\geq 0.$
$V'(t).$
${\frac {\partial V}{\partial x}}={\frac {2}{3}}xh$; ${\frac {\partial V}{\partial h}}={\frac {1}{3}}x^{2}$; $x'(t)={\frac {(1)(t+1)-t(1)}{(t+1)^{2}}}={\frac {1}{(t+1)^{2}}}$; and $h'(t)={\frac {-1}{(t+1)^{2}}}$
The chain rule gives: $V'(t)={\frac {\partial V}{\partial x}}x'(t)+{\frac {\partial V}{\partial h}}h'(t)$ $={\frac {2}{3}}xh{\frac {1}{(t+1)^{2}}}+{\frac {1}{3}}x^{2}{\frac {-1}{(t+1)^{2}}}$ $={\frac {2t}{3(t+1)^{4}}}-{\frac {t^{2}}{3(t+1)^{4}}}$
$={\frac {2t-t^{2}}{3(t+1)^{4}}}$
${\frac {\partial V}{\partial x}}={\frac {2}{3}}xh$; ${\frac {\partial V}{\partial h}}={\frac {1}{3}}x^{2}$; $x'(t)={\frac {(1)(t+1)-t(1)}{(t+1)^{2}}}={\frac {1}{(t+1)^{2}}}$; 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).
$xy\sin(z)=1,\ (1,2,\pi /6),\ (-1,-2,5\pi /6).$
$(x_{0},y_{0},z_{0})$
$x$
$y$
$z$
$dx$
$dy$
$dz$
$xy\sin(z)$
$d(xy\sin(z))=(dx)y_{0}\sin(z_{0})+x_{0}(dy)\sin(z_{0})+x_{0}y_{0}d(\sin(z))=(y_{0}\sin(z_{0}))dx+(x_{0}\sin(z_{0}))dy+(x_{0}y_{0}\cos(z_{0}))dz$
$1$
$d(1)=0$
$d(xy\sin(z))=d(1)$
$\iff (y_{0}\sin(z_{0}))dx+(x_{0}\sin(z_{0}))dy+(x_{0}y_{0}\cos(z_{0}))dz=0$
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:
$(x-1)+{\frac {1}{2}}(y-2)+{\sqrt {3}}(z-\pi /6)=0,\ (x+1)+{\frac {1}{2}}(y+2)+{\sqrt {3}}(z-5\pi /6)=0$
$(x_{0},y_{0},z_{0})$
$x$
$y$
$z$
$dx$
$dy$
$dz$
$xy\sin(z)$
$d(xy\sin(z))=(dx)y_{0}\sin(z_{0})+x_{0}(dy)\sin(z_{0})+x_{0}y_{0}d(\sin(z))=(y_{0}\sin(z_{0}))dx+(x_{0}\sin(z_{0}))dy+(x_{0}y_{0}\cos(z_{0}))dz$
$1$
$d(1)=0$
$d(xy\sin(z))=d(1)$
$\iff (y_{0}\sin(z_{0}))dx+(x_{0}\sin(z_{0}))dy+(x_{0}y_{0}\cos(z_{0}))dz=0$
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:
$(x-1)+{\frac {1}{2}}(y-2)+{\sqrt {3}}(z-\pi /6)=0,\ (x+1)+{\frac {1}{2}}(y+2)+{\sqrt {3}}(z-5\pi /6)=0$
$z=x^{2}e^{x-y},\ (2,2,4),\ (-1,-1,1).$
$(x_{0},y_{0},z_{0})$
$x$
$y$
$z$
$dx$
$dy$
$dz$
$z$
$dz$
$x^{2}e^{x-y}$
$d(x^{2}e^{x-y})=d(x^{2})e^{x_{0}-y_{0}}+x_{0}^{2}d(e^{x-y})$
$=(2x_{0}dx)e^{x_{0}-y_{0}}+x_{0}^{2}(e^{x_{0}-y_{0}}d(x-y))$
$=(2x_{0}e^{x_{0}-y_{0}})dx+x_{0}^{2}(e^{x_{0}-y_{0}}(dx-dy))$
$=(2x_{0}+x_{0}^{2})e^{x_{0}-y_{0}}dx-x_{0}^{2}e^{x_{0}-y_{0}}dy$
$dz=d(x^{2}e^{x-y})$
$\iff dz=(2x_{0}+x_{0}^{2})e^{x_{0}-y_{0}}dx-x_{0}^{2}e^{x_{0}-y_{0}}dy$
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:
$-8(x-2)+4(y-2)+z-4=0,\ x+y+z+1=0$
$(x_{0},y_{0},z_{0})$
$x$
$y$
$z$
$dx$
$dy$
$dz$
$z$
$dz$
$x^{2}e^{x-y}$
$d(x^{2}e^{x-y})=d(x^{2})e^{x_{0}-y_{0}}+x_{0}^{2}d(e^{x-y})$
$=(2x_{0}dx)e^{x_{0}-y_{0}}+x_{0}^{2}(e^{x_{0}-y_{0}}d(x-y))$
$=(2x_{0}e^{x_{0}-y_{0}})dx+x_{0}^{2}(e^{x_{0}-y_{0}}(dx-dy))$
$=(2x_{0}+x_{0}^{2})e^{x_{0}-y_{0}}dx-x_{0}^{2}e^{x_{0}-y_{0}}dy$
$dz=d(x^{2}e^{x-y})$
$\iff dz=(2x_{0}+x_{0}^{2})e^{x_{0}-y_{0}}dx-x_{0}^{2}e^{x_{0}-y_{0}}dy$
