Pick the one you are sitting. You get the sections it covers and
nothing else — the full guide material, figures and all — then questions
on exactly those sections. Coverage follows the department's own Fall 2026
lesson calendar.
Vectors themselves: components, length, dot, cross — plus parametric curves and the equations of a line in space.
STEP 01
The material
7 sections
§13.1 · §13.2 · §13.3 · §13.3b · §13.4 · §12.1 · §13.5 — and nothing else. Read them, then do the check-yourself questions inside each one with the answer folded away.
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 constantlya=⟨a1,a2,a3⟩∣a∣=a12+a22+a32a+b=⟨a1+b1,a2+b2,a3+b3⟩ca=⟨ca1,ca2,ca3⟩
unit vector — same direction, length exactly 1u=∣a∣a
Vector from P to Q is PQ=⟨q1−p1,q2−p2,q3−p3⟩ — 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 a pointing that way.
Normalise: divide by its own length, u=a/∣a∣. Now it has length 1.
Scale: multiply by the length you want. v=Lu.
Normalise first, scale second. Doing it in the other order gives the wrong length.
Check yourself
What is ∣⟨3,−4⟩∣?
Answer
9+16=5.
Write the unit vector in the direction of ⟨0,3,4⟩.
Answer
Length is 5, so ⟨0,53,54⟩. Normalise first, always.
What is the vector from P(1,2) to Q(4,6)?
Answer
PQ=⟨3,4⟩ — head minus tail, Q−P. Backwards is the classic slip.
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
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∣P1P2∣=(x2−x1)2+(y2−y1)2+(z2−z1)2
sphere · centre (h,k,l), radius r(x−h)2+(y−k)2+(z−l)2=r2
Why Pythagoras twice: walk along x, then along y — that diagonal on the floor is Δx2+Δy2. Now go up by Δz. 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: x2+ax→(x+2a)2−4a2.
Balance. Every −4a2 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 right side. If the right side is negative there is no sphere; if it is zero you get a single point.
Worked example
x2+y2+z2−4x+6y−2=0.
group
(x2−4x)+(y2+6y)+z2=2
complete
(x−2)2−4+(y+3)2−9+z2=2
balance
(x−2)2+(y+3)2+z2=15
Centre (2,−3,0), radius 15.
One equation in 3-D is a surface, not a curve.x=3 is a whole plane. x2+y2=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
0+16+9=5. Pythagoras, twice.
In 3-D, what does x2+y2=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+y2+(z+2)2=9?
Answer
Centre (1,0,−2), radius 3. The sign inside the bracket flips: (z+2)2 means z=−2.
Shine a light straight down on b. The shadow it casts along 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 numbera⋅b=a1b1+a2b2+a3b3anda⋅b=∣a∣∣b∣cosθ
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 mattercosθ=∣a∣∣b∣a⋅ba⊥b⟺a⋅b=0a⋅a=∣a∣2
Algorithm · angle between two vectors
Dot them componentwise: a⋅b.
Find both lengths∣a∣ and ∣b∣.
Divide:cosθ=∣a∣∣b∣a⋅b.
Arccos.θ=cos−1(⋅), always in [0,π].
Sign check before you compute: positive dot ⇒ acute, zero ⇒ right angle, negative ⇒ obtuse.
direction cosines — the angles a vector makes with each axiscosα=∣a∣a1,cosβ=∣a∣a2,cosγ=∣a∣a3,cos2α+cos2β+cos2γ=1
Those three cosines are the components of the unit vectora/∣a∣ — which is exactly why they square to 1.
work — the physics reason dot products existW=F⋅d=∣F∣∣d∣cosθ
Check yourself
Compute ⟨1,2,3⟩⋅⟨4,−1,0⟩.
Answer
4−2+0=2.
If a⋅b<0, is the angle acute or obtuse?
Answer
Obtuse. Negative dot means the shadow points backwards.
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 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"compba=∣b∣a⋅ba number
vector projection of a onto bprojba=(∣b∣2a⋅b)b=(compba)∣b∣ba vector
How to never mix them up. The vector projection is literally the scalar projection times the unit vector of b. So: get the number first, then attach a direction to it. One ∣b∣ in the denominator = number. Two = vector, because the second one is busy normalising b.
Algorithm · project a onto b
Dot. Compute a⋅b.
Decide which you want. Does the answer need a direction? If not, stop after step 3.
Scalar: divide by ∣b∣ once. Done — that is compba, and it may be negative.
Vector: multiply the scalar answer by the unit vector b/∣b∣. Equivalently, divide by ∣b∣2 and multiply by b — never by ∣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
a=⟨3,4,0⟩, b=⟨4,0,3⟩. Both have length 5.
dot
a⋅b=12+0+0=12
scalar
compba=512=2.4
vector
projba=2512⟨4,0,3⟩=⟨2548,0,2536⟩
check
its length is (48/25)2+(36/25)2=2560=2.4 ✓ matches the scalar answer
orthogonal decomposition — split a into "along b" plus "perpendicular to b"a=parallel to bprojba+perpendicular to b(a−projba)
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 aT and aN (14.3), and the flux/circulation split of a vector field (Unit 4).
Trap
projba=projab. Read the subscript as "onto". Swapping them gives a vector pointing in a completely different direction. And the scalar projection keeps its sign — if θ is obtuse, compba is negative, and that negative is the answer, not an error.
Check yourself
a⋅b=10 and ∣b∣=5. What is compba?
Answer
10/5=2. Divide by ∣b∣once — that gives the number.
Same numbers. Is projba a number or a vector?
Answer
A vector: 2510b. Divide twice — the second ∣b∣ is busy normalising.
If a⊥b, what is projba?
Answer
The zero vector. No shadow — the light is shining straight along b.
scalar triple product — volume of the boxV=∣a⋅(b×c)∣=a1b1c1a2b2c2a3b3c3
Reading it:b×c gives the base area as a vector standing on the base. Dotting with a takes 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
∣a∣=3, ∣b∣=4, and they meet at 90°. What is ∣a×b∣?
Answer
3⋅4⋅sin90°=12 — the biggest the tile can get.
What is a×a?
Answer
0. Zero angle, flat tile, no area.
You computed a×b. How do you check it in ten seconds?
Answer
Dot it with a. You must get 0. This catches the missing minus on the j component every time.
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, parametrizedx=f(t),y=g(t),a≤t≤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 coldsegment P→Q:circle, centre (h,k), radius R:ellipse, semi-axes a,b:r(t)=P+t(Q−P),0≤t≤1x=h+Rcost,y=k+Rsint,0≤t≤2πx=h+acost,y=k+bsint,0≤t≤2π
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 cost and sint, solve each
equation for the trig function and use cos2t+sin2t=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=2cost,y=3sint, 0≤t≤π.
identity
cost=2x and sint=3y, so
4x2+9y2=1 — an ellipse with semi-axes 2 and 3.
range
On 0≤t≤π, sint≥0, so y≥0.
Only the upper half of the ellipse.
direction
t=0⇒(2,0); t=2π⇒(0,3);
t=π⇒(−2,0). Right, up, left — counterclockwise.
answer
The upper half of 4x2+9y2=1,
traced once counterclockwise from (2,0) to (−2,0).
Trap
“Eliminate the parameter” is a lossy operation.x=t,y=t2 and
x=t2,y=t4 both give y=x2, 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 r(t)=r0+tv,
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≤t≤1.
Answer
x=−1+4t,y=4−2t. Start, plus t times
the displacement ⟨4,−2⟩.
What curve is x=1+3cost,y=−2+3sint, 0≤t≤2π?
Answer
The circle of radius 3 centred at
(1,−2), once counterclockwise from (4,−2): (x−1)2+(y+2)2=9.
x=sint,y=sin2t. Cartesian equation, and how much of it?
Answer
y=x2, but only the arc with
−1≤x≤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.
Quiz 1 covers the lines half of 13.5 only. Planes are Lesson 3, on 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 P0 with direction v=⟨a,b,c⟩r(t)=r0+tv⎩⎨⎧x=x0+aty=y0+btz=z0+ctax−x0=by−y0=cz−z0
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. 2x−1=5z−4,y=3.
plane through P0 with normal n=⟨a,b,c⟩n⋅(r−r0)=0⟺a(x−x0)+b(y−y0)+c(z−z0)=0⟺ax+by+cz=d
Free information: in ax+by+cz=d, the coefficients are the normal vector. You can read n=⟨a,b,c⟩ straight off any plane equation without doing anything.
