233 Final

1 Space Coordinates, Regions, and Surfaces

1.1 The Right-Hand Rule for Orienting Coordinate Axes

Before anything else in this course, you need to know which direction is which in 3D space. The right-hand rule establishes the standard orientation of the coordinate axes:

KEY CONCEPT

Hold your right hand so your fingers point along the positive $x$ -axis. Curl your fingers toward the positive $y$ -axis. Your thumb points in the direction of the positive $z$ -axis.

This is the convention used throughout multivariable calculus. If you set up your axes with the left hand instead, every cross product, every normal vector, and every orientation-dependent formula will give you the wrong sign.

The same right-hand rule reappears later for the cross product direction (Stage 3), but here it's specifically about knowing which way your axes point. On an exam, if you're unsure about the sign of a coordinate, hold up your right hand.

1.2 Distance and Midpoint Formulas in 3D

The distance formula is just the Pythagorean theorem applied three times — once for each axis direction:

FORMULA

Distance between $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ :

d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}

Midpoint:

M = \left(\frac{x_1+x_2}{2},\; \frac{y_1+y_2}{2},\; \frac{z_1+z_2}{2}\right)

Try It Yourself

Find the distance between $A(3, -1, 4)$ and $B(7, 2, -2)$ , and find the midpoint of $\overline{AB}$ .

Reveal Solution

Distance: I need the differences in each coordinate: $\Delta x = 7-3=4$ , $\Delta y = 2-(-1)=3$ , $\Delta z = -2-4=-6$ . Then:

d = \sqrt{4^2 + 3^2 + (-6)^2} = \sqrt{16+9+36} = \sqrt{61}

Midpoint: Average each coordinate:

M = \left(\frac{3+7}{2},\; \frac{-1+2}{2},\; \frac{4+(-2)}{2}\right) = (5,\; \tfrac{1}{2},\; 1)

1.3 Equations of Spheres

A sphere is the set of all points equidistant from a center. If the center is $(a,b,c)$ and the radius is $r$ , the equation is:

(x-a)^2 + (y-b)^2 + (z-c)^2 = r^2

Completing the square converts the expanded form back to standard form.

Try It Yourself

Find the center and radius of the sphere $x^2+y^2+z^2 - 8x + 2y + 6z + 1 = 0$ .

Reveal Solution

Group and complete the square for each variable:

$(x^2 - 8x + 16) + (y^2 + 2y + 1) + (z^2 + 6z + 9) = -1 + 16 + 1 + 9$

$(x-4)^2 + (y+1)^2 + (z+3)^2 = 25$

Center $(4, -1, -3)$ , radius $5$ .

1.4 Simple Surfaces, Regions, and Traces

KEY CONCEPT

Planes parallel to coordinate planes: $x = c$ (parallel to $yz$ -plane), $y = c$ (parallel to $xz$ -plane), $z = c$ (parallel to $xy$ -plane).

Cylinders: If a variable is absent from an equation, the surface extends infinitely in that direction. For example, $x^2+y^2=4$ is a circular cylinder of radius 2 extending along the $z$ -axis.

Traces: Fix one variable (e.g., set $z=c$ ) to get cross-sectional curves. Traces help you visualize unfamiliar surfaces.

Describing regions: Inequalities define volumes. For example, $1 \le x^2+y^2+z^2 \le 4$ describes the region between two concentric spheres (a spherical shell). The equation $z \ge 0$ restricts to the upper half-space.

Try It Yourself

Describe in words the region defined by $x^2 + y^2 \le 9$ , $0 \le z \le 5$ .

Reveal Solution

The inequality $x^2+y^2 \le 9$ defines a disk of radius 3 in the $xy$ -plane. Extending this with $0 \le z \le 5$ gives a solid cylinder of radius 3 and height 5, standing on the $xy$ -plane.

2 Vectors: Operations, Magnitude, and Applications

2.1 Representing Vectors

A vector has magnitude and direction. It can be represented as:

Arrow: a directed line segment from tail to tip
Components: $\vec{v} = \langle v_1, v_2, v_3 \rangle$
$\vec{i}, \vec{j}, \vec{k}$ form: $\vec{v} = v_1\vec{i} + v_2\vec{j} + v_3\vec{k}$

FORMULA

From points: $\overrightarrow{PQ} = \langle x_2-x_1,\; y_2-y_1,\; z_2-z_1\rangle$ (terminal minus initial)

From magnitude and angle (2D): $\vec{v} = \langle |\vec{v}|\cos\theta,\; |\vec{v}|\sin\theta\rangle$

Magnitude: $|\vec{v}| = \sqrt{v_1^2 + v_2^2 + v_3^2}$

Unit vector: $\hat{u} = \dfrac{\vec{v}}{|\vec{v}|}$

2.2 Vector Arithmetic

Addition, subtraction, and scalar multiplication are component-wise:

$\vec{u}+\vec{v} = \langle u_1+v_1,\; u_2+v_2,\; u_3+v_3 \rangle$
$c\vec{v} = \langle cv_1,\; cv_2,\; cv_3 \rangle$

Geometric interpretation: The triangle rule (tip-to-tail) and parallelogram rule both give vector addition geometrically. Scaling by $c \lt 0$ reverses direction.

KEY CONCEPT

Properties (tested in True/False):

$\vec{u}+\vec{v} = \vec{v}+\vec{u}$ (commutative) — TRUE
$c(\vec{u}+\vec{v}) = c\vec{u}+c\vec{v}$ (distributive) — TRUE
$|\vec{u}+\vec{v}| = |\vec{u}|+|\vec{v}|$ — FALSE in general (triangle inequality; equality only when parallel, same direction)

2.3 Force, Tension, and Velocity Word Problems

Vectors model physical quantities with both magnitude and direction. On the exam, you'll see problems where forces act on an object and you need to find individual tensions, resultant forces, or velocity components.

RECOGNITION PATTERN

Look for: "A weight is suspended by two ropes at angles...", "A boat is moving at speed $v$ across a river with current...", "Two forces act on a particle..."

Strategy: (1) Represent each force/velocity as a vector using magnitude and angle. (2) Set up equilibrium ( $\sum \vec{F} = \vec{0}$ ) or resultant ( $\vec{F}_{\text{net}} = \vec{F}_1 + \vec{F}_2$ ). (3) Solve component equations.

Try It Yourself

A 100-lb weight hangs from two cables. Cable 1 makes a 30° angle with the horizontal to the left; Cable 2 makes a 45° angle with the horizontal to the right. Find the tension in each cable.

Reveal Solution

Set up coordinates with $x$ horizontal (right) and $y$ vertical (up). The weight pulls down: $\vec{W} = \langle 0, -100\rangle$ .

Cable 1 pulls up-left: $\vec{T}_1 = \langle -T_1\cos 30°,\; T_1\sin 30° \rangle = \langle -\frac{\sqrt{3}}{2}T_1,\; \frac{1}{2}T_1\rangle$

Cable 2 pulls up-right: $\vec{T}_2 = \langle T_2\cos 45°,\; T_2\sin 45° \rangle = \langle \frac{\sqrt{2}}{2}T_2,\; \frac{\sqrt{2}}{2}T_2\rangle$

Equilibrium: $\vec{T}_1 + \vec{T}_2 + \vec{W} = \vec{0}$

$x$ : $-\frac{\sqrt{3}}{2}T_1 + \frac{\sqrt{2}}{2}T_2 = 0 \implies T_2 = \frac{\sqrt{3}}{\sqrt{2}}T_1 = \frac{\sqrt{6}}{2}T_1$

$y$ : $\frac{1}{2}T_1 + \frac{\sqrt{2}}{2}T_2 = 100$

Substituting: $\frac{1}{2}T_1 + \frac{\sqrt{2}}{2}\cdot\frac{\sqrt{6}}{2}T_1 = 100$ , so $\frac{1}{2}T_1 + \frac{\sqrt{3}}{2}T_1 = 100$ , giving $T_1(1+\sqrt{3}) = 200$ .

$T_1 = \frac{200}{1+\sqrt{3}} = \frac{200({\sqrt{3}-1})}{2} = 100(\sqrt{3}-1) \approx 73.2$ lb

$T_2 = \frac{\sqrt{6}}{2}\cdot 100(\sqrt{3}-1) = 50\sqrt{6}(\sqrt{3}-1) \approx 89.7$ lb

3 Dot Product, Cross Product, and Projections

3.1 The Dot Product

The dot product takes two vectors and produces a scalar. It measures how much two vectors "agree" in direction.

FORMULA

Algebraic: $\vec{u}\cdot\vec{v} = u_1v_1 + u_2v_2 + u_3v_3$

Geometric: $\vec{u}\cdot\vec{v} = |\vec{u}||\vec{v}|\cos\theta$

Angle between vectors: $\cos\theta = \dfrac{\vec{u}\cdot\vec{v}}{|\vec{u}||\vec{v}|}$

Orthogonality test: $\vec{u}\perp\vec{v} \iff \vec{u}\cdot\vec{v} = 0$

Work: $W = \vec{F}\cdot\vec{d}$

KEY CONCEPT

Dot product properties (True/False favorites):

$\vec{u}\cdot\vec{v} = \vec{v}\cdot\vec{u}$ — TRUE (commutative)
$\vec{u}\cdot(\vec{v}+\vec{w}) = \vec{u}\cdot\vec{v}+\vec{u}\cdot\vec{w}$ — TRUE (distributive)
$\vec{v}\cdot\vec{v} = |\vec{v}|^2$ — TRUE
$|\vec{u}\cdot\vec{v}| \le |\vec{u}||\vec{v}|$ — TRUE (Cauchy-Schwarz)
$|\vec{u}\cdot\vec{v}| = |\vec{u}||\vec{v}|$ — FALSE (equality only when $\theta = 0$ or $\pi$ )
The dot product is a scalar, not a vector

3.2 Orthogonal Projection

The intuition: Imagine shining a flashlight perpendicular to $\vec{v}$ . The shadow of $\vec{u}$ on the line containing $\vec{v}$ is the projection. It tells you "how much of $\vec{u}$ lies in the $\vec{v}$ direction."

