I am currently teaching a course in "vector analysis", following Colley's book.

So far we have reviewed multivariable calculus (a prereq for the course), and discussed: the derivative in general; the gradient, vector fields, divergence and curl; flow lines and conservative fields; arc length and the TNB frame; Kepler's laws (with complete proofs).

Logically speaking, I would like to move on to the change of variables formula for integration; but we have one more class meeting before spring break, and I'd like to present some "lighter" fare. Nothing suitable comes to mind.

Is there any "light and easy" and "fun" topic I could present to my students, involving some of the background I mentioned above?

  • $\begingroup$ I'm a big fan of Gauss's linking integral: en.wikipedia.org/wiki/… It may be more fun than easy. $\endgroup$
    – Adam
    Mar 5 '20 at 23:40
  • 7
    $\begingroup$ A different path, you might spend a whole class showing the role the determinant plays in modifying the area of the rectangle under a linear map. That would make the later discussion more concrete when you do the same at the level of the linearization. Munkrese Analysis on Manifolds text has a section on determinants which might give you ideas. $\endgroup$ Mar 6 '20 at 1:59
  • $\begingroup$ Post this as an answer! @JamesS.Cook $\endgroup$
    – Chris Cunningham
    Mar 6 '20 at 19:26
  • $\begingroup$ Post this as an answer! @Adam $\endgroup$
    – Chris Cunningham
    Mar 6 '20 at 19:27
  • 1
    $\begingroup$ @Adam: I've tried for years without much success to come up with a decent vector calculus exercise based on the Gauss integral formula for the linking number. It's a good example of an explicit formula that is not generally easy to evaluate by hand. Even for a torus knot the calculations are far more complicated than an ordinary student can handle. $\endgroup$
    – Dan Fox
    Mar 8 '20 at 12:06

If we study the tangent line to $y=f(x)$ at $(a,f(a))$ as given by: $$ y = L^a_f(x) = f(a)+f'(a)(x-a) $$ then you can verify that a little input interval $(a-h,a+h)$ is transferred to the output interval $(f(a)-f'(a)h,f(a)+f'(a)h)$ by the linearization $L^a_f$ of $f$ at $x=a$. What is the significance of the derivative in this viewpoint ? Observe the length of the input interval $(a-h,a+h)$ is $2h$ whereas the length of the output interval is $2f'(a)h$.

The value of the derivative tells us how the function stretches out the input. It gives a scale factor by which the input length is scaled to the output length.

A simple example, $f(x)=2x$ has $f([0,1]) = [0,2]$. The unit-interval $[0,1]$ is stretched by a factor of $2$ to give the length two interval $[0,2]$.

enter image description here

Most functions do not allow us to think about length in this global sense, but it is clear in this linear case. For a function with a variable derivative we have to think infinitesimally to see the scaling. This discussion ultimately leads to the formulation of arclength, so we can either return to it later or tie into that now (depending on your class).

Ok, great. Functions of one variable are pretty easy to visualize since two dimensions we can draw etc. But, what about a function which takes in two variables and outputs two variables ? That gives us 4 possibly independent degrees of freedom. We will not be able to view it in terms of graphs as it would require four dimensions (which I personally cannot directly visualize). So what to do ? One natural solution is to look at two separate planes where one plane is the arena of inputs and the other plane is the arena of outputs for the map. Let's say $T: dom(T) \subseteq \mathbb{R}^2_{xy} \rightarrow \mathbb{R}^2_{uv}$ meaning that we agree to use $x,y$ as Cartesian coordinates for the inputs and $u,v$ as Cartesian coordinates for the outputs. Previously we used $y=f(x)$ which has $x$ as independent and $y$ as dependent. Now, for $T$, we think of $x,y$ as independent and $u,v$ as dependendent. We could describe $T$ in terms of two equations: $$ u = T_1(x,y) \qquad \& \qquad v = T_2(x,y) $$ for all $(x,y) \in dom(T)$. In such a case we can write $T=(T_1,T_2)$ to indicate that the map $T$ has component functions $T_1$ and $T_2$ (this sentence should be skipped if kids look bewildered already).

