Calculus III, drawn · MA 26100 Fall 2026

UNIT 01

Vectors & curves

Ch. 12–14 · Quizzes 1–4

Everything here is about arrows. An arrow with a length and a direction, what happens when you multiply two of them together two different ways, and what happens when you let one arrow’s tip move around. Lessons 1–8.

13.1

Vectors

Quiz 1

A vector is a displacement: how far and which way. It does not care where it starts. Two arrows of the same length pointing the same way are the same vector.

the four things you do constantly

\mathbf a = \langle a_1,a_2,a_3\rangle \qquad |\mathbf a| = \sqrt{a_1^2+a_2^2+a_3^2}

\mathbf a + \mathbf b = \langle a_1+b_1,\ a_2+b_2,\ a_3+b_3\rangle \qquad c\,\mathbf a = \langle ca_1, ca_2, ca_3\rangle

unit vector — same direction, length exactly 1

\mathbf u = \frac{\mathbf a}{|\mathbf a|}

Vector from $P$ to $Q$ is $\overrightarrow{PQ} = \langle q_1-p_1,\ q_2-p_2,\ q_3-p_3\rangle$ — head minus tail. Getting this backwards is the single most common Unit 1 slip.

Algorithm · build a vector of prescribed length in a given direction

Get the direction as any vector $\mathbf a$ pointing that way.
Normalise: divide by its own length, $\mathbf u = \mathbf a/|\mathbf a|$ . Now it has length 1.
Scale: multiply by the length you want. $\mathbf v = L\,\mathbf u$ .

Normalise first, scale second. Doing it in the other order gives the wrong length.

Check yourself

What is $|\langle 3,-4\rangle|$ ?
Answer
$\sqrt{9+16}=5$ .
Write the unit vector in the direction of $\langle 0,3,4\rangle$ .
Answer
Length is 5, so $\left\langle 0,\tfrac35,\tfrac45\right\rangle$ . Normalise first, always.
What is the vector from $P(1,2)$ to $Q(4,6)$ ?
Answer
$\overrightarrow{PQ}=\langle 3,4\rangle$ — head minus tail, $Q-P$ . Backwards is the classic slip.

13.2

Space, distance, spheres

Quiz 1

Three number lines at right angles. A point needs three numbers instead of two — and distance is still just Pythagoras, applied twice.

drag to spin

toward yourightup

Spin it. The three brown steps and the one magenta arrow are the same information — the arrow is just the shortcut.

distance between two points

|P_1P_2| = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}

sphere · centre

(h,k,l)

, radius

r

(x-h)^2 + (y-k)^2 + (z-l)^2 = r^2

Why Pythagoras twice: walk along $x$ , then along $y$ — that diagonal on the floor is $\sqrt{\Delta x^2+\Delta y^2}$ . Now go up by $\Delta z$ . That floor-diagonal and the vertical rise are a right triangle too, so square-and-add again.

Algorithm · recognise a sphere from a messy equation

Group. Put all the $x$ terms together, all the $y$ terms, all the $z$ terms. Move the constant to the right.
Complete the square in each variable: $x^2+ax \to \left(x+\tfrac a2\right)^2 - \tfrac{a^2}{4}$ .
Balance. Every $-\tfrac{a^2}{4}$ you created must be added to the right-hand side too.
Read it off. Centre is $(h,k,l)$ from the three brackets, radius is $\sqrt{\text{right side}}$ . If the right side is negative there is no sphere; if it is zero you get a single point.

Worked example

$x^2+y^2+z^2 - 4x + 6y - 2 = 0$ .

group

$(x^2-4x)+(y^2+6y)+z^2 = 2$

complete

$(x-2)^2-4+(y+3)^2-9+z^2 = 2$

balance

$(x-2)^2+(y+3)^2+z^2 = 15$

Centre $(2,-3,0)$ , radius $\sqrt{15}$ .

One equation in 3-D is a surface, not a curve. $x=3$ is a whole plane. $x^2+y^2=9$ is an infinite cylinder — the $z$ is free, so the circle gets extruded forever.

Check yourself

Distance from $(1,0,2)$ to $(1,4,5)$ ?
Answer
$\sqrt{0+16+9}=5$ . Pythagoras, twice.
In 3-D, what does $x^2+y^2=4$ describe?
Answer
An infinite cylinder of radius 2 around the $z$ -axis — $z$ is free, so the circle gets extruded forever. One equation in 3-D is a surface, not a curve.
Centre and radius of $(x-1)^2+y^2+(z+2)^2=9$ ?
Answer
Centre $(1,0,-2)$ , radius 3. The sign inside the bracket flips: $(z+2)^2$ means $z=-2$ .

13.3

Dot product

Quiz 1

Shine a light straight down on $\mathbf b$ . The shadow it casts along $\mathbf a$ is the whole idea. Long shadow → big number. No shadow → zero. Shadow pointing backwards → negative.

Drag the angle. a·b = (length of a) × (length of the shadow). Same direction → positive. Square corner → zero. Opposite → negative.

two formulas, one number

\mathbf a\cdot\mathbf b = a_1b_1+a_2b_2+a_3b_3 \qquad\text{and}\qquad \mathbf a\cdot\mathbf b = |\mathbf a||\mathbf b|\cos\theta

The left one is how you compute it. The right one is what it means. Setting them equal is how every angle problem gets solved.

the three facts that matter

\cos\theta = \frac{\mathbf a\cdot\mathbf b}{|\mathbf a||\mathbf b|} \qquad \mathbf a\perp\mathbf b \iff \mathbf a\cdot\mathbf b = 0 \qquad \mathbf a\cdot\mathbf a = |\mathbf a|^2

Algorithm · angle between two vectors

Dot them componentwise: $\mathbf a\cdot\mathbf b$ .
Find both lengths $|\mathbf a|$ and $|\mathbf b|$ .
Divide: $\cos\theta = \dfrac{\mathbf a\cdot\mathbf b}{|\mathbf a||\mathbf b|}$ .
Arccos. $\theta = \cos^{-1}(\cdot)$ , always in $[0,\pi]$ .

Sign check before you compute: positive dot ⇒ acute, zero ⇒ right angle, negative ⇒ obtuse.

direction cosines — the angles a vector makes with each axis

\cos\alpha = \frac{a_1}{|\mathbf a|},\quad \cos\beta = \frac{a_2}{|\mathbf a|},\quad \cos\gamma = \frac{a_3}{|\mathbf a|}, \qquad \cos^2\alpha+\cos^2\beta+\cos^2\gamma = 1

Those three cosines are the components of the unit vector $\mathbf a/|\mathbf a|$ — which is exactly why they square to 1.

work — the physics reason dot products exist

W = \mathbf F\cdot\mathbf d = |\mathbf F||\mathbf d|\cos\theta

Check yourself

Compute $\langle1,2,3\rangle\cdot\langle4,-1,0\rangle$ .
Answer
$4-2+0=2$ .
If $\mathbf a\cdot\mathbf b \lt 0$ , is the angle acute or obtuse?
Answer
Obtuse. Negative dot means the shadow points backwards.
What is $\mathbf a\cdot\mathbf a$ ?
Answer
$|\mathbf a|^2$ . A vector agrees with itself completely.

13.3b

Scalar vs vector projection

Quiz 1

These two get confused constantly, and the difference is simple: one is a number (how long the shadow is) and the other is an arrow (the shadow itself, lying along $\mathbf b$ ).

Same picture, two different answers. comp is the tick mark on a ruler. proj is the arrow you could actually add to another vector.

scalar projection of a onto b — "component of a along b"

\operatorname{comp}_{\mathbf b}\mathbf a = \frac{\mathbf a\cdot\mathbf b}{|\mathbf b|} \qquad \textsf{a number}

vector projection of a onto b

\operatorname{proj}_{\mathbf b}\mathbf a = \left(\frac{\mathbf a\cdot\mathbf b}{|\mathbf b|^2}\right)\mathbf b = \bigl(\operatorname{comp}_{\mathbf b}\mathbf a\bigr)\frac{\mathbf b}{|\mathbf b|} \qquad \textsf{a vector}

How to never mix them up. The vector projection is literally the scalar projection times the unit vector of $\mathbf b$ . So: get the number first, then attach a direction to it. One $|\mathbf b|$ in the denominator = number. Two = vector, because the second one is busy normalising $\mathbf b$ .

Algorithm · project a onto b

Dot. Compute $\mathbf a\cdot\mathbf b$ .
Decide which you want. Does the answer need a direction? If not, stop after step 3.
Scalar: divide by $|\mathbf b|$ once. Done — that is $\operatorname{comp}_{\mathbf b}\mathbf a$ , and it may be negative.
Vector: multiply the scalar answer by the unit vector $\mathbf b/|\mathbf b|$ . Equivalently, divide by $|\mathbf b|^2$ and multiply by $\mathbf b$ — never by $|\mathbf b|$ .

Projecting onto b always lands on the b line. If your answer is not a multiple of b, you projected the wrong way round.

Worked example

$\mathbf a = \langle 3,4,0\rangle$ , $\mathbf b = \langle 4,0,3\rangle$ . Both have length 5.

dot

$\mathbf a\cdot\mathbf b = 12+0+0 = 12$

scalar

$\operatorname{comp}_{\mathbf b}\mathbf a = \dfrac{12}{5} = 2.4$

vector

$\operatorname{proj}_{\mathbf b}\mathbf a = \dfrac{12}{25}\langle4,0,3\rangle = \left\langle \tfrac{48}{25},\,0,\,\tfrac{36}{25}\right\rangle$

check

its length is $\sqrt{(48/25)^2+(36/25)^2} = \tfrac{60}{25} = 2.4$ ✓ matches the scalar answer

orthogonal decomposition — split a into "along b" plus "perpendicular to b"

\mathbf a = \underbrace{\operatorname{proj}_{\mathbf b}\mathbf a}_{\text{parallel to }\mathbf b} + \underbrace{\bigl(\mathbf a - \operatorname{proj}_{\mathbf b}\mathbf a\bigr)}_{\text{perpendicular to }\mathbf b}

This split is not a curiosity — it is reused three more times in this course: distance from a point to a line (13.5b), splitting acceleration into $a_T$ and $a_N$ (14.3), and the flux/circulation split of a vector field (Unit 4).

Trap

$\operatorname{proj}_{\mathbf b}\mathbf a \ne \operatorname{proj}_{\mathbf a}\mathbf b$ . Read the subscript as "onto". Swapping them gives a vector pointing in a completely different direction. And the scalar projection keeps its sign — if $\theta$ is obtuse, $\operatorname{comp}_{\mathbf b}\mathbf a$ is negative, and that negative is the answer, not an error.

Check yourself

$\mathbf a\cdot\mathbf b=10$ and $|\mathbf b|=5$ . What is $\operatorname{comp}_{\mathbf b}\mathbf a$ ?
Answer
$10/5=2$ . Divide by $|\mathbf b|$ once — that gives the number.
Same numbers. Is $\operatorname{proj}_{\mathbf b}\mathbf a$ a number or a vector?
Answer
A vector: $\dfrac{10}{25}\mathbf b$ . Divide twice — the second $|\mathbf b|$ is busy normalising.
If $\mathbf a\perp\mathbf b$ , what is $\operatorname{proj}_{\mathbf b}\mathbf a$ ?
Answer
The zero vector. No shadow — the light is shining straight along $\mathbf b$ .

13.4

Cross product

Quiz 1

Two arrows make a slanted tile. The cross product is a new arrow that stands straight out of that tile, and its length is exactly the tile's area.

drag to spin

angle between a and b

Flatten the angle to 0° and the tile vanishes — so does a × b. Open it to 90° and both are as big as they get. Swap the order and the arrow flips.

compute it as a 3×3 determinant

\mathbf a\times\mathbf b = \begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\ a_1&a_2&a_3\\ b_1&b_2&b_3\end{vmatrix} = \langle a_2b_3-a_3b_2,\ \ a_3b_1-a_1b_3,\ \ a_1b_2-a_2b_1\rangle

what it means

|\mathbf a\times\mathbf b| = |\mathbf a||\mathbf b|\sin\theta \qquad \mathbf a\times\mathbf b \perp \mathbf a \ \text{ and } \ \perp\mathbf b \qquad \mathbf b\times\mathbf a = -(\mathbf a\times\mathbf b)

Algorithm · the determinant, without mistakes

Write the grid. Row 1 is $\mathbf i,\mathbf j,\mathbf k$ ; row 2 is $\mathbf a$ ; row 3 is $\mathbf b$ . Order matters — $\mathbf a$ goes on top.
$\mathbf i$ component: cover column 1, cross-multiply what's left: $a_2b_3-a_3b_2$ .
$\mathbf j$ component: cover column 2, cross-multiply, then negate: $-(a_1b_3-a_3b_1)$ .
$\mathbf k$ component: cover column 3, cross-multiply: $a_1b_2-a_2b_1$ .
Verify for free. Dot your answer with $\mathbf a$ . You must get exactly 0. Same with $\mathbf b$ .

The minus sign on $\mathbf j$ is where nearly every lost point comes from. Step 5 catches it in ten seconds.

Area of a parallelogram

Spanned by $\mathbf a,\mathbf b$ : $\;A = |\mathbf a\times\mathbf b|$ .

Area of a triangle

Half of it: $\;A = \tfrac12|\overrightarrow{PQ}\times\overrightarrow{PR}|$ .

Are they parallel?

$\mathbf a\times\mathbf b=\mathbf 0$ ⟺ parallel. (Compare: $\mathbf a\cdot\mathbf b=0$ ⟺ perpendicular.)

scalar triple product — volume of the box

V = |\mathbf a\cdot(\mathbf b\times\mathbf c)| = \left|\begin{vmatrix}a_1&a_2&a_3\\ b_1&b_2&b_3\\ c_1&c_2&c_3\end{vmatrix}\right|

Reading it: $\mathbf b\times\mathbf c$ gives the base area as a vector standing on the base. Dotting with $\mathbf a$ takes $\mathbf a$ 's shadow on that stand-up direction — which is the height. Area × height = volume. If the triple product is zero, the box is flat, so the three vectors are coplanar.