FORMULA

Vector projection of $\vec{u}$ onto $\vec{v}$ :

\text{proj}_{\vec{v}}\vec{u} = \frac{\vec{u}\cdot\vec{v}}{|\vec{v}|^2}\,\vec{v}

Scalar projection (component):

\text{scal}_{\vec{v}}\vec{u} = \frac{\vec{u}\cdot\vec{v}}{|\vec{v}|}

The scalar projection is the signed length of the shadow. The vector projection points along $\vec{v}$ with that length.

Try It Yourself

Let $\vec{u} = \langle 3, 4 \rangle$ and $\vec{v} = \langle 5, 0 \rangle$ . Find $\text{proj}_{\vec{v}}\vec{u}$ and $\text{scal}_{\vec{v}}\vec{u}$ . Interpret geometrically.

Reveal Solution

$\vec{u}\cdot\vec{v} = 15+0 = 15$ , $|\vec{v}|^2 = 25$ , $|\vec{v}| = 5$ .

$\text{proj}_{\vec{v}}\vec{u} = \frac{15}{25}\langle 5,0\rangle = \langle 3, 0\rangle$ . The projection lies along the $x$ -axis with length 3 — it's the $x$ -component of $\vec{u}$ .

$\text{scal}_{\vec{v}}\vec{u} = \frac{15}{5} = 3$ . This is the signed length of the shadow. It's positive, so $\vec{u}$ has a component in the same direction as $\vec{v}$ .

3.3 The Cross Product

The cross product takes two vectors in $\mathbb{R}^3$ and produces a vector perpendicular to both.

FORMULA

Component formula:

\vec{u}\times\vec{v} = \langle u_2v_3-u_3v_2,\; u_3v_1-u_1v_3,\; u_1v_2-u_2v_1\rangle

Determinant method:

\vec{u}\times\vec{v} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \end{vmatrix}

Note: the middle component ( $\vec{j}$ ) gets a minus sign in cofactor expansion.

Magnitude: $|\vec{u}\times\vec{v}| = |\vec{u}||\vec{v}|\sin\theta$

Direction: Right-hand rule — point fingers along $\vec{u}$ , curl toward $\vec{v}$ , thumb = $\vec{u}\times\vec{v}$ .

KEY CONCEPT

Applications:

Area of parallelogram: $A = |\vec{u}\times\vec{v}|$
Area of triangle: $A = \frac{1}{2}|\vec{u}\times\vec{v}|$
Normal to a plane: $\vec{n} = \vec{v}_1 \times \vec{v}_2$ from two direction vectors in the plane
Parallel test: $\vec{u}\times\vec{v} = \vec{0}$ iff $\vec{u}$ and $\vec{v}$ are parallel

COMMON MISTAKE

Cross product properties (True/False):

$\vec{u}\times\vec{v} = -(\vec{v}\times\vec{u})$ — TRUE (anticommutative, NOT commutative!)
$\vec{u}\times\vec{u} = \vec{0}$ — TRUE
$(\vec{u}\times\vec{v})\cdot\vec{u} = 0$ — TRUE (cross product is perpendicular to both)
$|\vec{u}\times\vec{v}| = |\vec{v}\times\vec{u}|$ — TRUE (magnitudes are equal)
$(\vec{u}+\vec{v})\times\vec{v} = \vec{u}\times\vec{v}$ — TRUE (since $\vec{v}\times\vec{v}=\vec{0}$ )
Cross product is NOT associative

Try It Yourself

Find $\vec{a}\times\vec{b}$ for $\vec{a}=\langle 1,1,-2\rangle$ and $\vec{b}=\langle 3,-2,1\rangle$ . Then find the area of the triangle spanned by $\vec{a}$ and $\vec{b}$ .

Reveal Solution

Using the component formula (or determinant expansion):

$\vec{a}\times\vec{b} = \langle (1)(1)-(-2)(-2),\; (-2)(3)-(1)(1),\; (1)(-2)-(1)(3)\rangle = \langle 1-4,\; -6-1,\; -2-3\rangle = \langle -3, -7, -5\rangle$

Magnitude: $|\vec{a}\times\vec{b}| = \sqrt{9+49+25} = \sqrt{83}$

Area of triangle: $\frac{1}{2}\sqrt{83}$ .

4 Lines and Planes in 3D

4.1 Equations of Lines

A line in 3D is determined by a point on the line and a direction vector.

FORMULA

Vector form: $\vec{r}(t) = \vec{r}_0 + t\vec{v}$

Parametric equations: $x = x_0+at,\quad y = y_0+bt,\quad z = z_0+ct$

Symmetric equations: $\dfrac{x-x_0}{a} = \dfrac{y-y_0}{b} = \dfrac{z-z_0}{c}$

(where $\langle a,b,c\rangle$ is the direction vector and $(x_0,y_0,z_0)$ is a point on the line)

Line segment from $A$ to $B$ : $\vec{r}(t) = (1-t)A + tB,\quad 0\le t\le 1$

GOTCHA

Symmetric equations require all three direction components to be nonzero. If one component (say $a$ ) is zero, write $x = x_0$ separately and use symmetric form for the other two: $\frac{y-y_0}{b} = \frac{z-z_0}{c},\; x = x_0$ .

4.2 Equations of Planes

A plane is determined by a point and a normal vector $\vec{n} = \langle a,b,c\rangle$ .

FORMULA

Point-normal form: $\vec{n}\cdot(\vec{r}-\vec{r}_0)=0$

General form: $ax+by+cz = d$

Plane through three points $P,Q,R$ : compute $\vec{n} = \overrightarrow{PQ}\times\overrightarrow{PR}$ , then use point-normal form with any of the three points.

4.3 Relationships Between Lines and Planes

Lines parallel: direction vectors are scalar multiples
Lines orthogonal: direction vectors have zero dot product
Lines intersecting: solve parametric equations simultaneously for $t,s$ ; if no solution, they're parallel or skew
Planes parallel: normal vectors are scalar multiples
Planes orthogonal: normal vectors have zero dot product
Angle between planes: $\cos\theta = \frac{|\vec{n}_1\cdot\vec{n}_2|}{|\vec{n}_1||\vec{n}_2|}$
Point on a line/plane: substitute coordinates into the equation and check

4.4 Distance Formulas

These are two of the most important formulas in the course and both are testable.

FORMULA

Distance from point $(x_0,y_0,z_0)$ to plane $ax+by+cz=d$ :

D = \frac{|ax_0+by_0+cz_0-d|}{\sqrt{a^2+b^2+c^2}}

Why it works: The numerator is the scalar projection of the vector from any point on the plane to $(x_0,y_0,z_0)$ onto the normal $\langle a,b,c\rangle$ . The denominator normalizes that normal vector.

Try It Yourself

Find the distance from the point $(1, 2, -1)$ to the plane $2x - y + 2z = 4$ .

Reveal Solution

Here $a=2, b=-1, c=2, d=4$ . Plug in the point:

$D = \frac{|2(1)+(-1)(2)+2(-1)-4|}{\sqrt{4+1+4}} = \frac{|2-2-2-4|}{3} = \frac{|-6|}{3} = 2$

FORMULA

Distance from point $P$ to line through $P_0$ with direction $\vec{v}$ :

D = \frac{|\overrightarrow{P_0P}\times\vec{v}|}{|\vec{v}|}

Why it works: $|\overrightarrow{P_0P}\times\vec{v}|$ is the area of the parallelogram formed by $\overrightarrow{P_0P}$ and $\vec{v}$ . Dividing by the base length $|\vec{v}|$ gives the height, which is the perpendicular distance.

Try It Yourself

Find the distance from the point $P(1,0,1)$ to the line $x=1+t,\; y=2-t,\; z=3t$ .

Reveal Solution

The line passes through $P_0 = (1,2,0)$ with direction $\vec{v}=\langle 1,-1,3\rangle$ .

$\overrightarrow{P_0P} = \langle 1-1,\; 0-2,\; 1-0\rangle = \langle 0, -2, 1\rangle$

$\overrightarrow{P_0P}\times\vec{v} = \langle (-2)(3)-(1)(-1),\; (1)(1)-(0)(3),\; (0)(-1)-(-2)(1)\rangle = \langle -5, 1, 2\rangle$

$|\overrightarrow{P_0P}\times\vec{v}| = \sqrt{25+1+4} = \sqrt{30}$ , $|\vec{v}| = \sqrt{1+1+9} = \sqrt{11}$

$D = \frac{\sqrt{30}}{\sqrt{11}} = \sqrt{\frac{30}{11}}$

Try It Yourself

Find parametric and symmetric equations for the line through $P(2,1,3)$ and $Q(4,-1,5)$ .

Reveal Solution

Direction: $\overrightarrow{PQ} = \langle 2,-2,2\rangle$ . We can simplify to $\langle 1,-1,1\rangle$ .

Parametric: $x = 2+t,\; y = 1-t,\; z = 3+t$

Symmetric: $\frac{x-2}{1} = \frac{y-1}{-1} = \frac{z-3}{1}$ , or equivalently $x-2 = -(y-1) = z-3$ .

5 Vector-Valued Functions, Derivatives, and Arc Length

5.1 Domain of a Vector Function

For $\vec{r}(t) = \langle f(t), g(t), h(t)\rangle$ , the domain is the intersection of the domains of all component functions.

RECOGNITION PATTERN

Look for restrictions from: square roots (argument $\ge 0$ ), logarithms (argument $\gt 0$ ), denominators (cannot be zero). Take the intersection of all restrictions.

Try It Yourself

Find the domain of $\vec{r}(t) = \left\langle \sqrt{t-1},\; \ln(4-t),\; \dfrac{1}{t}\right\rangle$ .

Reveal Solution

Component 1: $t-1\ge 0 \implies t\ge 1$ .

Component 2: $4-t\gt 0 \implies t \lt 4$ .

Component 3: $t\neq 0$ .

Intersection: $[1, 4)$ . (The $t\neq 0$ constraint is automatically satisfied since $t\ge 1$ .)