Question: If $S \subset dom(T)$ is mapped by $T$ to $T(S)$ then how does the area of $S$ relate to the area of $T(S)$ ?

We should expect that this question cannot be nicely answered for nonlinear maps. Also, the area of $S$ is tough to calculate for general subsets. It follows that a good starting point to make progress on the above question is the most simple nontrivial case. We ought to study $S$ being a rectangle and $T$ being a linear map. To say $T$ is linear means we can write: $$ T(x,y) = \left[\begin{array}{cc} a & b \\ c & d \end{array} \right]\left[ \begin{array}{c} x \\ y \end{array} \right] = \left[ \begin{array}{c} ax+by \\ cx+dy \end{array} \right] = (ax+by, cx+dy). $$ the matrix $\left[\begin{array}{cc} a & b \\ c & d \end{array} \right]$ is known as the standard matrix of $T$ and a common shorthand notation is simply $[T] = \left[\begin{array}{cc} a & b \\ c & d \end{array} \right]$. Each choice of $2 \times 2$ matrix uniquely determines a linear map and vice-versa. We ought to expect the entries of $[T]$ somehow inform how the area of a rectangle $S$ is modified to give the area of $T(S)$.

At this point some student will complain that we've moved past rectangles and we care only about parallelograms since we are all itching to use the cross-product to calculate area. Furthermore, the students demand concrete definitions of parallelograms at this point so this whole exercise need not be vague. Since we have to give the audience what it wants let us define the parallelogram with sides $\vec{A},\vec{B}$ based at $\vec{r}_o$ by: $$ \mathcal{P}_{\vec{r}_o}(\vec{A},\vec{B}) = \vec{r}_o+ \{ s\vec{A}+t\vec{B} \ | \ 0 \leq s,t \leq 1 \} $$ Here we allow parallelograms to collapse to lines or even points. In other words, our parallelograms are possibly degenerate. If the parallelogram is based at the origin then we omit $\vec{r}_o$. For example, the unit-square $$ \mathcal{P}( (1,0),(0,1) ) = \{ (s(1,0)+t(0,1) \ | \ 0 \leq s,t \leq 1 \} = [0,1]^2 $$ Next, we ought to draw a picture to show that $\vec{A}$ and $\vec{B}$ align with the sides with common vertex at $\vec{r}_o$. It follows that the area of $\mathcal{P}_{\vec{r}_o}(\vec{A},\vec{B})$ is given by: $$ \text{area}( \mathcal{P}_{\vec{r}_o}(\vec{A},\vec{B})) = \| \vec{A} \times \vec{B} \| $$ Since this is getting a bit long, let me merely outline the remainder for the time being:

  1. Show that $ \mathcal{P}_{\vec{r}_o}(\vec{A},\vec{B})$ maps under $T$ to a new parallelogram $ \mathcal{P}_{T(\vec{r}_o)}(T(\vec{A}),T(\vec{B}))$. I strongly encourage the use of vector notation here!
  2. Notice that $T(\mathcal{P}_{\vec{r}_o}(\vec{A},\vec{B})$ has sides $T(\vec{A})$ and $T(\vec{B})$. Thus to calculate the area of $T(\mathcal{P}_{\vec{r}_o}(\vec{A},\vec{B}))$ we ought to calculate the magnitude of $ T(\vec{A}) \times T(\vec{B}) $.
  3. Compare the result above with $\| \vec{A} \times \vec{B} \|$. You should be able to find a number $m$ such that $\| T(\vec{A}) \times T(\vec{B}) \| = m\| \vec{A} \times \vec{B} \|$.

Spoiler Alert: $m = | ad-bc|$ where $$\text{det}([T]) = \text{det}\left[\begin{array}{cc} a & b \\ c & d \end{array} \right] = ad-bc. $$ is the determinant of the linear transformation $T$. Notice, if the determinant of $T$ is zero then the area of the mapped parallelogram is likewise zero. Linear maps only preserve the full dimension of their domain if they have nonzero determinant.