Check yourself

$|\mathbf a|=3$ , $|\mathbf b|=4$ , and they meet at $90°$ . What is $|\mathbf a\times\mathbf b|$ ?
Answer
$3\cdot4\cdot\sin90°=12$ — the biggest the tile can get.
What is $\mathbf a\times\mathbf a$ ?
Answer
$\mathbf 0$ . Zero angle, flat tile, no area.
You computed $\mathbf a\times\mathbf b$ . How do you check it in ten seconds?
Answer
Dot it with $\mathbf a$ . You must get 0. This catches the missing minus on the $\mathbf j$ component every time.

12.1

Parametric equations

Quiz 1

A Cartesian equation tells you where a curve is. A parametrization tells you how a point moves along it — where it starts, which way it goes, how fast, and how many times round. Those extra facts are the whole reason the rest of this course uses parametrizations for everything: curves in space, surfaces, line integrals, flux. This is where they start.

a plane curve, parametrized

x=f(t),\qquad y=g(t),\qquad a\le t\le b

Think of $t$ as time and $(f(t),g(t))$ as the position of a bug. Sweep $t$ from $a$ to $b$ and the bug draws the curve. Different bugs can draw the same picture.

Both bugs draw the same ellipse. The orange one goes counterclockwise once; the green one goes clockwise, twice as fast, twice around. They disagree about direction, speed, and how many laps. Eliminating the parameter keeps the picture and throws all three away — which is exactly the information line integrals care about.

the three you must know cold

\begin{aligned} \textsf{segment } P\to Q:&\quad \mathbf r(t)=P+t\,(Q-P),\qquad 0\le t\le 1\\[4pt] \textsf{circle, centre }(h,k)\textsf{, radius }R:&\quad x=h+R\cos t,\quad y=k+R\sin t,\qquad 0\le t\le 2\pi\\[4pt] \textsf{ellipse, semi-axes }a,b:&\quad x=h+a\cos t,\quad y=k+b\sin t,\qquad 0\le t\le 2\pi \end{aligned}

All three run counterclockwise as written, and all three start at the rightmost point when $t=0$ . Swap $\sin$ and $\cos$ , or negate one of them, and you change the starting point or the direction — never the picture.

Algorithm · eliminate the parameter

Look for the trig identity first. If you see $\cos t$ and $\sin t$ , solve each equation for the trig function and use $\cos^2t+\sin^2t=1$ . Do not try to solve for $t$ .
Otherwise solve the easier equation for $t$ — usually the linear one — and substitute into the other.
Carry the range across. Ask what values $x$ and $y$ actually take as $t$ runs over its interval. The Cartesian equation on its own will usually describe more curve than you have.
Say the direction out loud. Plug in $t=a$ , the midpoint, and $t=b$ , and mark the three points in order. That is the orientation, and no Cartesian equation records it.

Steps 3 and 4 are where the marks are. Step 1 and 2 are algebra you already have.

Worked example

Describe the curve $x=2\cos t,\;y=3\sin t$ , $0\le t\le\pi$ .

identity

$\cos t=\dfrac{x}{2}$ and $\sin t=\dfrac{y}{3}$ , so $\dfrac{x^2}{4}+\dfrac{y^2}{9}=1$ — an ellipse with semi-axes $2$ and $3$ .

range

On $0\le t\le\pi$ , $\sin t\ge 0$ , so $y\ge 0$ . Only the upper half of the ellipse.

direction

$t=0\Rightarrow(2,0)$ ; $t=\tfrac{\pi}{2}\Rightarrow(0,3)$ ; $t=\pi\Rightarrow(-2,0)$ . Right, up, left — counterclockwise.

answer

The upper half of $\dfrac{x^2}{4}+\dfrac{y^2}{9}=1$ , traced once counterclockwise from $(2,0)$ to $(-2,0)$ .

Trap

“Eliminate the parameter” is a lossy operation. $x=t,\;y=t^2$ and $x=t^2,\;y=t^4$ both give $y=x^2$ , but the first is the whole parabola and the second is only its right half, traced in and back out. If a question mentions direction, a starting point, or a number of laps, the Cartesian equation cannot answer it — keep the parametrization.

Where this reappears: 13.5 writes a line as $\mathbf r(t)=\mathbf r_0+t\mathbf v$ , which is the segment recipe with the range removed. 14.1 is this section with a third coordinate bolted on. 17.2 integrates along one of these, and the orientation you were told to keep track of here is what decides the sign of the answer.

Check yourself

Parametrize the segment from $(-1,4)$ to $(3,2)$ on $0\le t\le1$ .

Answer
$x=-1+4t,\;y=4-2t$ . Start, plus $t$ times the displacement $\langle 4,-2\rangle$ .
What curve is $x=1+3\cos t,\;y=-2+3\sin t$ , $0\le t\le 2\pi$ ?

Answer
The circle of radius $3$ centred at $(1,-2)$ , once counterclockwise from $(4,-2)$ : $(x-1)^2+(y+2)^2=9$ .
$x=\sin t,\;y=\sin^2 t$ . Cartesian equation, and how much of it?

Answer
$y=x^2$ , but only the arc with $-1\le x\le 1$ , and the point retraces it back and forth forever.
Two parametrizations give the same Cartesian equation. Name two things that can still differ.

Answer
Orientation, speed, starting point, and how much of the curve is covered — any of these. Only the point set is shared.

13.5

Lines & planes

Quiz 1 · Quiz 2

A line needs a point and a direction to walk. A plane needs a point and a direction to avoid. That is the entire difference.

Tilt n and the whole plane tilts with it. That single arrow is the plane's identity.

line through

P_0

with direction

\mathbf v = \langle a,b,c\rangle

\mathbf r(t) = \mathbf r_0 + t\mathbf v \qquad \begin{cases}x = x_0+at\\ y=y_0+bt\\ z=z_0+ct\end{cases} \qquad \frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}

Vector form, parametric form, symmetric form — the same line, written three ways. Symmetric form is just the parametric equations each solved for $t$ . If a direction component is 0 you cannot divide by it: write that coordinate as a separate condition, e.g. $\frac{x-1}{2}=\frac{z-4}{5},\ y=3$ .

plane through

P_0

with normal

\mathbf n = \langle a,b,c\rangle

\mathbf n\cdot(\mathbf r - \mathbf r_0) = 0 \quad\Longleftrightarrow\quad a(x-x_0)+b(y-y_0)+c(z-z_0)=0 \quad\Longleftrightarrow\quad ax+by+cz=d

Free information: in $ax+by+cz=d$ , the coefficients are the normal vector. You can read $\mathbf n = \langle a,b,c\rangle$ straight off any plane equation without doing anything.

Algorithm · plane through three points $P$ , $Q$ , $R$

Make two vectors in the plane: $\overrightarrow{PQ}$ and $\overrightarrow{PR}$ .
Cross them: $\mathbf n = \overrightarrow{PQ}\times\overrightarrow{PR}$ . It sticks out of the plane by construction.
Pick any one of the three points and write $\mathbf n\cdot(\mathbf r-\mathbf r_0)=0$ .
Expand and check that all three original points satisfy it.

Algorithm · classify two lines in space

Compare directions. If $\mathbf v_1 \parallel \mathbf v_2$ (one is a multiple of the other) the lines are parallel — identical if a point of one satisfies the other, otherwise distinct.
Otherwise set them equal — and use different parameters, $t$ for one and $s$ for the other. Same-letter parameters is the classic wrong answer.
Solve two of the three equations for $t$ and $s$ .
Test in the third. Consistent ⇒ they intersect (plug back in for the point). Contradiction ⇒ they are skew — never meeting, not parallel.

Angle between two planes

Angle between their normals: $\cos\theta = \dfrac{|\mathbf n_1\cdot\mathbf n_2|}{|\mathbf n_1||\mathbf n_2|}$ .

Line where two planes meet

Direction is $\mathbf n_1\times\mathbf n_2$ . For a point, set one variable to 0 and solve the two equations.

Parallel planes

Normals are multiples of each other. Same normal + different $d$ ⇒ parallel and distinct.

Check yourself

What is the normal vector of the plane $2x-y+3z=7$ ?
Answer
$\langle 2,-1,3\rangle$ — read straight off the coefficients, no work needed.
Line through $(1,0,0)$ with direction $\langle2,3,-1\rangle$ ?
Answer
$\mathbf r(t)=\langle 1+2t,\;3t,\;-t\rangle$ .
Two lines have non-parallel directions and no common point. What are they?
Answer
Skew. They never meet and they are not parallel — only possible in 3-D.

13.5b

Every distance formula, and why

Quiz 2

All four of these are the same move: build a vector from the thing to the point, then keep only the part that is perpendicular to whatever you're measuring from.

point

P_1

to the plane

ax+by+cz+d=0

D = \frac{|ax_1+by_1+cz_1+d|}{\sqrt{a^2+b^2+c^2}} = \bigl|\operatorname{comp}_{\mathbf n}\overrightarrow{P_0P_1}\bigr|

Read it as a scalar projection onto the normal. Take any point $P_0$ on the plane, form $\overrightarrow{P_0P_1}$ , and measure its shadow on $\mathbf n$ — the part of the trip that actually left the plane.

point

P

to the line through

Q

with direction

\mathbf v

D = \frac{|\overrightarrow{QP}\times\mathbf v|}{|\mathbf v|}

Why the cross product: $|\overrightarrow{QP}\times\mathbf v| = |\overrightarrow{QP}||\mathbf v|\sin\theta$ is the area of the parallelogram they span. Area ÷ base = height, and the height is the perpendicular distance. Same as $|\overrightarrow{QP} - \operatorname{proj}_{\mathbf v}\overrightarrow{QP}|$ — the leftover from 13.3b.

two skew lines

D = \frac{\bigl|(\mathbf r_2-\mathbf r_1)\cdot(\mathbf v_1\times\mathbf v_2)\bigr|}{|\mathbf v_1\times\mathbf v_2|}

Algorithm · distance between two skew lines

Find the common perpendicular: $\mathbf n = \mathbf v_1\times\mathbf v_2$ . It is at right angles to both lines at once.
Pick one point on each line, $P_1$ and $P_2$ , and form $\overrightarrow{P_1P_2}$ .
Project onto $\mathbf n$ : the distance is $\bigl|\operatorname{comp}_{\mathbf n}\overrightarrow{P_1P_2}\bigr|$ . Everything not along $\mathbf n$ can be walked off along the lines, so it doesn't count.

Equivalent framing: slide the two lines into two parallel planes with normal $\mathbf n$ , then measure plane-to-plane.

Two parallel planes

Take any point on one, use the point-to-plane formula on the other.

Line parallel to a plane

Take any point on the line, use point-to-plane.

Sanity check

Every distance is $\ge 0$ . Absolute value bars are not decoration.

Check yourself

Distance from the origin to $x+2y+2z=9$ ?
Answer
$\dfrac{|0+0+0-9|}{\sqrt{1+4+4}}=\dfrac93=3$ .
Why does point-to-line use a cross product?
Answer
$|\overrightarrow{QP}\times\mathbf v|$ is the parallelogram's area. Area ÷ base = height, and the height is the perpendicular distance.

13.6

Cylinders & quadric surfaces

Quiz 2

Section 13.6 has two halves, and the first one is nearly free. A cylinder is what you get when a variable is missing from the equation. Missing means unconstrained: that coordinate may be anything at all, so the curve drawn in the other two variables is dragged along the whole missing axis. A cylinder in this course is not necessarily round — it is any curve times a line.

a missing variable is a swept axis

x^{2}+z^{2}=9\qquad\text{(no }y\text{)}

In the $xz$ -plane that is a circle of radius $3$ . In space, nothing constrains $y$ , so every height $y=k$ gives the same circle. Stack them and you get an infinite tube whose axis is the $y$ -axis.

The equation never mentions

y

, so

y

is free. The circle repeats at every

y

— the straight lines along the tube are the rulings, and they run parallel to the axis of the variable that went missing.

Algorithm · name a cylinder

Spot the missing variable. One of $x$ , $y$ , $z$ does not appear. If all three appear, it is not a cylinder — go to the quadric test below.
Read the curve formed by the two that remain, in their own coordinate plane: circle, ellipse, parabola, hyperbola, or a pair of lines.
Sweep it along the missing variable's axis. That axis is the direction of the rulings.
Name it after the curve. Circular, elliptic, parabolic or hyperbolic cylinder — then say which axis the rulings are parallel to. Both halves are expected in the answer.

$y=-z^{2}$ has no $x$ : a parabola in the $yz$ -plane opening toward $-y$ , swept along the $x$ -axis. A parabolic cylinder, rulings parallel to the $x$ -axis.

Trap

$x^{2}+z^{2}=9$ is a circle in two dimensions and a tube in three. The question decides which by telling you the ambient space, and 13.6 is a chapter about surfaces in $\mathbb{R}^{3}$ — so the answer is the tube. Answering “a circle of radius 3” is the single most common way to drop this question.

Now the other half. Every quadric is decided by signs and squares. Slice it with flat planes and see what falls out — circles, parabolas, or hyperbolas.

drag to spin

Fast test: count the minus signs. Zero → ellipsoid. One → one sheet. Two → two sheets. Right side is 0 instead of 1 → cone. One variable unsquared → paraboloid.

Algorithm · identify a quadric by its traces

Standard form first. Get it to $\dfrac{x^2}{a^2}\pm\dfrac{y^2}{b^2}\pm\dfrac{z^2}{c^2} = 1,\ 0,$ or $=$ a single linear variable.
Count the minus signs among the squared terms: $0\Rightarrow$ ellipsoid, $1\Rightarrow$ hyperboloid of one sheet, $2\Rightarrow$ hyperboloid of two sheets.
Right side is 0? It's a cone.
One variable not squared? It's a paraboloid — elliptic if the two squared terms have the same sign, hyperbolic (a saddle) if opposite.
Confirm with traces. Set $z=k$ and see what curve you get, then $x=k$ , then $y=k$ . Ellipse / parabola / hyperbola tells you the rest.