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:
$-8(x-2)+4(y-2)+z-4=0,\ x+y+z+1=0$
$z=\tan ^{-1}(x+y),\ (0,0,0).$
$(x_{0},y_{0},z_{0})$
$x$
$y$
$z$
$dx$
$dy$
$dz$
$z$
$dz$
$\tan ^{-1}(x+y)$
$d(\tan ^{-1}(x+y))={\frac {1}{1+(x_{0}+y_{0})^{2}}}d(x+y)={\frac {dx+dy}{1+(x_{0}+y_{0})^{2}}}$
$dz=d(\tan ^{-1}(x+y))$
$\iff dz={\frac {dx+dy}{1+(x_{0}+y_{0})^{2}}}$
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 $(z-z_{0})={\frac {(x-x_{0})+(y-y_{0})}{1+(x_{0}+y_{0})^{2}}}$.
$(x_{0},y_{0},z_{0})=(0,0,0)$
$z=x+y$
$\iff x+y-z=0$
$(x_{0},y_{0},z_{0})$
$x$
$y$
$z$
$dx$
$dy$
$dz$
$z$
$dz$
$\tan ^{-1}(x+y)$
$d(\tan ^{-1}(x+y))={\frac {1}{1+(x_{0}+y_{0})^{2}}}d(x+y)={\frac {dx+dy}{1+(x_{0}+y_{0})^{2}}}$
$dz=d(\tan ^{-1}(x+y))$
$\iff dz={\frac {dx+dy}{1+(x_{0}+y_{0})^{2}}}$
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 $(z-z_{0})={\frac {(x-x_{0})+(y-y_{0})}{1+(x_{0}+y_{0})^{2}}}$.
$(x_{0},y_{0},z_{0})=(0,0,0)$
$z=x+y$
$\iff x+y-z=0$
$\sin(xyz)=1/2,\ (\pi ,1,1/6).$
$(x_{0},y_{0},z_{0})$
$x$
$y$
$z$
$dx$
$dy$
$dz$
$\sin(xyz)$
$d(\sin(xyz))=\cos(x_{0}y_{0}z_{0})d(xyz)=\cos(x_{0}y_{0}z_{0})((dx)y_{0}z_{0}+x_{0}(dy)z_{0}+x_{0}y_{0}(dz))=(y_{0}z_{0}\cos(x_{0}y_{0}z_{0}))dx+(x_{0}z_{0}\cos(x_{0}y_{0}z_{0}))dy+(x_{0}y_{0}\cos(x_{0}y_{0}z_{0}))dz$
$1/2$
$d(1/2)=0$
$d(\sin(xyz))=d(1/2)$
$\iff (y_{0}z_{0}\cos(x_{0}y_{0}z_{0}))dx+(x_{0}z_{0}\cos(x_{0}y_{0}z_{0}))dy+(x_{0}y_{0}\cos(x_{0}y_{0}z_{0}))dz=0$
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 $(y_{0}z_{0}\cos(x_{0}y_{0}z_{0}))(x-x_{0})+(x_{0}z_{0}\cos(x_{0}y_{0}z_{0}))(y-y_{0})+(x_{0}y_{0}\cos(x_{0}y_{0}z_{0}))(z-z_{0})=0$.
$(x_{0},y_{0},z_{0})=(\pi ,1,1/6)$
${\frac {\sqrt {3}}{12}}(x-\pi )+{\frac {\pi {\sqrt {3}}}{12}}(y-1)+{\frac {\pi {\sqrt {3}}}{2}}(z-1/6)=0$
$\iff (x-\pi )+\pi (y-1)+6\pi (z-1/6)=0$
$\iff x+\pi y+6\pi z=3\pi$
$(x_{0},y_{0},z_{0})$
$x$
$y$
$z$
$dx$
$dy$
$dz$
$\sin(xyz)$
$d(\sin(xyz))=\cos(x_{0}y_{0}z_{0})d(xyz)=\cos(x_{0}y_{0}z_{0})((dx)y_{0}z_{0}+x_{0}(dy)z_{0}+x_{0}y_{0}(dz))=(y_{0}z_{0}\cos(x_{0}y_{0}z_{0}))dx+(x_{0}z_{0}\cos(x_{0}y_{0}z_{0}))dy+(x_{0}y_{0}\cos(x_{0}y_{0}z_{0}))dz$
$1/2$
$d(1/2)=0$
$d(\sin(xyz))=d(1/2)$
$\iff (y_{0}z_{0}\cos(x_{0}y_{0}z_{0}))dx+(x_{0}z_{0}\cos(x_{0}y_{0}z_{0}))dy+(x_{0}y_{0}\cos(x_{0}y_{0}z_{0}))dz=0$
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 $(y_{0}z_{0}\cos(x_{0}y_{0}z_{0}))(x-x_{0})+(x_{0}z_{0}\cos(x_{0}y_{0}z_{0}))(y-y_{0})+(x_{0}y_{0}\cos(x_{0}y_{0}z_{0}))(z-z_{0})=0$.
$(x_{0},y_{0},z_{0})=(\pi ,1,1/6)$
${\frac {\sqrt {3}}{12}}(x-\pi )+{\frac {\pi {\sqrt {3}}}{12}}(y-1)+{\frac {\pi {\sqrt {3}}}{2}}(z-1/6)=0$
$\iff (x-\pi )+\pi (y-1)+6\pi (z-1/6)=0$
$\iff x+\pi y+6\pi z=3\pi$