The absolute value in $m = |ad-bc|$ is needed since the determinant can be negative. If you study it then you'll see the significance of the sign is that positive determinant maps preserve the orientation. Orientation for a pair of vectors is geometrically understood in terms of CCW rotation; $\{ \vec{A}, \vec{B} \}$ is positively oriented if vector $\vec{B}$ can be reached by rotating $\vec{A}$ in a CCW fashion (less than $180^o$). You can check, if $\text{det}(T) <0$ and $\{ \vec{A}, \vec{B} \}$ is positively oriented then $\{ T(\vec{A}), T(\vec{B}) \}$ will be negatively oriented in the sense that the vector $T(\vec{B})$ is reached from a CW rotation of $T(\vec{A})$ by some $\theta \leq 180^o$. The map pictured below has positive determinant: enter image description here

For a classroom talk, you probably should suggest notation for components of $\vec{A}$ and $\vec{B}$ so the students can calculate and share progress with one another. Alternatively, you could have them calculate a special case rather than attacking the derivation in full generality.

Finally, this leads us to the next question: what is the analog of the linearization of $y=f(x)$ ? How can we linearize something like $F(x,y) = (x^2+y^2, xy)$ at some point in the plane ? We'll see that an affine map $L_F^a = \vec{a}+d_aF$ where $d_aF$ is the differential of $F$ at $a$ gives the linearization of $F$. Moreover, $[d_aF]=J_F(a)$ is the Jacobian matrix. The determinant of this Jacobian matrix tells us how an infinitesimal area scales under the map $F$ just as $f'(a)$ tells us how the length of a small line-segment is stretched under the map $f$. Good news: the story for maps on $\mathbb{R}^n$ for $n>2$ is no different. But, we'll have to learn how to calculate determinants of bigger matrices... or... calculate some wedge products :)


As the students are by now familiar with the cross product, it is a natural question of how it might generalize to higher dimensions. It is surprising that the cross product only exists in dimensions $3$ and $7$: The $7$-dimenstional cross-product is the only "bilinear product of two vectors that is vector-valued, orthogonal, and has the same magnitude as in the 3D case."1 This could take you into very interesting territory.

  • Massey, W. S. "Cross products of vectors in higher dimensional Euclidean spaces." American Mathematical Monthly (1983): 697-701. (JSTOR link)

  • Seven-dimensional cross product: Wikipedia link.1 "Just as the $3$-dimensional cross product can be expressed in terms of the quaternions, the $7$-dimensional cross product can be expressed in terms of the octonions."

1Wikipedia on the $7$-dimensional cross product.

  • 1
    $\begingroup$ Shameless self-promotion: billcookmath.com/papers/2012-06_nD_pythag.pdf $\endgroup$
    – Bill Cook
    Mar 9 '20 at 15:34
  • $\begingroup$ Published version: "An n-dimensional Pythagorean theorem" College Math. J. 44 (2013), no. 2, 98—101. DOI 10.4169/college.math.j.44.2.098 $\endgroup$
    – Bill Cook
    Mar 9 '20 at 15:34
  • $\begingroup$ I think Colley's book has a problem or two about the (n-1)-ary cross product in R^n. :) $\endgroup$
    – Bill Cook
    Mar 9 '20 at 15:36

Based on a suggestion I once found here on MESE, my class last term proved (via Green’s theorem) the existence of a linear planimeter, showing how it could mechanically measure the area of a shape.

When we were done with the proof, I brought out the one I built at home from scrap wood:

Homemade linear planimeter made of wood

Figure 1: The red arm freely pivots on a block that is constrained to move linearly (sliding between two fixed pieces of wood, shown on the left). At the end of the arm is a piece of wire used to trace around a closed, oriented curve. On the arm is a wheel (also wood) that turns when the arm pivots so that the wire moves parallel to the oriented curve.