Algorithm · plane through three points P, Q, R
Make two vectors in the plane:PQ and PR.
Cross them:n=PQ×PR. It sticks out of the plane by construction.
Pick any one of the three points and write n⋅(r−r0)=0.
Expand and check that all three original points satisfy it.
Algorithm · classify two lines in space
Compare directions. If v1∥v2 (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: cosθ=∣n1∣∣n2∣∣n1⋅n2∣.
Line where two planes meet
Direction is n1×n2. 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
⟨2,−1,3⟩ — read straight off the coefficients, no work needed.
Line through (1,0,0) with direction ⟨2,3,−1⟩?
Answer
r(t)=⟨1+2t,3t,−t⟩.
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.
STEP 02
The questions
24 questions
On those sections only. Untimed, no calculator — the quiz doesn't allow one either.
0/24 answered
112.1 · Eliminating the parameternew
The curve x=2cost,y=3sint, 0≤t≤2π, is
Answer A
Solve each equation for the trig function: cost=2x and sint=3y.
Then cos2t+sin2t=1 gives 4x2+9y2=1.
The denominators are the squares of the amplitudes: 22=4 under x, 32=9 under y.
TrapSwapping the denominators. The number multiplying cost belongs under x, squared — not under y. Option B is that swap.
Which parametrization traces the line segment from (1,2) to (5,−4) exactly once as t runs from 0 to 1, starting at (1,2)?
Answer A
The straight-line recipe is r(t)=r0+t(r1−r0): start, plus t times the displacement.
r1−r0=⟨5−1,−4−2⟩=⟨4,−6⟩.
So x=1+4t, y=2−6t. At t=0 that is (1,2); at t=1 it is (5,−4). ✓
TrapOption D traces the same segment, correctly and exactly once — but backwards, from (5,−4) to (1,2). Direction is part of the answer. Option E covers the right points at the wrong speed.
cost=2x−3 and sint=−2y+1. Squaring and adding, (x−3)2+(y+1)2=4 — a circle of radius 2 centred at (3,−1).
Now the direction. t=0⇒(5,−1), the rightmost point. t=2π⇒(3,−3), below the centre.
Right, then down, then left: that is clockwise.
TrapReading the minus sign as a change of shape. Negating sint leaves the circle identical and only reverses the orientation. Radius is 2, not 4 — the 4 is r2.
TrapSubtracting the wrong way round. P−Q has the same length, so the error hides here — but it reverses the direction of every line you build from it later.
The equation x2+y2+z2−6x+4y−12=0 describes a sphere. Find its centre and radius.
Answer A
Group and complete the square, balancing the right side each time:
(x2−6x)+(y2+4y)+z2=12⇒(x−3)2−9+(y+2)2−4+z2=12.
(x−3)2+(y+2)2+z2=25. Centre (3,−2,0), radius 5.
TrapSign flip on the centre. (x−3)² means the centre coordinate is +3, not −3.
a⋅b=3+2+4=9 and ∣b∣2=1+4+4=9.
projba=∣b∣2a⋅bb=99⟨1,2,2⟩=⟨1,2,2⟩.
The scalar projection is the different object compba=∣b∣9=39=3 — option A, the planted distractor.
TrapDividing by |b| once (a number) when the question asked for the vector. Once = number, twice = vector.
For a=⟨4,3,0⟩ and b=⟨0,1,1⟩, find the scalar projection compba.
Answer A
a⋅b=(4)(0)+(3)(1)+(0)(1)=3 and ∣b∣=2.
compba=∣b∣a⋅b=23=232.
TrapOption C is projba — the vector projection, which divides by ∣b∣2 and then multiplies by b. The question asked for a number. Option B is that same computation with the b forgotten.
L1:x=1+t,y=2−t,z=3+2t and L2:x=2+2s,y=1−2s,z=6+4s. These two lines are
Answer A
Directions: v1=⟨1,−1,2⟩ and v2=⟨2,−2,4⟩=2v1. Parallel directions, so the lines cannot be skew or intersecting.
Now decide parallel-and-distinct versus identical: is L2's point (2,1,6) on L1? From x: 1+t=2⇒t=1, giving (2,1,5) on L1 — not (2,1,6).
Same direction, different point ⇒ parallel and distinct.
TrapStopping at “the directions are proportional, so it's the same line.” Parallel lines share a direction; identical lines also share a point. You must test a point.
Find symmetric equations for the line through (2,−1,4) parallel to v=⟨3,5,−2⟩.
Answer A
Start from the parametric form x=2+3t,y=−1+5t,z=4−2t and solve each for t:
t=3x−2, t=5y−(−1)=5y+1, t=−2z−4.
Setting them equal gives the symmetric form. The point is subtracted on top; the direction goes underneath.
TrapThe point's coordinate is −1, so the numerator is y−(−1)=y+1 — option E drops that. Option D swaps the roles: it puts the direction on top and the point underneath.
L1:x=1+t,y=2+3t,z=−1+2t and L2:x=2+s,y=−1+2s,z=3−s. These two lines are
Answer A
Directions ⟨1,3,2⟩ and ⟨1,2,−1⟩ are not multiples of each other, so the lines are not parallel. Either they meet or they are skew.
Set the x's equal: 1+t=2+s⇒t=1+s. Set the y's equal: 2+3t=−1+2s, so 3(1+s)−2s=−3, giving s=−6 and t=−5.
Now test the third equation: z1=−1+2(−5)=−11 but z2=3−(−6)=9. They disagree, so no single point is on both lines.
Not parallel and never meeting ⇒ skew.
TrapSolving the first two equations, finding a solution, and declaring the lines intersect. Two equations in two unknowns almost always have a solution — the third equation is the actual test.
Quiz 2 covers the planes half of 13.5. Lines were Lesson 2, on Quiz 1.
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 P0 with direction v=⟨a,b,c⟩r(t)=r0+tv⎩⎨⎧x=x0+aty=y0+btz=z0+ctax−x0=by−y0=cz−z0
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. 2x−1=5z−4,y=3.
plane through P0 with normal n=⟨a,b,c⟩n⋅(r−r0)=0⟺a(x−x0)+b(y−y0)+c(z−z0)=0⟺ax+by+cz=d
Free information: in ax+by+cz=d, the coefficients are the normal vector. You can read n=⟨a,b,c⟩ straight off any plane equation without doing anything.
Algorithm · plane through three points P, Q, R
Make two vectors in the plane:PQ and PR.
Cross them:n=PQ×PR. It sticks out of the plane by construction.
Pick any one of the three points and write n⋅(r−r0)=0.
Expand and check that all three original points satisfy it.
Algorithm · classify two lines in space
Compare directions. If v1∥v2 (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: cosθ=∣n1∣∣n2∣∣n1⋅n2∣.
Line where two planes meet
Direction is n1×n2. 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
⟨2,−1,3⟩ — read straight off the coefficients, no work needed.
Line through (1,0,0) with direction ⟨2,3,−1⟩?
Answer
r(t)=⟨1+2t,3t,−t⟩.
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.
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 P1 to the plane ax+by+cz+d=0D=a2+b2+c2∣ax1+by1+cz1+d∣=compnP0P1
Read it as a scalar projection onto the normal. Take any point P0 on the plane, form P0P1, and measure its shadow on n — the part of the trip that actually left the plane.
point P to the line through Q with direction vD=∣v∣∣QP×v∣
Why the cross product: ∣QP×v∣=∣QP∣∣v∣sinθ is the area of the parallelogram they span. Area ÷ base = height, and the height is the perpendicular distance. Same as ∣QP−projvQP∣ — the leftover from 13.3b.
two skew linesD=∣v1×v2∣(r2−r1)⋅(v1×v2)
Algorithm · distance between two skew lines
Find the common perpendicular:n=v1×v2. It is at right angles to both lines at once.
Pick one point on each line, P1 and P2, and form P1P2.
Project onto n: the distance is compnP1P2. Everything not along 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 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 ≥0. Absolute value bars are not decoration.
Check yourself
Distance from the origin to x+2y+2z=9?
Answer
1+4+4∣0+0+0−9∣=39=3.
Why does point-to-line use a cross product?
Answer
∣QP×v∣ is the parallelogram's area. Area ÷ base = height, and the height is the perpendicular distance.