5.2 Limits of Vector Functions

Limits are computed component-wise: $\lim_{t\to a}\vec{r}(t) = \langle \lim f(t),\; \lim g(t),\; \lim h(t)\rangle$ . Use all your Calc 1/2 techniques (L'Hopital, factoring, conjugates) on each component independently.

5.3 Curves, Eliminating the Parameter, and Graph Matching

A vector function's graph is a curve in the plane or in space. Common parameterizations:

Line segment $A\to B$ : $\vec{r}(t) = (1-t)A+tB$ , $0\le t\le 1$
Circle of radius $a$ : $\vec{r}(t)=\langle a\cos t, a\sin t\rangle$
Ellipse: $\vec{r}(t)=\langle a\cos t, b\sin t\rangle$
Part of graph $y=f(x)$ : $\vec{r}(t)=\langle t, f(t)\rangle$

KEY CONCEPT

Eliminating the parameter: Solve one component for $t$ , substitute into the others to get a Cartesian equation. This helps you sketch the curve.

Example: $\vec{r}(t)=\langle t^2, t^3\rangle$ . From $x=t^2$ , we get $t=\pm\sqrt{x}$ , so $y = t^3 = (\pm\sqrt{x})^3 = \pm x^{3/2}$ , giving the curve $y^2 = x^3$ .

KEY CONCEPT

Matching 3D vector functions to graphs: Project the curve onto coordinate planes. If $\vec{r}(t)=\langle \cos t, \sin t, t\rangle$ , the $xy$ -projection is a circle, the $xz$ -projection is a cosine wave, and the curve spirals upward — it's a helix.

Use projections to identify: circles vs. ellipses, upward vs. downward spirals, straight vs. oscillating paths.

5.4 Derivatives and Integrals of Vector Functions

FORMULA

Derivative: $\vec{r}'(t) = \langle x'(t), y'(t), z'(t)\rangle$

Integral: $\int\vec{r}(t)\,dt = \left\langle \int x(t)\,dt,\; \int y(t)\,dt,\; \int z(t)\,dt\right\rangle + \vec{C}$ — the constant of integration is a vector

Speed: $|\vec{r}'(t)|$ | Velocity: $\vec{r}'(t)$ | Unit tangent: $\vec{T}(t) = \frac{\vec{r}'(t)}{|\vec{r}'(t)|}$

Tangent line at $t_0$ : $\vec{L}(t) = \vec{r}(t_0) + (t-t_0)\vec{r}'(t_0)$

Finding $\vec{r}(t)$ from $\vec{r}'(t)$ and an initial condition: Integrate component-wise, then use the initial condition to determine $\vec{C}$ .

KEY CONCEPT

Differentiation rules (Theorem 3.3):

$\frac{d}{dt}(f\vec{u}) = f'\vec{u}+f\vec{u}'$ (product rule with scalar)
$\frac{d}{dt}(\vec{u}\cdot\vec{v}) = \vec{u}'\cdot\vec{v}+\vec{u}\cdot\vec{v}'$ (product rule for dot)
$\frac{d}{dt}(\vec{u}\times\vec{v}) = \vec{u}'\times\vec{v}+\vec{u}\times\vec{v}'$ (product rule for cross — order matters!)
$\frac{d}{dt}\vec{u}(f(t)) = f'(t)\vec{u}'(f(t))$ (chain rule)

Critical True/False: $\frac{d}{dt}|\vec{r}(t)| = |\vec{r}'(t)|$ is FALSE. The correct formula is $\frac{d}{dt}|\vec{r}| = \frac{\vec{r}\cdot\vec{r}'}{|\vec{r}|}$ .

5.5 Arc Length and Arc Length Parameterization

FORMULA

Arc length: $L = \int_a^b |\vec{r}'(t)|\,dt$

Arc length function: $s(t) = \int_{t_0}^t |\vec{r}'(u)|\,du$

KEY CONCEPT

Is the curve parameterized by arc length? Check if $|\vec{r}'(t)| = 1$ (speed equals 1) for all $t$ . If yes, the parameter is arc length — the length of the curve from $t_0$ to $t$ equals $t - t_0$ .

How to reparameterize by arc length:

Compute $s(t) = \int_{t_0}^t |\vec{r}'(u)|\,du$
Solve for $t$ in terms of $s$
Substitute back into $\vec{r}(t)$

Try It Yourself

Determine whether $\vec{r}(t) = \langle \frac{1}{\sqrt{2}}\cos t,\; \frac{1}{\sqrt{2}}\sin t,\; \frac{1}{\sqrt{2}}t\rangle$ is parameterized by arc length.

Reveal Solution

$\vec{r}'(t) = \langle -\frac{1}{\sqrt{2}}\sin t,\; \frac{1}{\sqrt{2}}\cos t,\; \frac{1}{\sqrt{2}}\rangle$

$|\vec{r}'(t)| = \sqrt{\frac{1}{2}\sin^2 t + \frac{1}{2}\cos^2 t + \frac{1}{2}} = \sqrt{\frac{1}{2}+\frac{1}{2}} = 1$

Since the speed is 1 for all $t$ , yes, this curve is parameterized by arc length.

6 Functions of Several Variables, Limits, and Partial Derivatives

6.1 Functions of Two Variables

"Is $z$ a function of $x$ and $y$ ?"

A surface $z = f(x,y)$ defines $z$ as a function of $x$ and $y$ if for each point $(x,y)$ in the domain, there is exactly one value of $z$ . This is the 3D analogue of the vertical line test: imagine a vertical line (parallel to the $z$ -axis) at each $(x,y)$ — it must hit the surface at most once.

KEY CONCEPT

$x^2+y^2+z^2 = 1$ does not define $z$ as a function of $x,y$ because for $(x,y)$ inside the unit disk, there are two $z$ -values (top and bottom hemisphere). But $z = \sqrt{1-x^2-y^2}$ does — it picks only the top hemisphere.

Domain and Level Curves

FORMULA

Domain restrictions: Denominators $\neq 0$ , radicands $\ge 0$ , log arguments $\gt 0$ . Sketch the domain as a region in the $xy$ -plane.

Level curves: Set $f(x,y) = c$ for various constants $c$ and sketch the resulting curves. Closely spaced contours = steep surface; widely spaced = gradual.

Level surfaces: For $f(x,y,z) = c$ , the level sets are surfaces in 3D.

Sketching Surfaces via Traces

KEY CONCEPT

Trace method: To sketch $z = f(x,y)$ , fix each variable in turn and sketch the resulting cross-sectional curve:

Fix $x = c$ : curve in the $yz$ -plane
Fix $y = c$ : curve in the $xz$ -plane
Fix $z = c$ : curve in the $xy$ -plane (level curve)

Combine these cross-sections to build a mental picture of the full surface. Remember: if a variable is absent from the equation, the surface extends uniformly in that direction.

Worked Example 1 — Parabolic Trough

Sketch $z = x^2$ .

Traces:

Fix $y = c$ : $z = x^2$ (same parabola in the $xz$ -plane for every $c$ — the variable $y$ is absent)
Fix $x = c$ : $z = c^2$ (horizontal line — constant height)
Fix $z = c$ : $x = \pm\sqrt{c}$ (two vertical lines in the $xy$ -plane)

The surface is a parabolic trough — imagine the parabola $z=x^2$ extruded infinitely along the $y$ -axis. Every cross-section parallel to the $xz$ -plane is the same upward-opening parabola.

Worked Example 2 — Bowl and Saddle

Compare $z = x^2 + y^2$ and $z = x^2 - y^2$ .

$z = x^2 + y^2$ (bowl-shaped surface opening upward):

Fix $z = c$ ( $c \ge 0$ ): $x^2+y^2 = c$ — circles of radius $\sqrt{c}$ . Level curves are concentric circles.
Fix $y = 0$ : $z = x^2$ — upward parabola. Fix $x = 0$ : $z = y^2$ — also upward parabola.
Shape: the surface curves upward in every direction from the origin. Minimum at the origin.

$z = x^2 - y^2$ (saddle-shaped surface):

Fix $z = 0$ : $x^2 = y^2$ , so $y = \pm x$ — two lines through the origin (the saddle's "cross").
Fix $y = 0$ : $z = x^2$ — curves upward. Fix $x = 0$ : $z = -y^2$ — curves downward.
Shape: curves up along the $x$ -axis and down along the $y$ -axis, like a horse saddle or mountain pass. The origin is a saddle point (neither max nor min).

GOTCHA

The exam info sheet says you do not need to know the standard names of these surfaces. Focus on the trace method: identify the shape of cross-sections, determine whether the surface opens up/down/sideways, and describe it in plain terms.

6.2 Limits and Continuity

For continuous functions: evaluate by direct substitution. Polynomials and rational functions (where defined) are continuous.

KEY CONCEPT

Two-path method: To show a limit does NOT exist, find two paths to the point that give different limiting values.

Example: For $\lim_{(x,y)\to(0,0)}\frac{xy}{x^2+y^2}$ , try $y=0$ (gives 0) and $y=x$ (gives $\frac{1}{2}$ ). Different values $\implies$ limit DNE.

GOTCHA

True/False: "If $f(x,y)\to L$ along every straight line through $(a,b)$ , then $\lim_{(x,y)\to(a,b)} f = L$ ." This is FALSE. A limit can exist along all lines but fail along a curved path (e.g., $y = x^2$ ).

KEY CONCEPT

Making a function continuous: If $f$ is piecewise-defined and you need to find a value of $k$ that makes it continuous at a point, the strategy is: compute the limit of the formula piece as you approach the point, then set that limit equal to the piecewise value $k$ .

Try It Yourself

Let $f(x,y) = \begin{cases} \dfrac{x^2-y^2}{x-y} & x \neq y \\[6pt] k & x = y \end{cases}$ . Find the value of $k$ that makes $f$ continuous at every point on the line $x=y$ .

