Arc Length Parameterization of Curves in Three-Dimensional Space

Arc Length Parameterization of Curves in Three-Dimensional Space

We recently looked at some examples of parameterizing various curves in $\mathbb{R}^3$ on the Parameterization of Curves in Three-Dimensional Space page. We also recently derived a formula for computing the arc length of a continuous bounded curve $C$ by the vector-valued function $\vec{r}(t) = (x(t), y(t), z(t))$ from $t = a$ to $t = b$ on the Arc Length of Curves in Three-Dimensional Space page with the following formula:

(1)
\begin{align} \mathrm{arc \: length} = \int_a^b \sqrt{ \left ( \frac{dx}{dt} \right)^2 + \left ( \frac{dy}{dt} \right)^2 + \left ( \frac{dz}{dt} \right)^2} \: dt \end{align}

We also noted that the Arc Length Function for $C$ Starting at $a$ is $s(t) = \int_a^t \| \vec{r'}(u) \| \: du = \int_a^t \sqrt{ \left ( \frac{dx}{du} \right )^2 + \left ( \frac{dy}{du} \right )^2 + \left ( \frac{dz}{du} \right )^2} \: du$.

We will now look at some examples of parameterizing a curve $C$ in terms of its arc length. Parameterizing a curve in this manner is often very useful, since as $t$ increases, $C$ is traced out at unit speed. Furthermore, many of the subsequent topics we will look are sometimes easier to work with provided that a curve can easily be parameterized in terms of its arc length.

Example 1

Parameterize the vector-valued function $\vec{r}(t) = (At, Bt, Ct)$ where $A^2 + B^2 + C^2 > 0$ starting at $t = 0$.

We first differentiate $\vec{r}(t)$ to get that $\vec{r'}(t) = (A, B, C)$. Therefore we have that $\| \vec{r'}(t) \| = \sqrt{A^2 + B^2 + C^2}$. Now:

(2)
\begin{align} \quad s= \int_0^t \| \vec{r'}(u) \| \: du = \int_0^t \sqrt{A^2 + B^2 + C^2} \: du = \sqrt{A^2 + B^2 + C^2}u \biggr \rvert_0^t = \sqrt{A^2 + B^2 + C^2}t \end{align}

Therefore we have that $s = \sqrt{A^2 + B^2 + C^2}t$ so $t = \frac{s}{\sqrt{A^2 + B^2 + C^2}}$ and hence the arc length parameterization starting at $0$ is:

(3)
\begin{align} \vec{r}(s) = \left ( \frac{As}{\sqrt{A^2 + B^2 + C^2}}, \frac{Bs}{\sqrt{A^2 + B^2 + C^2}}, \frac{Cs}{\sqrt{A^2 + B^2 + C^2}} \right ) \end{align}

Example 2

Parameterize the vector-valued function $\vec{r}(t) = (a \cos t, a \sin t, bt)$ starting at $t = 0$.

We first differentiate $\vec{r}(t)$ to get $\vec{r'}(t) = (-a \sin t, a \cos t, b)$. We will now find the norm/magnitude of $\vec{r'}(t)$ as follows:

(4)
\begin{align} \quad \| \vec{r'}(t) \| = \sqrt{a^2 \sin^2 t + a^2 \cos^2 t + b^2} = \sqrt{a^2 + b^2} \end{align}

Now we will find the arc length $s$ starting at $0$:

(5)
\begin{align} \quad s = \int_0^t \sqrt{a^2 + b^2} \: du = \sqrt{a^2 + b^2} \int_0^t 1 \: du = \sqrt{a^2 + b^2} u \biggr \rvert_0^t = \sqrt{a^2 + b^2}t \end{align}

Therefore we have that $s = \sqrt{a^2 + b^2}t$ and so $t = \frac{s}{\sqrt{a^2 + b^2}}$. Therefore our arc length parameterization of $\vec{r}$ is:

(6)
\begin{align} \vec{r}(s) = \left ( a \cos \left ( \frac{s}{\sqrt{a^2 + b^2}} \right) , a \sin \left ( \frac{s}{\sqrt{a^2 + b^2}} \right ) , \frac{bs}{\sqrt{a^2 + b^2}} \right ) \end{align}

Example 3

Parameterize the vector-valued function $\vec{r}(t) = (a \cos^3 t, a \sin^3 t, b \cos 2t )$ starting at $t = 0$.