Linear planimeter measuring wheel

Figure 2: Blow-up of the wheel, which I marked-off in tenths of a rotation. Note the red mark used to count whole rotations.

To test it, I drew a circle with a compass, and had a student measure it and determine its area with $\pi r^2$. I used the planimeter, having another student count the net change in number of rotations of the wheel. Students were surprised that our answers for the area were within 2% of each other. Then, we tested it with a print-out of a region between the x-axis and a complicated curve. [I prepared this sheet in advance, scaling it nicely so the planimeter would be able to trace around it.] We all agreed that computing the area by hand would be very difficult, so we used Wolfram|Alpha. Again, the planimeter's value of the area was close to Wolfram|Alpha's.

Finally, I brought out a polar planimeter I got on Ebay and the class oohed-and-ahed at the improved precision and accuracy when we used it to find the area of the circle and the "complicated curve" area. We did not go through the details of proving why the pivot could be on a fixed circle for the polar planimeter (as opposed to a fixed line in the linear case), but I suggested this as an interesting exploration for those who wanted to further explore the topic.

Polar planimeter Figure 3: $10 vintage planimeter from Ebay

Sketch of the proof that students worked through (with some help):

  • First, we drew an oriented curve that passed near a straight axis. [The axis represents the straight path that the planimeter's pivot slides on.] From the axis, segments of fixed length $L$ are drawn to points on the curve. These segments represent the fixed distance from the planimeter's pivot to the point of the wire at the end of the arm. This wire will trace along every point $(x,y)$ on the curve.

Oriented curve with vector V connecting point from x axis to point x,y on the curve. Vector F is perpendicular to vector V.

Figure 4: Set-up for creating a vector field

  • Now, we created our vector field by first considering a vector $\vec{v}$ that points from some arbitrary point $(a, 0)$ to the point $(x, y)$ on the curve. Our vector $\vec{F}(x,y)$ will be a unit vector, perpendicular to $\vec{v}$, and oriented in the same direction as the curve. Since $(a, 0)$ is assumed to be on a straight line, with the Pythagorean theorem, students found that $\vec{F}(x,y) = \langle \frac{-y}{L}, \frac{\sqrt{L^2-y^2}}{L} \rangle$.

  • Next, they determine that $curl \vec{F}$ is constant.

  • Finally, they use Green's Theorem to compute $L \int_C \vec{F}\cdot d\vec{r}$, seeing that it represents the area of the region enclosed by the curve $C$, and is calculated by: $L \times ($how far the wheel rolled$)$

Conclusion: I liked this application as it was right at their ability level skill-wise (using Green's Theorem), and because it pushed them conceptually (considering and writing the equation of a vector field that measures something physical, other than the typical examples explored in class: electricity/magnetism, gravity).

  • $\begingroup$ Very cool. I want one. $\endgroup$ Mar 12 '20 at 1:38

I have couple point five suggestions that are different ideas from the previous comment/answers:

  1. Do an application that uses the techniques they have already learned. This won't open a new topic math-wise before the break. And gives natural review. And can be made self-contained and defined in scope. Note, when I say "application", I don't mean a derivation into another math topic. I mean something in engineering or physics or chemistry or business or the like. In particular, look for something in data science or computers or the like (a Google search shows entire texts around this topic). Note, the point is really just a review and self contained. It's not like you have to turn them into radio engineers...but a little practice in applied word problems (regardless of exact field) is a good change of mental frame.

  2. Do a general review/drill day. Make it interactive by having an exercise sheet and a plan for how to march through the hour. Note that if you say (and mean it!) that the hour will help them with upcoming exams, that will get their attention.

2.5. (Related to 2, but little funner but less didactic.) "Vector Calc Jeopardy": very easy on the chalkboard--draw the grid, have questions ready, divide the room into teams, appoint a scorekeeper and answer-spotter (from the teams raising hands), insist that they answer "in the form of a question", etc. It's a bit more of a social interaction and a nice head out to the holidays...


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