The axis of a cone, paraboloid, or hyperboloid is the variable that behaves differently from the other two.

Algorithm · complete the square into standard form

Group by variable and move the bare constant to the right-hand side.
Factor out the coefficient of each squared term, so the bracket opens with a bare $x^{2}$ , $y^{2}$ or $z^{2}$ .
Complete each square by adding $(b/2)^{2}$ inside the bracket. Because the bracket is multiplied by that factored-out coefficient, what you must add to the other side is $a\cdot(b/2)^{2}$ , not $(b/2)^{2}$ . This is where the arithmetic goes wrong.
Divide so the right side is $1$ (or $0$ ).
Now count the minus signs with the algorithm above. Shifting a surface never changes what it is — only where it sits.

worked ·

4x^{2}-y^{2}+z^{2}+8x=0

\begin{aligned} 4(x^{2}+2x)-y^{2}+z^{2}&=0\\[2pt] 4(x^{2}+2x+1)-y^{2}+z^{2}&=0+4\cdot 1\\[2pt] 4(x+1)^{2}-y^{2}+z^{2}&=4\\[2pt] (x+1)^{2}-\tfrac{y^{2}}{4}+\tfrac{z^{2}}{4}&=1 \end{aligned}

One minus sign, right side $1$ : a hyperboloid of one sheet, axis the $y$ -axis, centred at $(-1,0,0)$ . The $+8x$ only moved it one unit along $x$ ; the surface was never anything else.

Trap

The one-sheet hyperboloid has one minus and is connected — you can walk around the waist. The two-sheet has two minuses and a gap: there is no surface at all for small $|z|$ . Test it by setting $z=0$ and seeing whether you get a real curve or an impossibility.

Check yourself

In space, what is $x^{2}+z^{2}=9$ ?
Answer
No $y$ , so $y$ is free: a circular cylinder of radius $3$ with rulings parallel to the $y$ -axis. In the plane it would be a circle; in $\mathbb{R}^{3}$ it is a tube.
Identify $y=-z^{2}$ in three dimensions.
Answer
No $x$ : a parabolic cylinder, the parabola $y=-z^{2}$ in the $yz$ -plane opening toward $-y$ , swept along the $x$ -axis.
Put $9x^{2}+4y^{2}+36z^{2}-18x=27$ into standard form and identify it.
Answer
$9(x^{2}-2x)=9(x-1)^{2}-9$ , so $9(x-1)^{2}+4y^{2}+36z^{2}=36$ , i.e. $\tfrac{(x-1)^{2}}{4}+\tfrac{y^{2}}{9}+z^{2}=1$ . No minus signs → an ellipsoid centred at $(1,0,0)$ .
Identify $x^2+y^2-z^2=1$ .
Answer
One minus sign → hyperboloid of one sheet (connected, like a cooling tower).
Identify $z=x^2-y^2$ .
Answer
One variable unsquared, opposite signs → hyperbolic paraboloid, the saddle.
Identify $z^2=x^2+y^2$ .
Answer
Right side is 0 when rearranged → cone.

14.1

Vector functions & space curves

Quiz 3

Feed in time, get back a position. The tip of the arrow traces a curve — a bead sliding along a bent wire.

a curve, as a moving arrow

\mathbf r(t) = \langle f(t),\ g(t),\ h(t)\rangle, \qquad \lim_{t\to a}\mathbf r(t) = \left\langle \lim f,\ \lim g,\ \lim h \right\rangle

Everything happens slot by slot. Limits, derivatives, integrals, continuity — do the one-variable thing to each component and reassemble.

Algorithm · parametrize the curve where two surfaces meet

Find the shadow. Eliminate one variable between the two equations to get a curve in a coordinate plane.
Parametrize that shadow. A circle $x^2+y^2=a^2$ becomes $x=a\cos t,\ y=a\sin t$ . A parabola becomes $x=t,\ y=t^2$ .
Recover the missing variable by plugging back into whichever surface equation solves for it most easily.
State the range of $t$ , and check the curve actually lies on both surfaces.

Worked example

Where the cylinder $x^2+y^2=1$ meets the plane $y+z=2$ .

shadow

The cylinder already gives the shadow: the unit circle.

parametrize

$x=\cos t,\ y=\sin t,\ 0\le t\le 2\pi$

recover

z

$z = 2-y = 2-\sin t$

$\mathbf r(t) = \langle \cos t,\ \sin t,\ 2-\sin t\rangle$ — a tilted ellipse riding on the cylinder.

Check yourself

Does $\mathbf r(t)=\langle\cos t,\;\sin t,\;3\rangle$ lie on a cylinder or a sphere?
Answer
Neither, quite — $x^2+y^2=1$ and $z=3$ , so it is a single circle: the cylinder cut by a horizontal plane.
What is the domain of $\langle\ln t,\;1/t,\;\sqrt t\rangle$ ?
Answer
$t \gt 0$ . Take the intersection of every component's domain.

14.2

Derivatives & integrals of vector functions

Quiz 3

Differentiate every slot. That is the whole rule. The result is tangent to the curve, pointing the way the bead is heading.

Derivative of position is velocity. Derivative of velocity is acceleration. Exactly Calc I — just done to all three coordinates at once.

derivative and unit tangent

\mathbf r'(t) = \langle f'(t),\ g'(t),\ h'(t)\rangle \qquad \mathbf T(t) = \frac{\mathbf r'(t)}{|\mathbf r'(t)|}

product rules — note which product each one keeps

\frac{d}{dt}\bigl[\mathbf u\cdot\mathbf v\bigr] = \mathbf u'\cdot\mathbf v + \mathbf u\cdot\mathbf v' \qquad \frac{d}{dt}\bigl[\mathbf u\times\mathbf v\bigr] = \mathbf u'\times\mathbf v + \mathbf u\times\mathbf v'

Trap

The cross-product rule must keep its order: $\mathbf u'\times\mathbf v$ , never $\mathbf v\times\mathbf u'$ . Cross products anticommute, so swapping flips the sign of your whole answer.

A fact worth memorising: if $|\mathbf r(t)|$ is constant, then $\mathbf r\cdot\mathbf r' = 0$ — the velocity is perpendicular to the position. That is why anything moving on a sphere has velocity tangent to the sphere, and it's the reason $\mathbf T' \perp \mathbf T$ in 14.3.

integral — antidifferentiate each slot, one vector constant

\int \mathbf r(t)\,dt = \left\langle \int f,\ \int g,\ \int h \right\rangle + \mathbf C

Check yourself

$\mathbf r(t)=\langle t^2,\;t^3\rangle$ . Find $\mathbf r\,\!\prime(1)$ .
Answer
$\langle 2t,3t^2\rangle$ at $t=1$ is $\langle2,3\rangle$ .
If $|\mathbf r(t)|$ is constant, what is $\mathbf r\cdot\mathbf r\,\!\prime$ ?
Answer
$0$ . Constant length means the velocity is perpendicular to the position — why anything moving on a sphere stays tangent to it.

14.3 · 14.5

Motion in space · T, N, B

Quiz 3

Sit on a roller coaster. T is where your nose points, N is the way you're pressed into your seat, and B is out through your shoulder. All three travel with you.

drag to spin

The frame rides the curve instead of standing still. Acceleration always splits into a T part (changing speed) and an N part (changing direction).

the moving frame

\mathbf T = \frac{\mathbf r'}{|\mathbf r'|} \qquad \mathbf N = \frac{\mathbf T'}{|\mathbf T'|} \qquad \mathbf B = \mathbf T\times\mathbf N

$\mathbf N$ is guaranteed perpendicular to $\mathbf T$ because $\mathbf T$ has constant length 1 — the fact from 14.2. $\mathbf B$ is perpendicular to both because that's what a cross product does.

acceleration always splits two ways

\mathbf a = a_T\,\mathbf T + a_N\,\mathbf N, \qquad a_T = \frac{d}{dt}|\mathbf v| = \frac{\mathbf r'\cdot\mathbf r''}{|\mathbf r'|}, \qquad a_N = \kappa|\mathbf v|^2 = \frac{|\mathbf r'\times\mathbf r''|}{|\mathbf r'|}

Read it physically. $a_T$ is the gas pedal — it changes how fast you go. $a_N$ is the steering wheel — it changes where you're pointed. There is never a $\mathbf B$ component: you are never accelerated sideways out of your own turn.

Algorithm · decompose acceleration

Compute $\mathbf r'$ and $\mathbf r''$ .
Speed: $|\mathbf r'|$ .
$a_T$ : dot them and divide by the speed, $\dfrac{\mathbf r'\cdot\mathbf r''}{|\mathbf r'|}$ .
$a_N$ : cross them and divide by the speed, $\dfrac{|\mathbf r'\times\mathbf r''|}{|\mathbf r'|}$ .
Check: $a_T^2 + a_N^2 = |\mathbf r''|^2$ . If that fails, one of them is wrong.

Dot for the gas pedal, cross for the steering wheel. Same two products, one divisor.

Projectile motion is this with $\mathbf a = \langle 0,0,-g\rangle$ : integrate once for $\mathbf v$ (constant of integration = initial velocity), again for $\mathbf r$ (constant = initial position).

Check yourself

What does $a_T$ measure?
Answer
The gas pedal — how fast your speed is changing. It is zero at constant speed, even in a hard turn.
Which product gives $a_N$ , dot or cross?
Answer
Cross. $a_N=\dfrac{|\mathbf r\,\!\prime\times\mathbf r\,\!\prime\prime|}{|\mathbf r\,\!\prime|}$ . Dot for the gas pedal, cross for the steering wheel.

14.4–14.5

Arc length & curvature

Quiz 4

A curve is hard to measure; a straight stick is easy. Chop the curve into a million tiny straight sticks, measure each, add them up.

Distance = speed × time, repeated for every instant. That's all the integral is doing.

arc length = speed integrated over time

L = \int_a^b |\mathbf r'(t)|\,dt = \int_a^b\sqrt{\left(\tfrac{dx}{dt}\right)^2+\left(\tfrac{dy}{dt}\right)^2+\left(\tfrac{dz}{dt}\right)^2}\,dt

Algorithm · reparametrize by arc length

Compute the speed $|\mathbf r'(t)|$ .
Build the arc-length function $s(t) = \displaystyle\int_0^t |\mathbf r'(u)|\,du$ — how far you've walked by time $t$ .
Invert it: solve $s = s(t)$ for $t$ in terms of $s$ .
Substitute back: $\mathbf r(t(s))$ . Now the parameter is distance, so the speed is exactly 1.

Only clean when the speed is constant or very simple — which is exactly when this gets assigned.

Curvature asks how small is the circle you're hugging. A tight little circle means a sharp turn; a huge circle looks almost straight.

A straight line hugs a circle of infinite radius, so its curvature is 0. A circle of radius R has curvature 1/R everywhere.

three formulas for the same κ — pick by what you're given

\kappa = \left|\frac{d\mathbf T}{ds}\right| = \frac{|\mathbf T'(t)|}{|\mathbf r'(t)|} = \frac{|\mathbf r'(t)\times\mathbf r''(t)|}{|\mathbf r'(t)|^3}

Use the third one

Almost always. It needs only $\mathbf r'$ and $\mathbf r''$ — no messy normalising.

Use the second one

When the problem already handed you $\mathbf T(t)$ .

For a plane curve $y=f(x)$

$\kappa = \dfrac{|f''|}{\bigl(1+(f')^2\bigr)^{3/2}}$

Sanity values: a straight line has $\kappa=0$ . A circle of radius $a$ has $\kappa = 1/a$ everywhere. The radius of curvature is $\rho = 1/\kappa$ — the radius of that best-fitting circle.

Check yourself

Speed of $\langle 3\cos t,\;3\sin t,\;4t\rangle$ ?
Answer
$\sqrt{9+16}=5$ , constant. So length over $[0,2\pi]$ is just $10\pi$ .
Curvature of a circle of radius 2?
Answer
$\kappa=1/2$ . Curvature is one over the radius of the hugging circle.
Curvature of a straight line?
Answer
$0$ — the hugging circle has infinite radius.

UNIT 02

Partial derivatives

Ch. 15 · Quizzes 4–6 · Midterm 1 ends at 15.7

Here is the trick that unlocks the rest of the course: a function of two variables is a hill. Feed it a spot on the ground, it hands you back a height. Lessons 9–16 are the rest of Exam 1; Lagrange (16.5) is Lesson 17 — the first thing on Exam 2.

15.1

Surfaces & level curves

Quiz 4

Slice the hill with flat sheets of glass at heights 1, 2, 3, 4. Each slice leaves a loop. Draw those loops flat on paper and you've made a topographic map. Nothing was lost.

drag to spin

height of the glass

Slide the glass up and down. Every height gives one loop, and the loops printed on the floor are the topographic map. Loops crowded = steep.

level curves (2 inputs) and level surfaces (3 inputs)

f(x,y) = k \qquad\qquad F(x,y,z) = k

Reading a contour map: loops crowded together = steep. Loops far apart = gentle. Closed rings shrinking inward = a peak or a pit. Contours that cross in an X = a saddle. You will use every one of these again in 15.7.

Algorithm · find and sketch the domain

List what can break: denominators (≠ 0), even roots (≥ 0), logs (> 0), arcsin/arccos (input in $[-1,1]$ ).
Write each as an inequality in $x$ and $y$ .
Sketch each boundary curve and shade the region satisfying all of them at once.
Mark boundaries solid or dashed — solid if the inequality is $\ge$ , dashed if strict.

Check yourself

What do the level curves of $f(x,y)=x+y$ look like?
Answer
Parallel straight lines $x+y=k$ , evenly spaced — a constant-slope ramp, not a hill.
Contours crowded close together mean what?
Answer
Steep. Big height change over little ground.

15.2

Limits & continuity

Quiz 4

On a number line you can only come in from the left or the right. On a plane you can spiral, zig-zag, arrive on any curve. Every road must agree or the limit doesn't exist.