## 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.
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".
$f(x,y)=x^{4}+2y^{2}-4xy$
${\frac {\partial f}{\partial x}}=4x^{3}-4y$
${\frac {\partial f}{\partial y}}=4y-4x$
$\left\{{\begin{array}{c}4x^{3}-4y=0\\4y-4x=0\end{array}}\right.$
$4y-4x=0\iff y=x$. 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.
${\frac {\partial f}{\partial x}}=4x^{3}-4y$
${\frac {\partial f}{\partial y}}=4y-4x$
$\left\{{\begin{array}{c}4x^{3}-4y=0\\4y-4x=0\end{array}}\right.$
$4y-4x=0\iff y=x$. 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.
$f(x,y)=\tan ^{-1}(xy)$
${\frac {\partial f}{\partial x}}={\frac {1}{1+(xy)^{2}}}y={\frac {y}{1+(xy)^{2}}}$
${\frac {\partial f}{\partial y}}={\frac {1}{1+(xy)^{2}}}x={\frac {x}{1+(xy)^{2}}}$
$\left\{{\begin{array}{c}y/(1+(xy)^{2})=0\\x/(1+(xy)^{2})=0\end{array}}\right.$
${\frac {y}{1+(xy)^{2}}}=0\iff y=0$ 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.
${\frac {\partial f}{\partial x}}={\frac {1}{1+(xy)^{2}}}y={\frac {y}{1+(xy)^{2}}}$
${\frac {\partial f}{\partial y}}={\frac {1}{1+(xy)^{2}}}x={\frac {x}{1+(xy)^{2}}}$
$\left\{{\begin{array}{c}y/(1+(xy)^{2})=0\\x/(1+(xy)^{2})=0\end{array}}\right.$
${\frac {y}{1+(xy)^{2}}}=0\iff y=0$ 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.
$f(x,y)=2xye^{-x^{2}-y^{2}}$
${\frac {\partial f}{\partial x}}=2ye^{-x^{2}-y^{2}}+(2xy)e^{-x^{2}-y^{2}}(-2x)=(2y-4x^{2}y)e^{-x^{2}-y^{2}}$
${\frac {\partial f}{\partial y}}=2xe^{-x^{2}-y^{2}}+(2xy)e^{-x^{2}-y^{2}}(-2y)=(2x-4xy^{2})e^{-x^{2}-y^{2}}$
$\left\{{\begin{array}{c}(2y-4x^{2}y)e^{-x^{2}-y^{2}}=0\\(2x-4xy^{2})e^{-x^{2}-y^{2}}=0\end{array}}\right.$
$\left\{{\begin{array}{c}(2y-4x^{2}y)e^{-x^{2}-y^{2}}=0\\(2x-4xy^{2})e^{-x^{2}-y^{2}}=0\end{array}}\right.\iff \left\{{\begin{array}{c}2y(1-2x^{2})=0\\2x(1-2y^{2})=0\end{array}}\right.$ $\iff \left\{{\begin{array}{c}y=0\;{\text{or}}\;x=-1/{\sqrt {2}}\;{\text{or}}\;x=1/{\sqrt {2}}\\x=0\;{\text{or}}\;y=-1/{\sqrt {2}}\;{\text{or}}\;y=1/{\sqrt {2}}\end{array}}\right.$. 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.
$(\pm 1/{\sqrt {2}},\pm 1/{\sqrt {2}}),$
$(\pm 1/{\sqrt {2}},\mp 1/{\sqrt {2}})$
${\frac {\partial f}{\partial x}}=2ye^{-x^{2}-y^{2}}+(2xy)e^{-x^{2}-y^{2}}(-2x)=(2y-4x^{2}y)e^{-x^{2}-y^{2}}$
