Calculating arc length of a curve by pythagorean theorem

Question

I was asked to explain the parametric curve arc length to a fellow student, only to find out that I don't completely understand it myself to be able to explain it. I've read multiple posts here about the subject but I still feel like something is missing for me.

My understanding is,

Let $C \rightarrow \mathbf{R_3}$ be an arbitrary curve and let $\textbf{x(s)}=$($x_1(s), x_2(s), x_3(s)$) for $s\in [a,b]$ be a parametric representation of $C$.

In order to derive the arc length $ds$ - where $ds$ is an infinitesimal change in arc length - when we're dealing with parametric equations, we break it down to how much change happened in each of the directions

(1) $$dx_i = \frac{dx_i}{ds}\cdot ds = x_i^ {'} ds \quad \quad i\in {1, 2, 3}$$

Based on this, the length of $ds$ becomes

(2)

\begin{equation} ds = \sqrt{\Sigma_{i=1}^{3}[x_i^{'}\cdot ds]^2}= \sqrt{\Sigma_{i=1}^{3}[x_i^{'}]^2}\cdot ds \end{equation}

By adding up $ds$ in the interval $s=a$ to $s=b$

(3) \begin{equation} arc \,\, length = \int_{s=a}^{s=b} ds = \int_{s=a}^{s=b} \sqrt{\Sigma_{i=1}^{3}[x_i^{'}]^2}\cdot ds \end{equation}

In case we wish to find the arc length on curve $C$ from $x_1=a_0$ to $x_1=a_1$, we would need to integrate over $dx_1$ instead of $ds$ \begin{equation*} ds = \sqrt{\Sigma_{i=1}^{3}[x_i^{'}]^2}\cdot ds = \sqrt{x_1^{'2}+x_2^{'2}+x_3^{'2}}\cdot ds = \sqrt{x_1^{'2}\cdot (1 + \frac{x_2^{'2}}{x_1^{'2}}+\frac{x_3^{'2}}{x_1^{'2}})} \cdot ds \end{equation*} \begin{equation*} ds = \sqrt{1 + \frac{x_2^{'2}}{x_1^{'2}}+\frac{x_3^{'2}}{x_1^{'2}}}\cdot {x_1^{'2}}\cdot ds = \sqrt{1 + \frac{x_2^{'2}}{x_1^{'2}}+\frac{x_3^{'2}}{x_1^{'2}}} \cdot \frac{dx_1}{ds}\cdot ds \end{equation*} \begin{equation*} arc \,\, length = \int_{x_1=a_0}^{x_1=a_1} \sqrt{1 + \frac{x_2^{'2}}{x_1^{'2}}+\frac{x_3^{'2}}{x_1^{'2}}} \cdot \frac{dx_1}{ds}\cdot ds \end{equation*}

Using Equation (1) above \begin{equation*} arc \,\, length = \int_{x_1=a_0}^{x_1=a_1} \sqrt{1 + \frac{x_2^{'2}}{x_1^{'2}}+\frac{x_3^{'2}}{x_1^{'2}}} \cdot dx_1 \end{equation*}

• First of all, I hope the above equations are right, as I feel like I'm messing up with the different notations.

• By taking a look at Equation (1), $ds$ can be visualized as distance along the curve $C$. $dx_i$ can be visualized as well as the distance along the $x_i$ axis. But I'm not quite sure what does $\frac{dx_i}{ds}$ represent here. I know it means the change in $x_i$ with regards to $ds$, but there's something here that I'm still missing. Is there a way to visualize that value, perhaps that could help me understand it.

  Or is it possible to look at it as $(\frac{dx_i}{ds})$ represents the speed in the $x_i$ direction and $ds$ being time so their multiplication $\frac{dx_i}{ds}\cdot ds$ is the distance in that direction? But then how can that be interpreted in Equation (2) as $time = speed \cdot time$ ?

• One last thing, looking at equation (2), wouldn't $\sqrt{\Sigma_{i=1}^{3}[x_i^{'}]^2}$ be just equal to 1?

Answer 1

Let us work with finite increments, and in 2D.

If the parametric equations of the curve are

$$x=x(t),y=y(t),$$

an increment $\Delta t$ in the parameter corresponds to increments $\Delta x, \Delta y$ in the coordinates.

By Pythagoras, the distance traveled is

$$\Delta s=\sqrt{\Delta x^2+\Delta y^2}$$

and the total curve length,

$$S=\sum \Delta s.$$

Now passing to the limit, we have, in differential terms

$$S=\oint\sqrt{dx^2+dy^2}=\int_{t_0}^{t_1}\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\,dt=\int_{t_0}^{t_1}\sqrt{\dot x^2+\dot y^2}\,dt.$$

If $t$ is time, we can also write the expression of the instantaneous speed,

$$v=\frac{ds}{dt}=\sqrt{\dot x^2+\dot y^2}$$ i.e. the Pythagorean combination of the horizontal and vertical speeds.

Now if the curve is described as $y=y(x)$, it suffices to replace $t$ by $x$ and

$$S=\int_{x_0}^{x_1}\sqrt{1+y'^2}\,dx.$$

The quantities $\dfrac{dx}{ds}$ and $\dfrac{dy}{ds}$ are the so-called direction cosines, i.e. the components of the unit tangent vector.

$$\vec t=\frac{\vec v}v=\frac{\left(\dot x,\dot y\right)}{\dfrac{ds}{dt}}=\left(\dfrac{dx}{ds},\dfrac{dy}{ds}\right)$$

and

$$\left(\dfrac{dx}{ds}\right)^2+\left(\dfrac{dy}{ds}\right)^2=1.$$

The generalization to 3D is immediate.