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Student explore the properties of an orthonormal basis using the Cartesian and \(S_z\) bases as examples.

In rectangular coordinates, the natural unit vectors are \(\{\boldsymbol{\hat x},\boldsymbol{\hat y}\}\), which point in the direction of increasing \(x\) and \(y\), respectively. Similarly, in polar coordinates the natural unit vectors are \(\boldsymbol{\hat r}\), which points in the direction of increasing \(r\), and \(\boldsymbol{\hat\phi}\), which points in the direction of increasing \(\phi\).

The unit tangent vector to a parametric curve is the unit vector tangent to the curve which points in the direction of increasing parameter. The principal unit normal vector to a parametric curve is the unit vector perpendicular to the curve “in the direction of bending”, which is the direction of the derivative of the unit tangent vector.

1. Consider the parametric curve \(\boldsymbol{\vec r} = 3\cos\phi\,\boldsymbol{\hat x} + 3\sin\phi\,\boldsymbol{\hat y}\) with \(\phi\in[0,2\pi]\). Calculate the unit tangent vector \(\boldsymbol{\hat T}\) and the principal unit normal vector \(\boldsymbol{\hat N}\) for this curve in terms of \(\boldsymbol{\hat x}\) and \(\boldsymbol{\hat y}\).
2. Consider a circle of radius \(3\) centered at the origin. Determine the unit tangent vector \(\boldsymbol{\hat T}\) and the principal unit normal vector \(\boldsymbol{\hat N}\) for this curve in terms of \(\boldsymbol{\hat r}\) and \(\boldsymbol{\hat\phi}\).
3. Compare your answers.

Instructor's Guide

Main ideas

• Geometric introduction of \(\boldsymbol{\hat r}\) and \(\boldsymbol{\hat\phi}\).
• Geometric introduction of unit tangent and normal vectors.

Prerequisites

• The position vector \(\vec{r}\).
• The derivative of the position vector is tangent to the curve.

Warmup

See the prerequisites. It is possible to briefly introduce these ideas immediately preceding this activity.

Props

• whiteboards and pens

Wrapup

Emphasize that \(\boldsymbol{\hat r}\) and \(\boldsymbol{\hat\phi}\) do not live at the origin! Encourage students to use the figure provided, which may help alleviate this confusion.

Point out to the students that \(\boldsymbol{\hat r}\) and \(\boldsymbol{\hat\phi}\) are defined everywhere (except at the origin), whereas \(\boldsymbol{\hat{T}}\) and \(\boldsymbol{\hat{N}}\) are properties of the curve. It is only on circles that these two notions coincide; \(\boldsymbol{\hat r}\) and \(\boldsymbol{\hat\phi}\) are adapted to round problems, and circles are round! Symmetry is important.

Emphasize that \(\{\boldsymbol{\hat r},\boldsymbol{\hat\phi}\}\) can be used as a basis (except at the origin). Point out to the students that their answer to the last problem gives them a formula expressing \(\boldsymbol{\hat r}\) and \(\boldsymbol{\hat\phi}\) in terms of \(\boldsymbol{\hat{x}}\) and \(\boldsymbol{\hat{y}}\). When comparing these basis vectors, they should all be drawn with their tails at the same point.

Details

We have had success helping students master the idea of “direction of bending” by describing the curve as part of a pickle jar; the principal unit normal vector points at the pickles!

In the Classroom

The easiest way to find \(\boldsymbol{\hat{N}}\) is to use the dot product to find vectors orthogonal to \(\boldsymbol{\hat{T}}\), then normalize. Students must then use the “direction of bending” criterion to choose between the two possible orientations.

Finding \(\boldsymbol{\hat{N}}\) in this way requires the student to give names to the its unknown components. This is a nontrivial skill; many students will have trouble with this.

It may be important to draw some examples. Despite that, students still feel wary of embracing \(\boldsymbol{\hat{r}}\) and \(\boldsymbol{\hat{\phi}}\). People will feel more comfortable over the next few classes but emphasize the geometry: the circle is still the circle and the unit tangent remains the same regardless.