Reveal Solution

Step 1: Simplify the formula piece. When $x\neq y$ :

$\dfrac{x^2-y^2}{x-y} = \dfrac{(x+y)(x-y)}{x-y} = x+y$

Step 2: Compute the limit as $(x,y)$ approaches a point on $x=y$ , say $(a,a)$ :

$\displaystyle\lim_{(x,y)\to(a,a)} f(x,y) = \lim_{(x,y)\to(a,a)}(x+y) = 2a$

Step 3: For continuity, the limit must equal the function value: $k = 2a$ . Since the piecewise value is a single constant $k$ and we need it to work at every point $(a,a)$ , this is only possible if the problem specifies a particular point. At any specific point $(a,a)$ , we need $k = 2a$ . For example, at $(1,1)$ : $k = 2$ . At $(3,3)$ : $k = 6$ .

General strategy: Factor/simplify to cancel the problematic term, evaluate the limit, set it equal to the piecewise value, solve for $k$ .

6.3 Partial Derivatives

Partial derivatives measure the rate of change of $f$ with respect to one variable while holding all others constant. Geometrically, $f_x(a,b)$ is the slope of the curve obtained by slicing the surface $z=f(x,y)$ with the plane $y=b$ .

FORMULA

Definition: $\frac{\partial f}{\partial x} = \lim_{h\to 0} \frac{f(x+h,y)-f(x,y)}{h}$

Computation: Hold $y$ constant and differentiate with respect to $x$ using standard Calc 1 rules.

Higher-order: $f_{xx}$ , $f_{yy}$ , $f_{xy}$ , $f_{yx}$ . Notation $\frac{\partial^2 f}{\partial y\partial x}$ means differentiate w.r.t. $x$ first, then $y$ .

Clairaut's Theorem: If $f$ has continuous second-order partials, then $f_{xy} = f_{yx}$ .

Limit Definition of Partial Derivatives

RECOGNITION PATTERN

When is the limit definition needed? For piecewise-defined functions where the formula changes at the point in question. You cannot just "take the derivative" because the function expression itself switches there.

Try It Yourself

Let $f(x,y) = \begin{cases} \frac{x^2 y}{x^2+y^2} & (x,y)\neq(0,0) \\ 0 & (x,y)=(0,0)\end{cases}$ . Find $f_x(0,0)$ and $f_y(0,0)$ .

Reveal Solution

We must use the limit definition because the formula changes at $(0,0)$ :

$f_x(0,0) = \lim_{h\to 0}\frac{f(h,0)-f(0,0)}{h} = \lim_{h\to 0}\frac{\frac{h^2\cdot 0}{h^2+0} - 0}{h} = \lim_{h\to 0}\frac{0}{h} = 0$

$f_y(0,0) = \lim_{h\to 0}\frac{f(0,h)-f(0,0)}{h} = \lim_{h\to 0}\frac{\frac{0\cdot h}{0+h^2} - 0}{h} = \lim_{h\to 0}\frac{0}{h} = 0$

Both partial derivatives are 0 at the origin.

6.4 PDE Verification

A common problem type: "Show that $u$ satisfies the PDE." You compute the required partial derivatives and substitute.

RECOGNITION PATTERN

Look for: "Verify that...", "Show that $f$ satisfies...", "Show that $u$ is a solution to..."

Common PDEs on this exam:

Wave equation: $u_{tt} = c^2 u_{xx}$
Laplace's equation: $u_{xx} + u_{yy} = 0$ (or with $u_{zz}$ in 3D)
Heat equation: $u_t = k\, u_{xx}$

Try It Yourself

Show that $u(x,t) = \sin(x - ct)$ satisfies the wave equation $u_{tt} = c^2 u_{xx}$ .

Reveal Solution

Compute the required partial derivatives:

$u_x = \cos(x-ct)$ , $u_{xx} = -\sin(x-ct)$

$u_t = -c\cos(x-ct)$ , $u_{tt} = -c^2\sin(x-ct) \cdot (-1) \cdot (-1) = -c^2\sin(x-ct)$

Wait, let me be careful: $u_t = \cos(x-ct)\cdot(-c) = -c\cos(x-ct)$

$u_{tt} = -c\cdot(-\sin(x-ct))\cdot(-c) = -c^2\sin(x-ct)$

And $c^2 u_{xx} = c^2\cdot(-\sin(x-ct)) = -c^2\sin(x-ct)$

Since $u_{tt} = -c^2\sin(x-ct) = c^2 u_{xx}$ , the wave equation is satisfied.

Try It Yourself

Show that $u(x,y) = e^x\sin y$ satisfies Laplace's equation $u_{xx}+u_{yy}=0$ .

Reveal Solution

$u_x = e^x\sin y$ , $u_{xx} = e^x\sin y$

$u_y = e^x\cos y$ , $u_{yy} = -e^x\sin y$

$u_{xx}+u_{yy} = e^x\sin y - e^x\sin y = 0$ . Laplace's equation is satisfied.

6.5 The Chain Rule for Multivariable Functions

FORMULA

Case 1: $z = f(x,y)$ , $x=x(t)$ , $y=y(t)$ :

\frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt}

Case 2: $z = f(x,y)$ , $x=x(s,t)$ , $y=y(s,t)$ :

\frac{\partial z}{\partial s} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial s} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial s}

Implicit differentiation: If $F(x,y,z)=0$ , then $\frac{\partial z}{\partial x} = -\frac{F_x}{F_z}$

Use a tree diagram: Draw the dependency tree. Multiply along each branch, add across paths.

7 Gradient, Tangent Planes, Linear Approximation, and Extrema

7.1 Directional Derivatives and Gradient

FORMULA

Gradient: $\nabla f = \langle f_x, f_y\rangle$ or $\langle f_x, f_y, f_z\rangle$

Directional derivative (must use a unit vector $\hat{u}$ ):

D_{\hat{u}}f = \nabla f\cdot\hat{u}

Maximum rate of change: $|\nabla f|$ , in the direction of $\nabla f$

Minimum rate of change: $-|\nabla f|$ , in the direction of $-\nabla f$

Zero rate of change: directions perpendicular to $\nabla f$

COMMON MISTAKE

Forgetting to normalize the direction vector. If the problem says "in the direction of $\langle 2,1,-2\rangle$ ", you must divide by $|\langle 2,1,-2\rangle| = 3$ to get the unit vector $\hat{u} = \langle\frac{2}{3},\frac{1}{3},-\frac{2}{3}\rangle$ .

7.2 Tangent Planes and Normal Lines

FORMULA

Tangent plane to $z = f(x,y)$ at $(x_0,y_0)$ :

z = f(x_0,y_0) + f_x(x_0,y_0)(x-x_0) + f_y(x_0,y_0)(y-y_0)

Tangent plane to level surface $F(x,y,z) = k$ at $(x_0,y_0,z_0)$ :

F_x(x_0,y_0,z_0)(x-x_0) + F_y(y-y_0) + F_z(z-z_0) = 0

Normal line to a level surface: direction = $\nabla F$ at the point:

\vec{r}(t) = \langle x_0, y_0, z_0\rangle + t\,\nabla F(x_0,y_0,z_0)

Tangent Line to a Level Curve (2D)

This is a separate skill from the tangent plane. For a level curve $f(x,y) = c$ in the plane, the gradient $\nabla f$ is normal to the curve.

FORMULA

Tangent line to $f(x,y) = c$ at $(x_0,y_0)$ :

f_x(x_0,y_0)(x-x_0) + f_y(x_0,y_0)(y-y_0) = 0

This is $\nabla f\cdot\langle x-x_0,\; y-y_0\rangle = 0$ — the tangent is perpendicular to the gradient.

Normal Line to a Level Curve (2D)

FORMULA

Normal line to $f(x,y) = c$ at $(x_0,y_0)$ : the direction is $\nabla f(x_0,y_0)$ :

\vec{r}(t) = \langle x_0,y_0\rangle + t\,\nabla f(x_0,y_0)

Or in symmetric form: $\frac{x-x_0}{f_x} = \frac{y-y_0}{f_y}$

Try It Yourself

Find the tangent line and normal line to the level curve $x^2+xy+y^2=7$ at the point $(1,2)$ .

Reveal Solution

Let $f(x,y) = x^2+xy+y^2$ . Then $f_x = 2x+y$ and $f_y = x+2y$ . At $(1,2)$ : $f_x = 4$ , $f_y = 5$ .

Tangent line: $4(x-1)+5(y-2)=0 \implies 4x+5y = 14$ .

Normal line: $\vec{r}(t)=\langle 1+4t,\; 2+5t\rangle$ , or in Cartesian: $\frac{x-1}{4}=\frac{y-2}{5}$ .

7.3 Linear Approximation and Differentials

The tangent plane is the linear approximation. For a point near $(a,b)$ :

FORMULA

Linear approximation:

f(x,y) \approx L(x,y) = f(a,b) + f_x(a,b)(x-a) + f_y(a,b)(y-b)

Total differential:

df = f_x\,dx + f_y\,dy

Use $dx = \Delta x$ , $dy = \Delta y$ to approximate $\Delta f \approx df$ .

Try It Yourself

Use linear approximation to estimate $\sqrt{(3.02)^2 + (3.97)^2}$ .

Reveal Solution

Let $f(x,y) = \sqrt{x^2+y^2}$ . We use the base point $(a,b) = (3,4)$ where $f(3,4) = 5$ .

$f_x = \frac{x}{\sqrt{x^2+y^2}}$ , so $f_x(3,4) = \frac{3}{5}$ .

$f_y = \frac{y}{\sqrt{x^2+y^2}}$ , so $f_y(3,4) = \frac{4}{5}$ .

$dx = 3.02-3 = 0.02$ , $dy = 3.97-4 = -0.03$ .

$f(3.02,3.97) \approx 5 + \frac{3}{5}(0.02) + \frac{4}{5}(-0.03) = 5 + 0.012 - 0.024 = 4.988$

(Exact value: $\sqrt{9.1204 + 15.7609} = \sqrt{24.8813} \approx 4.9881$ . Linear approximation is accurate.)

Try It Yourself

The radius of a cylinder is measured as 5 cm with error $\pm 0.1$ cm, and the height is 12 cm with error $\pm 0.2$ cm. Use differentials to approximate the maximum error in the computed volume $V = \pi r^2 h$ .

