While teaching multi-variable calculus for the first time in a while, I came across a tricky notational point in our textbook (Thomas' calculus - I'm not sure how widespread this notation is).

When $\mathbf{r}(t)$ is a vector-valued function, our book writes the arc length parametrization as $\mathbf{r}(s)$, with the same name for the function, changing only the variable. From the right perspective, this makes a lot of sense: one thinks of $\mathbf{r}$ as a polymorphic function that interprets its inputs in the uniquely sensible way, so when passed a value of $t$ it returns $\mathbf{r}(t)$, and when passed an arc length returns $\mathbf{r}(s)$.

On the other hand, this isn't the perspective my students (mostly first-year college students) have on functions. It violates the way functions normally behave---if $\mathbf{r}(t)=\langle t,t^2\rangle$ then $\mathbf{r}(s)$ doesn't equal $\langle s,s^2\rangle$. It creates some ambiguities---what does $\mathbf{r}(2)$ mean?

Assuming I'm stuck with this notation (it's the one our textbook uses, and I share a final with other people using the same book), what are good practices to work with and explain this to minimize how confusing it is? (And maybe help my students get the most out of this new perspective on functions.)

  • $\begingroup$ students are not so logical by in large. For example, try writing $f = f(x)$ and $g = g(y)$. I don't think many of them will complain you conflated the function with its value. Instead, they will usually understand what is meant by such shorthand. The same with $\vec{r}$. One sneaky way around it, just say $\vec{r}(s)$ is really just an abbreviation for $\vec{r}(s(t))$. Or, just be honest, it is an abuse of notation. That said, I really think most students do not notice if you don't draw attention to it. $\endgroup$ Commented Jan 31, 2018 at 3:50
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    $\begingroup$ @TheChef I like the idea of telling them it's an abbreviation. But I don't agree that student won't be confused. My students don't complain about conflating f with f(x), but they definitely find it confusing. They don't complain because they don't understand either concept well enough to articulate why it's confusing, but they definitely don't understand that shorthand. $\endgroup$ Commented Jan 31, 2018 at 5:17
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    $\begingroup$ @TheChef I doubt any mathematician really understands what is meant by that shorthand. They all think they understand it, but I haven't seen anyone who is able to formalise it (in the sense of implementing a proof checker on a computer that can consistently figure out what is going on). $\endgroup$ Commented Jan 31, 2018 at 9:46
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    $\begingroup$ Maybe address the ambiguity, give examples, and from then on use $\mathbf{r}_t(t)$ and $\mathbf{r}_s(s)$ on the blackboard and in exercises. And alert them to spots in the text where the ambiguity makes a difference. $\endgroup$ Commented Jan 31, 2018 at 13:06
  • $\begingroup$ @MichaelBächtold it an abuse of notation. That means two different ideas are being described by the same symbol in this case. I would also be unable to code it, yet, I think it has meaning. When we write f = f(x) and g = g(y) it likely means I have two variables $x$ and $y$ and for whatever reason $f(x,y)$ has partial derivative w.r.t. $y$ of zero and likewise $g(x,y)$ has partial derivative w.r.t. $x$ of zero. That said, Henry Towsner is correct, many students don't understand it. $\endgroup$ Commented Feb 1, 2018 at 0:38

3 Answers 3


By coincidence, I'm teaching multi-variable Calculus for the first time this semester, and have given some thought to how to handle this precise issue.

This seems to me to be closely related to an example I first encountered at http://math.oregonstate.edu/bridge/ideas/functions/, which I strongly encourage you to visit before you read the rest of this answer.

This example is discussed in some detail in a paper by Edward F. Redish and Eric Kuo, published at https://link.springer.com/article/10.1007/s11191-015-9749-7 (free preprint version at https://arxiv.org/pdf/1409.6272). The key insight concerns what the authors call Corinne's Shibboleth:

One of your colleagues is measuring the temperature of a plate of metal placed above an outlet pipe that emits cool air. The result can be well described in Cartesian coordinates by the function $$T(x,y) = k(x^2 + y^2)$$ where $k$ is a constant. If you were asked to give the following function, what would you write? $$T(r,\theta) = \textrm{ ?}$$ The context of the problem encourages you to think in terms of a particular physical system. Physicists tend to think of $T$ as a physical function – one that represents the temperature (in whatever units) at a particular point in space (in whatever coordinates). Mathematicians tend to consider $T$ as a mathematical function – one that represents a particular functional dependence relating a value to a pair of given numbers.