Some students are natural geometors and will realize what the desired vectors are. This is terrific. Certainly, it is worthwhile to explicitly demonstrate this, but the point is the geometry can often do the work for you. This will convince students of the value of smart coordinates.

Subsidiary ideas

• Dividing any vector by its length yields a unit vector.
• Using the dot product to find vectors perpendicular to a given vector.

Homework

• Some students will not be comfortable unless they work out the components of \(\boldsymbol{\hat r}\) and \(\boldsymbol{\hat\phi}\) with respect to \(\boldsymbol{\hat{x}}\) and \(\boldsymbol{\hat{y}}\). Let them.

Enrichment

• What units does a unit vector have? Do \(\boldsymbol{\hat r}\) and \(\boldsymbol{\hat\phi}\) have the same units?

Students hold rulers and meter sticks to represent a vector field. The instructor holds a hula hoop to represent a small area element. Students are asked to describe the flux of the vector field through the area element.

Students are shown a topographic map of an oval hill and imagine that the classroom is on the hill. They are asked to point in the direction of the gradient vector appropriate to the point on the hill where they are "standing".

Students learn about the geometric meaning of the amplitude and period parameters in the sine function. They also practice sketching the sum of two functions by hand.

Students learn how to express Angular Momentum as a vector quantity in polar coordinates, and then in Cylindrical and Spherical Coordinates

Write something you know about angular momentum.

Students are placed into small groups and asked to calculate the total differential of a function of two variables, each of which is in turn expressed in terms of two other variables.

This activity starts with a brief lecture introduction to power series and a short derivation of the formula for calculating the power series coefficients.

\[c_n={1\over n!}\, f^{(n)}(z_0)\]

Students use this formula to compute the power series coefficients for a \(\sin\theta\) (around both the origin and (if time allows) \(\frac{\pi}{6}\)). The meaning of these coefficients and the convergence behavior for each approximation is discussed in the whole-class wrap-up and in the follow-up activity: Visualization of Power Series Approximations.

This small group activity is designed to provide practice with the chain rule and to develop familiarity with polar coordinates. Students work in small groups to relate partial derivatives in rectangular and polar coordinates. The whole class wrap-up discussion emphasizes the importance of specifying what quantities are being held constant.

This small group activity using surfaces combines practice with the multivariable chain rule while emphasizing numerical representations of derivatives. Students work in small groups to measure partial derivatives in both rectangular and polar coordinates, then verify their results using the chain rule. The whole class wrap-up discussion emphasizes the relationship between a directional derivative in the \(r\)-direction and derivatives in \(x\)- and \(y\)-directions using the chain rule.

Consider the region \(D\) in the \(xy\)-plane shown below, which is bounded by \[u=9 \qquad u=36 \qquad v=1 \qquad v=4\] where \[u=xy \qquad v={y\over x}\] If you want to determine \(x\) and \(y\) as functions of \(u\) and \(v\), consider \(uv\) and \(u/v\).

• List as many methods as you can think of for finding the area of the given region.

  It is enough to refer to the methods by name or describe them briefly.

• For at least 3 of these methods, give explicitly the formulas you would use to find the area.

  You must put limits on your integrals, but you do not need to evaluate them.

• Using any 2 of these methods, find the area.

  One of these should be a method we have learned recently.

• Now consider the following integral over the same region \(D\):

  \(\int\!\!\int_D {y\over x} \>\>dA\)

• Which of the above methods can you use to do this integral?
• Do the integral.

Main ideas

• There are many ways to solve this problem!
• Using Jacobians (and inverse Jacobians)

Prerequisites

• Surface integrals
• Jacobians
• Green's/Stokes' Theorem

Warmup

Perhaps a discussion of single and double integral techniques for solving this problem.

Props

• whiteboards and pens

Wrapup

This is a good conclusion to the course, as it reviews many integration techniques. We emphasize that (2-dimensional) change-of-variable problems are a special case of surface integrals.

