$dx=\frac {dx}{dt}dt $. Why is this equality true and what does it mean?

Question

$dx=\frac {dx}{dt}dt $. I know that this deduction is obvious from the chain rule, given that we treat our dx and dt as just numbers. But I find it quite unsatisfactory to think of it in that sense. Is there a better / more "calculus-inclined" way of thinking about this equality. Can you please explain both the LHS and RHS individually.

Answer 1

There are two good ways to think about this.

First - integration (which is usually what this type of notation is shorthand for).

If $x=x(t)$, then $$dx=\frac{dx}{dt}dt=x'(t)\;dt$$ is shorthand for $$x(T)-x(0)=\int_{0}^{T}x'(t)\;dt,$$ which is something you can readily verify in a fully rigorous way. Let $\Pi=\{0=t_{0},t_{1},\ldots,t_{n-1},t_{n}=T\}$ be some partition of the interval $[0,T].$ Then, $$\begin{align*} x(T)-x(0)&=\sum_{k=1}^{n}x(t_{k})-x(t_{k-1})\\ &=\sum_{k=1}^{n}x'(t_{k}^{*})(t_{k}-t_{k-1})\;\;\;\;(t_{k}^{*}\in(t_{k-1},t_{k}))\\ &\to\int_{0}^{T}x'(t)\;dt\;\text{as}\;n\to\infty. \end{align*}$$ (Note how the first expression is unaffected by the limit $n\to\infty$.)

Intuitively, you can get $\Delta_{[0,T]} x$ by adding up a bunch of $dx$'s or a bunch of $\frac{dx}{dt}dt$'s.

In the second line we use the mean value theorem (or linear version of Taylor's theorem), which says $$x(t+\delta t)-x(t)=x'(t^{*})\delta t$$ for some $t^{*}_{k}\in(t,t+\delta t)$.

This leads us into the second interpretation, which is a fully computable/rigorous approximation to the expression using only finite quantities/concepts you are already familiar with.

If we switch to Leibniz notation and evaluate $dx/dt$ at $t^{*}$ then we have $$\delta x=\frac{dx}{dt}\delta t,$$ and you can see that this is basically the expression you are trying to verify the validity of. However, we do not typically know $t^{*}$ and anyway, the expression is suggesting we can evaluate $dx/dt$ at $t$. But if we do that we only get something slightly different: $$\delta x=\frac{dx}{dt}\delta t+o(|\delta t|).$$ The expression above is computable (since we know all of the quantities involved), except the error term which obeys the bound $\leq C(\delta t)^{2}.$

Essentially, your expression $dx=\frac{dx}{dt}dt$ is the above when $\delta t\to0$ (note that the error term goes to $0$ and the remaining expression should be true). Literally though, both sides ends up going to $0$ in the limit (this is obvious). What is actually important is how the ratio behaves, and that is what the expression is really trying to communicate. Indeed, dividing by $\delta t$ we get $$\frac{\delta x}{\delta t}=\frac{dx}{dt}+\frac{o(|\delta t|)}{\delta t}.$$

The error term still goes to $0$ as $\delta t\to0$, and the result is $$\frac{dx}{dt}=\frac{dx}{dt}.$$

Answer 2

Here's how I would think about it. Imagine you are at the blue dot in the graph below and you are moving along the curve one unit to the right (ΔX = 1). Where will this leave you? You can find out where you will be by taking the slope at the blue point (ΔT/ΔX) and multiplying by ΔT

Answer 3

Statements like $dx = \frac{dx}{dt} dt$ are best viewed as mnemonic devices, and nothing more (until way down the mathematical road, potentially, if at all, maybe).

They are helpful for remembering techniques when it comes to separable differential equations, for example, or the first derivative of a parametric equation:

$$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}.$$

But, do not buy into this notion too seriously. This analogy will break down. For example,

$$\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}} \neq \frac{d^2y/dt^2}{dx^2/dt^2}.$$