What is the geometrical difference between $\frac{dr}{dt}$ and $dr$?

Question

I am currently studying multivariate calculus and came across a section in my book which perplexes me. It states that a field $F(x, y) = (P(x, y), Q(x, y))$ can have the parameter $r(t) = (x(t), y(t))$. I follow thus far. The book then continues with explaining that the formal calculation with differentials is

$$ \frac{dr}{dt}=(\frac{dx}{dt},\frac{dy}{dt}) \Rightarrow dr = (dx, dy) $$

The book mentioned it simplified by multiplying with $dt$ on each side. I follow the math but struggle with understanding what this is actually saying geometrically. what does $dr$ and $\frac{dr}{dt}$ mean geometrically? I have never fully grasped the implication of multiplying and moving around the "delta" of equations and when it is nonsensical. My guess would be that $dr$ is a vector and $\frac{dr}{dt}$ is the rate of change the curve has at a given point. However, the book does not mention these things so I cannot verify this.

For context: The parametrisation above, in the book, is used one section down to explain the formula for curve integrals.

Answer 1

Let us work in one dimension for simplicity. Suppose you have a coordinate $x(t)$ parametrized by time telling you the location of (say) a particle on the real line. Then geometrically $dx$ exactly means an infinitesimal increment in the position of a particle due to an increment in time $dt$. Basically $dx = \dfrac{dx}{dt} dt$. And indeed $\dfrac{dx}{dt}$ is the velocity of the particle at a particular time. In higher dimensions, it's the same thing with $\vec{r}$ replacing $x$.

Regarding manipulating infinitesimals, it can indeed be a subtle issue in certain cases, but let me mention this: Everything could be done without manipulating infinitesimals because we are never actually interested in the infinitesimal expressions themselves; the infinitesimals are just a means to an end; a means of ultimately arriving at a finite expression. Let me give you a simple example of how it works in practice.

This is a really simple example but should give you the general idea. Suppose you have $\dfrac{dx(t)}{dt} = f(t)$ and you want to solve for $x(t)$. A common thing you would see in introductory calculus and physics textbooks is that you first multiply both sides by $dt$ and then integrate them as if they're separate variables: $$dx=f(t)dt \Rightarrow \int dx = \int_{0}^{t}f(t)dt \Rightarrow x(t)-x(0)=\int_{t_0}^{t}f(t)dt$$

But notice all these steps could be done in a rigorous manner without resorting to infinitesimals:

$$\dfrac{dx}{dt} = f(t)$$

$$\int_{0}^{t}\dfrac{dx}{dt}dt = \int_{0}^{t}f(t)dt$$

$$x(t)-x(0) = \int_{0}^{t}f(t)dt$$

Where we have used the Fundamental Theorem of calculus in going from step 2 to 3. Notice that nowhere did we use infinitesimals since they are not needed. They are just a convenient way of manipulating stuff, and if you think about it, the expression $dx = \dfrac{dx}{dt} dt$ is really just an infinitesimal Fundamental Theorem of calculus. That should give you some comfort in knowing that all these infinitesimal expressions actually have their roots in rigorous theorems, otherwise they won't work in the first place.

Now I should mention that infinitesimal expressions such as $dx$ can in fact be given a rigorous meaning, namely as covectors in the context of differential geometry. Even though in differential geometry they're not technically "infinitesimal" but they were constructed in such a way as to be able to think about them as infinitesimal expressions and manipulate them as we manipulate infinitesimal expressions. An actual approach to infinitesimal expressions has been constructed though in the field of mathematical logic (in particular, Model theory). But I must say that these approaches weren't particularly illuminating for me in understanding any new features of infinitesimals so I wouldn't worry about them for now if I were you.

Answer 2

The line integral of $F:\mathbb R^2\to\mathbb R^2$ over a curve $C$ parametrized by $r:[a,b]\to\mathbb R^2$ is defined by $$\int_a^b F(r(t))\cdot r'(t)\,dt$$

If we write $F(x,y)=(P(x,y),Q(x,y))$, $r(t)=(x(t),y(t))$ and (for simplicity) omit the arguments, the above integral becomes $$\int_a^b [Px'+Qy'] \,dt.$$

In Leibniz notation: $$\int_a^b \left[P\frac{dx}{dt}+Q\frac{dy}{dt}\right] \,dt.$$

In the words of C. H. Edwards, "a classical abbreviation" for this last integral is $$\int_C Pdx+Qdy.$$

Note that this integral can be written as a formal dot product: $$\int_C (P,Q)\cdot (dx,dy)$$

Thus, in the words of Williamson et al, the integral "can be still further shortened by writing" $dr=(dx,dy)$. Under this notation, the integral becomes $$\int_C F\cdot dr,$$ which is the classical notation for line integral.

From this point of view, $dr$ is just an abbreviation for $(dx,dy)$, which is just an abbreviation for $(x'(t),y'(t))=(\frac{dx}{dt},\frac{dy}{dt})$. So, the symbol $dr$ as well as the equality $dr=(dx,dy)$ have no meaning by themselves because they do not came from ordinary fractions. The formal calculation $\frac{dr}{dt}=\left(\frac{dx}{dt},\frac{dy}{dt}\right)\Rightarrow dr=(dx,dy)$ is just a method to help us remember the abbreviation. This is the basic meaning.

It seems (but I'm not sure) that the standard notation $$\int_C F\cdot dr=\int_a^b F(r(t))\cdot r'(t)\,dt$$ is just a generalization of the notation for line integral of scalar functions: $$\int_C f\cdot ds=\int_a^b f(r(t))\cdot |r'(t)|\,dt.$$

According to C. H. Edwards: "Historically this notation resulted from the expression $ds=|r'(t)|\,dt$ for the length of the 'infinitesimal' piece of path corresponding to the 'infinitesimal' time interval $[t,t+dt]$". Then, he says: "Later in this section we will provide the expression with an actual mathematical meaning (in contrast to its 'mythical' meaning here)". The "actual meaning" is given in terms of differential forms, but this is a more advanced point of view which I'm not sure you are interested in.