To disprove a limit: find two roads that disagree. To prove one: squeeze it, or switch to polar and show the answer doesn't depend on θ as r → 0.

the two-path test — this disproves, it never proves

\text{if } \lim_{(x,y)\to(a,b)}f \text{ along path 1} \ne \lim \text{ along path 2}, \quad\text{then the limit does not exist}

Algorithm · show a limit does NOT exist

Try the axes first. Set $y=0$ , take $x\to 0$ . Then $x=0$ , take $y\to 0$ . Cheapest possible test.
Try lines through the origin: $y=mx$ . If the answer still contains $m$ , you're done — different slopes give different limits.
Try matching powers. If the numerator has degree $p$ in one variable and the denominator degree $q$ , try $y = x^{p/q}$ or $y=x^2$ . This is what catches the sneaky ones that survive every straight line.
State it cleanly: "along $y=0$ the limit is $A$ ; along $y=x$ it is $B$ ; $A\ne B$ , so the limit does not exist."

Algorithm · show a limit DOES exist

Try direct substitution. If $f$ is built from polynomials, roots, exponentials, sines and cosines and nothing blows up — you're done, it's continuous.
Switch to polar: $x=r\cos\theta$ , $y=r\sin\theta$ , and let $r\to 0^+$ . If everything left over is bounded and the answer does not depend on $\theta$ , that's the limit.
Or squeeze: find $0\le |f - L| \le g(x,y)$ with $g\to 0$ . Useful bounds: $\left|\dfrac{x}{\sqrt{x^2+y^2}}\right|\le 1$ , $|\sin u|\le|u|$ , $x^2 \le x^2+y^2$ .

Checking a hundred paths proves nothing. Polar or squeeze is the only way to actually close it.

Worked example — polar closing the case

$\displaystyle\lim_{(x,y)\to(0,0)}\frac{x^3+y^3}{x^2+y^2}$ . Every straight line gives 0, but that isn't a proof.

polar

$\dfrac{r^3(\cos^3\theta+\sin^3\theta)}{r^2} = r\,(\cos^3\theta+\sin^3\theta)$

bound

$|\cos^3\theta+\sin^3\theta|\le 2$ , so the whole thing is $\le 2r$

conclude

as $r\to0$ this $\to 0$ regardless of $\theta$ . The limit is 0.

Check yourself

Along $y=0$ the limit is 3; along $y=x$ it is 5. What do you conclude?
Answer
The limit does not exist. Two roads disagree, and that is a complete proof.
You checked 100 straight lines and all gave 0. Does the limit exist?
Answer
No — you have proved nothing. Paths can only disprove. To prove, switch to polar and show the answer does not depend on $\theta$ , or squeeze it.

15.3

Partial derivatives

Quiz 5

Refuse to turn. Walk due east only and record your height — that's a plain 1-D curve, and its slope is $\partial f/\partial x$ . Turn 90° and walk due north for $\partial f/\partial y$ .

drag to spin

slide the wall

Move the wall and watch the slice curve change shape. A partial derivative is the slope of that one curve — nothing more exotic than Calc I.

definition — the other variable never moves

f_x(a,b) = \lim_{h\to 0}\frac{f(a+h,\,b)-f(a,b)}{h} \qquad f_y(a,b) = \lim_{h\to 0}\frac{f(a,\,b+h)-f(a,b)}{h}

Algorithm · compute a partial derivative

Cover up the other variables with your hand. Pretend they are numbers like 7.
Differentiate normally in the one remaining variable — product rule, chain rule, all of Calc I still applies.
Uncover. The frozen variables come back exactly as they were.
Then plug in the point, if a specific point was asked for. Never plug in first.

$\dfrac{\partial}{\partial x}\bigl(x^2 y^3\bigr) = 2x y^3$ — the $y^3$ was a constant, so it just rode along.

second and mixed partials

f_{xx} = \frac{\partial^2 f}{\partial x^2}\qquad f_{xy} = (f_x)_y = \frac{\partial^2 f}{\partial y\,\partial x} \qquad f_{yx} = (f_y)_x

Clairaut's theorem

If the mixed partials are continuous near a point, $f_{xy} = f_{yx}$ — the order you differentiate in doesn't matter. This is free error-checking: compute both and they must match.

Notation trap

$f_{xy}$ means " $x$ first, then $y$ ", but $\dfrac{\partial^2f}{\partial y\,\partial x}$ means the same thing read right-to-left. Subscripts run left-to-right; Leibniz notation runs inside-out. Clairaut makes them equal anyway, but exam wording depends on you knowing which is which.

Check yourself

Compute $\dfrac{\partial}{\partial x}\left(x^2y^3\right)$ .
Answer
$2xy^3$ . The $y^3$ was a constant, so it rode along untouched.
Is $f_{xy}$ the same as $f_{yx}$ ?
Answer
Yes, whenever the mixed partials are continuous (Clairaut). Compute both — it is free error-checking.

15.4

Chain rule

Quiz 5

Draw a tree. To find how fast $z$ changes when $t$ changes, multiply along each branch from $t$ up to $z$ , then add the branches together. Every chain rule in the course is this one picture.

Why add? Because t wiggling changes both x and y, and each of those independently pushes z. The effects stack.

Case 1 · one independent variable

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

Case 2 · two independent variables

\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} \qquad \frac{\partial z}{\partial t} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial t} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial t}

Algorithm · any chain rule, from the tree

Draw the tree. Output on top. Below it, every variable the output directly depends on. Below those, the variables they depend on. Keep going to the bottom.
Label every edge with the derivative of the upper node with respect to the lower one. Use $\partial$ when the node has more than one child, $d$ when it has exactly one.
Trace every path from the top down to the variable you're differentiating with respect to.
Multiply along each path, then add the paths.

Number of terms in your answer = number of distinct routes down the tree. Count the routes first, then check your answer has that many terms.

implicit differentiation, from

F(x,y)=0

and

F(x,y,z)=0

\frac{dy}{dx} = -\frac{F_x}{F_y} \qquad\qquad \frac{\partial z}{\partial x} = -\frac{F_x}{F_z}, \qquad \frac{\partial z}{\partial y} = -\frac{F_y}{F_z}

Where the minus sign comes from. Staying on the level set means $F$ never changes, so $dF = 0$ . Expand: $F_x\,dx + F_y\,dy = 0$ . Solve for $dy/dx$ and the minus falls out. Nothing memorised.

Algorithm · implicit partials without the formula

Differentiate the whole equation with respect to the variable you care about.
Treat the dependent variable as a function — every time you differentiate $z$ , write $\partial z/\partial x$ .
Collect all the $\partial z/\partial x$ terms on one side.
Solve for $\partial z/\partial x$ .

Check yourself

$z$ depends on $x$ and $y$ ; both depend on $t$ . How many terms in $dz/dt$ ?
Answer
Two — one per route down the tree. Count the routes before you compute, then check your answer has that many terms.

15.5

Gradient & directional derivatives

Quiz 5

Put the two partials side by side into one arrow. That arrow points straight uphill, it is always perpendicular to the contour lines, and its length is how steep that climb is.

Perpendicular is not a coincidence: walking along a contour keeps your height constant, so the height-change in that direction is zero — and zero change means zero dot product with ∇f.

the gradient

\nabla f = \left\langle f_x,\ f_y \right\rangle \qquad\text{or}\qquad \nabla f = \left\langle f_x,\ f_y,\ f_z \right\rangle

You don't have to walk straight uphill. Pick any direction $\mathbf u$ — your slope is the shadow of the gradient onto your direction. The dot product from 13.3, doing real work.

drag to spin

your direction u

Spin u around. Line it up with ∇f for the steepest climb; turn it 90° and you're walking the contour — dead flat. The green bar is the shadow.

directional derivative — u must be a UNIT vector

D_{\mathbf u}f = \nabla f\cdot\mathbf u = |\nabla f|\cos\theta

Algorithm · directional derivative

Compute $\nabla f$ symbolically, then evaluate at the point.
Normalise the direction. If you were given a vector, divide by its length. If you were given an angle $\theta$ from the $x$ -axis, use $\mathbf u = \langle\cos\theta,\sin\theta\rangle$ . If you were given "toward the point $Q$ ", use $\overrightarrow{PQ}/|\overrightarrow{PQ}|$ .
Dot them. $D_{\mathbf u}f = \nabla f\cdot\mathbf u$ .

Skipping step 2 is the most common lost point in Unit 2. A direction vector of length 3 inflates your answer by exactly 3×.

Question	Answer	Why
Fastest increase	direction $\nabla f$ , rate $\|\nabla f\|$	$\cos\theta$ is maximal at $\theta=0$
Fastest decrease	direction $-\nabla f$ , rate $-\|\nabla f\|$	$\cos\theta = -1$
No change at all	any $\mathbf u \perp \nabla f$	$\cos 90° = 0$ — you're walking the contour
Level set direction	tangent to the contour	same thing as the row above

The gradient is perpendicular to its own level set

Walking along a contour keeps your height constant, so the change in that direction is zero — and zero change means a zero dot product with $\nabla f$ . That single fact gives you tangent planes to any implicitly defined surface for free.

tangent plane & normal line to the surface

F(x,y,z)=k

F_x(P)(x-x_0)+F_y(P)(y-y_0)+F_z(P)(z-z_0) = 0 \qquad\quad \mathbf r(t) = \mathbf r_0 + t\,\nabla F(P)

Algorithm · tangent plane to any surface

Write it as a level surface. $z = f(x,y)$ becomes $F(x,y,z) = f(x,y)-z = 0$ .
Compute $\nabla F$ and evaluate at the point. That is the normal vector.
Use the point-normal plane equation from 13.5.
For the normal line, use the same $\nabla F$ as the direction vector.

Note step 1 reproduces $\langle f_x, f_y, -1\rangle$ from 15.6 — the two methods are the same method.

Check yourself

$\nabla f$ is perpendicular to what?
Answer
Its own level curve. Walking along a contour keeps height constant, so the change is zero — and zero change means zero dot product.
What is the maximum rate of increase at a point?
Answer
$|\nabla f|$ , the length of the gradient — a number, attained in the direction $\nabla f$ .
You walk along a contour line. What is $D_{\mathbf u}f$ ?
Answer
$0$ . Dead flat — that is what a contour means.

15.6

Tangent planes & linear approximation

Midterm 1

The Earth is round, but your backyard is flat. Zoom in far enough on any smooth surface and it becomes a plane — and using that plane instead of the real surface is what linear approximation means.

drag to spin

zoom in

The plane is fixed — it never moves. Zooming makes the surface come to it. That is exactly what linear approximation buys you.

tangent plane to

z=f(x,y)

(a,b)

z = f(a,b) + f_x(a,b)(x-a) + f_y(a,b)(y-b)

Read it as a sentence: start at the known height, then add "slope × how far you walked" for each direction. Its normal vector is $\langle f_x,\ f_y,\ -1\rangle$ .

differentials — the same statement, about small changes

dz = f_x\,dx + f_y\,dy \qquad \Delta z \approx dz

Algorithm · estimate a value or an error

Pick a nearby easy point $(a,b)$ where $f$ , $f_x$ , $f_y$ are trivial to evaluate.
Compute the three numbers $f(a,b)$ , $f_x(a,b)$ , $f_y(a,b)$ .
Set $dx = x-a$ and $dy = y-b$ — the small steps from easy point to real point.
Add it up: $f(x,y)\approx f(a,b)+f_x\,dx+f_y\,dy$ .
For maximum error, use $|dx|,|dy|$ at their worst and take absolute values of the coefficients: $|dz|\le |f_x||dx|+|f_y||dy|$ .

Differentiable is stronger than "both partials exist." A function can have both partials at a point and still not have a tangent plane there. In practice, if $f_x$ and $f_y$ are continuous near the point, $f$ is differentiable — that's the test you'll actually use.

Check yourself

What is the normal vector to the surface $z=f(x,y)$ ?
Answer
$\langle f_x,\;f_y,\;-1\rangle$ . (Or write it as $F=f-z=0$ and take $\nabla F$ — same thing.)
Write the differential $dz$ .
Answer
$dz=f_x\,dx+f_y\,dy$ . Nudge $x$ , nudge $y$ , add both effects.

15.7

Maximum & minimum values

Midterm 1

If $\nabla f = \mathbf 0$ you're on level ground. But level ground happens at a summit, at the bottom of a bowl, and at a mountain pass — where it goes up one way and down the other.

Contours that form closed rings around a point mean max or min. Contours that cross in an X mean saddle — the classic tell on an exam.

second derivative test

D = f_{xx}f_{yy} - \bigl(f_{xy}\bigr)^2 \quad\text{evaluated at the critical point}

If	Then	Picture
$D \gt 0$ and $f_{xx} \lt 0$	local maximum	both slices bend down — a summit
$D \gt 0$ and $f_{xx} \gt 0$	local minimum	both slices bend up — a bowl
$D \lt 0$	saddle point	slices bend opposite ways — a pass
$D=0$	test fails	look at the contours or test values directly

Algorithm · classify all critical points

Set both partials to zero: $f_x=0$ and $f_y=0$ . Solve the system.
Find every solution. Factor rather than divide — dividing by $x$ silently throws away the critical points with $x=0$ .
Compute $f_{xx}$ , $f_{yy}$ , $f_{xy}$ symbolically, once.
Evaluate $D$ at each point and read the table.
Report the function value $f$ at each max/min, not just the location.

Step 2 is where solutions get lost. If you ever divide by a variable, handle the "= 0" case separately.

Algorithm · absolute extrema on a closed bounded region

Interior: find critical points strictly inside $D$ , and list their $f$ values.
Boundary: break the boundary into pieces. Parametrize each piece so it becomes a one-variable problem.
Optimise each piece with ordinary Calc I: derivative zero, plus the endpoints of that piece.
Corners: evaluate $f$ at every corner of the region.
Compare the whole list. Biggest value is the absolute max, smallest is the absolute min. No second derivative test needed anywhere.

The extreme value theorem guarantees both exist when the region is closed and bounded — so if your list has no winner, your list is incomplete.

Trap

Forgetting the boundary. On a closed region the absolute maximum very often sits on the edge, where $\nabla f$ is not zero at all — so it never shows up as a critical point. Steps 2–4 are not optional.