${\frac {\partial f}{\partial y}}=2xe^{-x^{2}-y^{2}}+(2xy)e^{-x^{2}-y^{2}}(-2y)=(2x-4xy^{2})e^{-x^{2}-y^{2}}$
$\left\{{\begin{array}{c}(2y-4x^{2}y)e^{-x^{2}-y^{2}}=0\\(2x-4xy^{2})e^{-x^{2}-y^{2}}=0\end{array}}\right.$
$\left\{{\begin{array}{c}(2y-4x^{2}y)e^{-x^{2}-y^{2}}=0\\(2x-4xy^{2})e^{-x^{2}-y^{2}}=0\end{array}}\right.\iff \left\{{\begin{array}{c}2y(1-2x^{2})=0\\2x(1-2y^{2})=0\end{array}}\right.$ $\iff \left\{{\begin{array}{c}y=0\;{\text{or}}\;x=-1/{\sqrt {2}}\;{\text{or}}\;x=1/{\sqrt {2}}\\x=0\;{\text{or}}\;y=-1/{\sqrt {2}}\;{\text{or}}\;y=1/{\sqrt {2}}\end{array}}\right.$. 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.
$(\pm 1/{\sqrt {2}},\pm 1/{\sqrt {2}}),$
$(\pm 1/{\sqrt {2}},\mp 1/{\sqrt {2}})$
Find absolute maximum and minimum values of the function f on the set R.
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.
$f(x,y)=x^{2}+y^{2}-2y+1,\ R=\{(x,y)\mid x^{2}+y^{2}\leq 4\}$
$f(x,y)=x^{2}+y^{2}-2y+1$
$g(x,y)=x^{2}+y^{2}-4=0$
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 $(0,-2)$.
$f(x,y)=x^{2}+y^{2}-2y+1$
$g(x,y)=x^{2}+y^{2}-4=0$
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 $(0,-2)$.
$f(x,y)=x^{2}+y^{2}-2x-2y,$
$R$
$x\geq 0$
$y\geq 0$
$x+y\leq 2$
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 $(0,0),(2,0),(0,2)$.
$R$
$x\geq 0$
$y\geq 0$
$x+y\leq 2$
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 $(0,0),(2,0),(0,2)$.
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 $\lambda _{2}$.
$f(x,y,z)=x-y+z-2=0$
$\left\langle {\frac {\partial f}{\partial x}},{\frac {\partial f}{\partial y}},{\frac {\partial f}{\partial z}}\right\rangle =\langle 1,-1,1\rangle$
$(x,y,z)$
$(1,1,1)$
$\langle x-1,y-1,z-1\rangle$
$\langle 1,-1,1\rangle$
$\left\{{\begin{array}{c}x-y+z-2=0\\x-1=\lambda (1)\;{\text{and}}\;y-1=\lambda (-1)\;{\text{and}}\;z-1=\lambda (1)\end{array}}\right.$
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 $z=4/3$.
$(4/3,2/3,4/3)$
$x-y+z=2$
$(1,1,1)$
$(4/3,2/3,4/3)$
$f(x,y,z)=x-y+z-2=0$
$\left\langle {\frac {\partial f}{\partial x}},{\frac {\partial f}{\partial y}},{\frac {\partial f}{\partial z}}\right\rangle =\langle 1,-1,1\rangle$
$(x,y,z)$
$(1,1,1)$
$\langle x-1,y-1,z-1\rangle$
$\langle 1,-1,1\rangle$