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 axisx2+z2=9(no y)
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=−z2 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
x2+z2=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 R3 — 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 a2x2±b2y2±c2z2=1,0, or = a single linear variable.
Count the minus signs among the squared terms: 0⇒ ellipsoid, 1⇒ hyperboloid of one sheet, 2⇒ 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 x2, y2 or z2.
Complete each square by adding (b/2)2inside the bracket.
Because the bracket is multiplied by that factored-out coefficient, what you must
add to the other side is a⋅(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 · 4x2−y2+z2+8x=04(x2+2x)−y2+z24(x2+2x+1)−y2+z24(x+1)2−y2+z2(x+1)2−4y2+4z2=0=0+4⋅1=4=1
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 x2+z2=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 R3 it is a tube.
Identify y=−z2 in three dimensions.
Answer
No x: a parabolic cylinder, the parabola y=−z2 in the yz-plane opening toward −y, swept along the x-axis.
Put 9x2+4y2+36z2−18x=27 into standard form and identify it.
Answer
9(x2−2x)=9(x−1)2−9, so 9(x−1)2+4y2+36z2=36, i.e. 4(x−1)2+9y2+z2=1. No minus signs → an ellipsoid centred at (1,0,0).
Identify x2+y2−z2=1.
Answer
One minus sign → hyperboloid of one sheet (connected, like a cooling tower).
Identify z=x2−y2.
Answer
One variable unsquared, opposite signs → hyperbolic paraboloid, the saddle.
Identify z2=x2+y2.
Answer
Right side is 0 when rearranged → cone.
STEP 02
The questions
17 questions
On those sections only. Untimed, no calculator — the quiz doesn't allow one either.
0/17 answered
113.5 · Planes
Find an equation of the plane through (1,0,0), (0,2,0) and (0,0,4).
Answer B
PQ=⟨−1,2,0⟩,PR=⟨−1,0,4⟩.
n=PQ×PR=⟨8,4,2⟩∥⟨4,2,1⟩.
Through (1,0,0): 4(x−1)+2y+z=0⇒4x+2y+z=4.
Fast check: the intercept form 1x+2y+4z=1 clears to the same thing.
TrapSign slip on the j component of the cross product — verify n·PQ = 0.
Find an equation of the plane through A(1,0,0), B(0,2,0) and C(0,0,4).
Answer A
Two vectors in the plane: AB=⟨−1,2,0⟩ and AC=⟨−1,0,4⟩.
n=AB×AC=⟨2⋅4−0⋅0,0⋅(−1)−(−1)⋅4,0−(−2)⟩=⟨8,4,2⟩, or ⟨4,2,1⟩ after dividing by 2.
Through A(1,0,0): 4(x−1)+2y+z=0, so 4x+2y+z=4.
TrapReading the intercepts straight off as coefficients gives x+2y+4z=1 — option B. The intercept form is 1x+2y+4z=1; the denominators are the intercepts, not the numerators.
Two planes are orthogonal exactly when their normals are orthogonal.
n1=⟨2,1,−1⟩. For option A, n=⟨1,1,3⟩ and n1⋅n=2+1−3=0. ✓
The others: C gives 6−1−1=4, D gives 4+1−1=4, E gives 2+2−1=3 — none zero. B and F have normals proportional to n1, so those planes are parallel to the original, not perpendicular.
TrapOptions B and F look different but are the same plane family scaled by 2 and −1. Proportional normals mean parallel; orthogonal normals mean perpendicular.
Find the plane through (1,0,2) perpendicular to the line x=3−t,y=2+4t,z=1+t.
Answer A
A plane perpendicular to a line has the line's direction as its normal.
The direction is v=⟨−1,4,1⟩.
−1(x−1)+4(y−0)+1(z−2)=0, so −x+4y+z−1=0, that is −x+4y+z=1.
Check: −(1)+4(0)+2=1. ✓
TrapUsing the line's point(3,2,1) as the normal gives option C. The point tells you where the line is; the direction tells you which way it goes, and only the direction can be a normal here.
Count the minus signs among the squared terms: exactly one, and the right side is 1.
One minus ⇒hyperboloid of one sheet — the connected cooling-tower shape.
Check with a trace: setting z=0 gives the ellipse 4x2+9y2=1, a real curve,
so the surface exists at the waist and is not in two pieces.
TrapTwo minus signs would be two sheets; right side 0 would be a cone.
What is the trace of the cone z2=x2+y2 in the plane z=3?
Answer A
Substitute z=3: 9=x2+y2, a circle of radius 3.
The vertical traces are different: setting y=0 gives z2=x2, i.e. the pair of lines z=±x —
which is why the surface is a cone and not a paraboloid.
TrapReading the radius as z² = 9 instead of √9 = 3.
Two squared variables with opposite signs, plus one linear variable.
Along y=0: z=x2, a parabola opening up. Along x=0: z=−y2, a parabola opening down.
Up one way and down the other is exactly a saddle — the hyperbolic paraboloid.
TrapReading “hyperbolic” as “hyperboloid”. A hyperboloid has three squared terms and a constant; a hyperbolic paraboloid has two squared terms and one linear one.
For the surface z=x2−y2, what is the trace in the plane z=0?
Answer A
Set z=0: x2−y2=0, so (x−y)(x+y)=0.
That is y=x or y=−x — a pair of intersecting lines.
For z=k=0 the traces really are hyperbolas, opening along x when k>0 and along y when k<0. z=0 is the level where the hyperbola degenerates into its own asymptotes.
TrapAnswering “hyperbola” because every other horizontal trace is one. The one level that breaks the pattern is the one the question asks about.
One plus and two minuses ⇒ hyperboloid of two sheets, and the axis is the positive variable: z.
Sanity check on the traces: z=k needs 9k2−1≥0, so ∣k∣≥3. There is no surface at all between z=−3 and z=3 — that gap is the two separate sheets.
TrapConfusing this with 4x2+y2−z2=36, which has two pluses and is one sheet. Count the signs, then check whether some slab has no points in it: an empty gap means two sheets.
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 tangentr′(t)=⟨f′(t),g′(t),h′(t)⟩T(t)=∣r′(t)∣r′(t)
product rules — note which product each one keepsdtd[u⋅v]=u′⋅v+u⋅v′dtd[u×v]=u′×v+u×v′
Trap
The cross-product rule must keep its order: u′×v, never v×u′. Cross products anticommute, so swapping flips the sign of your whole answer.
A fact worth memorising: if ∣r(t)∣ is constant, then r⋅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 T′⊥T in 14.3.
integral — antidifferentiate each slot, one vector constant∫r(t)dt=⟨∫f,∫g,∫h⟩+C
Check yourself
r(t)=⟨t2,t3⟩. Find r′(1).
Answer
⟨2t,3t2⟩ at t=1 is ⟨2,3⟩.
If ∣r(t)∣ is constant, what is r⋅r′?
Answer
0. Constant length means the velocity is perpendicular to the position — why anything moving on a sphere stays tangent to it.
Quiz 3 covers 14.3 only — velocity, acceleration, circular and projectile motion. The T, N, B frame and the components of acceleration are 14.5, on Quiz 4 and later.
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 frameT=∣r′∣r′N=∣T′∣T′B=T×N
N is guaranteed perpendicular to T because T has constant length 1 — the fact from 14.2. B is perpendicular to both because that's what a cross product does.
acceleration always splits two waysa=aTT+aNN,aT=dtd∣v∣=∣r′∣r′⋅r′′,aN=κ∣v∣2=∣r′∣∣r′×r′′∣
Read it physically.aT is the gas pedal — it changes how fast you go. aN is the steering wheel — it changes where you're pointed. There is never a B component: you are never accelerated sideways out of your own turn.
Algorithm · decompose acceleration
Compute r′ and r′′.
Speed:∣r′∣.
aT: dot them and divide by the speed, ∣r′∣r′⋅r′′.
aN: cross them and divide by the speed, ∣r′∣∣r′×r′′∣.
Check:aT2+aN2=∣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 a=⟨0,0,−g⟩: integrate once for v (constant of integration = initial velocity), again for r (constant = initial position).
Check yourself
What does aT 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 aN, dot or cross?
Answer
Cross.aN=∣r′∣∣r′×r′′∣. Dot for the gas pedal, cross for the steering wheel.
STEP 02
The questions
16 questions
On those sections only. Untimed, no calculator — the quiz doesn't allow one either.
