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So, $\frac{dy}{dx}$ is defined as a limit. However, because of the

chain rule: $\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}$

And inverse function theorem: $\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}}$

It seems that $\frac{dy}{dx}$ has "fraction like" behavior.

Could somebody explain to me what justifies the second equality?

length ds = $\sqrt{(dx)^2+(dy)^2}=\sqrt{{1+(\frac{dy}{dx})^2}}dx$

Page 573 equation (1) of Gilbert Strang's calculus text

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    Where did you find that equality? That equality cannot be rigourously proven, since there is no definition of $dx$ and $dy$ by themselves. It can be justified heuristically, though. Is that what you want? – Jackozee Hakkiuz Dec 06 '21 at 22:34
  • sorry edited my post again with reference to Strangs calculus text –  Dec 06 '21 at 22:38
  • In the sense of Riemannian geometry maybe this can be justified as saying that the Riemannian metric $g$ is the usual inner product $$\mathrm dx^2+\mathrm dy^2$$ but then I am not sure what the second equality means (maybe $y$ depends parametrically on $x$ in some way?) – Maximilian Janisch Dec 06 '21 at 23:06

3 Answers3

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The expression $\sqrt{dx^2+dy^2}$ is shorthand - you can’t typically combine differentials like this and what would it even mean to integrate that? What it means is that if $x$ and $y$ are variables of a curve that are parametrized by some $t$, then to integrate $\sqrt{\frac{dx}{dt}^2+ \frac{dy}{dt}^2}dt$. If you factor out the $\frac{dx}{dt}$ term, you can apply the chain rule to the inside using $\frac{dy}{dt}/\frac{dx}{dt}= \frac{dy}{dx}$. On the outside, you can do u-substitution on your integral to get $\frac{dx}{dt}dt=dx$, giving you your alternative expression.

Eric
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$$ \sqrt{a^2+b^2} = \sqrt{a^2\left(1+\frac{b^2}{a^2}\right)}=a\sqrt{1+\frac{b^2}{a^2}} \quad \text{with}\ \ a=dx\ \ \text{and}\ \ b = dy.$$

You would need to justify why $\ dx\geq 0\ $ for the second equality above to hold, but that's not the focus of your question...

$$$$

Further to questions in the comments, let's say you want to calculate the (arc) length of $\ y=x^2\ $ between $\ x=0\ $ and $\ x=1.$ Then see here for the formula and how it is derived.

Adam Rubinson
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  • $dx$ and $dy$ aren't numbers – Maximilian Janisch Dec 06 '21 at 23:05
  • @MaximilianJanisch Just because they are not numbers does not mean you cannot treat them as numbers heuristically and fill in the correct details later. – Adam Rubinson Dec 06 '21 at 23:08
  • But what you wrote down in its current form is meaningless if one doesn't say what $\mathrm dx$ and $\mathrm dy$ are (to me they look like differential $1$-forms i.e. $1$-covariant tensor fields on some manifold) and how you multiply them and take square roots of them and so on. (I think specifying all of those "details" would be the main part of answering the above question.) (Admittedly it would be the duty of the exercise asker to provide a part of those details for us but oh well...) – Maximilian Janisch Dec 06 '21 at 23:09
  • @MaximilianJanisch Like I said, I am treating them as numbers, which is what the OP wants us to do. And it is not like this is entirely meaningless and non-adaptable to the strict definition of $ds,$ which is the infinitesimal length of a curve in 2d space. I'm not sure why you're bringing Riemann geometry or Tensor calculus into this when line integrals are a standard part of real analysis/calculus. e.g. when you want to compute the (arc) length of a curve like $y=x^2$ between two points... – Adam Rubinson Dec 06 '21 at 23:17
  • I had no clue what $\mathrm ds$ is supposed to mean (in fact I didn't know until looking into the text two minutes ago that we are dealing with a curve) so if you know it, please add that to your answer :) (I don't want to repeat myself but without any such added details, I am afraid that your answer is meaningless). – Maximilian Janisch Dec 06 '21 at 23:18
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I had a quick look at Gilbert's text and here is what I understood:

You are given a curve $\gamma=(x,y):I\to\mathbb R^2$, where $I\subset\mathbb R$ is some open interval.

Suppose for example that the curve is smooth and that $\gamma'(t)\neq 0$ for all $t\in I$.

Then as derived by me here, the length of $\gamma$ is given by $$\int_I \|\gamma'(t)\| \,\mathrm dt.$$ If you use the strange-looking notation $\gamma' = (\mathrm dx, \mathrm dy)$, then you obtain the "$\mathrm ds = \sqrt{\mathrm dx^2+\mathrm dy^2}$" part. (I will maybe give a justification for this notation down below in the future, but currently that part is incomplete.)

Furthermore, an application of the implicit function Theorem [2; Satz 169.1], details left as exercise to you, gives that there exists (after relabeling the axes $x$ and $y$ if necessary) for each $t\in I$ a neighborhood $J\subset I$ and a smooth function $f_J:J\to\mathbb R$ such that $\gamma(t) = (x(t),f_J(x(t)))$ for all $t\in J$.

Then the length of $\gamma\vert_J$ is $$\int_J \|\gamma'(t)\| \,\mathrm dt = \int_J \|(x(t),f_J(x(t)))'\| \,\mathrm dt=\int_J\lvert x'(t)\rvert\sqrt{1+f_J'(x(t))^2}\,\mathrm dt.$$ Now substitute $u=x=x(t)$ to obtain, in your notation, "$\mathrm ds = \sqrt{1+\left(\frac{\mathrm dy}{\mathrm dx}\right)^2}\,\mathrm dx$". (Here, $\frac{\mathrm dy}{\mathrm dx}$ is short for $f_J'$.) (And indeed you have to be wary of the sign of $x'$.)

This part is not finished yet.

Then every point $t\in I$ has a neighborhood $J\subset I$ such that the image of $\gamma\vert_J$ is an embedded $1$-dimensional smooth submanifold of $\mathbb R^2$. This follows from the Local Parametrization Theorem [1; Theorem 2.5]. Indeed, Such a $f_J$ provides the local parametrizations for [1; Theorem 2.5].

Literature

[2] Harro Heuser, Lehrbuch der Analysis. Teil 2. 11. Auflage. 2000.

Maximilian Janisch
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