Here are some of the methods one could use to do these integrals:

• change of variables (at least 2 ways)
• Area Corollary to Green's Theorem (at least 2 ways)
• ordinary single integral (at least 2 ways)
• ordinary double integral (at least 2 ways)
• surface integral

Details

In the Classroom

• Some students will want to simply use Jacobian formulas; encourage such students to try to solve this problem both by computing \(\frac{\partial(x,y)}{\partial(u,v)}\) and by computing \(\frac{\partial(u,v)}{\partial(x,y)}\).
• Other students will want to work directly with \(d\boldsymbol{\vec{r}}_1\) and \(d\boldsymbol{\vec{r}}_2\). This works fine if one first solves for \(x\) and \(y\) in terms of \(u\) and \(v\).

• Students who compute \(d\boldsymbol{\vec{r}}_1\) and \(d\boldsymbol{\vec{r}}_2\) directly can easily get confused, since they may try to eliminate \(x\) or \(y\), rather than \(u\) or \(v\).

  Along the curve \(v=\hbox{constant}\), one has \(dy=v\,dx\), so that \(d\boldsymbol{\vec{r}}_1 = dx\,\boldsymbol{\hat{x}} + dy\,\boldsymbol{\hat{y}} = (\boldsymbol{\hat{x}} + v\,\boldsymbol{\hat{y}})\,dx\), which some students will want to write in terms of \(x\) alone. But one needs to express this in terms of \(du\)! This can be done using \(du = x\,dy + y\,dx = x (v\,dx) + y\,dx = 2y\,dx\), so that \(d\boldsymbol{\vec{r}}_1 = (\boldsymbol{\hat{x}} + v\,\boldsymbol{\hat{y}}) \,\frac{du}{2y}\). A similar argument leads to \(d\boldsymbol{\vec{r}}_2 = (-\frac{1}{v}\,\boldsymbol{\hat{x}}+\boldsymbol{\hat{y}})\,\frac{x\,dv}{2}\) for \(u=\hbox{constant}\), so that \(d\boldsymbol{\vec{S}} = d\boldsymbol{\vec{r}}_1\times d\boldsymbol{\vec{r}}_2 = \boldsymbol{\hat{z}} \,\frac{x}{2y}\,du\,dv = \boldsymbol{\hat{z}} \,{du\,dv\over2v}\). This calculation can be done without solving for \(x\) and \(y\), provided one recognizes \(v\) in the penultimate expression.

  Emphasize that one must choose parameters, both on the region, and on each curve, and that \(u\) and \(v\) are chosen to make the limits easy.

• Take time before the activity to gauge students' recollection of single variable techniques and the Jacobian. After the activity, be sure to set up more than one approach. People will be fine after the first couple of steps but shouldn't leave class feeling stuck.

Subsidiary ideas

• Review of Green's Theorem
• Review of single integral techniques
• Review of double integral techniques

Enrichment

• Discuss the 3-dimensional case, perhaps relating it to volume integrals.

Students work in small groups to use completeness relations to change the basis of quantum states.

In this lecture, the instructor guides a discussion about translating between bra-ket notation and wavefunction notation for quantum systems.

Students use a completeness relations to write hydrogen atoms states in the energy and position bases.

Students are asked to draw lines of constant \(u\) and \(v\) in a \(u,v\) coordinate system. Then, in the same coordinate system, students must draw lines of constant \(x\) and constant \(y\) when

\[x(u,v)=u \] and \[y(u,v)=\frac{1}{2}u+3v. \]

Students consider how changing the volume of a system changes the internal energy of the system. Students use plastic graph models to explore these functions.

This small group activity is designed to help students visualize the cross product. Students work in small groups to determine the area of a triangle in space. The whole class wrap-up discussion emphasizes the geometric interpretation of the cross product.

Students use their arms to depict (sequentially) the different cylindrical and spherical basis vectors at the location of their shoulder (seen in relation to a specified origin of coordinates: either a set of axes hung from the ceiling of the room or perhaps a piece of furniture or a particular corner of the room).

First, students are shown diagrams of cylindrical and spherical coordinates. Common notation systems are discussed, especially that physicists and mathematicians use opposite conventions for the angles \(\theta\) and \(\phi\). Then students are asked to check their understanding by sketching several coordinate equals constant surfaces on their small whiteboards.

Students practice infinitesimal reasoning in cylindrical and spherical coordinates.

Write one thing you know about the derivative.