$\left\{{\begin{array}{c}x-y+z-2=0\\x-1=\lambda (1)\;{\text{and}}\;y-1=\lambda (-1)\;{\text{and}}\;z-1=\lambda (1)\end{array}}\right.$
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 $z=4/3$.
$(4/3,2/3,4/3)$
$x-y+z=2$
$(1,1,1)$
$(4/3,2/3,4/3)$
$z=x^{2}+y^{2}+10$
$x+2y-z=0.$
$f_{1}(x,y,z)=x^{2}+y^{2}-z+10=0$
$f_{2}(x,y,z)=x+2y-z=0$
$(x_{1},y_{1},z_{1})$
$(x_{2},y_{2},z_{2})$
$\langle x_{2}-x_{1},y_{2}-y_{1},z_{2}-z_{1}\rangle$
$\left\langle {\frac {\partial f_{1}}{\partial x}},{\frac {\partial f_{1}}{\partial y}},{\frac {\partial f_{1}}{\partial z}}\right\rangle =\langle 2x_{1},2y_{1},-1\rangle$
$\left\langle {\frac {\partial f_{2}}{\partial x}},{\frac {\partial f_{2}}{\partial y}},{\frac {\partial f_{2}}{\partial z}}\right\rangle =\langle 1,2,-1\rangle$
$\lambda _{1}$
$\lambda _{2}$
$\left\{{\begin{array}{c}x_{1}^{2}+y_{1}^{2}-z_{1}+10=0\\x_{2}+2y_{2}-z_{2}=0\\x_{2}-x_{1}=\lambda _{1}(2x_{1})\;{\text{and}}\;y_{2}-y_{1}=\lambda _{1}(2y_{1})\;{\text{and}}\;z_{2}-z_{1}=\lambda _{1}(-1)\\x_{2}-x_{1}=\lambda _{2}(1)\;{\text{and}}\;y_{2}-y_{1}=\lambda _{2}(2)\;{\text{and}}\;z_{2}-z_{1}=\lambda _{2}(-1)\end{array}}\right.$
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 $(x_{2},y_{2},z_{2})=(1.958{\overline {3}},3.91{\overline {6}},9.791{\overline {6}})$.
$(0.5,1,11.25)$
$f_{1}(x,y,z)=x^{2}+y^{2}-z+10=0$
$f_{2}(x,y,z)=x+2y-z=0$
$(x_{1},y_{1},z_{1})$
$(x_{2},y_{2},z_{2})$
$\langle x_{2}-x_{1},y_{2}-y_{1},z_{2}-z_{1}\rangle$
$\left\langle {\frac {\partial f_{1}}{\partial x}},{\frac {\partial f_{1}}{\partial y}},{\frac {\partial f_{1}}{\partial z}}\right\rangle =\langle 2x_{1},2y_{1},-1\rangle$
$\left\langle {\frac {\partial f_{2}}{\partial x}},{\frac {\partial f_{2}}{\partial y}},{\frac {\partial f_{2}}{\partial z}}\right\rangle =\langle 1,2,-1\rangle$
$\lambda _{1}$
$\lambda _{2}$
$\left\{{\begin{array}{c}x_{1}^{2}+y_{1}^{2}-z_{1}+10=0\\x_{2}+2y_{2}-z_{2}=0\\x_{2}-x_{1}=\lambda _{1}(2x_{1})\;{\text{and}}\;y_{2}-y_{1}=\lambda _{1}(2y_{1})\;{\text{and}}\;z_{2}-z_{1}=\lambda _{1}(-1)\\x_{2}-x_{1}=\lambda _{2}(1)\;{\text{and}}\;y_{2}-y_{1}=\lambda _{2}(2)\;{\text{and}}\;z_{2}-z_{1}=\lambda _{2}(-1)\end{array}}\right.$
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 $(x_{2},y_{2},z_{2})=(1.958{\overline {3}},3.91{\overline {6}},9.791{\overline {6}})$.
$(0.5,1,11.25)$

