Exam 1 study guide, worked

01

Review of Vectors

Lesson 1 · §13.1-13.4

Don't know this yet? §13.1 · §13.2 · §13.3 · §13.3 · §13.4 in the full guide →

From the official study guide

Vector Notation and Operations

A vector has both magnitude (length) and direction. Vector notation:

\vec{v} = \langle v_1, v_2, v_3 \rangle \quad \text{or} \quad \vec{v} = v_1\vec{i} + v_2\vec{j} + v_3\vec{k}

where $\vec{i} = \langle 1,0,0 \rangle$ , $\vec{j} = \langle 0,1,0 \rangle$ , $\vec{k} = \langle 0,0,1 \rangle$ are the standard basis vectors.

Basic Operations

Addition: $\vec{u} + \vec{v} = \langle u_1+v_1, u_2+v_2, u_3+v_3 \rangle$
Scalar multiplication: $c\vec{u} = \langle cu_1, cu_2, cu_3 \rangle$
Vector from P to Q: $\overrightarrow{PQ} = \langle x_2-x_1, y_2-y_1, z_2-z_1 \rangle$

Magnitude and Unit Vectors

Magnitude (Length) of vector $\vec{v} = \langle v_1, v_2, v_3 \rangle$ :

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

A unit vector is a vector of length 1. The unit vector in the direction of $\vec{v}$ is:

\hat{v} = \frac{\vec{v}}{|\vec{v}|}

Dot Product

Dot Product of $\vec{u} = \langle u_1, u_2, u_3 \rangle$ and $\vec{v} = \langle v_1, v_2, v_3 \rangle$ :

\vec{u} \cdot \vec{v} = u_1v_1 + u_2v_2 + u_3v_3 = |\vec{u}||\vec{v}|\cos\theta

where $\theta$ is the angle between the vectors. Result is a scalar.

$\vec{u} \cdot \vec{v} = 0 \Longleftrightarrow$ vectors are perpendicular (orthogonal)
Angle between vectors: $\cos\theta = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}||\vec{v}|}$

Cross Product

Cross Product of $\vec{u} = \langle u_1, u_2, u_3 \rangle$ and $\vec{v} = \langle v_1, v_2, v_3 \rangle$ :

\vec{u} \times \vec{v} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \end{vmatrix} = \langle u_2v_3-u_3v_2, u_3v_1-u_1v_3, u_1v_2-u_2v_1 \rangle

Result is a vector perpendicular to both $\vec{u}$ and $\vec{v}$ .

$|\vec{u} \times \vec{v}| = |\vec{u}||\vec{v}|\sin\theta$ = area of parallelogram formed by $\vec{u}$ and $\vec{v}$
$\vec{u} \times \vec{v} = \vec{0} \Longleftrightarrow$ vectors are parallel
$\vec{u} \times \vec{v} = -(\vec{v} \times \vec{u})$ (anti-commutative)

Practice

5 problems

1

Find the projection of $\vec{u} = \langle 2, -1, 3 \rangle$ onto $\vec{v} = \langle 1, 2, 2 \rangle$ .

no calculator, same as the exam
Answer $\left\langle \dfrac{2}{3},\, \dfrac{4}{3},\, \dfrac{4}{3}\right\rangle$ answer recomputed · sympy
1. The projection of $\vec u$ onto $\vec v$ is $\operatorname{proj}_{\vec v}\vec u=\dfrac{\vec u\cdot\vec v}{|\vec v|^2}\,\vec v$ — a vector *along* $\vec v$ .
2. $\vec u\cdot\vec v=(2)(1)+(-1)(2)+(3)(2)=2-2+6=6$ .
3. $|\vec v|^2=1+4+4=9$ .
4. $\operatorname{proj}_{\vec v}\vec u=\frac{6}{9}\langle1,2,2\rangle=\frac{2}{3}\langle1,2,2\rangle=\left\langle\frac23,\frac43,\frac43\right\rangle$ .
TrapDividing by $|\vec v|$ instead of $|\vec v|^2$ . That gives the scalar projection $\operatorname{comp}_{\vec v}\vec u=\frac{6}{3}=2$ , which is a number, not a vector. Read which one the question wants.
2

Find the area of the triangle with vertices $P(1,0,0)$ , $Q(0,2,0)$ , and $R(0,0,3)$ .

no calculator, same as the exam
Answer $\dfrac{7}{2}$ answer recomputed · sympy
1. Two edge vectors from $P$ : $\overrightarrow{PQ}=\langle-1,2,0\rangle$ and $\overrightarrow{PR}=\langle-1,0,3\rangle$ .
2. $\overrightarrow{PQ}\times\overrightarrow{PR}=\langle 2\cdot3-0\cdot0,\;0\cdot(-1)-(-1)\cdot3,\;0-(-2)\rangle=\langle6,3,2\rangle$ .
3. $|\langle6,3,2\rangle|=\sqrt{36+9+4}=7$ — that is the parallelogram area.
4. A triangle is half of it: $\text{area}=\frac72$ .
TrapForgetting the $\frac12$ . The cross product's length is the parallelogram, and $7$ is sitting right there looking like an answer.
3

If $\vec{u} = \langle 3, -1, 2 \rangle$ and $\vec{v} = \langle 1, 4, -1 \rangle$ , find $\vec{u} \times \vec{v}$ .

no calculator, same as the exam
Answer $\langle -7,\, 5,\, 13\rangle$ answer recomputed · sympy
1. $\vec u\times\vec v=\begin{vmatrix}\vec i&\vec j&\vec k\\3&-1&2\\1&4&-1\end{vmatrix}$ .
2. $\vec i$ : $(-1)(-1)-(2)(4)=1-8=-7$ .
3. $\vec j$ : $-\big[(3)(-1)-(2)(1)\big]=-(-5)=5$ — note the sign flip on the middle term.
4. $\vec k$ : $(3)(4)-(-1)(1)=12+1=13$ . So $\vec u\times\vec v=\langle-7,5,13\rangle$ .
TrapThe minus on the $\vec j$ component. Dropping it gives $\langle-7,-5,13\rangle$ , and the sanity check catches it instantly: $\vec u\cdot(\vec u\times\vec v)$ must be $0$ .
4

For what value of $c$ are the vectors $\langle 2, c, -1 \rangle$ and $\langle 3, -6, c \rangle$ orthogonal?

no calculator, same as the exam
Answer $c = \dfrac{6}{7}$ answer recomputed · sympy
1. Orthogonal means the dot product vanishes.
2. $\langle2,c,-1\rangle\cdot\langle3,-6,c\rangle=6-6c-c=6-7c$ .
3. $6-7c=0\Rightarrow c=\frac{6}{7}$ .
TrapMissing that $c$ appears in both vectors, so the last term $(-1)(c)$ also carries a $c$ . Treating it as a constant gives $c=1$ .
5

Find the angle between $\vec{u} = \langle 1, 1, 0 \rangle$ and $\vec{v} = \langle 0, 1, 1 \rangle$ .

no calculator, same as the exam
Answer $\theta = \dfrac{\pi}{3}$ answer recomputed · sympy
1. $\cos\theta=\dfrac{\vec u\cdot\vec v}{|\vec u||\vec v|}$ .
2. $\vec u\cdot\vec v=0+1+0=1$ , and $|\vec u|=|\vec v|=\sqrt2$ .
3. $\cos\theta=\frac{1}{2}\Rightarrow\theta=\frac{\pi}{3}$ .
TrapAnswering $\frac{\pi}{6}$ : $\cos^{-1}\!\frac12=\frac{\pi}{3}$ , not $\frac{\pi}{6}$ . $\frac{\pi}{6}$ is where the *sine* is $\frac12$ .

02

Lines and Planes in Space

Lesson 2 · §13.5

Don't know this yet? §13.5 in the full guide →

From the official study guide

Equations of a Line

Line through point $P_0(x_0, y_0, z_0)$ with direction vector $\vec{v} = \langle a, b, c \rangle$ :

Vector Equation

\vec{r}(t) = \vec{r}_0 + t\vec{v} = \langle x_0, y_0, z_0 \rangle + t\langle a, b, c \rangle

Parametric Equations

x = x_0 + at, \quad y = y_0 + bt, \quad z = z_0 + ct

Symmetric Equations (if $a, b, c \neq 0$ )

\frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c}

Equation of a Plane

Plane through point $P_0(x_0, y_0, z_0)$ with normal vector $\vec{n} = \langle a, b, c \rangle$ :

a(x-x_0) + b(y-y_0) + c(z-z_0) = 0

or equivalently: $ax + by + cz = d$ , where $d = ax_0 + by_0 + cz_0$

Finding Normal Vectors

If two direction vectors $\vec{v}_1$ and $\vec{v}_2$ lie in the plane, then $\vec{n} = \vec{v}_1 \times \vec{v}_2$
If given three points, use two vectors formed by the points

Note: Angle between two planes equals the angle between their normal vectors.