Reveal Solution

$dV = \frac{\partial V}{\partial r}\,dr + \frac{\partial V}{\partial h}\,dh = 2\pi rh\,dr + \pi r^2\,dh$

$dV = 2\pi(5)(12)(0.1) + \pi(25)(0.2) = 12\pi + 5\pi = 17\pi \approx 53.4 \text{ cm}^3$

7.4 Critical Points and the Second Derivatives Test

FORMULA

Critical points: Solve $f_x = 0$ and $f_y = 0$ simultaneously.

Discriminant:

D = f_{xx}f_{yy} - (f_{xy})^2

$D \gt 0$ and $f_{xx} \gt 0$ : local minimum
$D \gt 0$ and $f_{xx} \lt 0$ : local maximum
$D \lt 0$ : saddle point
$D = 0$ : test inconclusive

7.5 Absolute Extrema on Closed Bounded Regions

KEY CONCEPT

Procedure:

Find all interior critical points; evaluate $f$ there
Parameterize each boundary piece as a function of one variable; find its critical points
Evaluate $f$ at all boundary endpoints (corners)
Compare all values — largest is absolute max, smallest is absolute min

KEY CONCEPT

Extreme Value Theorem: If $f$ is continuous on a closed, bounded region $D$ , then $f$ attains both an absolute maximum and an absolute minimum on $D$ .

Why "closed" is needed: On an open region, $f$ might approach its supremum without ever reaching it. Example: $f(x,y)=x$ on the open disk $x^2+y^2\lt 1$ approaches 1 but never achieves it.

Why "bounded" is needed: On an unbounded region, $f$ might grow without bound. Example: $f(x,y)=x$ on $\{(x,y): y \ge 0\}$ has no maximum.

Both hypotheses are necessary. If either fails, extrema are not guaranteed.

8 Double and Triple Integrals

8.1 Double Integrals — Setup and Evaluation

FORMULA

Area: $A = \iint_R dA$ | Volume: $V = \iint_R [f(x,y)-g(x,y)]\,dA$

Mass: $m = \iint_R \rho(x,y)\,dA$ | Average value: $f_{\text{avg}} = \frac{1}{A}\iint_R f\,dA$

For iterated integrals, evaluate inside-out. The inner integral's variable matches the inner $d$ -variable.

8.2 Area of a Region via Double Integrals

FORMULA

\text{Area}(R) = \iint_R 1\,dA

To find the area of a flat region in the $xy$ -plane, integrate the constant function $1$ over the region. This is not the same as finding the volume under a surface — the integrand is simply $1$ , not $f(x,y)$ .

Try It Yourself

Use a double integral to find the area of the region bounded by $y = x^2$ and $y = 2x$ .

Reveal Solution

Step 1: Find intersection points. Set $x^2 = 2x$ , so $x^2 - 2x = 0$ , giving $x(x-2)=0$ . Intersections at $x=0$ and $x=2$ .

Step 2: Determine which curve is on top. For $0 \le x \le 2$ : $2x \ge x^2$ , so $y=2x$ is the upper bound and $y=x^2$ is the lower bound.

Step 3: Set up and evaluate.

\text{Area} = \iint_R 1\,dA = \int_0^2\int_{x^2}^{2x} 1\,dy\,dx = \int_0^2 (2x - x^2)\,dx

= \left[x^2 - \frac{x^3}{3}\right]_0^2 = 4 - \frac{8}{3} = \frac{4}{3}

8.3 Changing the Order of Integration

Why change order? Sometimes the integral is impossible (or extremely difficult) to evaluate in the given order, but becomes tractable in the reversed order.

KEY CONCEPT

Procedure:

Sketch the region using the given bounds
Re-describe the same region with the roles of $x$ and $y$ swapped
Write new bounds

Critical rule: If the limits depend on the other variable, you cannot simply swap the $d$ -variables. You must re-derive the bounds from the region sketch.

Try It Yourself

Evaluate by changing the order of integration: $\displaystyle\int_0^1\int_{\sqrt{y}}^{1}\frac{ye^{x^2}}{x^3}\,dx\,dy$

Reveal Solution

Step 1: Sketch the region. The inner integral goes from $x=\sqrt{y}$ to $x=1$ , the outer from $y=0$ to $y=1$ .

The curve $x = \sqrt{y}$ is the same as $y = x^2$ . So the region is bounded by $y=x^2$ on the left, $x=1$ on the right, $y=0$ on the bottom, and $y=1$ on the top.

Step 2: Reverse the order. Now let $x$ be the outer variable. For fixed $x$ , $y$ runs from $0$ to $x^2$ . And $x$ runs from $0$ to $1$ :

\int_0^1\int_0^{x^2}\frac{ye^{x^2}}{x^3}\,dy\,dx

Step 3: Evaluate. Inner integral: $\int_0^{x^2}\frac{ye^{x^2}}{x^3}\,dy = \frac{e^{x^2}}{x^3}\cdot\frac{y^2}{2}\Big|_0^{x^2} = \frac{e^{x^2}}{x^3}\cdot\frac{x^4}{2} = \frac{xe^{x^2}}{2}$

Outer integral: $\int_0^1\frac{xe^{x^2}}{2}\,dx = \frac{1}{4}e^{x^2}\Big|_0^1 = \frac{e-1}{4}$

8.4 Polar Coordinates

FORMULA

$x = r\cos\theta$ , $y = r\sin\theta$ , $r^2 = x^2+y^2$

dA = r\,dr\,d\theta

Use polar when: integrand contains $x^2+y^2$ ; region is a disk, annulus, or sector.

8.5 Triple Integrals

FORMULA

Volume: $V = \iiint_E dV$ | Mass: $m = \iiint_E \rho\,dV$ | Average value: $\frac{1}{V}\iiint_E f\,dV$

Cylindrical Coordinates

$x=r\cos\theta$ , $y=r\sin\theta$ , $z=z$ .

dV = r\,dz\,dr\,d\theta

Use for cylinders, bowl-shaped surfaces, and surfaces with circular cross-sections perpendicular to an axis.

Spherical Coordinates

$x = \rho\sin\phi\cos\theta$ , $y = \rho\sin\phi\sin\theta$ , $z = \rho\cos\phi$

dV = \rho^2\sin\phi\,d\rho\,d\theta\,d\phi

$\phi$ = angle from positive $z$ -axis ( $0\le\phi\le\pi$ ). $\theta$ = azimuthal angle ( $0\le\theta\le 2\pi$ ).

Use for spheres, hemispheres, and surfaces of the form $z^2 = x^2+y^2$ . Common: the surface $z = \sqrt{x^2+y^2}$ corresponds to $\phi = \pi/4$ .

KEY CONCEPT

Changing order in triple integrals: The same principle applies — sketch the region, then re-describe it with a different variable as the outermost integral. This is especially useful when switching from rectangular to cylindrical or spherical coordinates.

GOTCHA

True/False: $\int_0^{2\pi}\int_0^2\int_r^2 dz\,dr\,d\theta$ represents the volume between the surface $z = r$ and the plane $z=2$ — NOT the plane $x=2$ .

9 Vector Fields, Line Integrals, and Green's Theorem

9.1 Vector Fields

A vector field assigns a vector to each point: $\vec{F}(x,y,z) = \langle P, Q, R\rangle$ .

Sketching a Vector Field from a Formula

KEY CONCEPT

Procedure: Pick a grid of representative points. At each point, evaluate $\vec{F}$ , then draw an arrow starting at that point with the computed direction and proportional length.

Worked Example

Sketch $\vec{F}(x,y) = \langle -y,\; x\rangle$ by evaluating at a 3 x 3 grid of points.

$(x,y)$	$\vec{F}(x,y)$	Direction
$(1,0)$	$\langle 0, 1\rangle$	up
$(0,1)$	$\langle -1, 0\rangle$	left
$(-1,0)$	$\langle 0, -1\rangle$	down
$(0,-1)$	$\langle 1, 0\rangle$	right
$(1,1)$	$\langle -1, 1\rangle$	upper-left, length $\sqrt{2}$
$(-1,1)$	$\langle -1, -1\rangle$	lower-left, length $\sqrt{2}$
$(-1,-1)$	$\langle 1, -1\rangle$	lower-right, length $\sqrt{2}$
$(1,-1)$	$\langle 1, 1\rangle$	upper-right, length $\sqrt{2}$
$(0,0)$	$\langle 0, 0\rangle$	zero vector

The arrows rotate counterclockwise around the origin, all tangent to circles centered at the origin. Points farther from the origin have longer arrows (speed = $\sqrt{x^2+y^2}$ ).

KEY CONCEPT

Matching fields to graphs: Look for patterns — radial fields point away from/toward origin; rotational fields curl around origin; gradient fields have vectors perpendicular to level curves.

Conservative Vector Fields

FORMULA

Test for conservative: $\vec{F}$ is conservative iff $\nabla\times\vec{F} = \vec{0}$ (on simply-connected domain).

In 2D: $\vec{F}=\langle P,Q\rangle$ is conservative iff $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$

Finding potential function $f$ with $\nabla f = \vec{F}$ :

Integrate $P$ w.r.t. $x$ : $f = \int P\,dx + g(y,z)$
Differentiate w.r.t. $y$ , set equal to $Q$ ; solve for $g_y$ , integrate
Differentiate w.r.t. $z$ , set equal to $R$ ; solve for remaining constant

Divergence and Curl

FORMULA

Divergence (scalar): $\nabla\cdot\vec{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z}$

Interpretation: positive = expansion/source; negative = compression/sink.

Curl (vector): $\nabla\times\vec{F} = \left\langle \frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z},\; \frac{\partial P}{\partial z}-\frac{\partial R}{\partial x},\; \frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right\rangle$

Interpretation: local rotation axis and intensity.