0/16 answered
114.1 · Space curves
Which curve lies on the cylinder y2+z2=9 and advances at a constant rate along the cylinder\'s axis?
Answer B
The cylinder y2+z2=9 constrains y and z; its axis is the x-axis, so x must be the free coordinate.
Option B: y2+z2=9cos2t+9sin2t=9 ✓ and x=t advances at constant rate ✓.
A and E put the circle in x,y (wrong cylinder). D fails the identity. F has radius t, not 3.
TrapReading the cylinder’s axis off the wrong pair of variables.
Which function traces the circle of radius 3 centred on the z-axis in the plane z=5, counterclockwise as seen from above?
Answer A
Staying in the plane z=5 forces the third component to be the constant 5.
Radius 3 centred on the axis gives x=3cost, y=3sint; then x2+y2=9. ✓
(+cos,+sin) runs counterclockwise viewed from +z looking down.
TrapOption B has z=t — that is a helix, not a plane curve. Option D reverses the orientation. Option E shifts the centre to (3,3) and shrinks the radius to 1.
Find parametric equations for the tangent line to r(t)=⟨t,t2,t3⟩ at the point (2,4,8).
Answer A
First find the parameter value at that point. x=t=2, and indeed t2=4, t3=8. So t=2.
r′(t)=⟨1,2t,3t2⟩, so r′(2)=⟨1,4,12⟩ — the direction.
Line through (2,4,8) with direction ⟨1,4,12⟩: x=2+s,y=4+4s,z=8+12s.
TrapOption C uses r′ with t never substituted, effectively ⟨1,2,3⟩. Option B swaps the point and the direction. Option F evaluates 3t2 as 3⋅2=6 instead of 3⋅4=12.
r(t) is differentiable and ∣r(t)∣ is constant. What is r(t)⋅r′(t)?
Answer A
Use the product rule for the dot product: dtd[r⋅r]=r′⋅r+r⋅r′=2r⋅r′.
But r⋅r=∣r∣2, which is a constant, so its derivative is 0.
Hence 2r⋅r′=0, so r⋅r′=0.
Geometrically: a curve of constant distance from the origin lies on a sphere, and its velocity must be tangent to that sphere — perpendicular to the radius.
TrapThis is the fact behind uniform circular motion: constant speed means the acceleration is perpendicular to the velocity, which is the same statement applied to v instead of r.
A projectile leaves the origin with velocity ⟨20,0,20⟩ m/s under acceleration ⟨0,0,−10⟩ m/s². How far horizontally does it travel before returning to z=0?
Answer C
Integrate twice: z(t)=20t−5t2, x(t)=20t.
z=0⇒5t(4−t)=0⇒t=4 s (rejecting t=0, the launch).
Horizontal distance x(4)=20(4)=80 m.
TrapUsing t = 2 s (the time to the top) instead of the full flight time.
For r(t)=⟨4cos3t,4sin3t⟩, which statement is true?
Answer A
v=⟨−12sin3t,12cos3t⟩, so ∣v∣=12sin23t+cos23t=12 — constant.
a=⟨−36cos3t,−36sin3t⟩=−9r(t).
A negative multiple of the position vector points straight back at the origin. (This is centripetal acceleration.)
Also v⋅a=0, so the acceleration is perpendicular to the velocity, never parallel to it.
Trap“Constant speed” is not “no acceleration” — option F. The direction is changing the whole time, and changing direction is acceleration. Note also that the speed is 12, not the radius 4: the 3 inside the trig functions multiplies it.
A particle has acceleration a(t)=⟨0,2,6t⟩, with v(0)=⟨1,0,0⟩ and r(0)=⟨0,1,2⟩. Find r(1).
Answer A
Integrate once: v(t)=⟨0,2t,3t2⟩+C1. Since v(0)=⟨1,0,0⟩, C1=⟨1,0,0⟩, so v(t)=⟨1,2t,3t2⟩.
Integrate again: r(t)=⟨t,t2,t3⟩+C2. Since r(0)=⟨0,1,2⟩, C2=⟨0,1,2⟩.
r(t)=⟨t,t2+1,t3+2⟩, so r(1)=⟨1,2,3⟩.
TrapTwo integrations mean two vector constants, and each is fixed by its own initial condition. Losing C1 drops the x motion entirely; losing C2 gives ⟨1,1,1⟩.
A ball is launched from a height of 15 m with initial velocity ⟨20,0,10⟩ m/s, under acceleration ⟨0,0,−10⟩ m/s2. When does it hit the ground?
Answer A
Only the vertical component matters. Integrating twice, z(t)=15+10t−5t2 (initial height, plus initial vertical velocity, minus 21gt2).
Set z=0: 5t2−10t−15=0, so t2−2t−3=0, that is (t−3)(t+1)=0.
The positive root is t=3 s.
TrapDropping the initial height and solving 10t−5t2=0 gives t=2 — option B. The launch height is what keeps it in the air longer. And the coefficient is 21g=5, not g=10.
Lesson 8 runs 14.4 and 14.5 as far as Theorem 14.5. The later part of 14.5 is not on this quiz.
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 timeL=∫ab∣r′(t)∣dt=∫ab(dtdx)2+(dtdy)2+(dtdz)2dt
Algorithm · reparametrize by arc length
Compute the speed∣r′(t)∣.
Build the arc-length functions(t)=∫0t∣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: 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 givenκ=dsdT=∣r′(t)∣∣T′(t)∣=∣r′(t)∣3∣r′(t)×r′′(t)∣
Use the third one
Almost always. It needs only r′ and r′′ — no messy normalising.
Use the second one
When the problem already handed you T(t).
For a plane curve y=f(x)
κ=(1+(f′)2)3/2∣f′′∣
Sanity values: a straight line has κ=0. A circle of radius a has κ=1/a everywhere. The radius of curvature is ρ=1/κ — the radius of that best-fitting circle.
Check yourself
Speed of ⟨3cost,3sint,4t⟩?
Answer
9+16=5, constant. So length over [0,2π] is just 10π.
Curvature of a circle of radius 2?
Answer
κ=1/2. Curvature is one over the radius of the hugging circle.
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
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)=kF(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 ≥, dashed if strict.
Check yourself
What do the level curves of f(x,y)=x+y look like?
Answer
Parallel straight linesx+y=k, evenly spaced — a constant-slope ramp, not a hill.
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 provesif (x,y)→(a,b)limf along path 1=lim along path 2,then the limit does not exist
Algorithm · show a limit does NOT exist
Try the axes first. Set y=0, take x→0. Then x=0, take y→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=xp/q or y=x2. 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=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=rcosθ, y=rsinθ, and let r→0+. If everything left over is bounded and the answer does not depend on θ, that's the limit.
Or squeeze: find 0≤∣f−L∣≤g(x,y) with g→0. Useful bounds: x2+y2x≤1, ∣sinu∣≤∣u∣, x2≤x2+y2.
Checking a hundred paths proves nothing. Polar or squeeze is the only way to actually close it.
Worked example — polar closing the case
(x,y)→(0,0)limx2+y2x3+y3. Every straight line gives 0, but that isn't a proof.
polar
r2r3(cos3θ+sin3θ)=r(cos3θ+sin3θ)
bound
∣cos3θ+sin3θ∣≤2, so the whole thing is ≤2r
conclude
as r→0 this →0 regardless of θ. 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 θ, or squeeze it.
STEP 02
The questions
7 questions
On those sections only. Untimed, no calculator — the quiz doesn't allow one either.
0/7 answered
114.4 · Arc length
Find the length of the helix r(t)=⟨3cost,3sint,4t⟩ for 0≤t≤2π.
Answer B
r′=⟨−3sint,3cost,4⟩, so
∣r′∣=9sin2t+9cos2t+16=25=5 — constant speed.
L=∫02π5dt=10π. Distance = speed × time, exactly.
TrapForgetting to square the 4, or integrating |r| instead of |r′|.
For r(t)=⟨cost,sint,t⟩, find the binormal B at t=0.
Answer A
r′=⟨−sint,cost,1⟩, ∣r′∣=2, so
T(0)=21⟨0,1,1⟩.
T′=21⟨−cost,−sint,0⟩⇒T′(0)=21⟨−1,0,0⟩,
so N(0)=⟨−1,0,0⟩.
B=T×N=21⟨0,1,1⟩×⟨−1,0,0⟩=⟨0,−21,21⟩.