Practice

6 problems

1

The line through $(1, 3, -1)$ and $(3, 1, 2)$ intersects the plane $x + y + z = 6$ at what point?

no calculator, same as the exam
Answer $(3,\, 1,\, 2)$ answer recomputed · sympy
1. Direction: $\vec v=\langle3-1,\,1-3,\,2-(-1)\rangle=\langle2,-2,3\rangle$ .
2. Line: $\vec r(t)=\langle 1+2t,\;3-2t,\;-1+3t\rangle$ .
3. Substitute into $x+y+z=6$ : $(1+2t)+(3-2t)+(-1+3t)=3+3t=6\Rightarrow t=1$ .
4. $\vec r(1)=\langle3,1,2\rangle$ — the second given point happens to lie on the plane.
TrapSign errors in the direction vector. Notice the $x$ and $y$ parameters cancel, so the whole equation collapses to $3+3t$ ; if yours does not collapse, recheck $\vec v$ .
2

Find the equation of the plane containing $(2, 1, -1)$ and parallel to the vectors $\langle 1, 0, 3 \rangle$ and $\langle 2, 1, 0 \rangle$ .

no calculator, same as the exam
Answer $-3x + 6y + z + 1 = 0$ answer recomputed · sympy
1. Two vectors parallel to the plane give the normal through a cross product.
2. $\vec n=\langle1,0,3\rangle\times\langle2,1,0\rangle=\langle 0\cdot0-3\cdot1,\;3\cdot2-1\cdot0,\;1\cdot1-0\cdot2\rangle=\langle-3,6,1\rangle$ .
3. Point-normal form at $(2,1,-1)$ : $-3(x-2)+6(y-1)+1(z+1)=0$ .
4. Expand: $-3x+6+6y-6+z+1=0\Rightarrow -3x+6y+z+1=0$ .
Trap'Parallel to the vectors' does not mean the vectors are normals. They lie *in* the plane; their cross product is the normal.
2′restored

Find the distance from the point $(1,-2,3)$ to the plane $2x-y+2z=4$ .

no calculator, same as the exam
Official answer $\dfrac{5}{3}$ checked · official answer wrong

Correct answer $2$ recomputed · sympy
1. $D=\dfrac{|ax_0+by_0+cz_0-d|}{\sqrt{a^2+b^2+c^2}}$ with $\vec n=\langle2,-1,2\rangle$ .
2. Numerator: $|2(1)-(-2)+2(3)-4|=|2+2+6-4|=6$ .
3. Denominator: $\sqrt{4+1+4}=3$ .
4. $D=\dfrac63=2$ .
TrapForgetting to move $d$ to the same side. The formula measures $ax+by+cz-d$ , so a plane written as $2x-y+2z=4$ contributes $-4$ , not $+4$ .

The official answer is wrongThe commented-out source gives $\frac53$ . The numerator is $|2+2+6-4|=6$ , not $5$ , so the distance is $\frac63=2$ . This may be exactly why the problem was commented out.
Commented out on the official page, so it does not appear there. The skill is still on the syllabus, so it is restored here.
3

Find the line of intersection between the planes $x + y = 2$ and $y + z = 3$ .

no calculator, same as the exam
Answer $x = 2+t,\quad y = -t,\quad z = 3+t$ answer recomputed · sympy
1. Two equations, three unknowns — one free parameter, which is the line.
2. Let $y=-t$ . Then $x+y=2$ gives $x=2+t$ , and $y+z=3$ gives $z=3+t$ .
3. $x=2+t,\;y=-t,\;z=3+t$ .
4. Check the direction against $\vec n_1\times\vec n_2=\langle1,1,0\rangle\times\langle0,1,1\rangle=\langle1,-1,1\rangle$ — same direction, as it must be.
TrapAny parametrisation is correct; yours may look nothing like this one. Verify by plugging back into both plane equations rather than comparing to a printed answer.
4

Find the angle between the planes $x + y + z = 1$ and $x - y = 0$ .

no calculator, same as the exam
Answer $\theta = \dfrac{\pi}{2}$ (planes are perpendicular)answer recomputed · sympy
1. The angle between planes is the angle between normals: $\vec n_1=\langle1,1,1\rangle$ , $\vec n_2=\langle1,-1,0\rangle$ .
2. $\vec n_1\cdot\vec n_2=1-1+0=0$ .
3. Zero dot product $\Rightarrow\theta=\frac{\pi}{2}$ : the planes are perpendicular.
TrapReading coefficients off a plane written as $x-y=0$ and forgetting the missing $z$ is a coefficient of $0$ , not of $1$ .
5

For what value of $a$ do the lines $\vec{r}_1(t)=\langle 1+t,\, 2t,\, 3+t\rangle$ and $\vec{r}_2(s)=\langle 3-s,\, s+1,\, 2s+a\rangle$ intersect?

no calculator, same as the exam
Answer $a = 2$ answer recomputed · sympy
1. Intersection needs a single point, so use different parameters $t$ and $s$ and solve all three components.
2. $x$ : $1+t=3-s\Rightarrow t=2-s$ . $y$ : $2t=s+1$ .
3. Substitute: $2(2-s)=s+1\Rightarrow 4-2s=s+1\Rightarrow s=1$ , hence $t=1$ .
4. $z$ : $3+t=2s+a\Rightarrow 3+1=2+a\Rightarrow a=2$ .
TrapUsing $t$ for both lines. Two lines can cross at a point they reach at different times; forcing one parameter finds only the cases where they arrive together, and reports skew for lines that do meet.

03

Quadric Surfaces

Lessons 3-4 · §13.6

Don't know this yet? §13.6 in the full guide →

From the official study guide

A quadric surface is the graph of a second-degree equation in three variables.

Standard Quadric Surfaces
Surface	Standard Equation	Key Features
Ellipsoid	$\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$	All traces are ellipses
Elliptic Paraboloid	$z = \frac{x^2}{a^2} + \frac{y^2}{b^2}$	Opens upward (or downward if negative)
Hyperbolic Paraboloid	$z = \frac{x^2}{a^2} - \frac{y^2}{b^2}$	Saddle shape
Hyperboloid (1 sheet)	$\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2} = 1$	Connected, one piece
Hyperboloid (2 sheets)	$\frac{z^2}{c^2} - \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$	Two separate pieces
Elliptic Cone	$\frac{z^2}{c^2} = \frac{x^2}{a^2} + \frac{y^2}{b^2}$	Double cone, vertex at origin

Identifying Quadric Surfaces

Look at which variables are squared
Check the signs (all positive, mixed signs, or one isolated)
Check what the equation equals (0, 1, or a variable)
Sketch traces in the coordinate planes (xy, xz, yz)

Traces are curves of intersection with planes parallel to coordinate planes.

Practice

5 problems

1

Identify the surface: $\displaystyle -x^2 + \frac{y^2}{4} - \frac{z^2}{9} = 1$ .

no calculator, same as the exam
AnswerHyperboloid of two sheets (axis along $y$ )prose answer · not machine-checkable
1. Write it with the positive term first: $\dfrac{y^2}{4}-x^2-\dfrac{z^2}{9}=1$ .
2. One positive square, two negative, equals $+1$ — that is the two-sheet signature.
3. Trace at $y=0$ : $-x^2-\frac{z^2}{9}=1$ , impossible. Nothing at the origin, so it cannot be one connected sheet.
4. Hyperboloid of two sheets, axis along $y$ — the axis is always the variable whose square is positive.
TrapOne sheet vs two sheets is decided by the count of minus signs, not by which variable is isolated. One minus = one sheet (connected); two minuses = two sheets.
2

Identify the surface defined by $x^2 + 4x + y^2 - 2y + z^2 = -4$ . (Hint: complete the square.)

no calculator, same as the exam
AnswerSphere centered at $(-2,\, 1,\, 0)$ with radius $1$ answer recomputed · sympy
1. Complete the square in $x$ and $y$ : $x^2+4x=(x+2)^2-4$ and $y^2-2y=(y-1)^2-1$ .
2. $(x+2)^2-4+(y-1)^2-1+z^2=-4$ .
3. $(x+2)^2+(y-1)^2+z^2=1$ .
4. Sphere, centre $(-2,1,0)$ , radius $1$ .
TrapMoving the completion constants the wrong way. The $-4$ and $-1$ come out on the left, so they get added to the right: $-4+4+1=1$ .
3

Identify the surface: $z = 4x^2 + 9y^2$ .

no calculator, same as the exam
AnswerElliptic paraboloid opening upwardprose answer · not machine-checkable
1. One variable ( $z$ ) appears to the first power, the other two are squared with the same sign.
2. That is the paraboloid family. Both coefficients positive $\Rightarrow$ opens upward.
3. Traces $z=k>0$ are ellipses $4x^2+9y^2=k$ ; traces $x=0$ and $y=0$ are upward parabolas.
4. Elliptic paraboloid opening upward.
TrapCalling it a cone. A cone has $z^2$ , not $z$ ; the first power is what makes it a paraboloid.
4

Describe the trace of $\displaystyle \frac{x^2}{4} + \frac{y^2}{9} - z^2 = 1$ in the plane $z = 2$ .

no calculator, same as the exam
AnswerEllipse: $\dfrac{x^2}{20} + \dfrac{y^2}{45} = 1$ answer recomputed · sympy
1. Set $z=2$ : $\dfrac{x^2}{4}+\dfrac{y^2}{9}-4=1$ .
2. $\dfrac{x^2}{4}+\dfrac{y^2}{9}=5$ .
3. Divide by $5$ : $\dfrac{x^2}{20}+\dfrac{y^2}{45}=1$ — an ellipse.
TrapStopping at $\frac{x^2}{4}+\frac{y^2}{9}=5$ . A trace is named by its standard form, and the denominators change once you divide through.
5

Identify the surface: $\displaystyle x^2 + \frac{y^2}{4} = \frac{z^2}{9}$ .

no calculator, same as the exam
AnswerElliptic cone with axis along the $z$ -axisprose answer · not machine-checkable
1. Rewrite as $x^2+\dfrac{y^2}{4}-\dfrac{z^2}{9}=0$ — equal to zero, not $1$ .
2. Zero is the cone signature. Traces $z=k\ne0$ are ellipses that grow linearly with $|k|$ ; the trace $z=0$ is the single point at the origin (the vertex).
3. Elliptic cone with axis along the $z$ -axis.
TrapReading it as a hyperboloid. Same sign pattern — the right-hand side is the entire difference: $=1$ one sheet, $=-1$ two sheets, $=0$ the cone between them.