Estimating Divergence and Curl from a Graph

KEY CONCEPT

From a vector field's graph (arrows at various points):

Positive divergence: vectors spread outward from a point (more flow leaving than entering)
Negative divergence: vectors converge toward a point
Zero divergence: flow in equals flow out
Positive curl (2D): counterclockwise rotation tendency around the point
Negative curl (2D): clockwise rotation tendency
Zero curl: no rotation tendency — arrows don't swirl

Practice: place a tiny imaginary paddle wheel at the point. Would it spin? Which direction?

KEY CONCEPT

Identities tested in True/False:

$\text{curl}(\nabla f) = \vec{0}$ — always TRUE
$\nabla\cdot\vec{F}$ is a scalar, NOT a vector field — TRUE
$\nabla\times\vec{F}$ is a vector field — TRUE
$\nabla\cdot(\nabla\times\vec{F}) = 0$ always — TRUE (divergence of any curl is zero)
$\nabla\times(\vec{F}+\vec{G}) = \nabla\times\vec{F}+\nabla\times\vec{G}$ — TRUE (linearity)
$\nabla\cdot\vec{F}=\nabla\cdot\vec{G} \implies \vec{F}=\vec{G}$ — FALSE
$\nabla\times(\vec{F}\cdot\vec{G})$ — UNDEFINED ( $\vec{F}\cdot\vec{G}$ is scalar; curl of scalar is meaningless)
$\exists\vec{F}$ with $\nabla\times\vec{F}=\langle x,y,z\rangle$ — FALSE (since $\nabla\cdot\langle x,y,z\rangle = 3 \neq 0$ )

9.2 Line Integrals

FORMULA

Scalar line integral:

\int_C f\,ds = \int_a^b f(\vec{r}(t))|\vec{r}'(t)|\,dt

Vector field line integral (work):

\int_C \vec{F}\cdot d\vec{r} = \int_a^b \vec{F}(\vec{r}(t))\cdot\vec{r}'(t)\,dt

Applications of Line Integrals

KEY CONCEPT

Lateral surface area: If $f(x,y) \ge 0$ is a height function and $C$ is a curve in the $xy$ -plane, then $\int_C f\,ds$ gives the area of the "curtain" or "fence" hanging from $f$ down to the $xy$ -plane along $C$ .

Mass of a wire: If $\rho(x,y,z)$ is the linear density, the mass is $m = \int_C \rho\,ds$ .

Work: $W = \int_C \vec{F}\cdot d\vec{r}$ — work done by force $\vec{F}$ along path $C$ .

Circulation: $\oint_C \vec{F}\cdot d\vec{r}$ — the work integral around a closed curve. Measures the tendency of $\vec{F}$ to circulate around the curve.

Flux (2D): $\int_C \vec{F}\cdot\vec{n}\,ds = \int_C P\,dy - Q\,dx$ — flow across the curve.

Estimating Work, Circulation, and Flux from Graphs

KEY CONCEPT

From a vector field graph and a curve drawn on it:

Work/circulation positive: field vectors generally point in the same direction as the curve's orientation
Work/circulation negative: field vectors generally oppose the curve's direction
Work/circulation zero: field is perpendicular to the curve everywhere, or equal cancellation
Flux positive: field vectors generally cross the curve from left to right (outward for closed curves with CCW orientation)
Flux negative: field vectors cross from right to left (inward)
Flux zero: field is tangent to the curve everywhere, or equal cancellation

GOTCHA

True/False: $\int_{-C}f(x,y)\,ds = -\int_C f(x,y)\,ds$ is FALSE. The scalar integral $\int_C f\,ds$ uses $ds = |\vec{r}'(t)|\,dt$ which is always positive, so reversing orientation does NOT change it. This is TRUE for vector line integrals but FALSE for scalar line integrals.

9.3 Fundamental Theorem of Line Integrals

FORMULA

\int_C \nabla f\cdot d\vec{r} = f(B) - f(A)

Only depends on endpoints, not the path. Requires $\vec{F} = \nabla f$ (conservative).

Closed path corollary: $\oint_C \nabla f\cdot d\vec{r} = 0$ for any closed curve.

RECOGNITION PATTERN

Always check for conservative fields first. If a line integral problem gives you a complex parameterization, check whether $\vec{F}$ is conservative. If so, use FTLI instead of direct computation — much easier.

9.4 Green's Theorem

FORMULA

\oint_C P\,dx + Q\,dy = \iint_D \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right)dA

$C$ is the positively oriented (counterclockwise) boundary of $D$ .

Area via Green's Theorem

FORMULA

Choose $P$ and $Q$ so that $Q_x - P_y = 1$ . Three equivalent formulas:

A = \oint_C x\,dy = -\oint_C y\,dx = \frac{1}{2}\oint_C x\,dy - y\,dx

Try It Yourself

Use Green's Theorem to find the area enclosed by the ellipse $x = a\cos t$ , $y = b\sin t$ , $0 \le t \le 2\pi$ .

Reveal Solution

Using $A = \frac{1}{2}\oint_C x\,dy - y\,dx$ :

$dx = -a\sin t\,dt$ , $dy = b\cos t\,dt$

$x\,dy - y\,dx = a\cos t\cdot b\cos t\,dt - b\sin t\cdot(-a\sin t\,dt) = ab\cos^2 t\,dt + ab\sin^2 t\,dt = ab\,dt$

$A = \frac{1}{2}\int_0^{2\pi}ab\,dt = \frac{1}{2}\cdot ab\cdot 2\pi = \pi ab$

This elegantly gives the well-known ellipse area formula without any difficult integrals.

Using Green's Theorem to evaluate line integrals: Convert a line integral around a closed curve to a double integral over the enclosed region (or vice versa) — whichever is easier.

Using Green's Theorem to evaluate double integrals: If you have $\iint_D g\,dA$ and can find $P,Q$ with $Q_x - P_y = g$ , convert to a line integral around the boundary.

10 Surface Parameterization and Area

10.1 Parameterizing Surfaces

Express the surface as $\vec{r}(u,v) = \langle x(u,v),\; y(u,v),\; z(u,v)\rangle$ with bounds on $(u,v)$ .

Common parameterizations:

Explicit surface $z=f(x,y)$ : $\vec{r}(x,y)=\langle x, y, f(x,y)\rangle$
Sphere of radius $a$ : $\vec{r}(\theta,\phi) = \langle a\sin\phi\cos\theta,\; a\sin\phi\sin\theta,\; a\cos\phi\rangle$
Cylinder $x^2+y^2=a^2$ : $\vec{r}(\theta,z) = \langle a\cos\theta,\; a\sin\theta,\; z\rangle$

10.2 Normal Vector and Surface Area

FORMULA

Normal vector: $\vec{n} = \vec{r}_u\times\vec{r}_v$ (the cross product of the two tangent vectors)

Surface area (parametric):

A = \iint_D |\vec{r}_u\times\vec{r}_v|\,dA

Surface area (explicit $z=f(x,y)$ ):

A = \iint_R \sqrt{1+f_x^2+f_y^2}\,dA

Try It Yourself

Find the area of the surface $z = x^2+2y$ above the triangle with vertices $(0,0)$ , $(1,0)$ , $(1,2)$ .

Reveal Solution

$f_x = 2x$ , $f_y = 2$ . So $\sqrt{1+4x^2+4} = \sqrt{5+4x^2}$ .

Triangle region: $0\le x\le 1$ , $0\le y\le 2x$ .

$A = \int_0^1\int_0^{2x}\sqrt{5+4x^2}\,dy\,dx = \int_0^1 2x\sqrt{5+4x^2}\,dx$

Let $u=5+4x^2$ , $du=8x\,dx$ : $\int \frac{2}{8}\sqrt{u}\,du = \frac{1}{4}\cdot\frac{2}{3}u^{3/2}$

$= \frac{1}{6}(5+4x^2)^{3/2}\Big|_0^1 = \frac{1}{6}[9^{3/2}-5^{3/2}] = \frac{1}{6}[27-5\sqrt{5}]$

11 True/False Mastery

True/False questions make up a significant portion of the exam. They test conceptual understanding and edge-case reasoning. Work through these carefully — for each, decide your answer before revealing the solution.

11.1 Vectors and Dot/Cross Product

Try It Yourself

T/F 1: $|\vec{u}+\vec{v}| = |\vec{u}|+|\vec{v}|$ for all vectors $\vec{u},\vec{v}$ .

Reveal Answer

FALSE. Triangle inequality: $|\vec{u}+\vec{v}| \le |\vec{u}|+|\vec{v}|$ . Equality holds only when $\vec{u}$ and $\vec{v}$ are parallel and point in the same direction. Counterexample: $\vec{u}=\langle 1,0\rangle$ , $\vec{v}=\langle 0,1\rangle$ . Then $|\vec{u}+\vec{v}|=\sqrt{2}$ but $|\vec{u}|+|\vec{v}|=2$ .

Try It Yourself

T/F 2: $|\vec{u}\cdot\vec{v}| = |\vec{u}||\vec{v}|$ for all vectors.

Reveal Answer

FALSE. $|\vec{u}\cdot\vec{v}| = |\vec{u}||\vec{v}||\cos\theta|$ . Equality holds only when $\theta = 0$ or $\pi$ (parallel vectors). The correct inequality is Cauchy-Schwarz: $|\vec{u}\cdot\vec{v}| \le |\vec{u}||\vec{v}|$ .

Try It Yourself

T/F 3: $\vec{u}\times\vec{v} = \vec{v}\times\vec{u}$ for all vectors.

Reveal Answer

FALSE. Cross product is anticommutative: $\vec{u}\times\vec{v} = -(\vec{v}\times\vec{u})$ . The magnitudes are equal but directions are opposite.

Try It Yourself

T/F 4: $|\vec{u}\times\vec{v}| = |\vec{v}\times\vec{u}|$ for all vectors.

Reveal Answer

TRUE. $|\vec{u}\times\vec{v}| = |\vec{u}||\vec{v}|\sin\theta = |\vec{v}\times\vec{u}|$ . The magnitude is the same; only the direction reverses.

Try It Yourself

T/F 5: $(\vec{u}\times\vec{v})\cdot\vec{u} = 0$ for all vectors.