TrapComputing N × T instead of T × N — that flips every sign.
For f(x,y)=x2+4y2, the level curves f=k with k>0 are
Answer B
x2+4y2=k rearranges to kx2+k/4y2=1 — an ellipse
with semi-axes k and k/2, so it is twice as wide as it is tall.
They would be circles only if the two coefficients matched.
TrapSeeing "sum of squares" and answering circles. The unequal coefficients stretch it.
Consider (x,y)→(0,0)limx4+y2x2y. Which statement is correct?
Answer D
Along y=0: the quotient is 0. Along x=0: also 0. Every straight line y=mx also gives 0.
But along the parabola y=x2:
x4+x4x2⋅x2=2x4x4=21.
Two roads disagree, so the limit does not exist.
TrapTesting only straight lines. Matching the powers (here y ~ x²) is what exposes it.
Consider (x,y)→(0,0)limx2+y2x2−y2. Which is correct?
Answer D
Along y=0: x2x2=1. Along x=0: y2−y2=−1.
Two roads disagree, so the limit does not exist.
In polar the quotient is cos2θ−sin2θ=cos2θ — it depends on θ and never settles as r→0.
TrapConcluding "0/0 so it is 0". The axes alone already settle this one.
Refuse to turn. Walk due east only and record your height — that's a plain 1-D curve, and its slope is ∂f/∂x. Turn 90° and walk due north for ∂f/∂y.
drag to spin
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 movesfx(a,b)=h→0limhf(a+h,b)−f(a,b)fy(a,b)=h→0limhf(a,b+h)−f(a,b)
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.
∂x∂(x2y3)=2xy3 — the y3 was a constant, so it just rode along.
second and mixed partialsfxx=∂x2∂2ffxy=(fx)y=∂y∂x∂2ffyx=(fy)x
Clairaut's theorem
If the mixed partials are continuous near a point, fxy=fyx — the order you differentiate in doesn't matter. This is free error-checking: compute both and they must match.
Notation trap
fxy means "x first, then y", but ∂y∂x∂2f 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 ∂x∂(x2y3).
Answer
2xy3. The y3 was a constant, so it rode along untouched.
Is fxy the same as fyx?
Answer
Yes, whenever the mixed partials are continuous (Clairaut). Compute both — it is free error-checking.
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 variabledtdz=∂x∂zdtdx+∂y∂zdtdy
Case 2 · two independent variables∂s∂z=∂x∂z∂s∂x+∂y∂z∂s∂y∂t∂z=∂x∂z∂t∂x+∂y∂z∂t∂y
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 ∂ 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)=0dxdy=−FyFx∂x∂z=−FzFx,∂y∂z=−FzFy
Where the minus sign comes from. Staying on the level set means F never changes, so dF=0. Expand: Fxdx+Fydy=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 ∂z/∂x.
Collect all the ∂z/∂x terms on one side.
Solve for ∂z/∂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.
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∇f=⟨fx,fy⟩or∇f=⟨fx,fy,fz⟩
You don't have to walk straight uphill. Pick any direction u — your slope is the shadow of the gradient onto your direction. The dot product from 13.3, doing real work.
drag to spin
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 vectorDuf=∇f⋅u=∣∇f∣cosθ
Algorithm · directional derivative
Compute ∇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 θ from the x-axis, use u=⟨cosθ,sinθ⟩. If you were given "toward the point Q", use PQ/∣PQ∣.
Dot them.Duf=∇f⋅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 ∇f, rate ∣∇f∣
cosθ is maximal at θ=0
Fastest decrease
direction −∇f, rate −∣∇f∣
cosθ=−1
No change at all
any u⊥∇f
cos90°=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 ∇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)=kFx(P)(x−x0)+Fy(P)(y−y0)+Fz(P)(z−z0)=0r(t)=r0+t∇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 ∇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 ∇F as the direction vector.
Note step 1 reproduces ⟨fx,fy,−1⟩ from 15.6 — the two methods are the same method.
Check yourself
∇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
∣∇f∣, the length of the gradient — a number, attained in the direction ∇f.
You walk along a contour line. What is Duf?
Answer
0. Dead flat — that is what a contour means.
STEP 02
The questions
5 questions
On those sections only. Untimed, no calculator — the quiz doesn't allow one either.
0/5 answered
115.3 · Mixed partials
For f(x,y)=x2y3+x, find fxy at (1,2).
Answer B
fx=2xy3+1, then fxy=∂y∂(2xy3+1)=6xy2.
At (1,2): 6(1)(4)=24.
Clairaut check the other way round: fy=3x2y2⇒fyx=6xy2 ✓ same answer — free error-checking.
TrapDifferentiating twice in the same variable. f_xy means x first, then y.
Let z=x2y with x=s+t and y=st. Find ∂s∂z at (s,t)=(1,2).
Answer D
∂s∂z=zxxs+zyys=(2xy)(1)+(x2)(t).
At s=1,t=2: x=3,y=2, so =2(3)(2)(1)+9(2)=12+18=30.
Two routes down the tree, so two terms — count the routes before you compute.
TrapDropping one branch. Number of terms = number of paths from z down to s.
Mon Oct 5 · 8:00 pm · Lessons 1–16 · 13.1–15.7 (all of Ch. 13–14, Ch. 15 through max/min)
Midterm 1
Lessons 1–16, one hour, multiple choice, no calculator. Stops in the middle of Chapter 15 — Lagrange is NOT on it.
Worth knowing. Lessons 14–16 (tangent planes, max/min) are never quizzed. They arrive between Quiz 5 and this exam, so this is the only place they are ever tested before the final.
Only the sections no quiz ever covers are shown here. Everything from the earlier quizzes is on this exam too — work those tabs as well.
STEP 01
The material
2 sections
§15.6 · §15.7 — and nothing else. Read them, then do the check-yourself questions inside each one with the answer folded away.
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
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) at (a,b)z=f(a,b)+fx(a,b)(x−a)+fy(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 ⟨fx,fy,−1⟩.
differentials — the same statement, about small changesdz=fxdx+fydyΔz≈dz
Algorithm · estimate a value or an error
Pick a nearby easy point(a,b) where f, fx, fy are trivial to evaluate.
Compute the three numbersf(a,b), fx(a,b), fy(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)≈f(a,b)+fxdx+fydy.
For maximum error, use ∣dx∣,∣dy∣ at their worst and take absolute values of the coefficients: ∣dz∣≤∣fx∣∣dx∣+∣fy∣∣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 fx and fy 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
⟨fx,fy,−1⟩. (Or write it as F=f−z=0 and take ∇F — same thing.)
Write the differential dz.
Answer
dz=fxdx+fydy. Nudge x, nudge y, add both effects.
If ∇f=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 testD=fxxfyy−(fxy)2evaluated at the critical point
If
Then
Picture
D>0 and fxx<0
local maximum
both slices bend down — a summit
D>0 and fxx>0
local minimum
both slices bend up — a bowl
D<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:fx=0 and fy=0. Solve the system.
Find every solution. Factor rather than divide — dividing by x silently throws away the critical points with x=0.
Compute fxx, fyy, fxy symbolically, once.
Evaluate D at each point and read the table.
Report the function valuef 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 ∇f is not zero at all — so it never shows up as a critical point. Steps 2–4 are not optional.
Check yourself
D<0 at a critical point. What is it?
Answer
A saddle — and fxx 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 ∇f is not zero at all.
STEP 02
The questions
3 questions
On those sections only. Untimed, no calculator — the quiz doesn't allow one either.
0/3 answered
115.6 · Linear approximation
Use the linearization of f(x,y)=x2+y2 at (3,4) to estimate f(3.1,3.9).
Answer B
f(3,4)=5; fx=x2+y2x=53, fy=x2+y2y=54.
Δx=0.1,Δy=−0.1:
L=5+53(0.1)+54(−0.1)=5+0.06−0.08=4.98.
The move is mostly "sideways" along the level circle, so the value barely changes.
TrapSign of Δy. 3.9 is 0.1 below 4, so Δy is negative.
Find the absolute maximum of f(x,y)=xy on the closed disk x2+y2≤1.
Answer B
Interior:∇f=⟨y,x⟩=0 only at (0,0), where f=0.
Boundary: parametrize x=cost,y=sint, so f=costsint=21sin2t, whose maximum is 21.
Compare the whole list: {0,21,−21}. Absolute max =21, at (21,21).