Check yourself

$D \lt 0$ at a critical point. What is it?
Answer
A saddle — and $f_{xx}$ is irrelevant here.
You found every interior critical point on a closed region. Are you done?
Answer
No. Check every boundary piece and every corner. On a closed region the maximum usually lives on the edge, where $\nabla f$ is not zero at all.

15.8

Lagrange multipliers

Quiz 6

You must stay on a path. Walk along it and watch the contour lines you cross — as long as you're cutting across them you're still gaining height, so keep going. You've maxed out exactly where the path stops crossing and just touches one.

Touching = the two curves share a tangent line = their normals point the same way. The normals are ∇f and ∇g. So ∇f is a multiple of ∇g. The multiple is λ.

one constraint

\nabla f = \lambda\,\nabla g \qquad\text{and}\qquad g(x,y,z) = k

Touching = sharing a tangent line = normals pointing along the same line. The normals are $\nabla f$ and $\nabla g$ , so one must be a multiple of the other. That multiple is $\lambda$ — and you almost never need its value.

Algorithm · Lagrange, one constraint

Name $f$ and $g$ . $f$ is what you're optimising; $g=k$ is the restriction.
Write the system: $f_x=\lambda g_x$ , $f_y=\lambda g_y$ , $f_z=\lambda g_z$ , plus $g=k$ . Four equations, four unknowns.
Eliminate $\lambda$ first. Solve each equation for $\lambda$ and set them equal — that removes an unknown immediately.
Watch for zeros. Before dividing by anything, handle the case where it equals zero as its own branch. That's where the missing solutions hide.
Evaluate $f$ at every candidate point and just compare. Largest is the max, smallest is the min.

No second derivative test exists here. Comparison is the whole final step.

Worked example

Maximise $f = xy$ on the circle $g = x^2+y^2 = 8$ .

system

$y = 2\lambda x$ , $x = 2\lambda y$ , $x^2+y^2=8$

eliminate

$\dfrac{y}{2x} = \lambda = \dfrac{x}{2y} \Rightarrow y^2 = x^2 \Rightarrow y = \pm x$

constraint

$2x^2 = 8 \Rightarrow x=\pm 2$ , giving $(2,2),(-2,-2),(2,-2),(-2,2)$

compare

$f = 4,\ 4,\ -4,\ -4$ . Max 4, min −4.

zero branch

$x=0$ would force $y=0$ , which fails the constraint — so nothing was lost in step 2.

two constraints — the curve where two surfaces meet

\nabla f = \lambda\,\nabla g + \mu\,\nabla h, \qquad g = k_1, \qquad h = k_2

With two constraints you're stuck on a curve instead of a surface. $\nabla g$ and $\nabla h$ span the plane of directions perpendicular to that curve, and at an optimum $\nabla f$ has to lie in that plane — otherwise some leftover component would still be pushing you along the curve toward a better value.

When to use Lagrange vs. substitution

If the constraint solves cleanly for one variable, substitute and use 15.7 — it's faster. Reach for Lagrange when the constraint is symmetric, messy, or refuses to be solved for anything (a sphere, an ellipsoid, $x^2+y^2+z^2=1$ ).

Check yourself

What is the condition at a constrained optimum?
Answer
$\nabla f=\lambda\nabla g$ — the contour just touches the constraint, so their normals line up.
Do you need the second derivative test to finish?
Answer
No. Evaluate $f$ at every candidate point and compare. Biggest is the max, smallest is the min.

UNIT 03

Multiple integrals

Ch. 16 · Quizzes 6–8

Derivatives cut things apart; integrals pile things up. In 3-D the pile is made of tiny boxes — and the only real skill in this unit is choosing box shapes that match the region you were handed. Lessons 18–24.

16.1

Double integrals

Quiz 6

Tile the floor. On each tile, build a column as tall as the surface above it. Volume of one column = height × tile area. Add up all the columns.

drag to spin

tiles per side

Each column is height × tile area. Turn the tiles up and the staircase melts into the smooth solid. That limit is the double integral.

definition and Fubini's theorem

\iint_R f\,dA = \lim_{m,n\to\infty}\sum_{i=1}^m\sum_{j=1}^n f(x_{ij}^*,y_{ij}^*)\,\Delta A = \int_a^b\!\!\int_c^d f\,dy\,dx = \int_c^d\!\!\int_a^b f\,dx\,dy

Over a rectangle the order never matters. And if $f(x,y) = g(x)h(y)$ splits, the double integral splits too: $\iint = \left(\int_a^b g\right)\left(\int_c^d h\right)$ . Look for that before doing any work.

$f = 1$ gives the plain area of $R$ . That's the sanity check for every setup you write.

Check yourself

What does $\iint_R 1\,dA$ give you?
Answer
The plain area of $R$ . Use it to sanity-check any setup.
Over a rectangle, does the order of integration matter?
Answer
No (Fubini). And if $f(x,y)=g(x)h(y)$ the whole thing factors into two single integrals — look for that first.

16.2

General regions & setting up limits

Quiz 6

This is where the points get lost, and it's mechanical. Fire an arrow across the region. Where it enters and exits gives the inner limits. How far the arrow itself slides gives the outer limits — always plain numbers.

Same region, same answer, different work. Swap the order when the inner integral is impossible or the region needs splitting into pieces — that is the whole point of the "reverse the order" problems.

Type I · vertical arrows, y between two curves

\iint_R f\,dA = \int_a^b\!\!\int_{g_1(x)}^{g_2(x)} f(x,y)\,dy\,dx

Type II · horizontal arrows, x between two curves

\iint_R f\,dA = \int_c^d\!\!\int_{h_1(y)}^{h_2(y)} f(x,y)\,dx\,dy

Algorithm · set up the limits

Sketch the region. Always. Find where the boundary curves intersect by setting them equal.
Choose a sweep direction — vertical (Type I) or horizontal (Type II). Pick whichever needs fewer pieces.
Inner limits: enter → exit. Which curve does the arrow enter through, which does it leave through? Those are functions of the outer variable.
Outer limits: the arrow's travel range. These must be constants.
Check the shape: $\displaystyle\int_{\text{number}}^{\text{number}}\int_{\text{function}}^{\text{function}}$ . A variable in the outer limits means you set it up wrong.

Algorithm · reverse the order of integration

Read the given limits backwards into a sketch. Inner limits are curves; outer limits are the strip the arrow slides through. Draw the region they describe — do not skip to algebra.
Re-sweep the other way. Fire the arrow horizontally instead of vertically (or vice versa).
Solve the boundary curves for the other variable. $y = x^2$ becomes $x=\sqrt y$ .
Split into pieces if the enter-or-exit curve changes partway across.
Now integrate. The whole point is usually that the new inner integral is doable and the old one wasn't.

Classic tell: an inner integrand like $e^{y^2}$ or $\sin(x^2)$ that has no elementary antiderivative. That's the problem asking you to flip.

Worked example

$\displaystyle\int_0^1\!\!\int_x^1 e^{y^2}\,dy\,dx$ — the inner integral is impossible as written.

sketch

$0\le x\le 1$ and $x\le y\le 1$ : the triangle with corners $(0,0),(1,1),(0,1)$ .

re-sweep

Horizontally: for each $y$ in $[0,1]$ , $x$ runs $0 \to y$ .

rewrite

$\displaystyle\int_0^1\!\!\int_0^y e^{y^2}\,dx\,dy = \int_0^1 y\,e^{y^2}\,dy$

finish

$=\tfrac12 e^{y^2}\Big|_0^1 = \dfrac{e-1}{2}$

Check yourself

Your outer limits contain a variable. What went wrong?
Answer
The setup. Outer limits are always constants. Re-sweep the region.
The inner integrand is $e^{y^2}$ . What is the problem telling you?
Answer
Flip the order. No elementary antiderivative in that variable is the signal, every time.

16.3

Double integrals in polar

Quiz 7

Cut the region into wedges instead of tiles. A wedge is nearly a rectangle: one side is $dr$ long, the other is an arc of length $r\,d\theta$ — because the same angle sweeps a longer arc when you're farther out.

Use polar the moment you see x² + y², a circle, a disk, or an annulus. The messy algebra collapses into r².

the r is a stretch factor, not a rule to memorise

\iint_R f(x,y)\,dA = \int_\alpha^\beta\!\!\int_{r_1(\theta)}^{r_2(\theta)} f(r\cos\theta,\ r\sin\theta)\ \underbrace{r\,dr\,d\theta}_{dA}

Algorithm · convert to polar

Spot the trigger: $x^2+y^2$ , a circle, a disk, an annulus, a wedge, or limits with $\sqrt{a^2-x^2}$ .
Convert the integrand: $x\to r\cos\theta$ , $y\to r\sin\theta$ , and every $x^2+y^2 \to r^2$ .
Convert the region. $\theta$ sweeps the angular range; for each $\theta$ , $r$ runs from the inner boundary to the outer boundary.
Attach the $r$ . $dA = r\,dr\,d\theta$ . Write it before you do anything else.
Integrate — inner in $r$ , outer in $\theta$ .

Useful conversions: $\;x^2+y^2=a^2 \to r=a$ . $y=x \to \theta=\pi/4$ . $x^2+y^2=2ax \to r = 2a\cos\theta$ (a circle through the origin).

The single most expensive mistake in the course

Dropping the $r$ . $dA = r\,dr\,d\theta$ , not $dr\,d\theta$ . It comes from the wedge geometry above and it is also exactly what the Jacobian in 16.7 will produce. Write the $r$ the moment you write $dr\,d\theta$ .

Check yourself

What is $dA$ in polar?
Answer
$r\,dr\,d\theta$ .
Where does that $r$ come from?
Answer
The wedge. One side is $dr$ long; the other is an arc of length $r\,d\theta$ , because the same angle sweeps further when you are farther out.

16.4

Triple integrals

Quiz 7

Same idea, now the tiles are tiny cubes. To get limits: look at the solid's shadow on a coordinate plane, then ask how far the solid runs above each shadow point.

f = 1 gives volume. f = density gives mass. There are six possible orders (dz dy dx, dy dz dx, …) — pick whichever makes the inner surfaces simplest.

the shadow method

\iiint_E f\,dV = \iint_{\text{shadow}}\left[\int_{z_1(x,y)}^{z_2(x,y)} f\,dz\right]dA

Algorithm · set up a triple integral

Pick the sweep direction. Usually $z$ : fire an arrow straight up through the solid.
Inner limits — enter and exit surfaces. $z$ runs from the bottom surface $z_1(x,y)$ to the top surface $z_2(x,y)$ . Solve both for $z$ .
Project to get the shadow. Where the top and bottom surfaces meet is the shadow's boundary curve — set $z_1 = z_2$ and eliminate $z$ .
Set up that 2-D shadow region exactly as in 16.2 (or 16.3 if it's round).
Check the shape: function limits inside, then function limits, then numbers outermost.

Six orders exist ( $dz\,dy\,dx$ , $dy\,dz\,dx$ , …). If the $z$ -sweep gives an ugly shadow, try sweeping along $x$ or $y$ instead — the shadow lands on a different plane.

$f=1$

Volume of the solid.

$f=\rho$

Mass. Then $\bar x = \frac1m\iiint x\rho\,dV$ , and similarly for $\bar y,\bar z$ .

Inertia about the $z$ -axis

$I_z = \iiint (x^2+y^2)\,\rho\,dV$ — squared distance from the axis.

Check yourself

What does $\iiint_E 1\,dV$ give?
Answer
The volume of $E$ .
How do you get the two outer limits of a triple integral?
Answer
Project. The solid's shadow on a coordinate plane is a 2-D region — set it up exactly as in 16.2.

16.5a

Cylindrical coordinates

Quiz 7

Polar coordinates with a height bolted on. Cans, pipes, and anything round with a straight axis.

conversion and the volume element

x = r\cos\theta,\quad y=r\sin\theta,\quad z=z \qquad\qquad dV = r\,dz\,dr\,d\theta

Algorithm · when and how

Use it when the shadow on the $xy$ -plane is a disk, wedge, or annulus — or when you see $x^2+y^2$ in the surfaces.
Convert the surfaces: a cylinder $x^2+y^2=a^2$ becomes $r=a$ ; a cone $z=\sqrt{x^2+y^2}$ becomes $z=r$ ; a paraboloid $z=x^2+y^2$ becomes $z=r^2$ .
Order is usually $dz\,dr\,d\theta$ : $z$ between the two surfaces, $r$ across the shadow, $\theta$ around.
Attach the $r$ . Same $r$ , same reason, as in 16.3.

Check yourself

What is $dV$ in cylindrical?
Answer
$r\,dz\,dr\,d\theta$ — the same $r$ , for the same reason as polar.
The cone $z=\sqrt{x^2+y^2}$ becomes what?
Answer
$z=r$ . Clean — that is why cylindrical suits cones with flat tops.

16.5b

Spherical coordinates

Quiz 8

Distance from the origin, angle down from the north pole, angle around. Balls and cones. The famous $\rho^2\sin\varphi$ is not arbitrary — it's the little box's three edge lengths multiplied together.

drag to spin

latitude angle φ

Drag φ toward either pole and watch the box go skinny. That shrinking edge is ρ sin φ — the whole reason dV carries a sin φ.

conversion

x = \rho\sin\varphi\cos\theta,\qquad y=\rho\sin\varphi\sin\theta,\qquad z = \rho\cos\varphi,\qquad \rho^2 = x^2+y^2+z^2

the volume element, derived not memorised

dV = \underbrace{d\rho}_{\text{radial}}\times\underbrace{\rho\,d\varphi}_{\text{north–south}}\times\underbrace{\rho\sin\varphi\,d\theta}_{\text{east–west}} = \rho^2\sin\varphi\ d\rho\,d\varphi\,d\theta

Why $\sin\varphi$ : the circle of latitude at angle $\varphi$ has radius $\rho\sin\varphi$ , not $\rho$ . Near the poles it shrinks to nothing, so the same $d\theta$ sweeps almost no distance — and the volume element shrinks with it.