## Double Integrals over Rectangular Regions

Evaluate the given integral over the region R.
$\displaystyle \iint _{R}(x^{2}+xy)dA,\ R=\{(x,y)\mid x\in [1,2],\ y\in [-1,1]\}$
$14/3$
$14/3$
$\displaystyle \iint _{R}(xy\sin(x^{2}))dA,\ R=\{(x,y)\mid x\in [0,{\sqrt {\pi /2}}],\ y\in [0,1]\}$
$1/4$
$1/4$
$\displaystyle \iint _{R}{\frac {x}{(1+xy)^{2}}}dA,\ R=\{(x,y)\mid x\in [0,4],\ y\in [1,2]\}$
$\ln(5/3)$
$\ln(5/3)$
Evaluate the given iterated integrals.
$\displaystyle \int _{0}^{2}\int _{0}^{1}x^{5}y^{2}e^{x^{3}y^{3}}dydx$
$(e^{8}-9)/9$
$(e^{8}-9)/9$
$\displaystyle \int _{1}^{4}\int _{0}^{2}e^{y{\sqrt {x}}}dydx$
$e^{4}-e^{2}-2$
$e^{4}-e^{2}-2$

## Double Integrals over General Regions

Evaluate the following integrals.
$\displaystyle \iint _{R}xydA,$
$2$
$2$
$\displaystyle \iint _{R}(x+y)dA,$
$y=x^{2},$
$y=8-x^{2}.$
$152/3$
$152/3$
Use double integrals to compute the volume of the given region.
$z=8-x^{2}-2y^{2}.$
$4\pi {\sqrt {2}}$
$4\pi {\sqrt {2}}$
$z=y^{2}$
$R=\{(x,y)\mid y\in [0,1],\ x\in [y,1]\}.$
$1/12$
$1/12$
$z=x^{2}+y^{2}$
$z=50-x^{2}-y^{2}.$
$625\pi$
$625\pi$

## Double Integrals in Polar Coordinates

$\displaystyle \iint _{R}2xydA$
$R=\{(r,\theta )\mid r\in [1,3],\ \theta \in [0,\pi /2]\}$
$20$
$20$
$f(r,\theta )=1/r^{2}$
$\{(r,\theta )\mid r\in [2,4]\}.$
$\displaystyle {\frac {\ln 2}{6}}$
$\displaystyle {\frac {\ln 2}{6}}$
$\displaystyle \int _{0}^{3}\int _{0}^{\sqrt {9-x^{2}}}{\sqrt {x^{2}+y^{2}}}dydx.$
$9\pi /2$
$9\pi /2$
$\displaystyle \iint _{R}{\frac {x-y}{x^{2}+y^{2}+1}}dA$
$0$
$0$

## Triple Integrals

$\displaystyle \int _{1}^{\ln 8}\int _{0}^{\ln 4}\int _{0}^{\ln 2}e^{-x-y-2z}dxdydz.$
$\displaystyle {\frac {3}{16}}\left(e^{-2}-8^{-2}\right)$
$\displaystyle {\frac {3}{16}}\left(e^{-2}-8^{-2}\right)$
In the following exercises, sketching the region of integration may be helpful.
$8$
$8$
$z=\sin(y)$
$y\in [0,\pi ]$
$\pi$
$\pi$
$\displaystyle \int _{0}^{1}\int _{y}^{2-y}\int _{0}^{2-x-y}xydzdxdy.$
$2/15$
$2/15$
$\displaystyle \int _{0}^{1}\int _{-2}^{2}\int _{0}^{\sqrt {4-y^{2}}}dzdydx$
$\displaystyle \int _{0}^{1}\int _{0}^{2}\int _{-{\sqrt {4-z^{2}}}}^{\sqrt {4-z^{2}}}dydzdx$
$\displaystyle \int _{0}^{1}\int _{0}^{2}\int _{-{\sqrt {4-z^{2}}}}^{\sqrt {4-z^{2}}}dydzdx$

## Cylindrical And Spherical Coordinates

$\displaystyle \int _{0}^{3}\int _{0}^{\sqrt {9-x^{2}}}\int _{0}^{\sqrt {x^{2}+y^{2}}}{\frac {1}{\sqrt {x^{2}+y^{2}}}}dzdydx$
$9\pi /4$
$9\pi /4$
$D=\{(r,\theta ,z)\mid r\in [0,3],\ z\in [0,2]\}$
$\delta (r,\theta ,z)=5e^{-r^{2}}$
$10\pi (1-e^{-9})$
$10\pi (1-e^{-9})$
$z={\sqrt {17}}-{\sqrt {1+x^{2}+y^{2}}}$
$\displaystyle {\frac {2\pi (1+7{\sqrt {17}})}{3}}$
$\displaystyle {\frac {2\pi (1+7{\sqrt {17}})}{3}}$
$\iiint _{D}(x^{2}+y^{2}+z^{2})^{5/2}dV$
$\pi /2$
$\pi /2$
$\{(\rho ,\phi ,\theta )\mid \phi \leq \pi /3,\ z\in [0,4]\}$
$\delta (\rho ,\phi ,\theta )=5-z$
$128\pi$
$128\pi$
$x^{2}+z^{2}=1,\ y^{2}+z^{2}=1$
$16/3$
$16/3$

## Center of Mass and Centroid

${\frac {\langle 3,5,7\rangle }{4}}$
${\frac {\langle 3,5,7\rangle }{4}}$
$\rho (x)=1+\sin(x)$
$x\in [0,\pi ].$
$\pi /2$
$\pi /2$
$\rho (x)=2-x^{2}/16$
$x\in [0,4].$
$9/5$
$9/5$
$x^{2}+y^{2}=16.$
$\left({\frac {16}{3\pi }},{\frac {16}{3\pi }}\right)$
$\left({\frac {16}{3\pi }},{\frac {16}{3\pi }}\right)$
$y=\ln(x)$
$y=0$
$x=e$
$((e^{2}+1)/4,e/2-1)$
$((e^{2}+1)/4,e/2-1)$
$\{(x,y)\mid x\in [0,4],y\in [0,2]\}$
$\rho (x,y)=1+x/2.$
$(7/3,1)$
$(7/3,1)$
$\rho (x,y)=1+x+y.$
$(16/11,16/11)$
$(16/11,16/11)$

