Class prep · TJ Multivariable Calculus

See calculus in space,from vectors to Stokes' theorem.

A full multivariable calculus course, the standard first course, taught with worked examples and original practice. It covers vectors and space curves, partial derivatives and the gradient, optimization and Lagrange multipliers, multiple integrals, and vector calculus through Green's theorem, Stokes' theorem and the divergence theorem. The course is geometry first, and the problems are set deliberately harder and more abstract than a typical problem set. Every problem has a full worked solution. It is intended for students who have finished single-variable calculus through series, at the level of AP Calculus BC. Built by Neil Kothari, a TJ alumnus.

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$395 covers the full school year. The first lesson is free.

Try one. This is a test-level problem from the course.

Problem 1

Test level
Multiple choice
No calculator
4 pts

Let $\mathbf{a}$ and $\mathbf{b}$ be nonzero, non-parallel vectors in three-dimensional space and let $\mathbf{c}=\mathbf{a}\times(\mathbf{a}\times\mathbf{b})$ . Which statement is true for every such pair $\mathbf{a},\mathbf{b}$ ?

Answer

$\mathbf{c}$ lies in the plane of $\mathbf{a}$ and $\mathbf{b}$ and is perpendicular to $\mathbf{a}$ and nonzero, because a cross product is perpendicular to both factors; in fact $\mathbf{c}=(\mathbf{a}\cdot\mathbf{b})\mathbf{a}-|\mathbf{a}|^2\mathbf{b}$ .

Full solution

Work with directions first. Let $\mathbf{n}=\mathbf{a}\times\mathbf{b}$ . Because $\mathbf{a}$ and $\mathbf{b}$ are nonzero and not parallel, $\mathbf{n}\ne\mathbf{0}$ and $\mathbf{n}$ is perpendicular to the plane spanned by $\mathbf{a}$ and $\mathbf{b}$ .
Now $\mathbf{c}=\mathbf{a}\times\mathbf{n}$ is perpendicular to both of its factors: perpendicular to $\mathbf{a}$ and perpendicular to $\mathbf{n}$ . A vector perpendicular to $\mathbf{n}$ lies in the plane spanned by $\mathbf{a}$ and $\mathbf{b}$ , so $\mathbf{c}$ is in that plane and perpendicular to $\mathbf{a}$ .
It is not zero: $\mathbf{a}$ and $\mathbf{n}$ are perpendicular and both nonzero, so $|\mathbf{c}|=|\mathbf{a}||\mathbf{n}|\sin90^{\circ}=|\mathbf{a}||\mathbf{n}|>0$ . That rules out choice (C).
Algebra confirms it. The identity $\mathbf{a}\times(\mathbf{a}\times\mathbf{b})=\mathbf{a}(\mathbf{a}\cdot\mathbf{b})-\mathbf{b}(\mathbf{a}\cdot\mathbf{a})$ puts $\mathbf{c}$ visibly in the plane of $\mathbf{a}$ and $\mathbf{b}$ (choice A). Dotting with $\mathbf{a}$ gives $(\mathbf{a}\cdot\mathbf{b})|\mathbf{a}|^2-(\mathbf{a}\cdot\mathbf{b})|\mathbf{a}|^2=0$ , confirming perpendicularity.
Why the others are false: choice (B) says $\mathbf{c}\parallel\mathbf{n}$ , but $\mathbf{c}\perp\mathbf{n}$ , so they are never parallel. Choice (E) says $\mathbf{c}\parallel\mathbf{a}$ , but $\mathbf{c}\perp\mathbf{a}$ and $\mathbf{c}\ne\mathbf{0}$ . Choice (D) would need $(\mathbf{a}\cdot\mathbf{b})\mathbf{a}$ to be parallel to $\mathbf{b}$ , which happens only when $\mathbf{a}\cdot\mathbf{b}=0$ ; for $\mathbf{a}=\langle1,0,0\rangle$ and $\mathbf{b}=\langle1,1,0\rangle$ we get $\mathbf{c}=(1)\langle1,0,0\rangle-(1)\langle1,1,0\rangle=\langle0,-1,0\rangle$ , which is not parallel to $\mathbf{b}$ .

Where students lose points

Treating the cross product as if it were associative or as if repeating it returns to a familiar direction. Each cross product is perpendicular to both of its factors, so the second one turns the result back into the plane of the original two vectors. Always track which plane or line each result is perpendicular to.

original problems, each with a full worked solution: 390; original problems, each with a full worked solution
at test level or beyond: 270; at test level or beyond
short lessons, one for every idea: 30; short lessons, one for every idea
units, all available now: 5; units, all available now

Free lesson

Lesson 1.1: Vectors, the Dot Product and the Cross Product

How to read a vector as an arrow, measure agreement with the dot product, measure area and orientation with the cross product, and use the triple product for volume.