Lecture about finding \(\left|{\pm}\right\rangle _x\) and then \(\left|{\pm}\right\rangle _y\). There are two conventional choices to make: relative phase for \(_x\left\langle {+}\middle|{-}\right\rangle _x\) and \(_y\left\langle {+}\middle|{+}\right\rangle _x\).

So far, we've talked about how to calculate measurement probabilities if you know the input and output quantum states using the probability postulate:

\[\mathcal{P} = | \left\langle {\psi_{out}}\middle|{\psi_{in}}\right\rangle |^2 \]

Now we're going to do this process in reverse.

I want to be able to relate the output states of Stern-Gerlach analyzers oriented in different directions to each other (like \(\left|{\pm}\right\rangle _x\) and \(\left|{\pm}\right\rangle _x\) to \(\left|{\pm}\right\rangle \)). Since \(\left|{\pm}\right\rangle \) forms a basis, I can write any state for a spin-1/2 system as a linear combination of those states, including these special states.

I'll start with \(\left|{+}\right\rangle _x\) written in the \(S_z\) basis with general coefficients:

\[\left|{+}\right\rangle _x = a \left|{+}\right\rangle + be^{i\phi} \left|{-}\right\rangle \]

Notice that:

(1) \(a\), \(b\), and \(\phi\) are all real numbers; (2) the relative phase is loaded onto the second coefficient only.

My job is to use measurement probabilities to determine \(a\), \(b\), and \(\phi\).

I'll prepare a state \(\left|{+}\right\rangle _x\) and then send it through \(x\), \(y\), and \(z\) analyzers. When I do that, I see the following probabilities:

Input = \(\left|{+}\right\rangle _x\)	\(S_x\)	\(S_y\)	\(S_z\)
\(P(\hbar/2)\)	1	1/2	1/2
\(P(-\hbar/2)\)	0	1/2	1/2

First, looking at the probability for the \(S_z\) components:

\[\mathcal(S_z = +\hbar/2) = | \left\langle {+}\middle|{+}\right\rangle _x |^2 = 1/2\]

Plugging in the \(\left|{+}\right\rangle _x\) written in the \(S_z\) basis:

\[1/2 = \Big| \left\langle {+}\right|\Big( a\left|{+}\right\rangle + be^{i\phi} \left|{-}\right\rangle \Big) \Big|^2\]

Distributing the \(\left\langle {+}\right|\) through the parentheses and use orthonormality: \begin{align*} 1/2 &= \Big| a\cancelto{1}{\left\langle {+}\middle|{+}\right\rangle } + be^{i\phi} \cancelto{0}{\left\langle {+}\middle|{-}\right\rangle } \Big|^2 \\ &= |a|^2\\[12pt] \rightarrow a &= \frac{1}{\sqrt{2}} \end{align*}

Similarly, looking at \(S_z = -\hbar/2\): \begin{align*} \mathcal(S_z = +\hbar/2) &= | \left\langle {-}\middle|{+}\right\rangle _x |^2 = 1/2 \\ 1/2 = \Big| \left\langle {-}\right|\Big( a\left|{+}\right\rangle + be^{i\phi} \left|{-}\right\rangle \Big) \Big|^2\\ 1/2 &= \Big| a\cancelto{0}{\left\langle {-}\middle|{+}\right\rangle } + be^{i\phi} \cancelto{1}{\left\langle {-}\middle|{-}\right\rangle } \Big|^2 \\ &= |be^{i\phi}|^2\\ &= |b|^2 \cancelto{1}{(e^{i\phi})(e^{-i\phi})}\\[12pt] \rightarrow b &= \frac{1}{\sqrt{2}} \end{align*}

I can't yet solve for \(\phi\) but I can do similar calculations for \(\left|{-}\right\rangle _x\):

Input = \(\left|{-}\right\rangle _x\)	\(S_x\)	\(S_y\)	\(S_z\)
\(P(\hbar/2)\)	0	1/2	1/2
\(P(-\hbar/2)\)	1	1/2	1/2

\begin{align*} \left|{-}\right\rangle _x &= c \left|{+}\right\rangle + de^{i\gamma} \left|{-}\right\rangle \\ \mathcal(S_z = +\hbar/2) &= | \left\langle {+}\middle|{-}\right\rangle _x |^2 = 1/2\\ \rightarrow c = \frac{1}{\sqrt{2}}\\ \mathcal(S_z = +\hbar/2) &= | \left\langle {-}\middle|{-}\right\rangle _x |^2 = 1/2\\ \rightarrow d = \frac{1}{\sqrt{2}}\\ \end{align*}