04

Vector-Valued Functions

Lesson 5 · §14.1

Don't know this yet? §14.1 in the full guide →

From the official study guide

Definition

A vector-valued function (or vector function) has the form:

\vec{r}(t) = \langle f(t), g(t), h(t) \rangle = f(t)\vec{i} + g(t)\vec{j} + h(t)\vec{k}

The graph is a space curve traced out as t varies.

Limits and Continuity

The limit of a vector function is computed component-wise:

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

Practice

5 problems

1

Find the domain on which $\vec{r}(t) = \left\langle \sqrt{t-1},\, \ln(4-t),\, e^t \right\rangle$ is continuous.

no calculator, same as the exam
Answer $[1,\, 4)$ prose answer · not machine-checkable
1. A vector function is continuous exactly where every component is.
2. $\sqrt{t-1}$ needs $t\ge1$ .
3. $\ln(4-t)$ needs $4-t>0$ , i.e. $t<4$ .
4. $e^t$ is continuous everywhere. Intersect: $[1,4)$ .
TrapThe bracket types. $\sqrt{\;}$ allows equality (closed at $1$ ); $\ln$ never does (open at $4$ ).
2

Evaluate $\displaystyle \lim_{t \to 0} \left\langle \frac{\sin(3t)}{t},\, e^{2t},\, \frac{t^2+1}{t+1} \right\rangle$ .

no calculator, same as the exam
Answer $\langle 3,\, 1,\, 1\rangle$ answer recomputed · sympy
1. Componentwise.
2. $\lim_{t\to0}\frac{\sin 3t}{t}=3$ (the $\frac{\sin u}{u}$ limit with $u=3t$ ).
3. $\lim_{t\to0}e^{2t}=1$ .
4. $\lim_{t\to0}\frac{t^2+1}{t+1}=1$ . So the limit is $\langle3,1,1\rangle$ .
TrapWriting $1$ for the first component out of reflex. $\frac{\sin 3t}{t}=3\cdot\frac{\sin 3t}{3t}\to3$ .
3

Find the domain on which $\vec{r}(t) = \left\langle \frac{1}{t^2-4},\, \sqrt{t},\, \cos(t) \right\rangle$ is continuous.

no calculator, same as the exam
Answer $[0,2)\cup(2,\infty)$ prose answer · not machine-checkable
1. $\frac{1}{t^2-4}$ needs $t\ne\pm2$ .
2. $\sqrt t$ needs $t\ge0$ — which already kills $t=-2$ .
3. $\cos t$ is everywhere continuous.
4. $[0,\infty)$ with $t=2$ removed: $[0,2)\cup(2,\infty)$ .
TrapReporting both $\pm2$ as holes. $t=-2$ is not a hole, it is outside the domain the square root already imposed.
4

Describe the curve traced by $\vec{r}(t) = \langle 2\cos t,\, 2\sin t,\, 0 \rangle$ for $0 \le t \le 2\pi$ .

no calculator, same as the exam
AnswerCircle of radius 2 in the $xy$ -plane centered at the originprose answer · not machine-checkable
1. $x=2\cos t$ , $y=2\sin t$ gives $x^2+y^2=4$ , and $z=0$ throughout.
2. As $t$ runs $0\to2\pi$ the point goes once around, counterclockwise from $(2,0)$ .
3. A circle of radius 2 in the $xy$ -plane, centred at the origin.
TrapCalling it a helix. A helix needs $z$ to move; here $z\equiv0$ .
5

Evaluate $\displaystyle \lim_{t \to 0}\left\langle \frac{e^t - 1}{t},\, \frac{1-\cos t}{t^2},\, t\sin\!\left(\frac{1}{t}\right) \right\rangle$ .

no calculator, same as the exam
Answer $\left\langle 1,\, \dfrac{1}{2},\, 0\right\rangle$ answer recomputed · sympy
1. $\lim_{t\to0}\frac{e^t-1}{t}=1$ — this is $\frac{d}{dt}e^t$ at $0$ .
2. $\lim_{t\to0}\frac{1-\cos t}{t^2}=\frac12$ (two applications of l'Hôpital, or the series $1-\cos t\approx\frac{t^2}{2}$ ).
3. $\left|t\sin\frac1t\right|\le|t|\to0$ , so the third component is $0$ by the squeeze.
4. $\left\langle1,\frac12,0\right\rangle$ .
TrapDeclaring the third component undefined because $\sin\frac1t$ oscillates. Bounded times something going to zero still goes to zero.

05

Calculus of Vector-Valued Functions & Motion in Space

Lessons 6-7 · §14.2-14.3

Don't know this yet? §14.2 · §14.3 in the full guide →

From the official study guide

Derivative

\vec{r}'(t) = \lim_{h \to 0} \frac{\vec{r}(t+h) - \vec{r}(t)}{h} = \langle f'(t), g'(t), h'(t) \rangle

Integral

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

Differentiation Rules

For vector functions $\vec{u}(t)$ and $\vec{v}(t)$ :

$\frac{d}{dt}[\vec{u}(t) + \vec{v}(t)] = \vec{u}'(t) + \vec{v}'(t)$
$\frac{d}{dt}[c\vec{u}(t)] = c\vec{u}'(t)$
$\frac{d}{dt}[f(t)\vec{u}(t)] = f'(t)\vec{u}(t) + f(t)\vec{u}'(t)$
$\frac{d}{dt}[\vec{u}(t) \cdot \vec{v}(t)] = \vec{u}'(t) \cdot \vec{v}(t) + \vec{u}(t) \cdot \vec{v}'(t)$
$\frac{d}{dt}[\vec{u}(t) \times \vec{v}(t)] = \vec{u}'(t) \times \vec{v}(t) + \vec{u}(t) \times \vec{v}'(t)$

Motion in Space

If $\vec{r}(t)$ is the position vector at time t:

Velocity: $\vec{v}(t) = \vec{r}'(t)$
Speed: $|\vec{v}(t)| = |\vec{r}'(t)|$
Acceleration: $\vec{a}(t) = \vec{v}'(t) = \vec{r}''(t)$

Unit Tangent Vector

\vec{T}(t) = \frac{\vec{r}'(t)}{|\vec{r}'(t)|} = \frac{\vec{v}(t)}{|\vec{v}(t)|}

Practice

6 problems

1

A particle has velocity $\vec{v}(t) = \langle 2t,\, 3,\, -1 \rangle$ and at $t=0$ passes through $(1,-2,5)$ . Find $\vec{r}(t)$ .

no calculator, same as the exam
Answer $\vec{r}(t) = \langle t^2+1,\, 3t-2,\, -t+5\rangle$ answer recomputed · sympy
1. Position is the antiderivative of velocity: $\vec r(t)=\int\vec v(t)\,dt+\vec C$ .
2. $\int\langle2t,3,-1\rangle dt=\langle t^2,\,3t,\,-t\rangle+\vec C$ .
3. $\vec r(0)=\vec C=\langle1,-2,5\rangle$ .
4. $\vec r(t)=\langle t^2+1,\;3t-2,\;-t+5\rangle$ .
TrapForgetting that $\vec C$ is a vector of three separate constants, not one scalar.
2

The tangent line to $\vec{r}(t) = \langle t^2,\, 2t+1,\, t^3 \rangle$ at $t=1$ intersects the $xz$ -plane. Find the point of intersection.

no calculator, same as the exam
Answer $\left(-2,\, 0,\, -\dfrac{7}{2}\right)$ answer recomputed · sympy
1. $\vec r(1)=\langle1,3,1\rangle$ and $\vec r\,{}'(t)=\langle2t,2,3t^2\rangle$ , so $\vec r\,{}'(1)=\langle2,2,3\rangle$ .
2. Tangent line: $\langle1+2u,\;3+2u,\;1+3u\rangle$ .
3. The $xz$ -plane is $y=0$ : $3+2u=0\Rightarrow u=-\frac32$ .
4. $x=1+2(-\frac32)=-2$ , $z=1+3(-\frac32)=-\frac72$ . Point: $\left(-2,0,-\frac72\right)$ .
TrapUsing $t$ again for the tangent line. The line has its own parameter; reusing $t$ makes you solve on the curve instead of on its tangent.
3

Find the unit tangent vector $\vec{T}(0)$ for $\vec{r}(t) = \langle \cos t,\, \sin t,\, t \rangle$ .