This lesson starts with vectors as arrows in space and builds from there.

Problem 2

Core
True or false
No calculator
2 pts

True or False: two vectors $\mathbf a$ and $\mathbf b$ in space have lengths $|\mathbf a|=5$ and $|\mathbf b|=13$ and satisfy $\mathbf a\cdot\mathbf b=25$ . Then the triangle that has $\mathbf a$ and $\mathbf b$ as two of its sides, drawn from a common vertex, has area $60$ . Justify your answer.

Answer

False. The dot product gives $\cos\theta=\frac{25}{65}=\frac5{13}$ , so $\sin\theta=\frac{12}{13}$ and $|\mathbf a\times\mathbf b|=5\cdot13\cdot\frac{12}{13}=60$ . That number is the area of the parallelogram on $\mathbf a$ and $\mathbf b$ ; the triangle is half of it, so its area is $30$ .

Full solution

Start with what the dot product tells you about the angle. Since $\mathbf a\cdot\mathbf b=|\mathbf a||\mathbf b|\cos\theta$ , we get $\cos\theta=\dfrac{25}{5\cdot13}=\dfrac{5}{13}$ .
The cross product needs $\sin\theta$ . The angle between two vectors lies in $[0,\pi]$ , where sine is never negative, so $\sin\theta=+\sqrt{1-\frac{25}{169}}=\sqrt{\frac{144}{169}}=\frac{12}{13}$ .
Now $|\mathbf a\times\mathbf b|=|\mathbf a||\mathbf b|\sin\theta=5\cdot13\cdot\frac{12}{13}=60$ . This is the area of the parallelogram whose adjacent sides are $\mathbf a$ and $\mathbf b$ .
A diagonal cuts a parallelogram into two congruent triangles, so the triangle with sides $\mathbf a$ and $\mathbf b$ has area $\frac12|\mathbf a\times\mathbf b|=30$ , not $60$ . The statement is false.
Check with a concrete pair that has these lengths and this dot product: $\mathbf a=\langle5,0,0\rangle$ and $\mathbf b=\langle5,12,0\rangle$ (so $|\mathbf b|=13$ and $\mathbf a\cdot\mathbf b=25$ ). The triangle with vertices $(0,0)$ , $(5,0)$ , $(5,12)$ has base $5$ and height $12$ , so its area is $30$ , and $\mathbf a\times\mathbf b=\langle0,0,60\rangle$ has length $60$ .

Where students lose points

Stopping at $|\mathbf a\times\mathbf b|=60$ and calling that the triangle's area. The length of a cross product is the area of the parallelogram; a triangle is half of it. Also remember that $\sin\theta$ comes from $\cos\theta$ with a positive root, because $0\le\theta\le\pi$ .

Problem 3

Core
Multiple choice
No calculator
4 pts

Which of the following is a vector perpendicular to both $\mathbf{a}=\langle2,-1,3\rangle$ and $\mathbf{b}=\langle1,4,-2\rangle$ ?

Open the rest of this lesson, free

The full lesson and all 13 of its problems, with every solution.

How it works

Class prep that works alongside class

A full multivariable calculus course for grades 9 to 12. It works well alongside a school course and also stands on its own. Students who already know the material learn to draw the picture first and check every answer. Students who feel behind rebuild a topic with the lesson before they try the problems.

Every problem is original and set deliberately harder and more abstract than a typical problem set: true or false, multiple choice, multi-part and free response.
Each one is marked core, test level or beyond.
Solutions show every step and end with the mistake that most often costs points.
Worked examples in every lesson, and diagrams wherever a picture makes the idea clear.

What is inside

1Vectors, Lines, Planes and Curves in Space6 lessons · 78 problems
2Functions of Several Variables and Partial Derivatives6 lessons · 78 problems
3Optimization6 lessons · 78 problems
4Multiple Integrals6 lessons · 78 problems
5Vector Calculus6 lessons · 78 problems

The front entrance of Thomas Jefferson High School for Science and Technology

Who built it

Neil Kothari

TJHSST alumnus · Co-founder, EduAvenues · 10+ years in TJ preparation

Neil is a TJ alumnus and has spent more than a decade working with TJ students and their families. He built this course so students can practice the reasoning and the calculations together. Every problem is written for the course and every answer is checked before it goes in.

Tuition

$395for the full school year

One tuition covers every unit of the course. All 5 units are available now. The Vectors, the Dot Product and the Cross Product lesson stays free for everyone.

All 5 units: 30 lessons and 390 original problems
A full worked solution for every problem
Access through the end of the school year

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