## Vector Fields

One can sketch two-dimensional vector fields by plotting vector values, flow curves, and/or equipotential curves.
$\mathbf {F} =\nabla \phi$
$\phi (x,y)={\sqrt {x^{2}+y^{2}}}$
$\mathbf {F} =\left\langle {\frac {x}{\sqrt {x^{2}+y^{2}}}},{\frac {y}{\sqrt {x^{2}+y^{2}}}}\right\rangle$
$\mathbf {F} =\left\langle {\frac {x}{\sqrt {x^{2}+y^{2}}}},{\frac {y}{\sqrt {x^{2}+y^{2}}}}\right\rangle$
$\mathbf {F} =\nabla \phi$
$\phi (x,y)=\sin(x)\sin(y)$
$|x|\leq \pi$
$|y|\leq \pi$
$\nabla \phi (x,y)=\langle \cos(x)\sin(y),\sin(x)\cos(y)\rangle$
$\nabla \phi (x,y)=\langle \cos(x)\sin(y),\sin(x)\cos(y)\rangle$
$\mathbf {F} =\nabla \phi$
$\phi (x,y,z)=e^{-z}\sin(x+y)$
$\mathbf {F} =e^{-z}\left\langle \cos(x+y),\cos(x+y),-\sin(x+y)\right\rangle$
$\mathbf {F} =e^{-z}\left\langle \cos(x+y),\cos(x+y),-\sin(x+y)\right\rangle$

## Line Integrals

$\int _{C}(x^{2}+y^{2})ds$
${\frac {250{\sqrt {2}}}{3}}$
${\frac {250{\sqrt {2}}}{3}}$
$\int _{C}(x^{2}+y^{2})ds$
$128\pi$
$128\pi$
$\int _{C}(y-z)ds$
$\mathbf {r} (t)=\langle 3\cos(t),3\sin(t),t\rangle ,\ t\in [0,2\pi ]$
$-2{\sqrt {10}}\pi ^{2}$
$-2{\sqrt {10}}\pi ^{2}$
$\int _{C}\mathbf {F} \cdot d\mathbf {r}$
$\mathbf {F} =\langle x,y\rangle$
$\mathbf {r} (t)=\langle 4t,t^{2}\rangle ,\ t\in [0,1]$
$17/2$
$17/2$
$\displaystyle \mathbf {F} ={\frac {\langle x,y,z\rangle }{x^{2}+y^{2}+z^{2}}}$
$\ln(2{\sqrt {7}})$
$\ln(2{\sqrt {7}})$

## Conservative Vector Fields

Determine if the following vector fields are conservative on $\mathbb {R} ^{2}.$
$\langle -y,x+y\rangle$
$\langle 2x^{3}+xy^{2},2y^{3}+x^{2}y\rangle$
Determine if the following vector fields are conservative on their respective domains in $\mathbb {R} ^{3}.$ When possible, find the potential function.
$\langle y,x,1\rangle$
$\phi (x,y,z)=xy+z$
$\phi (x,y,z)=xy+z$
$\langle x^{3},2y,-z^{3}\rangle$
$\phi (x,y,z)=(x^{4}+4y^{2}-z^{4})/4$
$\phi (x,y,z)=(x^{4}+4y^{2}-z^{4})/4$