no calculator, same as the exam
Answer $\vec{T}(0) = \left\langle 0,\, \dfrac{1}{\sqrt{2}},\, \dfrac{1}{\sqrt{2}}\right\rangle$ answer recomputed · sympy
1. $\vec r\,{}'(t)=\langle-\sin t,\cos t,1\rangle$ , so $\vec r\,{}'(0)=\langle0,1,1\rangle$ .
2. $|\vec r\,{}'(0)|=\sqrt{0+1+1}=\sqrt2$ .
3. $\vec T(0)=\left\langle0,\frac{1}{\sqrt2},\frac{1}{\sqrt2}\right\rangle$ .
TrapSkipping the normalisation. $\vec T$ is a unit vector by definition; $\vec r\,{}'$ alone is not one here.
4

Find the speed of a particle with position $\vec{r}(t) = \langle e^t,\, e^{-t},\, \sqrt{2}\,t \rangle$ at $t = 0$ .

no calculator, same as the exam
AnswerSpeed $= 2$ answer recomputed · sympy
1. $\vec r\,{}'(t)=\langle e^t,\,-e^{-t},\,\sqrt2\rangle$ ; at $t=0$ this is $\langle1,-1,\sqrt2\rangle$ .
2. Speed $=\sqrt{1+1+2}=\sqrt4=2$ .
TrapDifferentiating $e^{-t}$ to $e^{-t}$ . The chain rule puts a minus in front, though here it squares away — the sign matters for direction, not for speed.
5

If $\vec{r}(t) = \langle t,\, t^2,\, t^3 \rangle$ , compute $\vec{r}'(t) \times \vec{r}''(t)$ .

no calculator, same as the exam
Answer $\langle 6t^2,\, -6t,\, 2\rangle$ answer recomputed · sympy
1. $\vec r\,{}'=\langle1,2t,3t^2\rangle$ and $\vec r\,{}''=\langle0,2,6t\rangle$ .
2. $\vec i$ : $(2t)(6t)-(3t^2)(2)=12t^2-6t^2=6t^2$ .
3. $\vec j$ : $-\big[(1)(6t)-(3t^2)(0)\big]=-6t$ .
4. $\vec k$ : $(1)(2)-(2t)(0)=2$ . So $\vec r\,{}'\times\vec r\,{}''=\langle6t^2,-6t,2\rangle$ .
TrapThis exact vector is the numerator of the curvature formula $\kappa=\frac{|\vec r\,{}'\times\vec r\,{}''|}{|\vec r\,{}'|^3}$ — recognise it and Lesson 8 gets shorter.
6

A particle travels along $\vec{r}(t) = \langle 3\cos t,\, 4\sin t,\, 0 \rangle$ . Find the speed at $t = \frac{\pi}{2}$ .

no calculator, same as the exam
AnswerSpeed $= 3$ answer recomputed · sympy
1. $\vec r\,{}'(t)=\langle-3\sin t,\;4\cos t,\;0\rangle$ .
2. At $t=\frac{\pi}{2}$ : $\langle-3,0,0\rangle$ .
3. Speed $=3$ .
TrapAssuming an ellipse is traversed at constant speed. It is not — the speed here is $\sqrt{9\sin^2t+16\cos^2t}$ , which runs between $3$ and $4$ . Only the circle has constant speed under this parametrisation.

06

Arc Length and Curvature

Lesson 8 · §14.4-14.5

Don't know this yet? §14.4 in the full guide →

From the official study guide

Arc Length

Arc Length of curve $\vec{r}(t)$ from $t = a$ to $t = b$ :

L = \int_a^b |\vec{r}'(t)|\,dt = \int_a^b \sqrt{[f'(t)]^2 + [g'(t)]^2 + [h'(t)]^2}\,dt

Arc Length Function

s(t) = \int_{t_0}^t |\vec{r}'(u)|\,du

Curvature

Curvature measures how quickly the curve changes direction:

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

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

\kappa(x) = \frac{|f''(x)|}{[1+(f'(x))^2]^{3/2}}

Practice

5 problems

1

Find the arc length of $\vec{r}(t) = \langle 3\cos t,\, 3\sin t,\, 4t \rangle$ for $0 \le t \le 2\pi$ .

no calculator, same as the exam
Answer $L = 10\pi$ answer recomputed · sympy
1. $\vec r\,{}'(t)=\langle-3\sin t,\;3\cos t,\;4\rangle$ .
2. $|\vec r\,{}'|=\sqrt{9\sin^2t+9\cos^2t+16}=\sqrt{9+16}=5$ — constant, the helix is traversed at constant speed.
3. $L=\int_0^{2\pi}5\,dt=10\pi$ .
TrapLosing the Pythagorean identity. $9\sin^2+9\cos^2=9$ ; if your integrand still contains $t$ , the collapse was missed.
2

Find the curvature $\kappa(t)$ of $\vec{r}(t) = \langle 2\cos t,\, 2\sin t,\, t \rangle$ .

no calculator, same as the exam
Answer $\kappa = \dfrac{2}{5}$ answer recomputed · sympy
1. For the helix $\langle a\cos t,a\sin t,bt\rangle$ , $\kappa=\dfrac{a}{a^2+b^2}$ — constant.
2. Here $a=2$ , $b=1$ : $\kappa=\dfrac{2}{4+1}=\dfrac25$ .
3. By the cross-product formula: $\vec r\,{}'\times\vec r\,{}''=\langle2\sin t,-2\cos t,4\rangle$ , whose length is $\sqrt{4+16}=2\sqrt5$ , and $|\vec r\,{}'|^3=(\sqrt5)^3=5\sqrt5$ . Ratio $=\frac{2\sqrt5}{5\sqrt5}=\frac25$ .
TrapUsing $\kappa=\frac1a$ — that is the circle of radius $a$ . Climbing in $z$ straightens the curve, so the helix is always less curved than its shadow.
3

Find the curvature of the plane curve $y = x^3$ at the point $(0,0)$ .

no calculator, same as the exam
Answer $\kappa = 0$ answer recomputed · sympy
1. For a graph, $\kappa(x)=\dfrac{|f''(x)|}{\left[1+(f'(x))^2\right]^{3/2}}$ .
2. $f'=3x^2$ , $f''=6x$ ; at $x=0$ both the numerator and $f'$ vanish.
3. $\kappa(0)=\dfrac{0}{1^{3/2}}=0$ .
TrapAssuming a curve that visibly bends must have $\kappa>0$ somewhere along it. $y=x^3$ is *flat to second order* at the origin — it is an inflection point, and curvature is exactly zero there.
4

Find the arc length of $\vec{r}(t) = \langle 3t,\, 4t,\, 0 \rangle$ for $1 \le t \le 5$ .

no calculator, same as the exam
Answer $L = 20$ answer recomputed · sympy
1. $\vec r\,{}'=\langle3,4,0\rangle$ , a constant vector: this is a straight line.
2. $|\vec r\,{}'|=5$ .
3. $L=\int_1^5 5\,dt=5\cdot4=20$ .
TrapIntegrating from $0$ to $5$ . The interval is $[1,5]$ , which has length $4$ , not $5$ .
5

A particle moves along $\vec{r}(t) = \langle 5\sin t,\, 3\sin t,\, 4\cos t \rangle$ . Find the distance traveled from $t=0$ to $t=\pi$ .

no calculator, same as the exam
Answer $L = \displaystyle\int_0^{\pi}\sqrt{16+18\cos^2 t}\,dt$ answer recomputed · sympy
1. $\vec r\,{}'(t)=\langle5\cos t,\;3\cos t,\;-4\sin t\rangle$ .
2. $|\vec r\,{}'|^2=25\cos^2t+9\cos^2t+16\sin^2t=34\cos^2t+16\sin^2t$ .
3. Write $\sin^2t=1-\cos^2t$ : $=16+18\cos^2t$ .
4. $L=\displaystyle\int_0^{\pi}\sqrt{16+18\cos^2t}\,dt$ — an elliptic integral, which has no elementary antiderivative. Setting it up correctly *is* the answer.
TrapHunting for a closed form. Most arc lengths do not have one; the exam asks you to set it up, so stop when the integral is right.

07

Functions of Several Variables

Lesson 9 · §15.1

Don't know this yet? §15.1 in the full guide →

From the official study guide

Definition

A function of two variables assigns to each ordered pair (x, y) in a set D a unique real number $z = f(x,y)$ .