TrapStopping at the interior critical point. On a closed region the max usually lives on the edge.
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∇f=λ∇gandg(x,y,z)=k
Touching = sharing a tangent line = normals pointing along the same line. The normals are ∇f and ∇g, so one must be a multiple of the other. That multiple is λ — 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:fx=λgx, fy=λgy, fz=λgz, plus g=k. Four equations, four unknowns.
Eliminate λ first. Solve each equation for λ 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=x2+y2=8.
system
y=2λx, x=2λy, x2+y2=8
eliminate
2xy=λ=2yx⇒y2=x2⇒y=±x
constraint
2x2=8⇒x=±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∇f=λ∇g+μ∇h,g=k1,h=k2
With two constraints you're stuck on a curve instead of a surface. ∇g and ∇h span the plane of directions perpendicular to that curve, and at an optimum ∇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, x2+y2+z2=1).
Check yourself
What is the condition at a constrained optimum?
Answer
∇f=λ∇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.
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
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∬RfdA=m,n→∞limi=1∑mj=1∑nf(xij∗,yij∗)ΔA=∫ab∫cdfdydx=∫cd∫abfdxdy
Over a rectangle the order never matters. And if f(x,y)=g(x)h(y) splits, the double integral splits too: ∬=(∫abg)(∫cdh). 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 ∬R1dA 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.
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∬RfdA=∫ab∫g1(x)g2(x)f(x,y)dydx
Type II · horizontal arrows, x between two curves∬RfdA=∫cd∫h1(y)h2(y)f(x,y)dxdy
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:∫numbernumber∫functionfunction. 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=x2 becomes x=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 ey2 or sin(x2) that has no elementary antiderivative. That's the problem asking you to flip.
Worked example
∫01∫x1ey2dydx — the inner integral is impossible as written.
sketch
0≤x≤1 and x≤y≤1: the triangle with corners (0,0),(1,1),(0,1).
re-sweep
Horizontally: for each y in [0,1], x runs 0→y.
rewrite
∫01∫0yey2dxdy=∫01yey2dy
finish
=21ey201=2e−1
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 ey2. What is the problem telling you?
Answer
Flip the order. No elementary antiderivative in that variable is the signal, every time.
STEP 02
The questions
4 questions
On those sections only. Untimed, no calculator — the quiz doesn't allow one either.
0/4 answered
115.8 · Lagrange multipliers
Find the maximum of f(x,y)=x+2y subject to x2+y2=5.
Answer C
∇f=⟨1,2⟩=λ⟨2x,2y⟩⇒y=2x.
Substituting: x2+4x2=5⇒x=±1, giving (1,2) and (−1,−2).
f(1,2)=5, f(−1,−2)=−5. Max =5, min =−5.
Cross-check: the maximum of a⋅r on a circle of radius R is ∣a∣R=5⋅5=5 ✓.
TrapReporting λ, or stopping at one of the two candidate points without comparing.
The integrand splits, so the double integral splits too:
∬RxydA=(∫02xdx)(∫03ydy)=(2)(29)=9.
Look for this factorisation before doing any work — over a rectangle it always applies when f=g(x)h(y).
TrapMultiplying the two upper limits (2 × 3 = 6). The antiderivatives are what multiply.
The region is 0≤y≤1,y≤x≤1, i.e. 0≤x≤1,0≤y≤x2. Sketch it, then re-sweep.
∫01∫0x2ex3dydx=∫01x2ex3dx=[31ex3]01=3e−1.
The x2 that the flip produces is exactly the factor that makes the substitution work.
TrapTrying to antidifferentiate e^(x³) in x. That is the signal to flip.
Region: 0≤x≤1,x≤y≤1 — the triangle above y=x. Re-swept horizontally: 0≤y≤1,0≤x≤y.
∫01∫0ysin(y2)dxdy=∫01ysin(y2)dy=[−21cos(y2)]01=21−cos1.
The y the flip produces is exactly what makes the substitution work.
TrapAttempting ∫sin(y²)dy directly. No elementary antiderivative — that is the signal to flip.
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 rdθ — 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∬Rf(x,y)dA=∫αβ∫r1(θ)r2(θ)f(rcosθ,rsinθ)dArdrdθ
Algorithm · convert to polar
Spot the trigger:x2+y2, a circle, a disk, an annulus, a wedge, or limits with a2−x2.
Convert the integrand:x→rcosθ, y→rsinθ, and every x2+y2→r2.
Convert the region.θ sweeps the angular range; for each θ, r runs from the inner boundary to the outer boundary.
Attach the r.dA=rdrdθ. Write it before you do anything else.
Integrate — inner in r, outer in θ.
Useful conversions: x2+y2=a2→r=a. y=x→θ=π/4. x2+y2=2ax→r=2acosθ (a circle through the origin).
The single most expensive mistake in the course
Dropping the r.dA=rdrdθ, not drdθ. 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 drdθ.
Check yourself
What is dA in polar?
Answer
rdrdθ.
Where does that r come from?
Answer
The wedge. One side is dr long; the other is an arc of length rdθ, because the same angle sweeps further when you are farther out.
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∭EfdV=∬shadow[∫z1(x,y)z2(x,y)fdz]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 z1(x,y) to the top surface z2(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 z1=z2 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 (dzdydx, dydzdx, …). 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=ρ
Mass. Then xˉ=m1∭xρdV, and similarly for yˉ,zˉ.
Inertia about the z-axis
Iz=∭(x2+y2)ρdV — squared distance from the axis.
Check yourself
What does ∭E1dV 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.
Find the volume of the tetrahedron bounded by x=0,y=0,z=0 and x+y+z=1.
Answer C
Sweep in z first: 0≤z≤1−x−y. The shadow on the xy-plane is the triangle 0≤y≤1−x,0≤x≤1.
V=∫01∫01−x∫01−x−ydzdydx=∫012(1−x)2dx=61.
Note the shape: function limits inside, then function, then numbers outermost.
TrapPutting a variable in the outermost limit. The outer limits are always constants.
Distance from the origin, angle down from the north pole, angle around. Balls and cones. The famous ρ2sinφ is not arbitrary — it's the little box's three edge lengths multiplied together.
drag to spin
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.
the volume element, derived not memoriseddV=radialdρ×north–southρdφ×east–westρsinφdθ=ρ2sinφdρdφdθ
Why sinφ: the circle of latitude at angle φ has radius ρsinφ, not ρ. Near the poles it shrinks to nothing, so the same dθ sweeps almost no distance — and the volume element shrinks with it.
Algorithm · set up in spherical
Use it when you see x2+y2+z2, a sphere, a ball, a cone, or a region between two spheres.
ρ limits: where the ray from the origin enters and leaves the solid. A sphere of radius a gives ρ=a; a plane z=c gives ρ=c/cosφ=csecφ.
φ limits: the cone angle. φ=0 is straight up, φ=π/2 is the xy-plane, φ=π is straight down. The cone z=x2+y2 is φ=π/4.
θ limits: how far around, usually 0 to 2π.
Attach ρ2sinφ and integrate, innermost ρ outward.
Ranges: ρ≥0, 0≤φ≤π, 0≤θ≤2π. φ never goes past π — 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
ρ2sinφdρdφdθ.
What is the range of φ?
Answer
0≤φ≤π — top to bottom. Only θ goes all the way round to 2π.
Why the sinφ?
Answer
The circle of latitude has radius ρsinφ, not ρ. Near the poles it shrinks to nothing, so the box goes skinny.
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 momentsm=∬RρdAMy=∬RxρdAMx=∬RyρdA
centre of massxˉ=mMy=m1∬RxρdAyˉ=mMx=m1∬RyρdA
moments of inertia — distance is SQUAREDIx=∬Ry2ρdAIy=∬Rx2ρdAI0=∬R(x2+y2)ρdA=Ix+Iy
Naming trap
My uses x, and Mx uses y. The subscript names the axis you are taking the moment about, not the variable in the integrand. Same crossover in Ix, which uses y2 — 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, x2+y2 for polar inertia.
Integrate ∬(weight)ρdA. 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?ρ is a constant, so it factors out of the top and bottom of xˉ and cancels completely. The answer is then called the centroid and depends only on the shape.
Check yourself
Write xˉ for a lamina.
Answer
xˉ=m1∬RxρdA — 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.