Reveal Answer

TRUE. The cross product $\vec{u}\times\vec{v}$ is perpendicular to both $\vec{u}$ and $\vec{v}$ . Two perpendicular vectors have zero dot product.

Try It Yourself

T/F 6: $(\vec{u}+\vec{v})\times\vec{v} = \vec{u}\times\vec{v}$ .

Reveal Answer

TRUE. Distribute: $(\vec{u}+\vec{v})\times\vec{v} = \vec{u}\times\vec{v} + \vec{v}\times\vec{v}$ . Since $\vec{v}\times\vec{v}=\vec{0}$ , the result is $\vec{u}\times\vec{v}$ .

Try It Yourself

T/F 7: If $\vec{u}\cdot\vec{v} = 0$ , then $\vec{u}=\vec{0}$ or $\vec{v}=\vec{0}$ .

Reveal Answer

FALSE. $\vec{u}\cdot\vec{v}=0$ means they could simply be orthogonal. Example: $\vec{u}=\langle 1,0\rangle$ , $\vec{v}=\langle 0,1\rangle$ . Neither is zero but $\vec{u}\cdot\vec{v}=0$ .

Try It Yourself

T/F 8: $\vec{u}\cdot\vec{v}\le|\vec{u}||\vec{v}|$ for all vectors.

Reveal Answer

TRUE. Since $\vec{u}\cdot\vec{v} = |\vec{u}||\vec{v}|\cos\theta$ and $\cos\theta \le 1$ , we have $\vec{u}\cdot\vec{v} \le |\vec{u}||\vec{v}|$ .

11.2 Lines and Planes

Try It Yourself

T/F 9: The vector $\langle 3,-1,2\rangle$ is parallel to the plane $6x-2y+4z=1$ .

Reveal Answer

FALSE. A vector is parallel to a plane iff it's perpendicular to the plane's normal. Normal = $\langle 6,-2,4\rangle$ . Check: $\langle 3,-1,2\rangle\cdot\langle 6,-2,4\rangle = 18+2+8 = 28 \neq 0$ . Not perpendicular to normal, so not parallel to plane. In fact, $\langle 3,-1,2\rangle = \frac{1}{2}\langle 6,-2,4\rangle$ , so this vector is parallel to the normal, hence perpendicular to the plane.

Try It Yourself

T/F 10: The equation $Ax+By+Cz=0$ represents a line in 3D.

Reveal Answer

FALSE. It represents a plane through the origin (with normal vector $\langle A,B,C\rangle$ ).

11.3 Vector Functions

Try It Yourself

T/F 11: $\vec{r}(t) = \langle t^3, 3t^3, 5t^3\rangle$ traces a line.

Reveal Answer

TRUE. Factor: $\vec{r}(t) = t^3\langle 1,3,5\rangle$ . All points lie on the line through the origin with direction $\langle 1,3,5\rangle$ . (The parameterization isn't constant-speed, but the curve is still a line.)

Try It Yourself

T/F 12: $\frac{d}{dt}|\vec{r}(t)| = |\vec{r}'(t)|$ .

Reveal Answer

FALSE. By the chain rule: $\frac{d}{dt}|\vec{r}| = \frac{\vec{r}\cdot\vec{r}'}{|\vec{r}|}$ . The left side measures how fast the distance from the origin changes; the right side is the speed (how fast the point moves along the curve). These are different — a point can move fast along a circle without getting farther from the origin.

11.4 Limits and Partial Derivatives

Try It Yourself

T/F 13: If $f(x,y)\to L$ along every straight line through $(a,b)$ , then $\lim_{(x,y)\to(a,b)}f(x,y)=L$ .

Reveal Answer

FALSE. The limit might fail along a curved path. Classic example: $f(x,y) = \frac{x^2 y}{x^4+y^2}$ . Along any line $y=mx$ , the limit is 0. But along $y=x^2$ , the limit is $\frac{1}{2}$ . So the limit DNE despite all straight-line limits agreeing.

Try It Yourself

T/F 14: If $f$ has continuous partial derivatives of all orders, then $f_{xxy} = f_{yxx}$ .

Reveal Answer

TRUE. By Clairaut's Theorem (applied repeatedly), the order of differentiation doesn't matter when all partials are continuous. $f_{xxy} = f_{xyx} = f_{yxx}$ .

11.5 Iterated Integrals

Try It Yourself

T/F 15: $\int_a^b\int_c^d f(x,y)\,dy\,dx = \int_c^d\int_a^b f(x,y)\,dx\,dy$ if $f$ is continuous.

Reveal Answer

TRUE. Fubini's Theorem guarantees this when the bounds are constants (rectangle) and $f$ is continuous.

Try It Yourself

T/F 16: $\int_0^1\int_0^{x^2}f(x,y)\,dy\,dx = \int_0^{x^2}\int_0^1 f(x,y)\,dx\,dy$ .

Reveal Answer

FALSE. When bounds depend on the other variable, you cannot simply swap. You must re-derive the bounds by sketching the region. The right side doesn't even make sense as written because the outer limits depend on $x$ , which would be the inner variable.

Try It Yourself

T/F 17: $\int_a^b\int_c^d g(x)h(y)\,dy\,dx = \left(\int_a^b g(x)\,dx\right)\left(\int_c^d h(y)\,dy\right)$ .

Reveal Answer

TRUE. When the integrand is separable ( $f(x,y)=g(x)h(y)$ ) and the region is a rectangle (constant bounds), the iterated integral factors into a product of single integrals.

Try It Yourself

T/F 18: $\int_{-1}^{1}\int_0^1 e^{x^2+y^2}\sin(y)\,dx\,dy = 0$ .

Reveal Answer

TRUE. The integrand $e^{x^2+y^2}\sin(y)$ is an odd function in $y$ (since $\sin(-y) = -\sin(y)$ and $e^{x^2+y^2}$ is even in $y$ ). The outer integral is over $y\in[-1,1]$ , a symmetric interval. The integral of an odd function over a symmetric interval is always 0.

11.6 Vector Calculus

Try It Yourself

T/F 19: If $f$ is a conservative scalar field, then $\oint_C \nabla f\cdot d\vec{r} = 0$ for any circle $C$ .

Reveal Answer

TRUE. By FTLI, the integral of a gradient over a closed curve equals $f(\text{end})-f(\text{start}) = 0$ since start = end.

Try It Yourself

T/F 20: If $\vec{F}=\langle P,Q\rangle$ and $P_y = Q_x$ on an open region $D$ , then $\vec{F}$ is conservative.

Reveal Answer

TRUE (for this course). Technically requires $D$ to be simply connected, but the course assumes this. The condition $P_y = Q_x$ is the 2D curl-free condition.

Try It Yourself

T/F 21: $\int_{-C}f(x,y)\,ds = -\int_C f(x,y)\,ds$ .

Reveal Answer

FALSE. For scalar line integrals, $ds$ is always positive. Reversing the curve does NOT change the sign. (This is TRUE for vector line integrals $\int_C\vec{F}\cdot d\vec{r}$ , but FALSE for scalar line integrals $\int_C f\,ds$ .)

Try It Yourself

T/F 22: The work done by a conservative force on a particle that moves along a closed path is 0.

Reveal Answer

TRUE. By FTLI, $\oint_C\nabla f\cdot d\vec{r} = f(\text{end})-f(\text{start})=0$ for any closed path.

Try It Yourself

T/F 23: $\nabla\cdot\vec{F}$ is a vector field.

Reveal Answer

FALSE. Divergence is a scalar-valued function, not a vector field.

Try It Yourself

T/F 24: $\text{curl}(\nabla f) = \vec{0}$ for any twice-differentiable scalar function $f$ .

Reveal Answer

TRUE. The curl of a gradient is always zero. This is equivalent to saying every gradient field is conservative (irrotational).

Try It Yourself

T/F 25: If $\nabla\cdot\vec{F}=\nabla\cdot\vec{G}$ , then $\vec{F}=\vec{G}$ .

Reveal Answer

FALSE. Many different vector fields can have the same divergence. For example, $\vec{F}=\langle x,0,0\rangle$ and $\vec{G}=\langle 0,y,0\rangle$ both have divergence 1, but $\vec{F}\neq\vec{G}$ .

Try It Yourself

T/F 26: There exists a vector field $\vec{F}$ such that $\nabla\times\vec{F} = \langle x, y, z\rangle$ .

Reveal Answer

FALSE. The divergence of any curl is always zero: $\nabla\cdot(\nabla\times\vec{F}) = 0$ . But $\nabla\cdot\langle x,y,z\rangle = 1+1+1 = 3 \neq 0$ . So no such $\vec{F}$ exists.

Try It Yourself

T/F 27: $\nabla\times(\vec{F}\cdot\vec{G}) = (\nabla\times\vec{F})\cdot(\nabla\times\vec{G})$ .

Reveal Answer

FALSE (not even well-defined). $\vec{F}\cdot\vec{G}$ is a scalar. Curl is only defined for vector fields. So $\nabla\times(\vec{F}\cdot\vec{G})$ is meaningless.

Try It Yourself

T/F 28: If $\vec{F}$ has path-independent line integrals on a connected domain $D$ , then $\vec{F}$ is conservative.

Reveal Answer

TRUE. Path independence on a connected domain is equivalent to being conservative.

12 Exam Patterns and Decision Trees

12.1 "Find the Equation of..." Decision Tree

12.2 "Evaluate the Line Integral" Decision Tree

RECOGNITION PATTERN

Step 1: Is the integrand $\vec{F}\cdot d\vec{r}$ (vector) or $f\,ds$ (scalar)?

Step 2 (if vector): Is $\vec{F}$ conservative? Check $\nabla\times\vec{F}=\vec{0}$ (or $P_y=Q_x$ in 2D).

Yes, conservative: Find potential function $f$ , use FTLI: $f(B)-f(A)$
No, closed curve: Consider Green's Theorem to convert to a double integral
No, not closed: Parameterize and compute directly

Step 2 (if scalar): Parameterize, compute $|\vec{r}'(t)|$ , integrate.