Visualization Methods

Graph: Surface in 3D space
Level Curves: Curves in xy-plane where $f(x,y) = k$ (constant)
Contour Map: Collection of level curves

Functions of Three Variables

For $w = f(x,y,z)$ :

Level Surfaces: Surfaces where $f(x,y,z) = k$

Practice

5 problems

1

Find the domain of $f(x, y) = \sqrt{9 - x^2 - y^2}$ .

no calculator, same as the exam
Answer $\{(x,y):\, x^2+y^2 \le 9\}$ prose answer · not machine-checkable
1. A square root needs a non-negative radicand: $9-x^2-y^2\ge0$ .
2. $x^2+y^2\le9$ .
3. The closed disk of radius $3$ centred at the origin, boundary included.
TrapWriting a strict inequality. The boundary circle is fine — it just makes $f=0$ there. Strictness belongs to logs and denominators.
2

Describe the nonempty level curves of $f(x, y) = x^2 + 4y^2$ .

no calculator, same as the exam
AnswerEllipses $\dfrac{x^2}{k}+\dfrac{y^2}{k/4}=1$ centered at origin ( $k>0$ )prose answer · not machine-checkable
1. Set $x^2+4y^2=k$ . For $k<0$ there are no points; $k=0$ is the single point at the origin.
2. For $k>0$ , divide: $\dfrac{x^2}{k}+\dfrac{y^2}{k/4}=1$ .
3. Ellipses centred at the origin, twice as wide as they are tall (semi-axes $\sqrt k$ and $\sqrt k/2$ ).
TrapCalling them circles. The $4$ squashes the $y$ -direction; equal denominators would be needed for circles.
3

Describe the level curves of $f(x, y) = y - x^2$ .

no calculator, same as the exam
AnswerParabolas $y = x^2 + k$ prose answer · not machine-checkable
1. $y-x^2=k\Rightarrow y=x^2+k$ .
2. Parabolas $y=x^2$ , shifted vertically by $k$ — one for every real $k$ , and they never intersect.
TrapExpecting closed curves. Level curves of a saddle-ish surface are unbounded; only bowl-shaped surfaces give loops.
4

Find the domain of $f(x, y) = \ln(x + y - 1)$ .

no calculator, same as the exam
Answer $\{(x,y):\, x+y > 1\}$ prose answer · not machine-checkable
1. $\ln$ requires a strictly positive argument: $x+y-1>0$ .
2. $x+y>1$ : the open half-plane strictly above the line $y=1-x$ , the line itself excluded.
TrapIncluding the line. At $x+y=1$ the log is $\ln 0$ , undefined — this is the one place the boundary is genuinely out.
5

Describe the level surfaces of $f(x, y, z) = x^2 + y^2 + z^2$ .

no calculator, same as the exam
AnswerSpheres of radius $\sqrt{k}$ centered at origin ( $k>0$ )prose answer · not machine-checkable
1. $x^2+y^2+z^2=k$ with $k>0$ is a sphere of radius $\sqrt k$ .
2. $k=0$ collapses to the origin; $k<0$ is empty.
3. Concentric spheres centred at the origin.
TrapA level *surface* of a three-variable function, not a level curve. One dimension up: $f(x,y,z)=k$ carves a surface out of space.

08

Limits and Continuity

Lesson 10 · §15.2

Don't know this yet? §15.2 in the full guide →

From the official study guide

Limits

We write $\displaystyle\lim_{(x,y) \to (a,b)} f(x,y) = L$ , if f(x,y) approaches L as (x,y) approaches (a,b) from any direction.

Proving a Limit Does Not Exist

Find two different paths that give different limits:

Along x-axis: $y = 0$
Along y-axis: $x = 0$
Along $y = x$ or $y = mx$
Along parabolic paths: $y = x^2$

Continuity

f is continuous at (a,b) if:

\lim_{(x,y) \to (a,b)} f(x,y) = f(a,b)

Practice

5 problems

1

Evaluate $\displaystyle\lim_{(x,y) \to (0,0)} \frac{x^3 y}{x^4 + y^2}$ , or show it does not exist.

no calculator, same as the exam
Official answerlimit does not existchecked · official answer wrong

Correct answer $0$ recomputed · sympy
1. Try paths first. Along $y=0$ : $0$ . Along $x=0$ : $0$ . Along $y=x^2$ : $\frac{x^3\cdot x^2}{x^4+x^4}=\frac{x}{2}\to0$ . Every path gives $0$ , which is a signal to stop hunting and start proving.
2. Squeeze. By AM–GM, $x^4+y^2\ge 2x^2|y|$ , so for $(x,y)\ne(0,0)$ :
3. $\left|\frac{x^3y}{x^4+y^2}\right|\le\frac{|x|^3|y|}{2x^2|y|}=\frac{|x|}{2}.$
4. $\frac{|x|}{2}\to0$ , so the limit exists and equals $0$ .
TrapA finite pile of paths never proves a limit exists — but it also never proves it fails. When every path you try agrees, switch to a bound.

The official answer is wrongThe official answer says the limit does not exist. It does: the AM–GM squeeze above bounds the expression by $\frac{|x|}{2}$ . (Compare $\frac{x^2y}{x^4+y^2}$ , which genuinely fails along $y=x^2$ — one power of $x$ away.)
2

Evaluate $\displaystyle\lim_{(x,y) \to (0,0)} \frac{3x^2 y}{x^2 + y^2}$ , or show it does not exist.

no calculator, same as the exam
Official answerlimit does not existchecked · official answer wrong

Correct answer $0$ recomputed · sympy
1. Every path gives $0$ : along $y=mx$ , $\frac{3x^2(mx)}{x^2(1+m^2)}=\frac{3mx}{1+m^2}\to0$ .
2. Squeeze, and this one is immediate: $x^2\le x^2+y^2$ , so $\dfrac{x^2}{x^2+y^2}\le1$ and
3. $\left|\frac{3x^2y}{x^2+y^2}\right|\le 3|y|.$
4. $3|y|\to0$ as $(x,y)\to(0,0)$ , so the limit exists and equals $0$ .
TrapDegree counting is the fast check: the numerator is degree $3$ , the denominator degree $2$ . Numerator degree strictly larger $\Rightarrow$ the quotient is squeezed to $0$ . Only equal degrees can produce a path-dependent limit.

The official answer is wrongThe official answer says the limit does not exist. It does: it is $0$ , by the one-line bound $\left|\frac{3x^2y}{x^2+y^2}\right|\le3|y|$ . The classic DNE example is $\frac{xy}{x^2+y^2}$ — numerator and denominator both degree $2$ .
3

Find the value of $k$ such that $\displaystyle\lim_{(x,y) \to (0,0)} \frac{x^4 - 2k(x^2+y^2) - y^4}{x^2+y^2} = 6$ .

no calculator, same as the exam
Answer $k = -3$ answer recomputed · sympy
1. Split it: $\dfrac{x^4-y^4}{x^2+y^2}-\dfrac{2k(x^2+y^2)}{x^2+y^2}$ .
2. First piece factors: $\dfrac{(x^2-y^2)(x^2+y^2)}{x^2+y^2}=x^2-y^2\to0$ .
3. Second piece is exactly $2k$ , for every $(x,y)\ne(0,0)$ .
4. Limit $=-2k$ . Set $-2k=6\Rightarrow k=-3$ .
TrapLosing the sign. The term enters as $-2k(x^2+y^2)$ , so the limit is $-2k$ , and $k=3$ would give $-6$ .
4

Evaluate $\displaystyle\lim_{(x,y) \to (0,0)} \frac{\sin(x^2 + y^2)}{x^2 + y^2}$ .

no calculator, same as the exam
Answer $1$ answer recomputed · sympy
1. Substitute $u=x^2+y^2$ . As $(x,y)\to(0,0)$ , $u\to0^+$ .
2. The expression is $\dfrac{\sin u}{u}$ .
3. $\lim_{u\to0}\frac{\sin u}{u}=1$ .
TrapThe substitution is legitimate here precisely because $x$ and $y$ only ever appear in the combination $x^2+y^2$ . It is not legitimate in general.
5

Evaluate $\displaystyle\lim_{(x,y) \to (0,0)} \frac{x^3 - y^3}{x^2 + y^2}$ , or show it does not exist.

no calculator, same as the exam
Answer $0$ answer recomputed · sympy
1. Polar coordinates: $x=r\cos\theta$ , $y=r\sin\theta$ , and $(x,y)\to(0,0)$ is exactly $r\to0^+$ .
2. $\dfrac{r^3(\cos^3\theta-\sin^3\theta)}{r^2}=r(\cos^3\theta-\sin^3\theta)$ .
3. $|\cos^3\theta-\sin^3\theta|\le2$ , so the whole thing is $\le 2r\to0$ uniformly in $\theta$ .
4. The limit is $0$ .
TrapPolar only settles it when the $\theta$ factor is bounded and the $r$ factor goes to zero on its own. If $\theta$ survives in the limit, polar has told you it does not exist.