Algorithm · set up in spherical

Use it when you see $x^2+y^2+z^2$ , a sphere, a ball, a cone, or a region between two spheres.
$\rho$ limits: where the ray from the origin enters and leaves the solid. A sphere of radius $a$ gives $\rho = a$ ; a plane $z=c$ gives $\rho = c/\cos\varphi = c\sec\varphi$ .
$\varphi$ limits: the cone angle. $\varphi=0$ is straight up, $\varphi=\pi/2$ is the $xy$ -plane, $\varphi=\pi$ is straight down. The cone $z=\sqrt{x^2+y^2}$ is $\varphi=\pi/4$ .
$\theta$ limits: how far around, usually $0$ to $2\pi$ .
Attach $\rho^2\sin\varphi$ and integrate, innermost $\rho$ outward.

Ranges: $\rho\ge 0$ , $0\le\varphi\le\pi$ , $0\le\theta\le 2\pi$ . $\varphi$ never goes past $\pi$ — that would be going round twice.

The ρ² sin φ is not arbitrary: dρ × (ρ dφ) × (ρ sin φ dθ). Two ρ's and one sin φ fall out on their own. φ ∈ [0, π] (top to bottom), θ ∈ [0, 2π] (all the way around).

Check yourself

What is $dV$ in spherical?
Answer
$\rho^2\sin\varphi\;d\rho\,d\varphi\,d\theta$ .
What is the range of $\varphi$ ?
Answer
$0\le\varphi\le\pi$ — top to bottom. Only $\theta$ goes all the way round to $2\pi$ .
Why the $\sin\varphi$ ?
Answer
The circle of latitude has radius $\rho\sin\varphi$ , not $\rho$ . Near the poles it shrinks to nothing, so the box goes skinny.

16.6

Mass, moments, centre of mass, inertia

Quiz 8

A metal plate with heavy and light spots. Its mass is density piled over the region. Its centre of mass is where you'd stick a pin so it sits level — the average position, with heavy bits voting harder.

Everything here is "∬ (some weight) × ρ dA". Weight = 1 → mass. Weight = x → moment. Weight = distance² → moment of inertia. Same integral, different sticker.

mass and moments

m = \iint_R \rho\,dA \qquad M_y = \iint_R x\,\rho\,dA \qquad M_x = \iint_R y\,\rho\,dA

centre of mass

\bar x = \frac{M_y}{m} = \frac{1}{m}\iint_R x\,\rho\,dA \qquad\qquad \bar y = \frac{M_x}{m} = \frac{1}{m}\iint_R y\,\rho\,dA

moments of inertia — distance is SQUARED

I_x = \iint_R y^2\rho\,dA \qquad I_y = \iint_R x^2\rho\,dA \qquad I_0 = \iint_R (x^2+y^2)\rho\,dA = I_x + I_y

Naming trap

$M_y$ uses $x$ , and $M_x$ uses $y$ . The subscript names the axis you are taking the moment about, not the variable in the integrand. Same crossover in $I_x$ , which uses $y^2$ — distance from the $x$ -axis is $|y|$ .

Algorithm · any of these quantities

Set up the region once — limits, and polar if it's round.
Pick the weight from the table: $1$ for mass, $x$ or $y$ for a moment, $x^2+y^2$ for polar inertia.
Integrate $\iint(\text{weight})\rho\,dA$ . Every quantity in this section is that one integral with a different sticker.
For a centre of mass, divide by $m$ at the very end. Moments are not divided; centres of mass are.

Symmetry shortcut: if the region and the density are both symmetric about a line, the centre of mass lies on that line. That can kill half the work.

Uniform density? $\rho$ is a constant, so it factors out of the top and bottom of $\bar x$ and cancels completely. The answer is then called the centroid and depends only on the shape.

Check yourself

Write $\bar x$ for a lamina.
Answer
$\bar x=\dfrac1m\displaystyle\iint_R x\,\rho\,dA$ — the moment, divided by the mass.
Why is moment of inertia weighted by distance squared?
Answer
Because it measures resistance to spinning, and mass far from the axis resists far more than mass near it.

16.7

Change of variables & the Jacobian

bonus · not on the plan

Purdue’s 37 lessons skip 16.7 — the plan jumps straight from mass calculations (16.6) to vector fields (17.1). Read this only if you want to see where the $r$ and $\rho^2\sin\varphi$ actually come from.

Bend the coordinate grid into whatever shape makes your region a plain rectangle. The bending stretches areas, so you owe a correction factor — and the Jacobian is exactly that factor.

Polar is the famous special case: for x = r cos θ, y = r sin θ the Jacobian works out to exactly r. Same for spherical: ρ² sin φ. The rule you memorised was always a Jacobian.

the Jacobian determinant

\frac{\partial(x,y)}{\partial(u,v)} = \begin{vmatrix}\dfrac{\partial x}{\partial u} & \dfrac{\partial x}{\partial v}\\[8pt] \dfrac{\partial y}{\partial u} & \dfrac{\partial y}{\partial v}\end{vmatrix} \qquad\qquad dA = \left|\frac{\partial(x,y)}{\partial(u,v)}\right|\,du\,dv

change of variables

\iint_R f(x,y)\,dA = \iint_S f\bigl(x(u,v),\,y(u,v)\bigr)\left|\frac{\partial(x,y)}{\partial(u,v)}\right|du\,dv

Algorithm · change of variables

Choose $u$ and $v$ from the boundary. If the region is bounded by $x+y=1$ , $x+y=3$ , $x-y=0$ , $x-y=2$ , then take $u=x+y$ , $v=x-y$ — the region becomes the rectangle $1\le u\le3$ , $0\le v\le 2$ .
Invert to get $x(u,v)$ and $y(u,v)$ .
Compute the Jacobian and take its absolute value.
Rewrite the integrand in $u$ and $v$ .
Integrate over the new, rectangular region.

Shortcut: if inverting is ugly, compute $\dfrac{\partial(u,v)}{\partial(x,y)}$ instead and use $\dfrac{\partial(x,y)}{\partial(u,v)} = 1\Big/\dfrac{\partial(u,v)}{\partial(x,y)}$ .

Polar and spherical were Jacobians all along

For $x=r\cos\theta,\ y=r\sin\theta$ the Jacobian works out to exactly $r$ . For spherical it works out to exactly $\rho^2\sin\varphi$ . The rules you were told to memorise in 16.3 and 16.5 are just this one theorem, already applied.

Check yourself

What is the Jacobian for polar coordinates?
Answer
Exactly $r$ — the rule you memorised in 16.3 was this theorem all along.

UNIT 04

Vector calculus

Ch. 17 · Quizzes 8–10 → Midterm 2 → Final

One last new object: instead of a number at every point, an arrow at every point. Then four theorems that all say the same sentence — what happens on the edge equals what the derivative does inside. Lessons 25–33 finish Exam 2; Stokes and the divergence theorem (34–37) appear only on the Final.

17.1

Vector fields

Quiz 8

Wind, current, gravity, magnetic pull. Two questions decide everything that follows: does the wind go around, and does the wind go out?

Written down: F(x,y) = ⟨P(x,y), Q(x,y)⟩. The left field is ⟨−y, x⟩. The right field is ⟨x, y⟩.

a vector field

\mathbf F(x,y) = P(x,y)\,\mathbf i + Q(x,y)\,\mathbf j \qquad\quad \mathbf F(x,y,z) = P\,\mathbf i + Q\,\mathbf j + R\,\mathbf k

Gradient fields are the special ones: $\mathbf F = \nabla f$ for some scalar $f$ . Physically, $f$ is a hill (or a potential) and $\mathbf F$ points the way water would run. Section 17.3 is entirely about spotting these.

Check yourself

What does divergence measure?
Answer
Net outflow from a tiny box: faucet if positive, drain if negative.
What does curl measure?
Answer
Local spin — whether a paddlewheel turns. Not whether the flow curves: straight flow that is faster on one side still spins it.

17.2

Line integrals

Quiz 9

Walk a curve and add something up as you go. If the something is a height you get a curtain's area. If it's a push you get work.

Slice the curtain into thin strips: each is height f × width ds. And ds = |r′(t)| dt, exactly the arc-length piece from Chapter 2.

scalar line integral — with respect to arc length

\int_C f\,ds = \int_a^b f\bigl(\mathbf r(t)\bigr)\,\bigl|\mathbf r'(t)\bigr|\,dt

Now the vector version. Wind straight at your back: full help. Wind from the side: does nothing. Wind in your face: fights you. Only the part of the push along your walk counts.

Dot product again, at every point along the curve. F·dr keeps only the along-the-path share of the push.

work — three notations for one integral

\int_C \mathbf F\cdot d\mathbf r = \int_a^b \mathbf F\bigl(\mathbf r(t)\bigr)\cdot\mathbf r'(t)\,dt = \int_C P\,dx + Q\,dy + R\,dz

Algorithm · evaluate any line integral

Parametrize $C$ . Line segment from $A$ to $B$ : $\mathbf r(t) = (1-t)A + tB$ , $0\le t\le1$ . Circle of radius $a$ : $\langle a\cos t, a\sin t\rangle$ . Graph $y=g(x)$ : $\langle t, g(t)\rangle$ .
Compute $\mathbf r'(t)$ . For a scalar integral you also need $|\mathbf r'(t)|$ .
Substitute the parametrization into $\mathbf F$ or $f$ so everything is in terms of $t$ alone.
Dot (vector version) or multiply by $|\mathbf r'|$ (scalar version).
Integrate over the parameter range. If $C$ is made of several pieces, do each separately and add.

Orientation trap

$\displaystyle\int_{-C}\mathbf F\cdot d\mathbf r = -\int_C \mathbf F\cdot d\mathbf r$ — reversing the walk flips the sign of work. But $\displaystyle\int_{-C} f\,ds = \int_C f\,ds$ — a curtain has the same area no matter which way you walk along its base. Vector integrals care about direction; scalar ones don't.

Check yourself

What does $\int_C f\,ds$ give when $f=1$ ?
Answer
The length of the curve.
You walk $C$ backwards. What happens to $\int_C\mathbf F\cdot d\mathbf r$ ?
Answer
The sign flips. But $\int_C f\,ds$ does not change — a curtain has the same area either way.

17.3

Conservative fields & the FTLI

Quiz 9

Climb from the car park to the summit. Take the switchbacks or take the ridge — you gained exactly the same height. Some fields are just "downhill from a hill", and for those the route is irrelevant.

Equivalent statements, all meaning the same thing: F = ∇f · the work is path-independent · every closed loop gives zero · curl F = 0 (on a region with no holes).

fundamental theorem for line integrals

\int_C \nabla f\cdot d\mathbf r = f\bigl(\mathbf r(b)\bigr) - f\bigl(\mathbf r(a)\bigr) = f(\text{end}) - f(\text{start})

This is the FTC from Calc I wearing a coat: integrate a derivative, get boundary values. The whole path collapses to two evaluations.

Four ways of saying the same thing

$\mathbf F$ is conservative ⟺ $\mathbf F = \nabla f$ for some potential $f$ ⟺ $\int_C\mathbf F\cdot d\mathbf r$ is path-independent ⟺ $\oint_C\mathbf F\cdot d\mathbf r = 0$ around every closed loop. On a simply connected region these are also equivalent to $\operatorname{curl}\mathbf F = \mathbf 0$ .

the test

\text{2-D: } \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} \qquad\qquad \text{3-D: } \operatorname{curl}\mathbf F = \nabla\times\mathbf F = \mathbf 0

Algorithm · find the potential function $f$

Test first. Check $P_y = Q_x$ (2-D) or $\operatorname{curl}\mathbf F=\mathbf 0$ (3-D). If it fails, stop — there is no $f$ .
Integrate $P$ with respect to $x$ : $f = \displaystyle\int P\,dx + g(y,z)$ . The constant of integration is a function of the other variables, not a number.
Differentiate your $f$ with respect to $y$ and set it equal to $Q$ . Solve for $g_y$ .
Integrate $g_y$ in $y$ , picking up a new constant $h(z)$ .
Repeat with $z$ and $R$ to pin down $h(z)$ . Add a genuine constant $C$ at the end.
Verify: $\nabla f$ should give back $\mathbf F$ exactly.

Then use it: $\int_C\mathbf F\cdot d\mathbf r = f(\text{end}) - f(\text{start})$ , and the parametrization never gets written down.

Worked example

$\mathbf F = \langle 2xy,\ x^2+2y\rangle$ , from $(0,0)$ to $(1,3)$ .

test

$P_y = 2x$ , $Q_x = 2x$ ✓ conservative

step 2

$f = \int 2xy\,dx = x^2y + g(y)$

step 3

$f_y = x^2 + g'(y) = x^2+2y \Rightarrow g'(y)=2y \Rightarrow g = y^2$

potential

$f = x^2y + y^2$

evaluate

$f(1,3)-f(0,0) = (3+9) - 0 = \mathbf{12}$ — no parametrization needed.

Trap

Curl zero is not enough on a region with a hole. The classic $\mathbf F = \left\langle \frac{-y}{x^2+y^2},\ \frac{x}{x^2+y^2}\right\rangle$ passes $P_y=Q_x$ everywhere it is defined, yet a loop around the origin gives $2\pi$ , not 0. The region must be simply connected — no punctures.

Check yourself

What is the 2-D test for a conservative field?
Answer
$\dfrac{\partial P}{\partial y}=\dfrac{\partial Q}{\partial x}$ — and the region must have no holes.
$\mathbf F$ is conservative. What is $\oint_C\mathbf F\cdot d\mathbf r$ around any closed loop?
Answer
$0$ . You end where you started, so the height change is zero.

17.4

Green's theorem

Midterm 2

Chop the region into cells and spin each one counterclockwise. Along any wall between two cells, one cell pushes up and its neighbour pushes down — dead even. The only walls with nobody on the other side are on the outside.