12.3 "Set Up the Integral" — Coordinate System Choice

RECOGNITION PATTERN

Use polar when: integrand contains $x^2+y^2$ ; region is a disk, annulus, or sector.

Use cylindrical when: region has circular symmetry in the $xy$ -plane (cylinders, bowl-shaped surfaces opening along an axis, surfaces with circular cross-sections); integrand involves $x^2+y^2$ .

Use spherical when: region involves spheres or surfaces of the form $z^2 = x^2+y^2$ ; integrand involves $x^2+y^2+z^2$ ; bounds are naturally expressed with $\rho$ and $\phi$ .

Stay rectangular when: region has flat/linear boundaries; no obvious symmetry.

12.4 Cross-Topic Integration Problems

Try It Yourself

Let $\vec{F}(x,y) = \langle 2xy+y^2,\; x^2+2xy\rangle$ . Evaluate $\int_C \vec{F}\cdot d\vec{r}$ where $C$ goes from $(0,0)$ to $(1,2)$ along any path.

Reveal Solution

Step 1: Check conservative. $P = 2xy+y^2$ , $Q = x^2+2xy$ . $P_y = 2x+2y$ , $Q_x = 2x+2y$ . Since $P_y = Q_x$ , $\vec{F}$ is conservative.

Step 2: Find potential. $f = \int(2xy+y^2)\,dx = x^2y+xy^2+g(y)$ . Then $f_y = x^2+2xy+g'(y) = x^2+2xy$ , so $g'(y)=0$ , $g(y)=C$ . Thus $f = x^2y+xy^2$ .

Step 3: FTLI. $f(1,2)-f(0,0) = (1)(2)+(1)(4)-0 = 6$ .

Try It Yourself

Find the absolute maximum and minimum of $f(x,y)=x^2+y^2-2x$ on the closed disk $x^2+y^2\le 4$ .

Reveal Solution

Interior critical points: $f_x = 2x-2=0$ , $f_y = 2y=0$ , so $(1,0)$ . $f(1,0) = 1+0-2 = -1$ .

Boundary: Parameterize: $x=2\cos t,\; y=2\sin t$ . Then $f = 4\cos^2 t+4\sin^2 t-4\cos t = 4-4\cos t$ . Critical: $f'(t) = 4\sin t = 0$ , so $t=0$ or $t=\pi$ .

At $t=0$ : $(2,0)$ , $f=0$ . At $t=\pi$ : $(-2,0)$ , $f=8$ .

Compare: Interior: $-1$ . Boundary: $0$ and $8$ .

Absolute minimum: $-1$ at $(1,0)$ . Absolute maximum: $8$ at $(-2,0)$ .

13 Exam Simulation Problems

Work these under timed conditions — roughly 8–10 minutes each, matching exam pacing.

Problem 1

Find the equation of the plane through $P(3,-1,1)$ , $Q(4,0,2)$ , and $R(6,3,1)$ .

Reveal Solution

$\overrightarrow{PQ} = \langle 1,1,1\rangle$ , $\overrightarrow{PR} = \langle 3,4,0\rangle$ .

$\vec{n} = \overrightarrow{PQ}\times\overrightarrow{PR} = \langle (1)(0)-(1)(4),\; (1)(3)-(1)(0),\; (1)(4)-(1)(3)\rangle = \langle -4, 3, 1\rangle$

Plane: $-4(x-3)+3(y+1)+1(z-1)=0$ , so $-4x+3y+z = -4(3)+3(-1)+1 = -12-3+1 = -14$ .

Or: $-4x+3y+z=-14$ , equivalently $4x-3y-z = 14$ .

Problem 2

Set up (but do not evaluate) a triple integral in spherical coordinates for the volume of the region inside the sphere $x^2+y^2+z^2=4$ and above the surface $z=\sqrt{x^2+y^2}$ .

Reveal Solution

The sphere is $\rho = 2$ . The surface $z = \sqrt{x^2+y^2}$ means $\rho\cos\phi = \rho\sin\phi$ , i.e., $\tan\phi = 1$ , so $\phi = \pi/4$ .

The region is above that surface ( $\phi \le \pi/4$ ) and inside the sphere ( $\rho \le 2$ ). Full revolution in $\theta$ .

$V = \int_0^{2\pi}\int_0^{\pi/4}\int_0^2 \rho^2\sin\phi\,d\rho\,d\phi\,d\theta$

Problem 3

Use Green's Theorem to evaluate $\oint_C (x^2y\,dx + xy^2\,dy)$ where $C$ is the circle $x^2+y^2=1$ oriented counterclockwise.

Reveal Solution

$P = x^2y$ , $Q = xy^2$ . $Q_x = y^2$ , $P_y = x^2$ .

$\oint_C = \iint_D (y^2 - x^2)\,dA$

Switch to polar: $\int_0^{2\pi}\int_0^1 (r^2\sin^2\theta - r^2\cos^2\theta)\cdot r\,dr\,d\theta = \int_0^{2\pi}\int_0^1 r^3(\sin^2\theta-\cos^2\theta)\,dr\,d\theta$

$= \int_0^{2\pi}(-\cos 2\theta)\,d\theta \cdot \int_0^1 r^3\,dr = \left[-\frac{\sin 2\theta}{2}\right]_0^{2\pi}\cdot\frac{1}{4} = 0\cdot\frac{1}{4} = 0$ .

Problem 4

Find the directional derivative of $f(x,y,z)=x^2y+x\sqrt{1+z}$ at $(1,2,3)$ in the direction of $\langle 2,1,-2\rangle$ . What is the maximum rate of change and in what direction?

Reveal Solution

$f_x = 2xy+\sqrt{1+z} = 2(1)(2)+\sqrt{4} = 6$

$f_y = x^2 = 1$

$f_z = \frac{x}{2\sqrt{1+z}} = \frac{1}{2\cdot 2} = \frac{1}{4}$

$\nabla f(1,2,3) = \langle 6, 1, \frac{1}{4}\rangle$

Unit vector: $\hat{u} = \frac{\langle 2,1,-2\rangle}{3} = \langle \frac{2}{3}, \frac{1}{3}, -\frac{2}{3}\rangle$

$D_{\hat{u}}f = 6\cdot\frac{2}{3}+1\cdot\frac{1}{3}+\frac{1}{4}\cdot(-\frac{2}{3}) = 4+\frac{1}{3}-\frac{1}{6} = \frac{24+2-1}{6} = \frac{25}{6}$

Max rate = $|\nabla f| = \sqrt{36+1+\frac{1}{16}} = \sqrt{\frac{593}{16}} = \frac{\sqrt{593}}{4}$ , in direction $\nabla f = \langle 6, 1, \frac{1}{4}\rangle$ .

Problem 5

Find the tangent plane and normal line to the surface $x^2+y^2+z^2 = 14$ at the point $(1, 2, 3)$ .

Reveal Solution

Let $F(x,y,z) = x^2+y^2+z^2$ . Then $\nabla F = \langle 2x, 2y, 2z\rangle$ . At $(1,2,3)$ : $\nabla F = \langle 2, 4, 6\rangle$ .

Tangent plane: $2(x-1)+4(y-2)+6(z-3) = 0$ , i.e., $2x+4y+6z = 28$ , or $x+2y+3z = 14$ .

Normal line: $\vec{r}(t) = \langle 1+2t,\; 2+4t,\; 3+6t\rangle$ , or $\frac{x-1}{2}=\frac{y-2}{4}=\frac{z-3}{6}$ .

Math 233 Final Exam — Comprehensive Study Guide

1.1 The Right-Hand Rule for Orienting Coordinate Axes

1.2 Distance and Midpoint Formulas in 3D

1.3 Equations of Spheres

1.4 Simple Surfaces, Regions, and Traces

2.1 Representing Vectors

2.2 Vector Arithmetic

2.3 Force, Tension, and Velocity Word Problems

3.1 The Dot Product

3.2 Orthogonal Projection

3.3 The Cross Product

4.1 Equations of Lines

4.2 Equations of Planes

4.3 Relationships Between Lines and Planes

4.4 Distance Formulas

5.1 Domain of a Vector Function

5.2 Limits of Vector Functions

5.3 Curves, Eliminating the Parameter, and Graph Matching

5.4 Derivatives and Integrals of Vector Functions

5.5 Arc Length and Arc Length Parameterization

6.1 Functions of Two Variables

"Is zzz a function of xxx and yyy?"

Domain and Level Curves

Sketching Surfaces via Traces

6.2 Limits and Continuity

6.3 Partial Derivatives

Limit Definition of Partial Derivatives

6.4 PDE Verification

6.5 The Chain Rule for Multivariable Functions

7.1 Directional Derivatives and Gradient

7.2 Tangent Planes and Normal Lines

Tangent Line to a Level Curve (2D)

Normal Line to a Level Curve (2D)

7.3 Linear Approximation and Differentials

7.4 Critical Points and the Second Derivatives Test

7.5 Absolute Extrema on Closed Bounded Regions

8.1 Double Integrals — Setup and Evaluation

8.2 Area of a Region via Double Integrals

8.3 Changing the Order of Integration

8.4 Polar Coordinates

8.5 Triple Integrals

Cylindrical Coordinates

Spherical Coordinates

9.1 Vector Fields

Sketching a Vector Field from a Formula

Conservative Vector Fields

Divergence and Curl

Estimating Divergence and Curl from a Graph

9.2 Line Integrals

Applications of Line Integrals

Estimating Work, Circulation, and Flux from Graphs

9.3 Fundamental Theorem of Line Integrals

9.4 Green's Theorem

Area via Green's Theorem

10.1 Parameterizing Surfaces

10.2 Normal Vector and Surface Area

11.1 Vectors and Dot/Cross Product

11.2 Lines and Planes

11.3 Vector Functions

11.4 Limits and Partial Derivatives

11.5 Iterated Integrals

11.6 Vector Calculus

12.1 "Find the Equation of..." Decision Tree

12.2 "Evaluate the Line Integral" Decision Tree

12.3 "Set Up the Integral" — Coordinate System Choice

12.4 Cross-Topic Integration Problems

"Is $z$ a function of $x$ and $y$ ?"