09

Partial Derivatives

Lesson 11 · §15.3

Don't know this yet? §15.3 in the full guide →

From the official study guide

First-Order Partial Derivatives

Partial derivative with respect to x (treat y as constant):

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

Partial derivative with respect to y (treat x as constant):

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

Higher-Order Partial Derivatives

$f_{xx} = \frac{\partial^2 f}{\partial x^2}$
$f_{yy} = \frac{\partial^2 f}{\partial y^2}$
$f_{xy} = \frac{\partial^2 f}{\partial y \partial x}$
$f_{yx} = \frac{\partial^2 f}{\partial x \partial y}$

Clairaut's Theorem

If $f_{xy}$ and $f_{yx}$ are both continuous:

f_{xy} = f_{yx}

Practice

5 problems

1

Let $f(x,y,z)=\cos(xy)+e^{y^2z}+\ln(z^3)$ . Compute $f_{yxz}(0,1,1)$ .

no calculator, same as the exam
Answer $f_{yxz}(0,1,1) = 0$ answer recomputed · sympy
1. Read the order right: $f_{yxz}$ means differentiate by $y$ , then $x$ , then $z$ .
2. $f_y=-x\sin(xy)+2yz\,e^{y^2z}$ .
3. $f_{yx}=-\sin(xy)-xy\cos(xy)$ — the $e^{y^2z}$ term has no $x$ and dies here.
4. $f_{yxz}=0$ : what is left has no $z$ in it at all. Evaluating at $(0,1,1)$ is not even needed.
TrapGrinding through the $\ln(z^3)$ term. It has no $x$ and no $y$ , so it is annihilated by the very first derivative.
2

If $f(x, y) = x^3 y^2 + \sin(xy)$ , find $f_{xy}(0, 0)$ .

no calculator, same as the exam
Answer $f_{xy}(0,0) = 1$ answer recomputed · sympy
1. $f_x=3x^2y^2+y\cos(xy)$ .
2. $f_{xy}=6x^2y+\cos(xy)-xy\sin(xy)$ (product rule on $y\cos(xy)$ ).
3. At $(0,0)$ : $0+\cos0-0=1$ .
TrapEvaluating at $(0,0)$ before differentiating. $f(x,0)=0$ for all $x$ , so that route reports $0$ — you must differentiate symbolically first, then substitute.
3

If $f(x,y) = e^{x^2 y}$ , find $f_x(1, 2)$ .

no calculator, same as the exam
Answer $f_x(1,2) = 4e^2$ answer recomputed · sympy
1. Hold $y$ constant: $f_x=e^{x^2y}\cdot 2xy$ .
2. At $(1,2)$ : $e^{2}\cdot2\cdot1\cdot2=4e^2$ .
TrapDifferentiating the exponent as if $y$ were a variable too. In $f_x$ , $y$ is a constant — that is the entire content of 'partial'.
4

Let $f(x,y)=x^2\ln(y)$ . Find $f_{xy}$ and verify Clairaut's theorem by also computing $f_{yx}$ .

no calculator, same as the exam
Answer $f_{xy} = f_{yx} = \dfrac{2x}{y}$ answer recomputed · sympy
1. $f_x=2x\ln y$ , so $f_{xy}=\dfrac{2x}{y}$ .
2. The other order: $f_y=\dfrac{x^2}{y}$ , so $f_{yx}=\dfrac{2x}{y}$ .
3. They agree — as Clairaut guarantees, since both mixed partials are continuous on $y>0$ .
TrapBelieving Clairaut always applies. It needs continuity of the mixed partials; the standard counterexample $\frac{xy(x^2-y^2)}{x^2+y^2}$ has $f_{xy}(0,0)\ne f_{yx}(0,0)$ .
5

If $f(x,y) = \arctan\!\left(\frac{y}{x}\right)$ , find $f_x + f_y$ at $(1, 1)$ .

no calculator, same as the exam
Answer $f_x + f_y = 0$ at $(1,1)$ answer recomputed · sympy
1. For $f=\arctan\frac{y}{x}$ : $f_x=\dfrac{-y}{x^2+y^2}$ and $f_y=\dfrac{x}{x^2+y^2}$ .
2. At $(1,1)$ : $f_x=-\frac12$ , $f_y=\frac12$ .
3. $f_x+f_y=0$ .
TrapDropping the inner derivative $\frac{\partial}{\partial x}\frac{y}{x}=-\frac{y}{x^2}$ . The $\frac{1}{1+(y/x)^2}$ factor then has to be cleared of its fractions to reach $x^2+y^2$ .

10

The Chain Rule

Lesson 12 · §15.4

Don't know this yet? §15.4 in the full guide →

From the official study guide

Chain Rule - Case 1

If $z = f(x,y)$ where $x = g(t)$ and $y = h(t)$ :

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

Chain Rule - Case 2

If $z = f(x,y)$ where $x = g(s,t)$ and $y = h(s,t)$ :

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

Implicit Differentiation

If $F(x,y,z) = 0$ defines z implicitly:

\frac{\partial z}{\partial x} = -\frac{F_x}{F_z}, \quad \frac{\partial z}{\partial y} = -\frac{F_y}{F_z}

Practice

6 problems

1

If $z = x^2 y$ , where $x = t^2$ and $y = t^3$ , find $\frac{dz}{dt}$ at $t = 1$ .

no calculator, same as the exam
Answer $\left.\dfrac{dz}{dt}\right|_{t=1} = 7$ answer recomputed · sympy
1. Chain rule: $\frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}=2xy(2t)+x^2(3t^2)$ .
2. At $t=1$ : $x=1$ , $y=1$ , so $=2(1)(1)(2)+1(3)=4+3=7$ .
3. Sanity check by substituting first: $z=(t^2)^2t^3=t^7$ , $\frac{dz}{dt}=7t^6=7$ at $t=1$ . Same.
4. $\dfrac{dz}{dt}\Big|_{t=1}=7$ .
TrapSubstituting first is fine here and often faster — but the exam wants the tree diagram, and for a general $f$ you cannot substitute at all.
2

Suppose $xz^2 + y^2z = 2$ defines $z$ implicitly as a function of $x$ and $y$ . Find $\frac{\partial z}{\partial x}$ at the point $(1, 1, 1)$ .

no calculator, same as the exam
Answer $\dfrac{\partial z}{\partial x} = -\dfrac{1}{3}$ answer recomputed · sympy
1. Write $F=xz^2+y^2z-2$ , so the surface is $F=0$ .
2. $F_x=z^2$ and $F_z=2xz+y^2$ .
3. $\dfrac{\partial z}{\partial x}=-\dfrac{F_x}{F_z}=-\dfrac{z^2}{2xz+y^2}$ .
4. At $(1,1,1)$ : $-\dfrac{1}{2+1}=-\dfrac13$ .
TrapThe minus sign in $-F_x/F_z$ . It is not optional and it is the single most-dropped symbol in 15.4.
3

If $w=xy+yz$ , $x=s+t$ , $y=st$ , $z=s-t$ , find $\frac{\partial w}{\partial s}$ at $s=1,\, t=0$ .

no calculator, same as the exam
Answer $\left.\dfrac{\partial w}{\partial s}\right|_{s=1,t=0} = 0$ answer recomputed · sympy
1. $\frac{\partial w}{\partial s}=w_x x_s+w_y y_s+w_z z_s=(y)(1)+(x+z)(t)+(y)(1)$ .
2. At $s=1,t=0$ : $x=1$ , $y=0$ , $z=1$ , so $=0+(2)(0)+0=0$ .
3. Or substitute first: $w=st(s+t)+st(s-t)=2s^2t$ , $w_s=4st=0$ at $(1,0)$ .
4. $\dfrac{\partial w}{\partial s}=0$ .
TrapMissing that $y$ multiplies two branches ( $w=xy+yz$ feeds $y$ into both terms), so $y_s$ appears once with coefficient $x+z$ .
4

If $\sin(xy) = e^{2z}$ , use implicit differentiation to find $\frac{\partial z}{\partial x}$ at $\left(1, \frac{\pi}{2}, 0\right)$ .

no calculator, same as the exam
Answer $\dfrac{\partial z}{\partial x} = 0$ answer recomputed · sympy
1. $F=\sin(xy)-e^{2z}$ , so $F_x=y\cos(xy)$ and $F_z=-2e^{2z}$ .
2. At $\left(1,\frac{\pi}{2},0\right)$ : $F_x=\frac{\pi}{2}\cos\frac{\pi}{2}=0$ and $F_z=-2$ .
3. $\dfrac{\partial z}{\partial x}=-\dfrac{0}{-2}=0$ .
Trap $\cos\frac{\pi}{2}=0$ is what kills this one. Check the point before doing any algebra — sometimes it collapses the whole expression.
5

If $z = f(x,y)$ with $x = r\cos\theta$ and $y = r\sin\theta$ , express $\frac{\partial z}{\partial r}$ in terms of $f_x$ and $f_y$ .

no calculator, same as the exam
Answer $\dfrac{\partial z}{\partial r} = f_x\cos\theta + f_y\sin\theta$ prose answer · not machine-checkable
1. $\frac{\partial z}{\partial r}=f_x\frac{\partial x}{\partial r}+f_y\frac{\partial y}{\partial r}$ .
2. $\frac{\partial x}{\partial r}=\cos\theta$ and $\frac{\partial y}{\partial r}=\sin\theta$ (with $\theta$ held fixed).
3. $\dfrac{\partial z}{\partial r}=f_x\cos\theta+f_y\sin\theta$ .
4. Worth noticing: that is $\nabla f\cdot\langle\cos\theta,\sin\theta\rangle$ — the directional derivative straight outward. Lesson 13 in disguise.
TrapWriting $\frac{\partial z}{\partial\theta}$ by mistake; that one is $-f_x r\sin\theta+f_y r\cos\theta$ , and the factors of $r$ are the tell.
6

If $F(x,y,z) = x^2+y^2+z^2-3xyz=0$ defines $z$ implicitly, find $\frac{\partial z}{\partial x}$ at $(1,1,1)$ , or explain why it does not exist.

no calculator, same as the exam
Answer $\left.\dfrac{\partial z}{\partial x}\right|_{(1,1,1)} = -1$ answer recomputed · sympy
1. $F=x^2+y^2+z^2-3xyz$ , so $F_x=2x-3yz$ and $F_z=2z-3xy$ .
2. At $(1,1,1)$ : $F_x=2-3=-1$ and $F_z=2-3=-1$ .
3. $F_z=-1\ne0$ , so the implicit function theorem applies and $z$ really is a function of $(x,y)$ near this point.
4. $\dfrac{\partial z}{\partial x}=-\dfrac{-1}{-1}=-1$ .
TrapSkipping the $F_z\ne0$ check. When $F_z=0$ the surface has a vertical tangent plane there and $\frac{\partial z}{\partial x}$ does not exist — which is exactly what the 'or explain why not' is fishing for.