Each cell's spin is (Q_ₓ − P_ᵧ) dA — the 2-D curl. Add them all up and you're left holding a single trip around the boundary.

circulation form

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

flux (divergence) form — same theorem, rotated 90°

\oint_C \mathbf F\cdot\mathbf n\,ds = \iint_D \left(\frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y}\right)dA = \iint_D \operatorname{div}\mathbf F\ dA

Algorithm · apply Green's theorem

Check the hypotheses. $C$ must be closed, simple (no self-crossings), positively oriented — counterclockwise, region on your left — and $\mathbf F$ smooth on all of $D$ .
Identify $P$ and $Q$ from $\oint P\,dx+Q\,dy$ .
Compute $Q_x - P_y$ and simplify. This is usually far smaller than either piece.
Set up $\iint_D$ over the enclosed region — polar if it's round.
If $C$ runs clockwise, apply the theorem anyway and negate the result.

Also runs backwards: a horrible double integral can become an easy loop integral. Read the problem for which direction is cheaper.

area for free — choose P and Q so that

Q_x-P_y=1

A = \oint_C x\,dy = -\oint_C y\,dx = \tfrac12\oint_C x\,dy - y\,dx

Algorithm · regions with a hole

Orient both boundaries positively: outer curve counterclockwise, inner curve clockwise — so the region always stays on your left.
Apply Green's theorem to the whole boundary $C = C_{\text{outer}} + C_{\text{inner}}$ .
If $Q_x-P_y = 0$ on the region between them, the two loop integrals are equal in magnitude — so you can replace a nasty outer curve with a tiny convenient circle around the hole.

That last trick is how the $2\pi$ in the 17.3 trap gets computed.

Check yourself

Which way must $C$ be oriented for Green's theorem?
Answer
Counterclockwise, region on your left. If it runs clockwise, apply the theorem anyway and negate.
Give a line integral that returns the enclosed area.
Answer
$A=\oint_C x\,dy$ (or $-\oint_C y\,dx$ ). Chosen so that $Q_x-P_y=1$ .

17.5

Curl & divergence

Midterm 2

Curl: drop a paddlewheel in and see if it spins. Not "is the water going in a circle" — "is one side pushed harder than the other".

Curl is a vector in 3D: it points along the axis the paddlewheel spins about, by the right-hand rule, and its length is how fast. In 2D only the z-component survives: Q_ₓ − P_ᵧ.

curl — a vector, computed as a determinant

\operatorname{curl}\mathbf F = \nabla\times\mathbf F = \begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\[2pt] \dfrac{\partial}{\partial x}&\dfrac{\partial}{\partial y}&\dfrac{\partial}{\partial z}\\[6pt] P&Q&R\end{vmatrix}

The direction of curl is the axis the paddlewheel spins about (right-hand rule); its length is how fast. In 2-D only the $\mathbf k$ -component survives, and it is exactly the $Q_x - P_y$ from Green's theorem.

Divergence: draw a tiny box. Count what flows in, count what flows out. More out than in means something inside is making it.

Divergence is a number, not a vector — there's no direction to "how much is being created here."

divergence — a scalar

\operatorname{div}\mathbf F = \nabla\cdot\mathbf F = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z}

Two identities worth memorising outright

$\operatorname{div}(\operatorname{curl}\mathbf F) = 0$ — a spin never creates anything. $\operatorname{curl}(\nabla f) = \mathbf 0$ — a hill never makes a whirlpool. Together they explain why a conservative field always has zero curl, and why a curl field always has zero flux out of a closed surface.

Type trap

curl of a vector field is a vector. Divergence of a vector field is a number. So $\operatorname{div}(\operatorname{curl}\mathbf F)$ makes sense and $\operatorname{curl}(\operatorname{div}\mathbf F)$ does not — you can't cross a scalar. Checking the type of your answer catches half of all algebra errors here.

Check yourself

What is $\operatorname{div}(\operatorname{curl}\mathbf F)$ ?
Answer
$0$ , always. A spin never creates anything.
What is $\operatorname{curl}(\nabla f)$ ?
Answer
$\mathbf 0$ , always. A hill never makes a whirlpool.
Is $\operatorname{div}\mathbf F$ a scalar or a vector?
Answer
A scalar. So $\operatorname{curl}(\operatorname{div}\mathbf F)$ is not even defined — checking types catches half of all errors here.

17.6a

Parametric surfaces

Quiz 10

A surface is a rubber sheet with its own two coordinates. Move $u$ a little and you slide one way; move $v$ and you slide the other. Those two slides span a tiny tile — and 13.4 already told you its area.

The cross product does double duty: its direction is the surface normal, its length is the tile's area. Both of the things a surface integral needs.

tangent vectors, normal, and area element

\mathbf r_u = \frac{\partial\mathbf r}{\partial u},\quad \mathbf r_v = \frac{\partial\mathbf r}{\partial v},\qquad \mathbf n\,dS = (\mathbf r_u\times\mathbf r_v)\,du\,dv, \qquad dS = |\mathbf r_u\times\mathbf r_v|\,du\,dv

The cross product does double duty: its direction is the surface normal, its length is the tile's area. Both things a surface integral needs, from one computation.

Surface	Parametrization	$\|\mathbf r_u\times\mathbf r_v\|$
Graph $z=f(x,y)$	$\langle x,\ y,\ f(x,y)\rangle$	$\sqrt{1+f_x^2+f_y^2}$
Sphere radius $a$	$\langle a\sin\varphi\cos\theta,\ a\sin\varphi\sin\theta,\ a\cos\varphi\rangle$	$a^2\sin\varphi$
Cylinder radius $a$	$\langle a\cos\theta,\ a\sin\theta,\ z\rangle$	$a$
Cone $z=\sqrt{x^2+y^2}$	$\langle r\cos\theta,\ r\sin\theta,\ r\rangle$	$r\sqrt2$
Surface of revolution	$\langle x,\ f(x)\cos\theta,\ f(x)\sin\theta\rangle$	$f\sqrt{1+(f')^2}$

Row 1 and row 2 are worth recognising instantly — they reproduce the surface-area formula from 17.6 and the $\rho^2\sin\varphi$ from 16.5.

Check yourself

What does $(\mathbf r_u\times\mathbf r_v)\,du\,dv$ equal?
Answer
$\mathbf n\,dS$ — both the direction and the area, from one computation. Never normalise it and then also multiply by $dS$ .

17.6b

Surface area

Quiz 10

Briggs files surface area under Surface Integrals (17.6) — which is why it sits here rather than back in Chapter 16.

A patch of hillside is always bigger than the flat patch beneath it. How much bigger depends on the tilt — and the tilt is exactly what the partial derivatives measure.

Same idea as arc length's √(1 + (dy/dx)²), one dimension up. Steeper means more surface hiding above the same footprint.

area of the graph of

z=f(x,y)

over the region

R

A(S) = \iint_R \sqrt{1 + \bigl(f_x\bigr)^2 + \bigl(f_y\bigr)^2}\ dA

Same shape as arc length, which was $\int\sqrt{1+(dy/dx)^2}\,dx$ — one dimension up. Flat ground gives $f_x=f_y=0$ , the root becomes 1, and surface area collapses to plain area. That's your check.

Algorithm · surface area of a graph

Solve for $z$ so you have $z=f(x,y)$ .
Compute $f_x$ and $f_y$ , then $1+f_x^2+f_y^2$ and simplify before integrating — this is usually where a clean root appears.
Find $R$ : the shadow of the surface on the $xy$ -plane.
Integrate over $R$ , going polar if the shadow is round (it usually is).

For a surface given parametrically instead, jump to 17.6: $A = \iint |\mathbf r_u\times\mathbf r_v|\,dA$ .

Check yourself

What is the stretch factor for the graph $z=f(x,y)$ ?
Answer
$\sqrt{1+f_x^2+f_y^2}$ .
On flat ground, what does that factor become?
Answer
$1$ , so surface area collapses to plain area. That is the check on every surface-area setup.

17.6c

Surface integrals & flux

Quiz 10

Hold a hoop in the wind. Facing it head-on you catch all of it. Turn it edge-on and you catch nothing — even though the wind hasn't changed. Only the part perpendicular to the surface gets through.

drag to spin

tilt the net

Tilt to 90° and the wind hasn't changed at all — but nothing gets through. Only the component along n crosses the surface.

scalar surface integral — mass of a sheet

\iint_S f\,dS = \iint_D f\bigl(\mathbf r(u,v)\bigr)\,\bigl|\mathbf r_u\times\mathbf r_v\bigr|\,du\,dv

flux of a vector field

\iint_S \mathbf F\cdot d\mathbf S = \iint_S \mathbf F\cdot\mathbf n\,dS = \iint_D \mathbf F\cdot(\mathbf r_u\times\mathbf r_v)\,du\,dv

Algorithm · compute a flux integral

Parametrize $S$ — use the table in 17.6 whenever it fits.
Compute $\mathbf r_u\times\mathbf r_v$ . Do not normalise it and do not compute $dS$ separately; the un-normalised cross product already carries both.
Check orientation. Does the cross product point the way the problem wants (outward, upward)? If not, negate it.
Substitute $\mathbf F$ at the parametrized points, so everything is in $u$ and $v$ .
Dot and integrate over the parameter region $D$ .

For a graph $z=g(x,y)$ oriented upward, this collapses to $\iint_D\bigl(-P g_x - Q g_y + R\bigr)dA$ .

Orientation trap

Swapping $u$ and $v$ flips the normal, and flipping the normal negates your flux. Always sanity-check by evaluating the normal at one concrete point: for a sphere with outward orientation it should point away from the origin.

Check yourself

You turn the net edge-on to the wind. How much gets through?
Answer
Nothing. $\mathbf F\perp\mathbf n$ makes $\mathbf F\cdot\mathbf n=0$ , even though the wind never changed.
Should you normalise $\mathbf r_u\times\mathbf r_v$ before dotting with $\mathbf F$ ?
Answer
No. The raw cross product already carries $dS$ inside it.

17.7

Stokes' theorem

Final Exam

Dip a bent wire loop in soap. The film can be flat or bulge out like a bubble. Both have the same rim, so both have the same total spin through them. Pick the lazy one.

drag to spin

morph the film

Morph the soap film however you like. The rim never moves, so the answer never changes — Green's theorem is this picture lying flat.

circulation around the rim = curl through any surface it bounds

\oint_C \mathbf F\cdot d\mathbf r = \iint_S (\operatorname{curl}\mathbf F)\cdot d\mathbf S

Algorithm · use Stokes' theorem

Decide which side is easier. Nasty curve, simple curl → go right. Nasty surface, simple curve → go left.
Choose the friendliest surface with that rim. A flat disk or the plane region inside $C$ almost always beats the surface you were handed.
Compute $\operatorname{curl}\mathbf F$ as a determinant.
Match orientations by the right-hand rule: curl the fingers of your right hand along $C$ , and your thumb must point the way $\mathbf n$ points. Mismatch = wrong sign.
Evaluate the surface integral using 17.6.

Green's theorem is exactly this with the surface lying flat in the $xy$ -plane and $\mathbf n = \mathbf k$ .

Check yourself

You are told $S$ is 'any smooth surface with boundary $C$ '. Does the choice matter?
Answer
No. Only the rim matters — pick the laziest surface, usually the flat disk.
When is Green's theorem just Stokes' theorem?
Answer
When the surface lies flat in the $xy$ -plane with $\mathbf n=\mathbf k$ . Same theorem, tilted.

17.8

The divergence theorem

Final Exam

Fill a solid with tiny cubes. Whatever leaks out the right face of one cube leaks into the left face of the next. Every internal face cancels with its partner. The only faces with nothing behind them are on the outer skin.

Each cube's surplus is div F · dV. Add them all up and you're left with the flux through the outside surface. Exactly Green's proof, in 3D.

flux out of a closed surface = divergence inside

\oiint_S \mathbf F\cdot d\mathbf S = \iiint_E \operatorname{div}\mathbf F\ dV

Algorithm · use the divergence theorem

Confirm $S$ is closed — a sealed bag with no openings — and oriented outward.
Compute $\operatorname{div}\mathbf F = P_x+Q_y+R_z$ . This is usually startlingly simple compared to $\mathbf F$ itself.
Set up $\iiint_E$ over the enclosed solid, in whichever coordinates match it (16.4–16.5).
Integrate. Often $\operatorname{div}\mathbf F$ is a constant, and the answer is just that constant times the volume.

Algorithm · when the surface is NOT closed

Cap it. Add a simple surface $S_2$ (usually a flat disk) to seal the opening.
Apply the theorem to the sealed union: $\displaystyle\oiint_{S\cup S_2} = \iiint_E \operatorname{div}\mathbf F\,dV$ .
Compute the cap's flux separately — it's flat, so it's easy.
Subtract: $\displaystyle\iint_S = \iiint_E \operatorname{div}\mathbf F\,dV - \iint_{S_2}$ .
Watch the cap's orientation: as part of the sealed bag it must point outward, which for a bottom cap means downward.

This capping trick is the single most-tested idea on the final.

Check yourself

What two conditions must $S$ satisfy?
Answer
It must be closed (a sealed bag) and oriented outward.
The surface is open. Now what?
Answer
Cap it, apply the theorem to the sealed union, compute the cap's flux separately, then subtract it. The cap must point outward too.

17 ·

The punchline: all five are one theorem

What happens on the edge equals what the derivative does inside. The only thing that changes between them is how many dimensions you're in.

Every row: integrate a derivative over the inside = integrate the original over the edge. Learn the sentence and the four formulas stop being four things.

Theorem	Statement	The "inside"	Its edge
FTC	$\int_a^b f'\,dx = f(b)-f(a)$	an interval	2 endpoints
FTLI	$\int_C \nabla f\cdot d\mathbf r = f(B)-f(A)$	a curve	2 endpoints
Green	$\oint_C \mathbf F\cdot d\mathbf r = \iint_D (Q_x-P_y)\,dA$	a flat region	one loop
Stokes	$\oint_C \mathbf F\cdot d\mathbf r = \iint_S \operatorname{curl}\mathbf F\cdot d\mathbf S$	a surface	its rim
Divergence	$\oiint_S \mathbf F\cdot d\mathbf S = \iiint_E \operatorname{div}\mathbf F\,dV$	a solid	its skin

If you remember one thing

Derivatives measure local change. Integrals add local change up. When you add up local change over a whole region, everything in the middle cancels with its neighbour and you're left standing on the boundary. That is the entire second half of Calc III.