As a result, physicists tend to answer that $T(r,\theta) = kr^2$ because they interpret $x^2 + y^2$ physically as the square of the distance from the origin. If $r$ and $\theta$ are the polar coordinates corresponding to the rectangular coordinates $x$ and $y$, the physicists' answer yields the same value for the temperature at the same physical point in both representations. In other words, physicists assign meaning to the variables $x$, $y$, $r$, and $\theta$ – the geometry of the physical situation relating the variables to one another.

Mathematicians, on the other hand, may regard $x$, $y$, $r$, and $\theta$ as dummy variables denoting two arbitrary independent variables. The variables $(r, \theta)$ or $(x,y)$ don't have any meaning constraining their relationship. Mathematicians focus on the mathematical grammar of the expression rather than any possible physical meaning. The function as defined instructs one to square the two independent variables, add them, and multiply the result by $k$. The result should therefore be $T(r,\theta) = k(r^2 + \theta^2)$.

Typically, a physicist will be upset at the mathematician's result. You might hear, "You can’t add $r^2$ and $\theta^2$! They have different units!" The mathematician is likely to be upset at the physicist's result. You might hear, "You can’t change the functional dependence without changing the name of the symbol! You have to write something like $$T(x,y) = S(r,\theta) = kr^2.$$ To which the physicist might respond, "You can’t write that the temperature equals the entropy! That will be too confusing." (Physicists often use $S$ to represent entropy.)

(Parenthetically, coming up soon in my multivariable Calculus course -- and presumably in the OPs course as well -- are functions expressed in polar, cylindrical, and spherical coordinates, so the particulars of this example may be salient in that context as well.)

The relationship between this example and the one in the OP is that the convention of using the notations $\mathbf{r}(t)$ and $\mathbf{r}(s)$ to refer to two different parametrizations of the same path seems (to me) to be in accordance with the "Physicist's interpretation": the variables $t$, $s$ and $\mathbf{r}$ are understood to have physical meaning (respectively: time, arc length, and position), so that $\mathbf{r}(t)$ means "the position corresponding to time $t$" while $\mathbf{r}(s)$ means "the position corresponding to arc length $s$". This is perfectly coherent and reasonable if the variables are understood as "physical variables", but it is at odds with the mathematician's understanding of a function as a mapping that assigns a value to an arbitrary input represented by a dummy variable.

What to do about it? Here I think one strategy that can be helpful is to use less functional notation, rather than more. Consider the example of an object moving in a circular path of radius $5$, i.e. $\mathbf{r}(t)=\left<5\cos(t),5\sin(t)\right>$. I would propose actually eliding the variable from the left-hand side entirely, and writing it this way:

Consider a particle whose position at time $t$ is given by $$\mathbf{r} = \left<5\cos(t),5\sin(t)\right>$$ Then the velocity is $$\mathbf{r}' = \left<-5\sin(t), 5\cos(t)\right>$$ so the speed of the object is $$|\mathbf{r}'| = \sqrt{25\cos^2(t) + 25\sin^2(t)} = 5 $$ and therefore the distance traveled between $t=0$ and $t=T$ is $$ \int_0^T 5 dt = 5T$$ This gives us the relationship $s = 5t$, where $s$ is the distance traveled by time $t$. Inverting this relationship we have $t = \frac{s}{5}$, which allows us to express the position of the object when it has traveled a distance $s$ as $$\mathbf{r} = \left<5\cos\left(\frac{s}{5}\right),5\sin\left(\frac{s}{5}\right)\right>$$

Notice that in this exposition functional notation has been almost completely avoided. An expression like $\mathbf{r}(2)$ (which, as the OP notes, is ambiguous) would be handled by using a descriptive phrase: either "$\mathbf{r}$ when $t=2$" or "$\mathbf{r}$ when $s=2$", depending on which is intended.

If you think about it, this is really no different from the common single-variable usage of writing something like $y = 5x^3 +2x$, which also avoids functional notation. And just as we do not hesitate to write $\frac{dy}{dx}$, I think it is perfectly reasonable to write $\frac{d\mathbf{r}}{dt}$ and $\frac{d\mathbf{r}}{ds}$ to refer to the respective derivatives.