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 fieldF(x,y)=P(x,y)i+Q(x,y)jF(x,y,z)=Pi+Qj+Rk
Gradient fields are the special ones: F=∇f for some scalar f. Physically, f is a hill (or a potential) and 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.
STEP 02
The questions
4 questions
On those sections only. Untimed, no calculator — the quiz doesn't allow one either.
0/4 answered
116.5 · Spherical triple integral
Evaluate ∭Ex2+y2+z2dV where E is the unit ball.
Answer B
In spherical the integrand is just ρ, and dV=ρ2sinφdρdφdθ:
∫02π∫0π∫01ρ⋅ρ2sinφdρdφdθ=∫ρ341⋅∫sinφ2⋅∫dθ2π=π.
TrapLetting φ run to 2π. φ ∈ [0, π]; only θ goes all the way round.
A lamina of constant density occupies the triangle 0≤x≤1,0≤y≤x. Find xˉ.
Answer C
Mass (density 1): m=∫01∫0xdydx=∫01xdx=21.
Moment: ∫01∫0xxdydx=∫01x2dx=31.
xˉ=1/21/3=32.
Sensible: the triangle carries more area at large x, so the balance point sits right of centre.
TrapForgetting to divide by the mass — that leaves the moment, 1/3, option B.
Find the polar moment of inertia I0=∬D(x2+y2)dA for the unit disk with constant density 1.
Answer B
In polar the integrand is r2 and dA=rdrdθ, so you integrate r3:
I0=∫02π∫01r2⋅rdrdθ=2π[4r4]01=2π.
Because distance is squared, mass far from the axis counts far more than mass near it.
TrapDropping the Jacobian r, which turns r³ into r² and gives 2π/3.
divF=∂x(x)+∂y(−y)=1−1=0.
2-D curl =Qx−Py=∂x(−y)−∂y(x)=0−0=0.
So the field neither creates nor spins: it stretches along x exactly as fast as it
compresses along y. Zero curl on all of R2 (simply connected) also makes it
conservative, with potential f=21x2−21y2.
TrapAssuming arrows that "spread out" must have positive divergence — here the two axes cancel.
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∫Cfds=∫abf(r(t))r′(t)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∫CF⋅dr=∫abF(r(t))⋅r′(t)dt=∫CPdx+Qdy+Rdz
Algorithm · evaluate any line integral
Parametrize C. Line segment from A to B: r(t)=(1−t)A+tB, 0≤t≤1. Circle of radius a: ⟨acost,asint⟩. Graph y=g(x): ⟨t,g(t)⟩.
Compute r′(t). For a scalar integral you also need ∣r′(t)∣.
Substitute the parametrization into F or f so everything is in terms of t alone.
Dot (vector version) or multiply by ∣r′∣ (scalar version).
Integrate over the parameter range. If C is made of several pieces, do each separately and add.
Orientation trap
∫−CF⋅dr=−∫CF⋅dr — reversing the walk flips the sign of work. But ∫−Cfds=∫Cfds — 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 ∫Cfds give when f=1?
Answer
The length of the curve.
You walk C backwards. What happens to ∫CF⋅dr?
Answer
The sign flips. But ∫Cfds does not change — a curtain has the same area either way.
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∫C∇f⋅dr=f(r(b))−f(r(a))=f(end)−f(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
F is conservative ⟺ F=∇f for some potential f ⟺ ∫CF⋅dr is path-independent ⟺ ∮CF⋅dr=0 around every closed loop. On a simply connected region these are also equivalent to curlF=0.
the test2-D: ∂y∂P=∂x∂Q3-D: curlF=∇×F=0
Algorithm · find the potential function f
Test first. Check Py=Qx (2-D) or curlF=0 (3-D). If it fails, stop — there is no f.
Integrate P with respect to x:f=∫Pdx+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 gy.
Integrate gy 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:∇f should give back F exactly.
Then use it: ∫CF⋅dr=f(end)−f(start), and the parametrization never gets written down.
Worked example
F=⟨2xy,x2+2y⟩, from (0,0) to (1,3).
test
Py=2x, Qx=2x ✓ conservative
step 2
f=∫2xydx=x2y+g(y)
step 3
fy=x2+g′(y)=x2+2y⇒g′(y)=2y⇒g=y2
potential
f=x2y+y2
evaluate
f(1,3)−f(0,0)=(3+9)−0=12 — no parametrization needed.
Trap
Curl zero is not enough on a region with a hole. The classic F=⟨x2+y2−y,x2+y2x⟩ passes Py=Qx everywhere it is defined, yet a loop around the origin gives 2π, not 0. The region must be simply connected — no punctures.
Check yourself
What is the 2-D test for a conservative field?
Answer
∂y∂P=∂x∂Q — and the region must have no holes.
F is conservative. What is ∮CF⋅dr around any closed loop?
Answer
0. You end where you started, so the height change is zero.
STEP 02
The questions
3 questions
On those sections only. Untimed, no calculator — the quiz doesn't allow one either.
0/3 answered
117.2 · Line integral · work
Evaluate ∫Cydx+xdy+zdz along the straight segment from (0,1,1) to (2,3,5).
Answer D
r(t)=⟨2t,1+2t,1+4t⟩,0≤t≤1, so dx=2dt,dy=2dt,dz=4dt.
∫01[(1+2t)(2)+(2t)(2)+(1+4t)(4)]dt=∫01(2+4t+4t+4+16t)dt=∫01(6+24t)dt=6+12=18.
TrapMismatching which component multiplies which differential.
F=⟨y+z,x+z,x+y⟩. Find ∫CF⋅dr along any path from (0,0,0) to (1,2,3).
Answer C
∇×F=⟨1−1,1−1,1−1⟩=0 on all of R3, so F is conservative.
Build the potential: f=∫(y+z)dx=xy+xz+g(y,z); matching fy=x+z gives gy=z⇒g=yz.
So f=xy+yz+xz, and ∫CF⋅dr=f(1,2,3)−f(0,0,0)=2+6+3=11.
TrapParametrizing a path. Test for conservative first — it collapses to two evaluations.
Wed Nov 18 · 8:00 pm · Lessons 17–31 · 15.8–17.6 (Lagrange through surface integrals)
Midterm 2
Lessons 17–31, one hour, multiple choice, no calculator. Starts at Lagrange.
Worth knowing. Green's theorem (17.4) and curl / divergence (17.5) are never quizzed — Lesson 29 and 30 fall between Quiz 9 and this exam. The ground rules say Lessons 17–31; the calendar teaches through Lesson 33 before the exam. Assume 17–31 and know 32–33.
Only the sections no quiz ever covers are shown here. Everything from the earlier quizzes is on this exam too — work those tabs as well.
STEP 01
The material
2 sections
§17.4 · §17.5 — and nothing else. Read them, then do the check-yourself questions inside each one with the answer folded away.
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 (Qₓ − Pᵧ) dA — the 2-D curl. Add them all up and you're left holding a single trip around the boundary.
circulation form∮CPdx+Qdy=∬D(∂x∂Q−∂y∂P)dA
flux (divergence) form — same theorem, rotated 90°∮CF⋅nds=∬D(∂x∂P+∂y∂Q)dA=∬DdivFdA
Algorithm · apply Green's theorem
Check the hypotheses.C must be closed, simple (no self-crossings), positively oriented — counterclockwise, region on your left — and F smooth on all of D.
Identify P and Q from ∮Pdx+Qdy.
Compute Qx−Py and simplify. This is usually far smaller than either piece.
Set up ∬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 Qx−Py=1A=∮Cxdy=−∮Cydx=21∮Cxdy−ydx
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 boundaryC=Couter+Cinner.
If Qx−Py=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 2π 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.
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: Qₓ − Pᵧ.
curl — a vector, computed as a determinantcurlF=∇×F=i∂x∂Pj∂y∂Qk∂z∂R
The direction of curl is the axis the paddlewheel spins about (right-hand rule); its length is how fast. In 2-D only the k-component survives, and it is exactly the Qx−Py 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 scalardivF=∇⋅F=∂x∂P+∂y∂Q+∂z∂R
Two identities worth memorising outright
div(curlF)=0 — a spin never creates anything. curl(∇f)=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 div(curlF) makes sense and curl(divF) 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 div(curlF)?
Answer
0, always. A spin never creates anything.
What is curl(∇f)?
Answer
0, always. A hill never makes a whirlpool.
Is divF a scalar or a vector?