Check yourself

Say all four theorems in one sentence.
Answer
What happens on the edge equals what the derivative does inside. Everything in the middle cancels with its neighbour.
You have a horrible closed-surface flux integral. First thought?
Answer
Divergence theorem — compute $\operatorname{div}\mathbf F$ first. It is usually startlingly simpler than $\mathbf F$ .

REFERENCE

One page

everything above, compressed

If you can rebuild the pictures from these lines, you're ready.

R.1

Fall 2026 map

Purdue splits MA 26100 into 37 lessons. Ten recitation quizzes, two evening midterms and a final are laid on top of them — and the boundaries do not fall on chapter breaks. Everything below is transcribed from the department's own Fall 2026 calendar, read 2026-09-01.

Stop	When	Lessons	Sections
Quiz 1	Tue Sep 1 · recitation	1, 2	13.1–13.4 · 12.1 · 13.5 lines	study it
Quiz 2	Tue Sep 8 · recitation	3, 4	13.5 planes · 13.6	study it
Quiz 3	Tue Sep 15 · recitation	5, 6, 7	14.1 · 14.2 · 14.3	study it
Quiz 4	Tue Sep 22 · recitation	8, 9, 10	14.4 · 14.5 · 15.1 · 15.2	study it
Quiz 5	Tue Sep 29 · recitation	11, 12, 13	15.3 · 15.4 · 15.5	study it
Midterm 1	Mon Oct 5 · 8:00 pm	1–16	13.1–15.7 (all of Ch. 13–14, Ch. 15 through max/min)	study it
Quiz 6	Tue Oct 20 · recitation	17, 18, 19	15.8 · 16.1 · 16.2	study it
Quiz 7	Tue Oct 27 · recitation	20, 21, 22	16.3 · 16.4 · 16.5 cylindrical	study it
Quiz 8	Tue Nov 3 · recitation	23, 24, 25	16.5 spherical · 16.6 · 17.1	study it
Quiz 9	Tue Nov 10 · recitation	26, 27, 28	17.2 · 17.3	study it
Midterm 2	Wed Nov 18 · 8:00 pm	17–31	15.8–17.6 (Lagrange through surface integrals)	study it
Quiz 10	Tue Dec 1 · recitation	31, 32, 33	17.6 · all three lessons	study it
Final Exam	Finals week · date TBD	1–37	12.1–17.8 · everything	study it

Three windows have no quiz. Lessons 14–16 (tangent planes, max/min) land between Quiz 5 and Midterm 1. Lessons 29–30 (Green's theorem, curl and divergence) land between Quiz 9 and Midterm 2. Lessons 34–37 (Stokes, divergence theorem) land after the last quiz and are on the final only. Nothing forces you to learn those on schedule, which is exactly why they are where grades go.

Weeks with no quiz

Tue Aug 25 — Quiz 0 — Calc I & II review
Tue Oct 6 — no quiz — Midterm 1 was the night before
Tue Oct 13 — no recitation — Fall Break
Tue Nov 17 — no quiz — Midterm 2 review
Tue Nov 24 — no recitation — Thanksgiving
Tue Dec 8 — no quiz — final review (Lessons 34–37)

The two lowest quiz scores are dropped, and there are no make-ups — the drops are the make-up policy.

16.7 (change of variables / the Jacobian) is not on the lesson plan at all — the calendar goes 16.6 → 17.1. It is kept here as background, badged as a bonus, and no quiz or exam requires it.

Component	Weight	Detail
Quizzes	15%	Every recitation, timed, no calculator. The 2 lowest are dropped.
Homework	15%	MyLab Math, generally due Tue and Thu 11:59 pm. The 3 lowest are dropped.
Midterm 1	20%	Mon Oct 5, 8:00 pm · Lessons 1–16
Midterm 2	20%	Wed Nov 18, 8:00 pm · Lessons 17–31
Final Exam	30%	Finals week · everything

Dates that are not about material

2026-09-04 — Last day to drop with no record on your transcript
2026-11-24 — Last day to drop with a W

Calculators are allowed on nothing — not quizzes, not midterms, not the final.

R.2

Which theorem?

You're never asked to evaluate one specific integral — you're asked to get an answer. Each theorem is a door from the integral you were handed to one you'd rather do.

Read it as: "I'm here, I'd rather be there, which door?" Almost every exam problem is chosen so that one direction is brutal and the other is three lines.

You're handed	Look for	Do this
$\int_C \mathbf F\cdot d\mathbf r$ , open curve	$P_y = Q_x$ ?	Find $f$ , answer is $f(B)-f(A)$ — 17.3
$\oint_C \mathbf F\cdot d\mathbf r$ , flat closed loop	region is easy	Green: $\iint (Q_x-P_y)\,dA$ — 17.4
$\oint_C \mathbf F\cdot d\mathbf r$ , closed loop in 3-D	curl is simple	Stokes over any capping surface — 17.7
$\iint_S \mathbf F\cdot d\mathbf S$ , closed surface	div is simple	Divergence theorem — 17.8
$\iint_S \mathbf F\cdot d\mathbf S$ , open surface	can you cap it?	Cap, apply divergence, subtract the cap — 17.8
$\iint_S(\operatorname{curl}\mathbf F)\cdot d\mathbf S$	the rim	Stokes backwards: just walk the rim — 17.7

R.3

Coordinates

System	Element	Reach for it when
Rectangular	$dA = dx\,dy$ · $dV = dx\,dy\,dz$	Boxes, planes, straight edges.
Polar	$dA = r\,dr\,d\theta$	$x^2+y^2$ , a circle, disk, ring, wedge.
Cylindrical	$dV = r\,dz\,dr\,d\theta$	Cans, pipes, cones with a flat top.
Spherical	$dV = \rho^2\sin\varphi\,d\rho\,d\varphi\,d\theta$	$x^2+y^2+z^2$ , balls, cones from the origin.
Your own	$dA = \left\|\dfrac{\partial(x,y)}{\partial(u,v)}\right\|du\,dv$	A slanted region you can flatten to a rectangle.

R.4

Formula sheet

Unit 1 · Vectors

$\mathbf a\cdot\mathbf b = |\mathbf a||\mathbf b|\cos\theta$ the shadow
$|\mathbf a\times\mathbf b| = |\mathbf a||\mathbf b|\sin\theta$ the tile's area
$\operatorname{comp}_{\mathbf b}\mathbf a = \dfrac{\mathbf a\cdot\mathbf b}{|\mathbf b|}$ a number
$\operatorname{proj}_{\mathbf b}\mathbf a = \dfrac{\mathbf a\cdot\mathbf b}{|\mathbf b|^2}\mathbf b$ an arrow along b
$V = |\mathbf a\cdot(\mathbf b\times\mathbf c)|$ volume of the box; 0 ⟹ coplanar
$\mathbf n\cdot(\mathbf r-\mathbf r_0)=0$ a plane
$D = \dfrac{|ax_1+by_1+cz_1+d|}{\sqrt{a^2+b^2+c^2}}$ point to plane

Unit 1 · Curves

$L = \int_a^b|\mathbf r'(t)|\,dt$ speed × time
$\mathbf T = \mathbf r'/|\mathbf r'|$ , $\mathbf N = \mathbf T'/|\mathbf T'|$ , $\mathbf B = \mathbf T\times\mathbf N$ the moving frame
$\kappa = \dfrac{|\mathbf r'\times\mathbf r''|}{|\mathbf r'|^3}$ 1 / the hugging circle
$a_T = \dfrac{\mathbf r'\cdot\mathbf r''}{|\mathbf r'|}$ , $a_N = \dfrac{|\mathbf r'\times\mathbf r''|}{|\mathbf r'|}$ gas pedal, steering wheel

Unit 2 · Partials

$\nabla f = \langle f_x, f_y, f_z\rangle$ compass pointing uphill
$D_{\mathbf u}f = \nabla f\cdot\mathbf u$ the gradient's shadow; u must be unit
$z = f(a,b)+f_x\Delta x+f_y\Delta y$ tangent plane
$\dfrac{dz}{dt} = z_x x'+z_y y'$ add the routes down the tree
$\dfrac{\partial z}{\partial x} = -\dfrac{F_x}{F_z}$ implicit
$D = f_{xx}f_{yy}-f_{xy}^2$ $D \lt 0$ ⟹ saddle
$\nabla f = \lambda\nabla g$ contour kisses the fence

Unit 3 · Integrals

$\iint_R f\,dA$ columns on tiles
$\iint f\,r\,dr\,d\theta$ never drop the r
$A=\iint\sqrt{1+f_x^2+f_y^2}\,dA$ surface area
$\bar x = \frac1m\iint x\rho\,dA$ balance point
$I_0 = \iint (x^2+y^2)\rho\,dA$ inertia — distance squared
$dV = \rho^2\sin\varphi\,d\rho\,d\varphi\,d\theta$ three edges, multiplied

Unit 4 · Fields

$\int_C f\,ds = \int f(\mathbf r(t))|\mathbf r'(t)|\,dt$ a curtain's area
$\int_C \mathbf F\cdot d\mathbf r = \int \mathbf F(\mathbf r(t))\cdot\mathbf r'(t)\,dt$ work
$\operatorname{curl}\mathbf F = \nabla\times\mathbf F$ does a paddlewheel spin?
$\operatorname{div}\mathbf F = \nabla\cdot\mathbf F$ faucet or drain?
$\mathbf n\,dS = (\mathbf r_u\times\mathbf r_v)\,du\,dv$ normal and area at once

Unit 4 · The theorems

$\int_C\nabla f\cdot d\mathbf r = f(B)-f(A)$ endpoints only
$\oint_C P\,dx+Q\,dy = \iint_D (Q_x-P_y)\,dA$ Green
$\oint_C\mathbf F\cdot d\mathbf r = \iint_S \operatorname{curl}\mathbf F\cdot d\mathbf S$ Stokes
$\oiint_S\mathbf F\cdot d\mathbf S = \iiint_E \operatorname{div}\mathbf F\,dV$ divergence

R.5

Traps that cost points

Where	The mistake	The fix
13.1	$\overrightarrow{PQ}$ written tail minus head	Head minus tail. Always $Q-P$ .
13.3b	Dividing by $\|\mathbf b\|$ once for the vector projection	Once = number. Twice = vector.
13.4	Forgetting the minus on the $\mathbf j$ component	Dot your answer with $\mathbf a$ ; it must be 0.
13.5	Same parameter $t$ for both lines	Use $t$ and $s$ , or you'll miss intersections.
15.5	Direction vector not normalised in $D_{\mathbf u}f$	Divide by its length first, every time.
15.7	Skipping the boundary on a closed region	The max usually lives on the edge.
15.8	Dividing by a variable that could be 0	Split into branches before dividing.
16.2	A variable left in the outer limits	Outer limits are numbers. Always.
16.3	Dropping the $r$ in $r\,dr\,d\theta$	Write the $r$ the moment you write $dr\,d\theta$ .
16.5	Letting $\varphi$ run to $2\pi$	$\varphi\in[0,\pi]$ . Only $\theta$ goes all the way round.
17.2	Ignoring orientation on $\int\mathbf F\cdot d\mathbf r$	Reversing the curve negates it (but not $\int f\,ds$ ).
17.4	Using Green on a clockwise curve	Apply it anyway, then negate.
17.6	Normalising $\mathbf r_u\times\mathbf r_v$ and also using $dS$	The raw cross product already is $\mathbf n\,dS$ .
17.8	Divergence theorem on an open surface	Cap it, apply, then subtract the cap.

R.6

Self-check

Don't tick a box because you recognise the words. Tick it when you can sketch the picture from memory and say the algorithm out loud.

Ready

0 / 46

Every formula in this course is a sentence about a picture. When a problem looks unfamiliar, don't reach for the formula — reach for the picture, and ask what it's doing. Where does the arrow point? What cancels? What's left on the edge?

Vectors & curves

Vectors

Algorithm · build a vector of prescribed length in a given direction

Check yourself

Space, distance, spheres

Algorithm · recognise a sphere from a messy equation

Worked example

Check yourself

Dot product

Algorithm · angle between two vectors

Check yourself

Scalar vs vector projection

Algorithm · project a onto b

Worked example

Trap

Check yourself

Cross product

Algorithm · the determinant, without mistakes

Area of a parallelogram

Area of a triangle

Are they parallel?

Check yourself

Parametric equations

Algorithm · eliminate the parameter

Worked example

Trap

Check yourself

Lines & planes

Algorithm · plane through three points PPP, QQQ, RRR

Algorithm · classify two lines in space

Angle between two planes

Line where two planes meet

Parallel planes

Check yourself

Every distance formula, and why

Algorithm · distance between two skew lines

Two parallel planes

Line parallel to a plane

Sanity check

Check yourself

Cylinders & quadric surfaces

Algorithm · name a cylinder

Trap

Algorithm · identify a quadric by its traces

Algorithm · complete the square into standard form

Trap

Check yourself

Vector functions & space curves

Algorithm · parametrize the curve where two surfaces meet

Worked example

Check yourself

Derivatives & integrals of vector functions

Trap

Check yourself

Motion in space · T, N, B

Algorithm · decompose acceleration

Check yourself

Arc length & curvature

Algorithm · reparametrize by arc length

Use the third one

Use the second one

For a plane curve y=f(x)y=f(x)y=f(x)

Check yourself

Partial derivatives

Surfaces & level curves

Algorithm · find and sketch the domain

Check yourself

Limits & continuity

Algorithm · show a limit does NOT exist

Algorithm · show a limit DOES exist

Worked example — polar closing the case

Check yourself

Partial derivatives

Algorithm · compute a partial derivative

Notation trap

Check yourself

Chain rule

Algorithm · any chain rule, from the tree

Algorithm · implicit partials without the formula

Check yourself

Algorithm · plane through three points $P$ , $Q$ , $R$

For a plane curve $y=f(x)$

$f=1$

$f=\rho$

Inertia about the $z$ -axis

Algorithm · find the potential function $f$