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


\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?


1 Answer 1


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

If the parametric equations of the curve are


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


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

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



The generalization to 3D is immediate.

  • $\begingroup$ Thank you for the answer, as I understand from the explanations, the equations in my post are correct in 3D space. So I do understand the general idea behind the math and how it's happening however, could you refer to the bullet points in my post as those little obstacles that are just in my way of linking everything together. $\endgroup$
    – Elia
    Oct 17, 2018 at 19:32
  • $\begingroup$ @Elia: see my update. $\endgroup$
    – user65203
    Oct 18, 2018 at 7:04

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