So now I have: \begin{align*} \left|{+}\right\rangle _x &= \frac{1}{\sqrt{2}} \left|{+}\right\rangle + \frac{1}{\sqrt{2}}e^{i\beta} \left|{-}\right\rangle \\ \left|{-}\right\rangle _x &= \frac{1}{\sqrt{2}} \left|{+}\right\rangle + \frac{1}{\sqrt{2}}e^{i\gamma} \left|{-}\right\rangle \\ \end{align*}

I know \(\beta \neq \gamma\) because these are not the same state - they are orthogonal to each other: \begin{align*} 0 &= \,_x\left\langle {+}\middle|{-}\right\rangle _x \\ &= \Big(\frac{1}{\sqrt{2}} \left\langle {+}\right| + \frac{1}{\sqrt{2}}e^{i\beta} \left\langle {-}\right| \Big)\Big( \frac{1}{\sqrt{2}} \left|{+}\right\rangle + \frac{1}{\sqrt{2}}e^{i\gamma} \left|{-}\right\rangle \Big)\\ \end{align*}

Now FOIL like mad and use orthonormality: \begin{align*} 0 &= \frac{1}{2}\Big(\cancelto{1}{\left\langle {+}\middle|{+}\right\rangle } + e^{i\gamma} \cancelto{0}{\left\langle {+}\middle|{-}\right\rangle } + e^{i\beta} \cancelto{0}{\left\langle {-}\middle|{+}\right\rangle } + e^{i(\gamma - \beta)}\cancelto{1}{\left\langle {-}\middle|{-}\right\rangle } \Big)\\ &= \frac{1}{2}\Big(1 + e^{i(\gamma - \beta} \Big) \\ \rightarrow & \quad e^{i(\gamma-\beta)} = -1 \end{align*}

This means that \(\gamma-\beta = \pi\). I don't have enough information to solve for \(\beta\) and \(\gamma\), but there is a one-time conventional choice made that \(\beta = 0\) and \(\gamma = 1\), so that: \begin{align*} \left|{+}\right\rangle _x &= \frac{1}{\sqrt{2}} \left|{+}\right\rangle + \frac{1}{\sqrt{2}}\cancelto{1}{e^{i0}} \left|{-}\right\rangle \\ \left|{-}\right\rangle _x &= \frac{1}{\sqrt{2}} \left|{+}\right\rangle + \frac{1}{\sqrt{2}}\cancelto{-1}{e^{i\pi}} \left|{-}\right\rangle \\[12pt] \rightarrow \left|{+}\right\rangle _x &= \frac{1}{\sqrt{2}} \left|{+}\right\rangle \color{red}{+} \frac{1}{\sqrt{2}}\left|{-}\right\rangle \\ \left|{-}\right\rangle _x &= \frac{1}{\sqrt{2}} \left|{+}\right\rangle \color{red}{-} \frac{1}{\sqrt{2}}\left|{-}\right\rangle \\[12pt] \end{align*}

When \(\left|{\pm}\right\rangle _y\) is the input state:

Input = \(\left|{+}\right\rangle _y\)	\(S_x\)	\(S_y\)	\(S_z\)
\(P(\hbar/2)\)	1/2	1	1/2
\(P(-\hbar/2)\)	1/2	0	1/2

Input = \(\left|{-}\right\rangle _y\)	\(S_x\)	\(S_y\)	\(S_z\)
\(P(\hbar/2)\)	1/2	0	1/2
\(P(-\hbar/2)\)	1/2	1	1/2

The calculations proceed in the same way. The \(S_z\) probabilities give me: \begin{align*} \left|{+}\right\rangle _y &= \frac{1}{\sqrt{2}} \left|{+}\right\rangle + \frac{1}{\sqrt{2}}\cancelto{1}{e^{i\alpha}} \left|{-}\right\rangle \\ \left|{-}\right\rangle _y &= \frac{1}{\sqrt{2}} \left|{+}\right\rangle + \frac{1}{\sqrt{2}}\cancelto{-1}{e^{i\theta}} \left|{-}\right\rangle \\ \end{align*}

The orthongality between \(\left|{+}\right\rangle _y\) and \(\left|{-}\right\rangle _y\) mean that \(\theta - \alpha = \pi\).

