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connected but not path connected?

I’ll start with an example.

In physics, $x(t)$ represents the $x$-position of a particle, and $v(t)$ its ($x$-)velocity. To determine the total displacement of a particle on the interval $[a, b]$, we can use the formula

$$\Delta x = \int_a^b{v(t)~dt}$$

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To me, this makes sense, because $v = \frac {dx}{dt}\\$, so the above equation is equivalent to:

$$\Delta x = \int_a^b{v(t)~dt} = \int_a^b{\frac {dx}{dt}~dt} = \int_a^b{dx} = x(b) – x(a) = \Delta x$$

However, I’ve been told that you can’t just “cancel” the $dt$ differential because it’s not “proper.”

Another example uses parametric arc length:

$$\ell = \int_a^b{\sqrt{\left ( \frac {dx}{dt} \right )^2 + \left ( \frac {dy}{dt} \right )^2} dt}$$

Now take a standard function, $y = f(x)$. We can define $x = t$, and then we have $y(t) = f(t)$, $x(t) = t$, and $\frac{dx}{dt} = 1$. Then, $\frac {dy}{dt} = \frac {{dy}~/~{dx}}{{dx}~/~{dt}} = \frac {{dy}~/~{dx}} 1 = \frac {dy}{dx}$, and our formula simplifies to

$$\ell = \int_a^b{\sqrt{1 + \left ( \frac {dy}{dx} \right )^2} dt}$$

This is indeed the correct formula for arc length of a function (also derivable with Pythagorean theorem), but it relies on being able to cancel the $dx$s in $\frac {{dy}~/~{dx}}{{dx}~/~{dt}}$.

So, my question is, **when, if ever, can you cancel differentials?**

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You’ve clearly demonstrated one of the dangers of a naive notion of “cancelling” differentials: recall from elementary calculus that

$$\int_a^b{dx} = b-a$$

In particular, both equalities in

$$\int_a^b{\frac {dx}{dt}~dt} = \int_a^b{dx} = x(b) – x(a)$$

are incorrect. The only reason you got the correct result at the end is because you remembered what you *meant* to do, and ignored what you *actually* did.

The “right” thing to do from an elementary calculus standpoint is the change of variable formula. In particular, you would have

$$ \int_a^b \frac{dx}{dt}~ dt = \int_{x(a)}^{x(b)}~dx$$

In a realistic problem, you might have actually believed yourself when you got to the expression $\int_a^b \, dx$ and gotten the wrong answer. A lot of beginning students get greatly confused over similar mistakes.

You’ve given yourself extra chances to get the wrong answer in your formula for arclength, since what you wrote isn’t even a differential; e.g. you probably meant to multiply the integrand by $dt$ in the first formula for $\ell$.

When you’re more comfortable with differenetial forms and a more differential-geometric-like notation (which I strongly suspect you *implicitly* are without knowing what that means), you will recognize that differential forms satisfy an identity

$$ \frac{dx}{dy} \, dy = dx $$

whenever $x$ is differentiable with respect to $y$ (and there is only one “degree of freedom”). And if $\gamma$ is some path through the “state space” that $x$ and $y$ measure, then

$$ \int_\gamma f \frac{dx}{dy} \, dy = \int_\gamma f \, dx $$

just as one might expect. $f$ is supposed to be any scalar field in the above. The important distinction between this and the calculus learned in elementary calculus is that we are thinking in terms of a path through “state space” rather than an integral between numeric limits, and that $x$ and $y$ implicitly have some meaning as (dependent) variables on “state space” rather than being indeterminate “dummy” variables.

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