Length of a curve: misunderstood

Question

I keep it simple: the length of a curve can be calculated as integral:

L = $ \int \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 } \cdot dt$

not $[L(f) = \int_{r_1}^{r_2}f'(t)\,\mathrm{dt}]$.

Looking at this formula it seemed self-evident to me: u're summing up all the parts of a curve as it moves in time. However by thinking about it a little more I'm flustered: how can an Integral spit out a length? Isn't it all about surface?

Answer 1

An integral is in effect "summing up tiny bits of a thing to get the whole thing." That thing can be an area, and when the thing you are integrating is graphed, the (signed) area below it is the result. So any integral can be drawn as an area, it's just that sometimes we don't want to use that kind of description.

You can add up lots of dm to get a mass $M$. Lots of bits of length to get a length $L$. Lots of bits of volume $dV$ to get a volume $V$. And so forth.

In your example, a tiny bit of arc length, $dL$, can be drawn as the hypotenuse of a tiny triangle with legs $dx$ and $dy$, so by the Pythagorean theorem, $$(dL)^2 = (dx)^2 + (dy)^2$$, or multiplying by $(dt/dt)^2$, we get $$ \left(\frac{dL}{dt}\right)^2(dt)^2 = \left[\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 \right] (dt)^2 $$ and finally $$ \left(\frac{dL}{dt}\right)(dt) = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 } \cdot dt$$ and $$ L = \int dL = \int \left(\frac{dL}{dt}\right)(dt) = \int \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 } \cdot dt$$

Answer 2

However by thinking about it a little more I'm flustered: how can an Integral spit out a length? Isn't it all about surface?

A familiar illustration of how integral spit out a length (distance traveled = area underneath the speed curve) from Physics class:

For constant speed $v=5 \frac{m}{s}$, $S=\int_0^{10} 5\,dt=50(m)$

For constant acceleration $a=\frac{1}{2} \frac{m}{s^2}$, $S=\int_0^{10} \frac{1}{2}t\,dt=25 (m)$

Answer 3

The expression of "an integral represents area under a line" is correct in most cases but this is often misunderstood to be its only meaning, Lets look at the general case: $$\int_a^bf(x)dx$$ Whilst the $dx$ is often thought of as telling you "what we integrate wrt" but also remember that whilst it is small, it is not $0$ and so we can think of it as a small spacial step.

Thinking of it as area gives us a height $f(x)$ multiplied by a width $dx$, giving the area of a small rectangle $f(x)dx$ then to find the area over $[a,b]$ we have to do this for the whole range. For this to be accurate (a smooth function does not have steps like this) we must make $dx$ very small so $f(x)$ is close to constant over the range of $dx$.

Another interesting thing you may notice is that the integral symbol $\int$ looks somewhat like an $S$, in the sense that the [Riemann integral][1] is defined as a summation where the size of the terms tend to zero and the number of terms tend to infinity.

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What is going on in this case is we are again looking at very small steps, $dx,dy$ and noting that as these tend to zero we can represent line length $dL$ the hypotenuse of the triangle formed, or: $$dL^2=dx^2+dy^2$$ where $x(t),y(t)$ parametrically represent the function $f$. Now we want these small lengths over the required domain, and so: $$L=\int dL=\int_{t_1}^{t_2}\sqrt{dx^2+dy^2}=\int_{t_1}^{t_2}\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}dt$$

[1]: https://en.wikipedia.org/wiki/Riemann_integral#:~:text=Loosely%20speaking%2C%20the%20Riemann%20integral,more%20specifically%20Riemann%2Dintegrable).