But I also know the \(S_x\) probabilities and how to write \(|ket{\pm}_x\) in the \(S_z\) basis. For an input of \(\left|{+}\right\rangle _y\): \begin{align*} \mathcal(S_x = +\hbar/2) &= | \,_x\left\langle {+}\middle|{+}\right\rangle _y |^2 = 1/2 \\ 1/2 &= \Big| \Big(\frac{1}{\sqrt{2}} \left\langle {+}\right| + \frac{1}{\sqrt{2}}\left\langle {-}\right|\Big) \Big( \frac{1}{\sqrt{2}}\left|{+}\right\rangle + \frac{1}{\sqrt{2}}e^{i\alpha} \left|{-}\right\rangle \Big) \Big|^2\\ 1/2 &= \Big| \frac{1}{\sqrt{2}} \frac{1}{\sqrt{2}} \cancelto{1}{\left\langle {+}\middle|{+}\right\rangle } + \frac{1}{\sqrt{2}} \frac{1}{\sqrt{2}}e^{i\alpha} \cancelto{1}{\left\langle {-}\middle|{-}\right\rangle } \Big|^2 \\ &= \frac{1}{4}|1+e^{i\alpha}|^2\\ &= \frac{1}{4} \Big( 1+e^{i\alpha}\Big) \Big( 1+e^{-i\alpha}\Big)\\ &= \frac{1}{4} \Big( 2+e^{i\alpha} + e^{-i\alpha}\Big)\\ &= \frac{1}{4} \Big( 2+2\cos\alpha\Big)\\ \frac{1}{2} &= \frac{1}{2} + \frac{1}{2}\cos\alpha \\ 0 &= \cos\alpha\\ \rightarrow \alpha = \pm \frac{\pi}{2} \end{align*}

Here, again, I can't solve exactly for alpha (or \(\theta\)), but the convention is to choose \(alpha = \frac{\pi}{2}\) and \(\theta = \frac{3\pi}{2}\), making \begin{align*} \left|{+}\right\rangle _y &= \frac{1}{\sqrt{2}} \left|{+}\right\rangle + \frac{1}{\sqrt{2}}\cancelto{i}{e^{i\pi/2}} \left|{-}\right\rangle \\ \left|{-}\right\rangle _y &= \frac{1}{\sqrt{2}} \left|{+}\right\rangle + \frac{1}{\sqrt{2}}\cancelto{-i}{e^{i3\pi/2}} \left|{-}\right\rangle \\ \rightarrow \left|{+}\right\rangle _y &= \frac{1}{\sqrt{2}} \left|{+}\right\rangle \color{red}{+} \frac{\color{red}{i}}{\sqrt{2}} \left|{-}\right\rangle \\ \left|{-}\right\rangle _y &= \frac{1}{\sqrt{2}} \left|{+}\right\rangle \color{red}{-} \frac{\color{red}{i}}{\sqrt{2}} \left|{-}\right\rangle \\ \end{align*}

If I use these two convenctions for the relative phases, then I can write down \(\left|{\pm}\right\rangle _n\) in an arbitrary direction described by the spherical coordinates \(\theta\) and \(\phi\) as:

Discuss the generalize eigenstates: \begin{align*}\ \left|{+}\right\rangle _n &= \cos \frac{\theta}{2} \left|{+}\right\rangle + \sin \frac{\theta}{2} e^{i\phi} \left|{-}\right\rangle \\ \left|{-}\right\rangle _n &= \sin \frac{\theta}{2} \left|{+}\right\rangle - \cos \frac{\theta}{2} e^{i\phi} \left|{-}\right\rangle \end{align*}

And how the \(\left|{\pm}\right\rangle _x\) and \(\left|{\pm}\right\rangle _y\) are consistent.