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    $\begingroup$ I very much like you answer and would have suggested the same for teaching. Let me just add that there is a well known notation for "$\mathbf{r}$ when $s=2$". Its $\mathbf{r}|_{s=2}$. But beware that many colleagues might object to it when it is not applied to a derivative. Also I'm not a 100% sure about its semantics. For example $d(x|_{x=2})$ cannot be the same as $(dx)|_{x=2}$, so it's not literally the same as logicians substitution. $\endgroup$ Commented Feb 1, 2018 at 8:12
  • $\begingroup$ That example is fantastic, and clarifying. $\endgroup$ Commented Feb 1, 2018 at 20:07
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    $\begingroup$ Note that it's not just reasonable to write $d\mathbf{r}/dt$ and $d\mathbf{r}/ds$ but preferable, since otherwise how is one to know which one $\mathbf{r}'$ is supposed to be? (If only two parametrizations are used, one with independent variable $t$ given first, and another with independent variable $s$ parametrized by arclength, then it's probably a safe bet that $d\mathbf{r}/dt$ is meant, and you should tell the students so, but I still always write $d\mathbf{r}/dt$ in class.) $\endgroup$ Commented Aug 20, 2018 at 8:07

I don't have much to add to mweiss' nice answer in terms of teaching suggestions. But I would like to add my own point of view on what the abuse of notation $\mathbf{r}=\mathbf{r}(s)=\mathbf{r}(t)$ means and where it comes from historically.

The first person to do this abuse of notation (implicitly) seems to have been Jacobi around 1830 (I suggest you read that link after reading the rest of what I'll say). Note that 1830 is more than 100 years after Bernoulli and Euler introduced the notation $y=f(x)$ and more than 140 years after Leibniz startet to talk of functions. In the period between Leibniz and Jacobi we find luminaries like Euler, Lagrange, Laplace, Fourier, Bolzano, Cauchy and Gauss who, as far as I can tell, never did this. So it's not the case, as some people believe, that mathematicians and physicist have been writing $y=y(x)$ ever since the invention of functions. Even after Jacobi there are many people, like Riemann, Peano or Planck who apparently didn't do it. So it's also not the case that physicists invented $y=y(x)$ or somehow need it more urgently than mathematicians.

So why did Jacobi start doing it? In the above link you can read what he himself had to say about it (I highly recommend it to anyone teaching multivariable calculus), but his own words actually don't explain it completely. To understand it better we first need to understand how the word function was used prior to about 1900. Most mathematicians seem unaware of the dramatic change of meaning this word underwent during the period 1900-1920. And it's a non-trivial task for a modern mathematician with a set theoretic perspective to make sense of what it meant prior to 1900. But let's try.

If you open any calculus textbook written after Leibniz and before at least 1910 (a ~200 year period) you'll find that the word "function" is always defined as function of something. Here is a typical example from the end of that period taken from Peano's Calcolo differenziale e principii di calcolo integrale 1884, p.3:

Among the variables there are those to which we can assign arbitrarily and successively different values, called independent variables, and others whose values depend on the values given to the first ones. These are called dependent variables or functions of the first ones.

We shall first treat functions of a single independent variable, and we shall say that: a function $y$ of $x$ is given in an interval $(a, b)$, if to any value of $x$ in between $a$ and $b$ corresponds a unique and determinate value for $y$ - whatever the means of determining it.

So for example $x^{2}$ is a definite function of $x$ for any value of $x$, and is hence given in any interval; $\sqrt{x}$ understood as the arithmetic root of $x$, is given for all positive values of $x$ ; while $\frac{1}{1} + \frac{1}{2} +\frac {1}{3} +\cdots + \frac{1}{x}$ is a function of $x$ defined only for integer and positive values of the variable, etc.

So a function, like $x^2$ or $y$, is a variable quantity, just like $x$, only that it satisfies some additional property. If you feel uncomfortable with "$y$ is a function of $x$" because you are so used to modern functions, maybe paraphrasing it as "$y$ depends on $x$" helps a bit. (Certainly if $f:\mathbb{R}\to \mathbb{R}$, no modern mathematician would say that $f$ depends on $x\in \mathbb{R}$.) Hence whenever they called something a function, they had to add of what that thing is a function. But often it was clear from the context or the notation, so they would soon drop the "of $x$" and simply talk of functions. So for example on p.13 we find Peano writing

One says that a function $f(x)$ becomes maximal relative to an interval $(a,b)$ when $x=x_0$...

To be correct he should have said something like: "One says that a function of $x$, $f(x)$, becomes ...". But it was clear form the notation. (Observe that nothing prevents $f(x)$ from being a function of something else too. For example when $x$ is itself a function of $t$, then $f(x)$ would also be a function of $t$.) We inherited the phrase "$f(x)$ is a function" —which we find in every modern calculus textbook (and btw. you used the question)— from this pre 1900 period, even though according to our modern convention we should correctly say "$f$ is a function".

To emphasise again the difference between "function of ..." and our modern "functions": when $y=f(x)$ they called $y$ and $f(x)$ a function, while we call $f$ a function.