11

Directional Derivatives & Gradient

Lesson 13 · §15.5

Don't know this yet? §15.5 in the full guide →

From the official study guide

Directional Derivative

The directional derivative $D_{\vec{u}}f(a,b)$ measures rate of change in direction of unit vector $\vec{u}$ :

D_{\vec{u}}f(a,b) = \nabla f(a,b) \cdot \vec{u} = f_x(a,b)u_1 + f_y(a,b)u_2

Gradient Vector

\nabla f(x,y) = \langle f_x(x,y), f_y(x,y) \rangle

Properties of the Gradient

$\nabla f$ points in direction of maximum rate of increase
$|\nabla f|$ gives the maximum rate of increase
$-\nabla f$ points in direction of maximum decrease
$\nabla f$ is orthogonal to level curves

Practice

5 problems

1

Find the directional derivative of $f(x,y)=x^2y-3y^2$ at the point $(2,1)$ in the direction of the vector $\vec{v}=\langle 3,4\rangle$ .

no calculator, same as the exam
Answer $D_{\hat{u}}f(2,1) = \dfrac{4}{5}$ answer recomputed · sympy
1. $\nabla f=\langle 2xy,\;x^2-6y\rangle$ ; at $(2,1)$ that is $\langle4,\,4-6\rangle=\langle4,-2\rangle$ .
2. Normalise the direction: $|\langle3,4\rangle|=5$ , so $\hat u=\left\langle\frac35,\frac45\right\rangle$ .
3. $D_{\hat u}f=\langle4,-2\rangle\cdot\left\langle\frac35,\frac45\right\rangle=\frac{12-8}{5}=\frac45$ .
TrapDotting with $\langle3,4\rangle$ unnormalised, which gives $4$ — five times too big. The directional derivative is only a rate if the direction has length $1$ .
2

Find the maximum rate of change of $f(x,y,z)=x\sqrt{yz}$ at the point $(4,1,4)$ , and state the direction in which it occurs.

no calculator, same as the exam
AnswerMax rate of change: $\sqrt{21}$ , in direction $\langle 2,\, 4,\, 1\rangle$ answer recomputed · sympy
1. $f=x\sqrt{yz}$ , so $f_x=\sqrt{yz}$ , $f_y=\dfrac{xz}{2\sqrt{yz}}$ , $f_z=\dfrac{xy}{2\sqrt{yz}}$ .
2. At $(4,1,4)$ : $\sqrt{yz}=2$ , so $\nabla f=\left\langle2,\;\frac{16}{4},\;\frac{4}{4}\right\rangle=\langle2,4,1\rangle$ .
3. Maximum rate $=|\nabla f|=\sqrt{4+16+1}=\sqrt{21}$ .
4. It occurs in the direction of $\nabla f$ itself, $\langle2,4,1\rangle$ .
TrapGiving the maximum rate as a vector or the direction as a number. The rate is $|\nabla f|$ ; the direction is $\nabla f$ (or its unit version).
3

At $(1,2)$ , in what direction does $f(x,y)=xe^{-y}$ decrease most rapidly? What is that rate of decrease?

no calculator, same as the exam
AnswerDirection: $\langle -1,\, 1\rangle$ ; rate of decrease: $\sqrt{2}\,e^{-2}$ answer recomputed · sympy
1. $\nabla f=\left\langle e^{-y},\;-xe^{-y}\right\rangle$ ; at $(1,2)$ : $\left\langle e^{-2},\,-e^{-2}\right\rangle$ .
2. Steepest decrease is $-\nabla f=\left\langle-e^{-2},\,e^{-2}\right\rangle$ , i.e. the direction $\langle-1,1\rangle$ .
3. The rate of decrease is $|\nabla f|=e^{-2}\sqrt{2}$ .
TrapReporting the rate as negative. 'Rate of decrease' is already signed by the word; the magnitude $\sqrt2\,e^{-2}$ is the number wanted.
4

Find the directional derivative of $f(x,y,z)=x^2+2y^2+3z^2$ at $(1,1,1)$ in the direction of $\langle 1,-1,0\rangle$ .

no calculator, same as the exam
Answer $D_{\hat{u}}f(1,1,1) = -\sqrt{2}$ answer recomputed · sympy
1. $\nabla f=\langle2x,4y,6z\rangle$ ; at $(1,1,1)$ : $\langle2,4,6\rangle$ .
2. $\hat u=\dfrac{\langle1,-1,0\rangle}{\sqrt2}$ .
3. $D_{\hat u}f=\dfrac{2-4+0}{\sqrt2}=\dfrac{-2}{\sqrt2}=-\sqrt2$ .
TrapRationalising carelessly: $\frac{-2}{\sqrt2}=-\sqrt2$ , not $-\frac{\sqrt2}{2}$ .
5

Suppose $\nabla f(2,3)=\langle 4,-3\rangle$ . Find $D_{\vec{u}}f(2,3)$ where $\vec{u}$ is the unit vector in the direction of $\langle 3,4\rangle$ .

no calculator, same as the exam
Answer $D_{\hat{u}}f(2,3) = 0$ answer recomputed · sympy
1. $\hat u=\left\langle\frac35,\frac45\right\rangle$ .
2. $D_{\hat u}f=\langle4,-3\rangle\cdot\left\langle\frac35,\frac45\right\rangle=\frac{12-12}{5}=0$ .
3. Zero is meaningful: $\langle3,4\rangle$ is perpendicular to $\nabla f$ , so this direction runs along the level curve through $(2,3)$ .
TrapTreating $0$ as a computational accident. A zero directional derivative always means you are moving tangent to the level curve.

12

Tangent Planes & Linear Approximation

Lesson 14 · §15.6

Don't know this yet? §15.6 in the full guide →

From the official study guide

Tangent Plane

Equation of tangent plane to $z = f(x,y)$ at $(a, b, f(a,b))$ :

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

Linear Approximation

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

Differential

dz = f_x\,dx + f_y\,dy

Normal Vector

For $z = f(x,y)$ : $\vec{n} = \langle f_x, f_y, -1 \rangle$

For $F(x,y,z) = 0$ : $\vec{n} = \nabla F = \langle F_x, F_y, F_z \rangle$

Practice

5 problems

1

Consider the surface $2x^2-4x+y^3+z+2=0$ . Find the point(s) on the surface where the tangent plane is parallel to the $xy$ -plane.

no calculator, same as the exam
Answer $(1,\, 0,\, 0)$ answer recomputed · sympy
1. Solve for $z$ : $z=-2x^2+4x-y^3-2$ .
2. Parallel to the $xy$ -plane means the tangent plane is horizontal: $z_x=0$ and $z_y=0$ .
3. $z_x=-4x+4=0\Rightarrow x=1$ ; $z_y=-3y^2=0\Rightarrow y=0$ .
4. $z=-2+4-0-2=0$ . The point is $(1,0,0)$ .
TrapReporting only $(x,y)$ . The question asks for points on the surface, so the $z$ -coordinate has to be computed.
2

Find the equation of the tangent plane to $z = x^2 + 2xy$ at the point $(1, 1, 3)$ .

no calculator, same as the exam
Answer $4x + 2y - z = 3$ answer recomputed · sympy
1. $z_x=2x+2y$ and $z_y=2x$ ; at $(1,1)$ : $z_x=4$ , $z_y=2$ .
2. $z=3+4(x-1)+2(y-1)$ .
3. Expand: $z=4x+2y-3$ , i.e. $4x+2y-z=3$ .
TrapLeaving off $f(a,b)=3$ . The tangent plane has to pass through the point of tangency; without the constant it passes through the origin instead.
3

Use the linear approximation of $f(x,y)=\sqrt{x^2+y^2}$ at $(3,4)$ to estimate $f(3.1, 3.9)$ .

no calculator, same as the exam
Answer $L(3.1, 3.9) \approx 4.98$ answer recomputed · sympy
1. $f(3,4)=5$ . $f_x=\dfrac{x}{\sqrt{x^2+y^2}}=\frac35$ , $f_y=\dfrac{y}{\sqrt{x^2+y^2}}=\frac45$ .
2. $L(x,y)=5+\frac35(x-3)+\frac45(y-4)$ .
3. At $(3.1,3.9)$ : $\Delta x=0.1$ , $\Delta y=-0.1$ .
4. $L=5+0.6(0.1)+0.8(-0.1)=5+0.06-0.08=4.98$ .
TrapSign of $\Delta y$ . Moving from $4$ to $3.9$ is $-0.1$ ; getting that backwards gives $5.02$ , and both look plausible.
4

If $z = x^3 y^2$ , find the differential $dz$ .

no calculator, same as the exam
Answer $dz = 3x^2y^2\,dx + 2x^3y\,dy$ answer recomputed · sympy
1. $dz=f_x\,dx+f_y\,dy$ .
2. $f_x=3x^2y^2$ and $f_y=2x^3y$ .
3. $dz=3x^2y^2\,dx+2x^3y\,dy$ .
TrapWriting $\Delta z$ for $dz$ . The differential is the linear estimate; $\Delta z$ is the true change, and they differ by higher-order terms.
5

Find the normal vector to the surface $x^2+y^2-z = 0$ at the point $(1, 2, 5)$ .

no calculator, same as the exam
Answer $\vec{n} = \langle 2,\, 4,\, -1\rangle$ answer recomputed · sympy
1. Write the surface as a level set: $F(x,y,z)=x^2+y^2-z=0$ .
2. $\nabla F=\langle2x,\;2y,\;-1\rangle$ .
3. At $(1,2,5)$ : $\langle2,4,-1\rangle$ .
TrapReading $\langle f_x,f_y,-1\rangle$ and the sign on the last slot. Both $\langle2,4,-1\rangle$ and $\langle-2,-4,1\rangle$ are normal; the form that comes out of $\nabla F$ is the one asked for.