We first differentiate $\vec{r}(t)$ to get $\vec{r'}(t) = (-3a \sin t \cos^2 t, 3a \sin^2 t \cos t, - 2b \sin (2t))$. We will also use the trigonometric substitution that $\sin (2t) = 2 \sin t \cos t$ and so:

(7)
\begin{align} \quad \vec{r'}(t) = (-3a \sin t \cos^2 t, 3a \sin^2 t \cos t, - 4b \sin t \cos t) \end{align}

Therefore we have that:

(8)
\begin{align} \quad \| \vec{r'}(t) \| = \sqrt{9a^2 \sin^2 \cos^4 t + 9a^2 \sin^4 t \cos^2 t + 16b^2 \sin^2 t \cos^2 t} \\ \quad \| \vec{r'}(t) \| = \sqrt{[\sin^2 t \cos^2 t][9a^2 \cos^2 t + 9a^2 \sin^2 t + 16b^2]} \\ \quad \| \vec{r'}(t) \| = \sin t cos t \sqrt{9a^2 \cos^2 t + 9a^2 \sin^2 t + 16b^2} \\ \quad \| \vec{r'}(t) \| = \sin t \cos t \sqrt{9a^2 + 16b^2} \end{align}

So let's calculate the arc length $s$ starting at $0$:

(9)
\begin{align} \quad s = \int_0^t \sqrt{9a^2 + 16b^2} \sin u \cos u \: du = \sqrt{9a^2 + 16b^2} \int_0^t \sin u \cos u \: du = -\frac{1}{2} \sqrt{9a^2 + 16b^2} \left ( \cos^2 u \right) \biggr \rvert_0^t \\ = -\frac{\sqrt{9a^2 + 16b^2}}{2} (\cos^2 t - 1) = \frac{\sqrt{9a^2 + 16b^2}}{2} (1 - \cos^2 t) = \frac{\sqrt{9a^2 + 16b^2}}{2} \sin^2 t \end{align}

Therefore we have that:

(10)
\begin{align} \quad s = \frac{\sqrt{9a^2 + 16b^2}}{2} \sin^2 t \\ \quad \sin^2 t = \frac{2s}{\sqrt{9a^2 + 16b^2}} \\ \quad \sin t = \sqrt{\frac{2s}{\sqrt{9a^2 + 16b^2}}} \end{align}

We also have that:

(11)
\begin{align} \quad s = \frac{\sqrt{9a^2 + 16b^2}}{2} (1 - \cos^2 t) \\ \quad (1 - \cos^2 t) = \frac{2s}{\sqrt{9a^2 + 16b^2}} \\ \quad \cos^2 t = 1 - \frac{2s}{\sqrt{9a^2 + 16b^2}} \\ \quad \cos t = \sqrt{1 - \frac{2s}{\sqrt{9a^2 + 16b^2}}} \end{align}

Lastly we will use the trigonometric identity where $\cos 2t = 1 - 2\sin^2 t$ and so:

(12)
\begin{align} \quad \cos 2t = 1 - \frac{4s}{\sqrt{9a^2 + 16b^2}} \end{align}

Therefore the arc length parameterization of $\vec{r}$ is:

(13)
\begin{align} \quad \vec{r'}(s) = \left ( a \left [ 1 - \frac{2s}{\sqrt{9a^2 + 16b^2}} \right ]^{3/2}, a \left [ \frac{2s}{\sqrt{9a^2 + 16b^2}} \right ]^{3/2}, b \left [ 1 - \frac{4s}{\sqrt{9a^2 + 16b^2}} \right ] \right ) \end{align}