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I was watching Khan Academy's video on 'Introduction to the line integral' when he does something interesting. Namely, he 'multiplies' a term by dt/dt:

$\frac{dt}{dt} * \sqrt{(dx^2 + dy^2)}$

to change it into the more workable

$dt \cdot \sqrt{\frac{dx}{dt}^2 + \frac{dy}{dt}^2}$

My question is what mathematical rational for how and why this 'abuse' of differential notation can be used, and why treating it as an algebraic value 'works' as such. I have read this thread: Is it mathematically valid to separate variables in a differential equation? which provided a really good explanation for this sort of thing with regards to that form of DE. However, I sometimes come across such treatments of differential operators and can never seem to find the more rigorous explanations.

Cheers and I hope my question is not too unclear.

Community
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masiewpao
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    It is more or less linearization for well behaved functions: $f(x + \Delta x) = f(x) + f'(x) \Delta x + O((\Delta x)^2)$ and $df / dx \approx \Delta f / \Delta x$.for small $\Delta x$, with $\Delta f = f(x + \Delta x) - f(x)$.. – mvw Apr 20 '16 at 17:52
  • Hmm, I don't quite understand how linearisation explains the usage of the differentials in this way, but I will see if I can work it out, thanks! – masiewpao Apr 20 '16 at 18:26
  • What is your understanding of $dx$ here? If you don't care you may as well be content with the observed manipulation of terms. – Christian Blatter Jun 19 '16 at 19:19

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There is a formulation of calculus in terms of the field of hyperreal numbers which provides a formalization of this type of reasoning. This text by H. Jerome Keisler discusses the concept in chapter 2, which is about derivatives.

As discussed on the linked wiki page, the hyperreal numbers are an extension of the real numbers which include nonzero infinitesimal numbers. A hyperreal number $\Delta x$ is called infinitesimal when $0\le\Delta x<r$ for any given real number $r$. Taking the hyperreals to be countably infinite sequences of real numbers as in the ultrapower construction of the hyperreals, we see that $\left(\frac{1}{n}\right)_{n\in\mathbb{N}}=\left(1,\frac{1}{2},\frac{1}{3},\dots\right)$ is not $0=(0,0,0,\dots)$ since the two sequences differ in a cofinite number of entries. We also see that $\left(\frac{1}{n}\right)_{n\in\mathbb{N}}<r=(r,r,r,\dots)$ for any $r\in\mathbb{R}$, since we have that $\frac{1}{n}<r$ for a cofinite number of $n\in\mathbb{N}$, so we have exhibited an infinitesimal number.

The advantage of this system is that the hyperreals, including infinitesimals, obey the usual rules of arithmetic. Thus, for example, we can compute the "infinitesimal secant" from $x$ to $x+\Delta x$ for $f(x)=x^2$ and a nonzero infinitesimal $\Delta x$ (like our number $\left(\frac{1}{n}\right)_{n\in\mathbb{N}}$) by computing $$\frac{(x+\Delta x)^2-x^2}{\Delta x}=\frac{x^2+2x\Delta x+(\Delta x)^2-x^2}{\Delta x}=\frac{2x\Delta x+(\Delta x)^2}{\Delta x}=2x+\Delta x.$$

We then define a standard part function $\operatorname{st}(x)$ which takes a hyperreal number and returns the unique real number $r$, if it exists, such that $x-r$ is infinitesimal. We then see that $\operatorname{st}(2x+\Delta x)=2x$ and go on to define the derivative as $$f'(x)=\operatorname{st}\left(\frac{f(x+\Delta x)-f(x)}{\Delta x}\right)$$ for any nonzero infinitesimal $\Delta x$.

In more generality, any time we are working with differentials, such as $dt$ in your example, we can always treat $dt$ as a nonzero infinitesimal and perform arithmetic as usual as long as we remember that we have to be able to take the standard part correctly at the end of our computation.

Charlotte Aten
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  • You should possibly edit your answer to explain some of the mentioned concepts. – Colbi Jun 19 '16 at 19:43
  • Hi there, thanks for your answer! I THINK I have some intuition of what you're saying, however why is it that in some texts this sort of treatment is considered 'wrong'? By that I mean dy/dx as a whole represents something; the derivative, which means that you cannot treat dy or dx as a variable, since dy/dx itself is not ''one variable over another''? – masiewpao Jul 19 '16 at 04:13
  • I can think of two reasons why you might be told that this sort of thing is "wrong". The first is that the author of your text, despite knowing that hyperreal numbers and so forth exist, does not want to get sidetracked by discussing infinitesimals and simply wants you to be able to perform the computation. The second is that the author is unaware that this work has been done, as infinitesimals were not on solid mathematical ground for a long time. Indeed, the study of calculus in this manner is still often called "nonstandard analysis". – Charlotte Aten Jul 24 '16 at 15:38