13

Maximum and Minimum Problems

Lessons 15-16 · §15.7

Don't know this yet? §15.7 in the full guide →

From the official study guide

Critical Points

Point (a,b) is a critical point if:

f_x(a,b) = 0 \text{ and } f_y(a,b) = 0 \quad (\nabla f(a,b) = \vec{0})

Second Derivative Test

Let the discriminant be:

D = f_{xx}(a,b)f_{yy}(a,b) - [f_{xy}(a,b)]^2

Classification

If $D > 0$ and $f_{xx} > 0$ : local minimum
If $D > 0$ and $f_{xx} < 0$ : local maximum
If $D < 0$ : saddle point
If $D = 0$ : inconclusive

Absolute Max/Min on Closed Region

Step 1: Find all critical points inside region D

Step 2: Find extreme values on the boundary of D

Step 3: Compare all values - largest is absolute max, smallest is absolute min

Practice

6 problems

1

Find the critical points of $f(x,y) = x^2y - y + 3y^2$ .

no calculator, same as the exam
Answer $(0,\,\tfrac{1}{6}),\;(1,0),\;(-1,0)$ answer recomputed · sympy
1. $f_x=2xy=0$ and $f_y=x^2-1+6y=0$ .
2. $f_x=0$ splits into two cases: $x=0$ or $y=0$ .
3. $x=0$ : $-1+6y=0\Rightarrow y=\frac16$ , giving $\left(0,\frac16\right)$ .
4. $y=0$ : $x^2-1=0\Rightarrow x=\pm1$ , giving $(1,0)$ and $(-1,0)$ .
TrapDividing $2xy=0$ by $x$ . That silently discards the $x=0$ branch — and here it is a whole critical point.
2

Classify the critical point of $f(x,y) = x^2+xy+y^2+3x-3y+4$ using the second derivative test.

no calculator, same as the exam
AnswerLocal minimum at $(-3,\, 3)$ answer recomputed · sympy
1. $f_x=2x+y+3=0$ and $f_y=x+2y-3=0$ .
2. Solve: $y=-3-2x$ , then $x+2(-3-2x)-3=0\Rightarrow-3x-9=0\Rightarrow x=-3$ , $y=3$ .
3. $f_{xx}=2$ , $f_{yy}=2$ , $f_{xy}=1$ , so $D=4-1=3>0$ .
4. $D>0$ and $f_{xx}=2>0$ : local minimum at $(-3,3)$ .
TrapSquaring $f_{xy}$ is required: $D=f_{xx}f_{yy}-(f_{xy})^2$ . Using $f_{xy}$ unsquared gives $3$ here too, by coincidence — the habit will fail you elsewhere.
3

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

no calculator, same as the exam
Answer $M + m = 15$ answer recomputed · sympy
1. Interior: $f_x=2x-2=0\Rightarrow x=1$ , $f_y=2y=0\Rightarrow y=0$ . Inside the disk, and $f(1,0)=1-2+0+4=3$ .
2. Boundary $x^2+y^2=4$ : substitute $y^2=4-x^2$ , so $f=x^2-2x+(4-x^2)+4=8-2x$ with $x\in[-2,2]$ .
3. That is linear and decreasing: max $12$ at $x=-2$ , min $4$ at $x=2$ .
4. Compare everything: $M=12$ , $m=3$ , so $M+m=15$ .
TrapIgnoring the interior critical point because the boundary already gave a max and a min. The absolute minimum here is *inside*, at $(1,0)$ .
4

Find and classify all critical points of $f(x,y) = xy - x^3 - y^2$ .

no calculator, same as the exam
AnswerSaddle point at $(0,0)$ ; local max at $(\tfrac{1}{6}, \tfrac{1}{12})$ answer recomputed · sympy
1. $f_x=y-3x^2=0$ and $f_y=x-2y=0$ , so $x=2y$ and $y=3x^2$ .
2. $y=3(2y)^2=12y^2\Rightarrow y(1-12y)=0\Rightarrow y=0$ or $y=\frac1{12}$ .
3. Critical points: $(0,0)$ and $\left(\frac16,\frac1{12}\right)$ .
4. $f_{xx}=-6x$ , $f_{yy}=-2$ , $f_{xy}=1$ , so $D=12x-1$ . At $(0,0)$ : $D=-1<0$ , saddle. At $\left(\frac16,\frac1{12}\right)$ : $D=1>0$ and $f_{xx}=-1<0$ , local maximum.
Trap $D$ depends on $x$ here, so it must be re-evaluated at each point. A single computed $D$ cannot classify two different critical points.
5

Show that $f(x,y) = x^2 - y^2$ has a saddle point at $(0,0)$ using the second derivative test.

no calculator, same as the exam
AnswerSaddle point at $(0,0)$ answer recomputed · sympy
1. $f_x=2x=0$ and $f_y=-2y=0$ , so the only critical point is $(0,0)$ .
2. $f_{xx}=2$ , $f_{yy}=-2$ , $f_{xy}=0$ .
3. $D=(2)(-2)-0=-4<0$ , so $(0,0)$ is a saddle point.
4. The geometry backs it up: along $y=0$ the surface is $x^2$ (a valley), along $x=0$ it is $-y^2$ (a ridge). This is the standard saddle.
Trap $D<0$ is the one case that is fully decisive with no second test needed. Do not go looking at $f_{xx}$ afterwards; its sign is irrelevant when $D<0$ .
6

Find the absolute maximum and absolute minimum of $f(x,y) = 4xy - x^2 - 2y^2 + 2$ on the region $0 \le x \le 2$ , $0 \le y \le 1$ .

no calculator, same as the exam
AnswerAbs. max $= 4$ at $(2,1)$ ; abs. min $= -2$ at $(2,0)$ answer recomputed · sympy
1. Interior: $f_x=4y-2x=0$ and $f_y=4x-4y=0$ give $x=y$ and $x=2y$ , so only $(0,0)$ — a corner, not interior. No interior critical point.
2. Edge $y=0$ : $f=2-x^2$ on $x\in[0,2]$ , running from $2$ down to $-2$ .
3. Edge $y=1$ : $f=4x-x^2$ on $[0,2]$ , running from $0$ up to $4$ at $x=2$ .
4. Edge $x=0$ : $f=2-2y^2\in[0,2]$ . Edge $x=2$ : $f=-2y^2+8y-2$ , from $-2$ at $y=0$ to $4$ at $y=1$ .
5. Absolute max $=4$ at $(2,1)$ ; absolute min $=-2$ at $(2,0)$ .
TrapForgetting the four corners. Both extremes here happen to be corner points, which the edge parametrisations only catch at their endpoints.

Review of Vectors

Vector Notation and Operations

Basic Operations

Magnitude and Unit Vectors

Dot Product

Cross Product

Practice

Lines and Planes in Space

Equations of a Line

Vector Equation

Parametric Equations

Symmetric Equations (if a,b,c≠0a, b, c \neq 0a,b,c=0)

Equation of a Plane

Finding Normal Vectors

Practice

Quadric Surfaces

Identifying Quadric Surfaces

Practice

Vector-Valued Functions

Definition

Limits and Continuity

Practice

Calculus of Vector-Valued Functions & Motion in Space

Derivative

Integral

Differentiation Rules

Motion in Space

Unit Tangent Vector

Practice

Arc Length and Curvature

Arc Length

Arc Length Function

Curvature

Practice

Functions of Several Variables

Definition

Visualization Methods

Functions of Three Variables

Practice

Limits and Continuity

Limits

Proving a Limit Does Not Exist

Continuity

Practice

Partial Derivatives

First-Order Partial Derivatives

Higher-Order Partial Derivatives

Clairaut's Theorem

Practice

The Chain Rule

Chain Rule - Case 1

Chain Rule - Case 2

Implicit Differentiation

Practice

Directional Derivatives & Gradient

Directional Derivative

Gradient Vector

Properties of the Gradient

Practice

Tangent Planes & Linear Approximation

Tangent Plane

Linear Approximation

Differential

Normal Vector

Practice

Maximum and Minimum Problems

Critical Points

Second Derivative Test

Classification

Absolute Max/Min on Closed Region

Practice

Symmetric Equations (if $a, b, c \neq 0$ )