Take-home messages from today's interactive lecture:

• You can think about the differential \(dx\) as a small amount of \(x\). While technically, it is an infinitesimally small amount, in practice it is almost always possible to think of it as just really, really small. By which, we mean small enough that any error you make by giving it a finite rather than infinitesmal size is too small to make a significant difference in your calculation. Much more about this in later activities.
• A differentials equation tells you how a small change in one variable is related to a small change in one or more other variables. You can interpret these equations geometrically with a figure.
• When doing algebra with differentials equations, you can think of each differential as a new variable.
• When you zap an equation with \(d\), the resulting differentials equation is linear in the differentials, i.e. each of the differentials appears to the first power. This is an open invitation to do linear algebra to rearrange the equation.
• The dimensions of \(dx\) are the same as the dimensions of \(x\). After all, \(dx\) is just a small amount of \(x\).
• Every equation involving differentials should have matching factors of smallness in every term. For example: \begin{align} dx&=2y^2\, dy&\hbox{one factor of \(d\) in every term}\\ dx\, dy&=r\, dr\, d\theta &\hbox{two factors of \(d\) in every term} \end{align} as opposed to: \[\cancel{dx=2y\, dy\, dz}\]

Instructor's Guide

Introduction

The vector differential \(d\vec{r}\) is a fundamental unifying feature of our approach to vector calculus and therefore to electrostatics and magnetostatics. It is used both to define derivatives like the gradient and directional derivatives and integrals in space, from integrals along curves to surfaces and volumes.

This interactive lecture is the first step in defining and evaluating the vector differential. It should introduce the scalar differential and how to calculate it by “zapping with \(d\)”, i.e. it is mainly procedural. Emphasize that zapping with \(d\) linearizes the equations so that it is easy to do substitution with differentials. Also emphasize the role of systems of equations.

An important non-procedural, geometric message is how the differentials (as opposed to differential) equation tells you how small changes in one variable are related to small changes in other variables.

The important takehome messages are listed in the summary handout for students.

You might want to have the student's build their understanding through a thoughtful series of SWBQ's (one at a time with class discussion after each one!)

Global prompt: “Zap the following expression with \(d\). Then draw a figure that shows you what the differentials equation is telling you about the relationship of small changes.”

• After they have done the following problem, remind them that most of them already know how to do this from \(u\)-substitution so that their knowledge about that is in their working memory. \[u=y^3\]
• Also a \(u\)-substitution problem, but they need to do chain rule. Mention that this is a great time for them to be reviewing basic calculus if they need to. Also, they must know, in any give expression, which letters are constants and which are variables. \[u=A e^{-(\frac{x}{a})^2}\]
• This example shows that you don't have to have an equation with an isolated variable on one side AND you can have more than two variables. Look at this example both for \(r=\) constant and \(r\) as a variable. Draw the figure for the case \(r=\) constant. How much harder is the figure if \(r\) is variable? \[x^2+y^2=r^2\]
• A system of equations. Ask the students to solve for \(dx\) and \(dy\) OR for \(dr\) and \(d\phi\). Which is easy and which is hard? Remind students that in \(u\)-substitution they probably know that they need to change ALL instances of the old variable to \(u\)'s, the same applies here. If they are changing to \(r\) and \(\phi\) from \(x\) and \(y\), they should, if possible, change ALL of the instances. Sometimes this is algebraically impossible to do, but they should at least keep trackin their minds that, if they still have \(x\)'s and \(y\)'s, they should now think of these as functions of \(r\) and \(\phi\) and not independent, i.e. \(x=x(r,\phi)\) and \(y=y(r,\phi)\) even if they can't write down explicitly what those functions are. These nuances are particularly important in thermodynamics. (Note: We like to use \(\phi\) for polar coordinates so that these coordinates agree with the physicists' use of spherical coordinates.) \begin{align} r^2&=x^2+y^2\\ \tan\phi&=\frac{y}{x} \end{align}
• (Optional:) Another system of equations. Ask the students to solve for \(dx\) and \(dy\) OR for \(dr\) and \(d\phi\). Which is easy and which is hard? \begin{align} x&=r\cos\phi\\ y&=r\sin\phi \end{align}