Answer
A scalar. So curl(divF) is not even defined — checking types catches half of all errors here.
STEP 02
The questions
4 questions
On those sections only. Untimed, no calculator — the quiz doesn't allow one either.
0/4 answered
117.4 · Green's theorem
C is the circle x2+y2=9 traversed counterclockwise. Find ∮C(y3)dx+(x3+3xy2)dy.
Answer E
P=y3,Q=x3+3xy2.
Py=3y2 and Qx=3x2+3y2, so the scalar curl is Qx−Py=3x2.
Green: ∮C=∬D3x2dA. In polar, x=rcosθ:
∫02π∫033r2cos2θrdrdθ=3⋅434∫02πcos2θdθ=4243⋅π=4243π.
TrapForgetting the r in r dr dθ, or dropping the −P_y term.
F has continuous second partials. Which expression is guaranteed to be 0 (the zero scalar or zero vector)?
Answer B
div(curlF)=0 always — a spin creates nothing.
(The companion identity is curl(∇f)=0.)
Option A is not even defined: divF is a scalar and you cannot take
the curl of a scalar. div(∇f)=∇2f is the Laplacian, generally nonzero.
TrapType errors. curl eats a vector and returns a vector; div eats a vector and returns a scalar.
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 elementru=∂u∂r,rv=∂v∂r,ndS=(ru×rv)dudv,dS=∣ru×rv∣dudv
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
∣ru×rv∣
Graph z=f(x,y)
⟨x,y,f(x,y)⟩
1+fx2+fy2
Sphere radius a
⟨asinφcosθ,asinφsinθ,acosφ⟩
a2sinφ
Cylinder radius a
⟨acosθ,asinθ,z⟩
a
Cone z=x2+y2
⟨rcosθ,rsinθ,r⟩
r2
Surface of revolution
⟨x,f(x)cosθ,f(x)sinθ⟩
f1+(f′)2
Row 1 and row 2 are worth recognising instantly — they reproduce the surface-area formula from 17.6 and the ρ2sinφ from 16.5.
Check yourself
What does (ru×rv)dudv equal?
Answer
ndS — both the direction and the area, from one computation. Never normalise it and then also multiply by dS.
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 RA(S)=∬R1+(fx)2+(fy)2dA
Same shape as arc length, which was ∫1+(dy/dx)2dx — one dimension up. Flat ground gives fx=fy=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 fx and fy, then 1+fx2+fy2 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=∬∣ru×rv∣dA.
Check yourself
What is the stretch factor for the graph z=f(x,y)?
Answer
1+fx2+fy2.
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.
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 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∬SfdS=∬Df(r(u,v))ru×rvdudv
flux of a vector field∬SF⋅dS=∬SF⋅ndS=∬DF⋅(ru×rv)dudv
Algorithm · compute a flux integral
Parametrize S — use the table in 17.6 whenever it fits.
Compute ru×rv. 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 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 ∬D(−Pgx−Qgy+R)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.F⊥n makes F⋅n=0, even though the wind never changed.
Should you normalise ru×rv before dotting with F?
Answer
No. The raw cross product already carries dS inside it.
STEP 02
The questions
4 questions
On those sections only. Untimed, no calculator — the quiz doesn't allow one either.
0/4 answered
117.6 · Surface area
Find the area of the part of the paraboloid z=x2+y2 lying below z=1.
Answer A
zx=2x,zy=2y⇒1+4x2+4y2=1+4r2; the shadow is the unit disk.
A=∫02π∫011+4r2rdrdθ=2π⋅121[(1+4r2)3/2]01=6π(55−1).
TrapDropping the r from dA, which kills the clean u-substitution.
S is the part of the plane z=2x+2y lying above the unit disk x2+y2≤1. Find ∬S1dS (the area of S).
Answer C
zx=2,zy=2⇒dS=1+4+4dA=3dA.
The shadow is the unit disk, of area π, so ∬SdS=3π.
The stretch factor is constant here because the surface is a plane — the tilt never changes.
TrapForgetting the √(1+zₓ²+z_y²) factor and reporting the shadow’s area, π.
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 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∮CF⋅dr=∬S(curlF)⋅dS
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 curlF 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 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 n=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 n=k. Same theorem, tilted.
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∬SF⋅dS=∭EdivFdV
Algorithm · use the divergence theorem
Confirm S is closed — a sealed bag with no openings — and oriented outward.
Compute divF=Px+Qy+Rz. This is usually startlingly simple compared to F itself.
Set up ∭E over the enclosed solid, in whichever coordinates match it (16.4–16.5).
Integrate. Often divF 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 S2 (usually a flat disk) to seal the opening.
Apply the theorem to the sealed union:∬S∪S2=∭EdivFdV.
Compute the cap's flux separately — it's flat, so it's easy.
Subtract:∬S=∭EdivFdV−∬S2.
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.
STEP 02
The questions
7 questions
On those sections only. Untimed, no calculator — the quiz doesn't allow one either.
0/7 answered
117.7 · Stokes · spatial triangle
C is the triangle with vertices (1,0,0),(0,1,0),(0,0,1), oriented counterclockwise viewed from the point (1,1,1). For F=⟨z,x,y⟩, find ∮CF⋅dr.
Answer B
∇×F=⟨1,1,1⟩. The triangle lies in x+y+z=1, whose unit normal
(pointing toward (1,1,1)) is n=31⟨1,1,1⟩.
(∇×F)⋅n=33=3, a constant.
The triangle's area is 23, so ∮=3⋅23=23.
TrapUsing the un-normalised ⟨1,1,1⟩ as n, or the projected area instead of the true area.
S is any upward-oriented smooth surface whose boundary is the circle x2+y2=4 in the plane z=1, oriented counterclockwise seen from above. For F=⟨−y,x,z2⟩, find ∬S(∇×F)⋅ndS.
Answer C
Stokes turns the surface integral into one lap around the rim, so the surface itself is irrelevant.
Parametrize C:r(t)=⟨2cost,2sint,1⟩, r′=⟨−2sint,2cost,0⟩.
F(r(t))=⟨−2sint,2cost,1⟩.
F⋅r′=4sin2t+4cos2t=4, so ∮=4(2π)=8π.
(Equivalently ∇×F=⟨0,0,2⟩ and the flux through the disk is 2⋅π(2)2=8π.)
TrapTrying to build the unknown surface. Stokes says only the rim matters.
C is the boundary of the first-octant part of the plane x+y+z=1, oriented counterclockwise viewed from above. For F=⟨y,z,x⟩, find ∮CF⋅dr.
Answer B
∇×F=⟨−1,−1,−1⟩. The triangle lies in x+y+z=1 with upward unit normal n=31⟨1,1,1⟩.
(∇×F)⋅n=−33=−3, constant.
Triangle area =23, so ∮=−3⋅23=−23.
TrapLosing the sign. A negative circulation is a real answer, not an error.
Find the outward flux of F=⟨x,y,z⟩ across the sphere of radius 2 centred at the origin.
Answer D
divF=1+1+1=3, a constant, so the triple integral is
3×(volume)=3⋅34π(2)3=32π.
Check directly: on the sphere F⋅n=∣r∣=2, and the area is 4π(2)2=16π, giving 2⋅16π=32π ✓.
TrapUsing radius 1, or forgetting that div is constant so no integration is needed.
Find the outward flux of F=⟨x,y,z⟩ through the closed surface of the solid cylinder x2+y2≤1, 0≤z≤2.
Answer D
divF=3, constant, so the flux is 3×volume.
Volume =π(1)2(2)=2π, giving 6π.
Doing it by faces would mean three separate surface integrals — the theorem replaces all of them.
TrapForgetting the two flat caps if you insist on computing faces directly.
S is the paraboloid z=4−x2−y2 with z≥0, oriented upward — it is open at the bottom. Find the flux of F=⟨x,y,z⟩ through S.
Answer D
Cap it with the disk D:z=0,x2+y2≤4, oriented downward so the union is a sealed outward-oriented bag.
Volume: ∫02π∫02(4−r2)rdrdθ=2π[2r2−4r4]02=8π.
divF=3, so the sealed flux is 3(8π)=24π.
On the cap z=0: n=⟨0,0,−1⟩ and F⋅n=−z=0, so the cap contributes nothing.
Therefore the flux through S alone is 24π−0=24π.
TrapApplying the divergence theorem to S by itself. It is not closed — you must cap it, then subtract the cap.