## Green's Theorem

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

## Divergence And Curl

$\langle 2x,4y,-3z\rangle$
$3$
$3$
$\displaystyle {\frac {\langle x,y,z\rangle }{1+x^{2}+y^{2}}}$
$\displaystyle {\frac {x^{2}+y^{2}+3}{(1+x^{2}+y^{2})^{2}}}$
$\displaystyle {\frac {x^{2}+y^{2}+3}{(1+x^{2}+y^{2})^{2}}}$
$\langle x^{2}-y^{2},xy,z\rangle$
$\langle 0,0,3y\rangle$
$\langle 0,0,3y\rangle$
$\langle z^{2}\sin(y),xz^{2}\cos(y),2xz\sin(y)\rangle$
$\mathbf {0}$
$\mathbf {0}$
$\mathbf {F} =\mathbf {a} \times \mathbf {r}$
$\mathbf {a}$
$\mathbf {r} =\langle x,y,z\rangle$
$\mathbf {F}$
$2\mathbf {a}$
$\mathbf {a} =\langle a_{1},a_{2},a_{3}\rangle$
$\mathbf {F} =\mathbf {a} \times \mathbf {r} =\langle a_{2}z-a_{3}y,a_{3}x-a_{1}z,a_{1}y-a_{2}x\rangle =\langle f,g,h\rangle$, and then
$\nabla \cdot \mathbf {F} =\mathbf {f} _{x}+\mathbf {g} _{y}+\mathbf {h} _{z}=0+0+0=0,$
$\nabla \times \mathbf {F} =\langle h_{y}-g_{z},f_{z}-h_{x},g_{x}-f_{y}\rangle =\langle a_{1}+a_{1},a_{2}+a_{2},a_{3}+a_{3}\rangle =2\mathbf {a} .$
$\mathbf {a} =\langle a_{1},a_{2},a_{3}\rangle$
$\mathbf {F} =\mathbf {a} \times \mathbf {r} =\langle a_{2}z-a_{3}y,a_{3}x-a_{1}z,a_{1}y-a_{2}x\rangle =\langle f,g,h\rangle$, 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

$2x-4y+3z=16.$
$\langle u,v,(16-2u+4v)/3\rangle ,\ u,v\in \mathbb {R}$
$\langle u,v,(16-2u+4v)/3\rangle ,\ u,v\in \mathbb {R}$
$z^{2}=1+x^{2}+y^{2}.$
$\langle {\sqrt {v^{2}-1}}\cos(u),{\sqrt {v^{2}-1}}\sin(u),v\rangle ,\ u\in [0,2\pi ],\ |v|\geq 1$
$\langle {\sqrt {v^{2}-1}}\cos(u),{\sqrt {v^{2}-1}}\sin(u),v\rangle ,\ u\in [0,2\pi ],\ |v|\geq 1$
$f(x,y,z)=xy$
$2/{\sqrt {3}}$
$2/{\sqrt {3}}$
$f(x,y,z)=x^{2}+y^{2}$
$z=x^{2}+y^{2},\ z\in [0,4].$
${\frac {(391{\sqrt {17}}+1)\pi }{60}}$
${\frac {(391{\sqrt {17}}+1)\pi }{60}}$
$\mathbf {F} =\langle x,y,z\rangle$
    $z^{2}=x^{2}+y^{2},\ z\in [0,1],$
$0$
$0$
$\mathbf {F} =\langle -y,z,1\rangle$
    $y=x^{2},\ z\in [0,4],\ x\in [0,1],$
$-10$
$-10$

## Stokes' Theorem

$\mathbf {F} =\langle x^{2}-z^{2},y,2xz\rangle$
$z=4-x-y$
${\frac {-128}{3}}$
${\frac {-128}{3}}$
$\mathbf {F} =\langle y^{2},-z^{2},x\rangle$
$\mathbf {r} (t)=\langle 3\cos(t),4\cos(t),5\sin(t)\rangle .$
$15\pi$
$15\pi$
$\iint _{S}(\nabla \times F)\cdot \mathbf {n} dS$
$\mathbf {F} =\langle x,y,z\rangle$
$S$
${\frac {x^{2}}{4}}+{\frac {y^{2}}{9}}+z^{2}=1$
$\mathbf {n}$
$0$
$0$
$\iint _{S}(\nabla \times F)\cdot \mathbf {n} dS$
$\mathbf {F} =\langle 2y,-z,x-y-z\rangle$
$S$
$x^{2}+y^{2}+z^{2}=25$
$3\leq z\leq 5$
$\mathbf {n}$
$-32\pi$
$-32\pi$

## Divergence Theorem

Compute the net outward flux of the given field across the given surface.
$\mathbf {F} =\langle x,-2y,3z\rangle$
$S$
${\sqrt {6}}$
$16{\sqrt {6}}\pi$
$16{\sqrt {6}}\pi$
$\mathbf {F} =\langle x,2y,z\rangle$
$S$
$x+y+z=1$
$2/3$
$2/3$
$\mathbf {F} =\langle y+z,x+z,x+y\rangle$
$S$
$\{(x,y,z)\mid |x|\leq 1,|y|\leq 1,|z|\leq 1\}$
$0$
$0$
$\mathbf {F} =\langle x,y,z\rangle$
$S$
$z=4-x^{2}-y^{2}$
$24\pi$
$24\pi$
$\mathbf {F} =\langle z-x,x-y,2y-z\rangle$
$S$
$-224\pi$
$-224\pi$
$\mathbf {F} =\langle x,2y,3z\rangle$
$S$
$x^{2}+y^{2}=1$
$x^{2}+y^{2}=4$
$z=0$
$z=8$
$144\pi$
$144\pi$
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