If they called $f(x)$ the function, how did they call $f$? After all the notation $f(x)$ existed since Bernoulli ~1718. Didn't they also call $f$ a function? No, not officially. The first ones (Bernoulli, Euler, Lagrange) called $f$ the characteristic of the function $f(x)$. I interpret this as saying: its just a character used to distinguish one function of $x$, say $f(x)$, from another, say $g(x)$. But mainly people before 1900 didn't call $f$ anything at all! Peano for example doesn't in his calculus book. In fact, no one treated $f$ as a mathematical object on its own, before Dedekind, Peano, Cantor and Frege (to a certain extent independently) changed that around 1890. (There are some surprising historical bits in that thread. For example: Dedekind, Peano and Cantor all gave $f$ a new name, calling it resp. "map", "prefix sign for a function" and "allocation". They all preserved the original use of "function" for $f(x)$! Moreover Dedekind formulated his notion of map nine years before realising that he could identify it with the $f$ appearing in their functions $f(x)$. Only Frege (unfortunately) suggested calling $f$ a function, and somehow it became standard. We would probably have less difficulty communicating today, if $f$ had not gotten the name function.)

Returning to $y=y(x)$ and why Jacobi started that. Besides the good reasons about partial derivatives he gives in De Determinantibus 1840, I suspect that he simply wanted a notation to indicate of what a certain variable is to be considered a function. I used to believe that in the equation $y=y(x)$ the $y$ on the left and on the right were objects of different types which by abuse where given the same name (the right one being of type $\mathbb{R}\to\mathbb{R}$, the left one of type $\mathbb{R}$.) But I now think that this is not what Jacobi intended and how physicist would like to use it. Instead, we should literally think of the $y$ on the right and left as the same object (of type "variable quantity"), and what is being abused is the notation for function application $f(x)$. In other words: had Jacobi chosen another notation like square brackets $y[x]$ to annotate that $y$ is considered as function of $x$, instead of using the already existing notation $f(x)$ for "application of a map to a variable", we might have less trouble with it now. I suspect one could formalise this idea similarly to how computer scientist formalise "ascription". See for example Pierce, Types and Programming Languages.

Having said that, I don't think that by doing that we would gain much. Writing $\mathbf{r}=\mathbf{r}[t]$ seems mainly an aide for the reader, and we might as well do without it, as mweiss suggested. The harder problem seems to be to correctly formalise the notion of "variable quantity" and surrounding things like the notation $dy/dx$.

  • $\begingroup$ This is an interesting history that I wasn't aware of. I agree that mathematicians are too glib about taking the modern set theoretic definition of a function for granted, but this is a facet of that I haven't seen before. $\endgroup$ Commented Feb 1, 2018 at 20:07
  • $\begingroup$ This is a wonderful answer. I probably am not careful enough in my teaching. I'm torn between being correct and making connections. I know if I am a bit fuzzy it allows me to see more students as correct. Anyway, one thing I think we can say about math before 1900 is that the focus was on equations rather than functions. Your point that the word function has been revised in our modern usage vs. the 19th century use is very helpful. $\endgroup$ Commented Mar 27, 2019 at 2:45

When I was a student learning the difference, I found the unit-speed interpretation to be more helpful. Try using an animation in Matlab, Python, or Mathematica to illustrate this for simple curves by simultaneously tracing out $r(t) = (\cos{2t}, \sin{2t})$ and $r(s)$ or $\tilde{r}(t) = (t, t^2)$ and $\tilde{r}(s)$. Seeing the curve being traced out in its "normal" parametrization and its arclength parametrization will help show that the parameters are describing the same curve but at different speeds (the two parameters for the curve corresponding to $\tilde{r}$ show this especially well since the standard parametrization is not constant speed).

I think students will understand that implicitly, the arclength parametrization is just hiding the composition with a function involving the inverse of the arclength integral and doing so allows you to travel at unit speed. In other words, $r(s)$ really just means $r(t(s))$. The technicalities may be better suited for a differential geometry course (see the first chapter of Elementary Differential Geometry by Andrew Pressley), as concepts like regularity of a curve and the Inverse Function Theorem are required.

  • $\begingroup$ If the $r$ is to have the same meaning in $r(s)$ as it has in $r(t)$, except that $r(s)$ is really an abbreviation, then $r(s)$ has to be an abbreviation of $r(t(s))$, not of $r(s(t))$. And if the $t$ is also to have the same meaning in $r(t)$ as in $r(t(s))$, then $r(t)$ also has to be an abbreviation for $r(t(s))$. There is some sense in that, to say that the fundamental parametrization is always by arclength. (Although note that this parametrization still depends on orientation, and possibly a choice of base point, at least for infinite curves.) $\endgroup$ Commented Aug 20, 2018 at 8:16
  • $\begingroup$ Thanks for the correction! $\endgroup$ Commented Mar 24, 2019 at 18:38

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