What happens when we divide two different rates of change?

Question

I was learning parametric equations and I saw this:

\begin{align} \frac{dy}{dx} &= \frac{\frac{dy}{dt}}{\frac{dx}{dt}}.\\\\ \end{align}

The situation is, for example, that there are two different parametric equations in terms of $t.$ And we want to find $dy/dx.$

\begin{align} y(t) &= t^2+1 \\\\ \frac{dy}{dt} &= 2t \\\\ x(t) &= t+9 \\\\ \frac{dx}{dt} &= 1 \\\\ \frac{dy}{dx} &= \frac{2t}{1} \\\\ \end{align}

I know mathematically (!), that $dt$ cancels out, and we get $dy/dx.$

\begin{align} \frac{dy}{dx} &= \frac{dy}{dt} \times \frac{dt}{dx}.\\\\ \end{align}

Question: 1 But I don't understand this intuitively. How can dividing two different rates of change give us the rate of change between two different equations?

And why we are even doing that? We could combine $y(t)$ and $x(t)$ in one equation and then calculate $dy/dx,$

\begin{align} x(t) &= t+9 \\\\ t &= x-9 \\\\ y(x) &= (x-9)^2+1 \\\\ \frac{dy}{dx} &= 2(x-9) \\\\ & OR \\\\ \frac{dy}{dx} &= 2((t+9)-9) \\\\ \end{align}

Question: 1.2 And not only that, but I have been taught that $dy/dt$ or $dx/dt$ or $dy/dx$ doesn't mean we are diving two quantities; it's just a notation and it's not exactly a two numbers division. So, how can we treat $dy/dx$ as a quotient in the above step, and do it many times, while in solving KVL equations, we treat $dy/dx$ as if it is a simple algebraic term and we move $dx$ or $dy$ around like it is a normal term?

\begin{align} \frac{dy}{dx} &= \frac{y(x+dx)-y(x)}{(x+dx)-(x)}\\\\ \end{align}

There must a deep intuitive understanding I am lacking about division, derivatives and parametric equations.

Answer 1

Question n°1. You ask why $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}y/\mathrm{d}t}{\mathrm{d}x/\mathrm{d}t}$ can be interpreted itself as a rate of change. I guess that a concrete example will be more meaningful. Let's say that $x(t)$ and $y(t)$ measure euros and dollars through time. Then, $\frac{\mathrm{d}y}{\mathrm{d}t}$ and $\frac{\mathrm{d}x}{\mathrm{d}t}$ correspond to their respective variation rates in time. As a consequence, $\frac{\mathrm{d}y}{\mathrm{d}x}$ will represent the change rate of dollars versus euros (through time).

But why to do so ? Because it sometimes allows to handle easier expressions than a straightforward computation of $\frac{\mathrm{d}y}{\mathrm{d}x}$. If $x = f(t)$ and $y = g(t)$, then one has $y = g(f^{-1}(x))$. Yet, $f$ and $g$ might be indeed easier to differentiate than $g \circ f^{-1}$, or even worse, you might be unable to find a closed form for $f^{-1}$ (in particular when $f$ is not bijective).

Question n°2. In fact, the expression $\frac{\mathrm{d}y}{\mathrm{d}x}$ may be treated as a quotient sometimes $-$ that is the purpose of Leibniz notation in the end $-$, but it needs to be done with caution, otherwise this kind of manipulations can lead to nonsensical results, that is why it is usually taught as a forbidden treatment, until students acquire enough maturity on this topic (analogously to square roots of negative numbers, in a way). In the present case, the chain rule turns to be one of such manipulations, initially taught formally from the definition of the derivative and then reinterpreted as a "rule of three" with infinitesimal quantities.