# Arc Length of Curves in Three-Dimensional Space

We will now look at computing the arc length specified by a vector-valued function $\vec{r}(t)$ on an interval for which $t$ is in $I = [a, b]$.

Let $C$ be a continuous and bounded curve specified by the vector-valued function $\vec{r}(t) = (x(t), y(t), z(t))$ for $a ≤ t ≤ b$, and subdivided the interval $I = [a, b]$ into $n$ subintervals with endpoints $a= t_0, t_1, t_2, ..., t_n = b$ where $t_0 < t_1 < ... < t_{n-1} < t_n$.

Now let the vector $\vec{r_i} = \vec{r}(t_i)$ for $1 ≤ i ≤ n$ be the position vector with terminal point defined by $t = t_i$. Then the vector $\vec{r_i} - \vec{r_{i-1}}$ is a chord-vector, that the length/magnitude of this chord vector $\| \vec{r_i} - \vec{r_{i-1}} \|$ approximates the length of $C$ between the points defined by $t = t_{i-1}$ and $t = t_{i}$. The sum of these approximate lengths approximates the total length of the curve $C$

(1)Of course, the length of $C$, call it $L$ is larger than $\sum_{i=1}^{n} \| \vec{r_i} - \vec{r_{i-j}} \|$ since these approximations will be underestimates of $L$ as we're summing the lengths of the shortest distances between successive $\vec{r}(t_i)$. Therefore for all $n \in \mathbb{N}$, $\sum_{i=1}^{n} \| \vec{r_i} - \vec{r_{i-j}} \| ≤ L$, however, it is not hard to see that we subdivide our intervals more and more, our approximation gets larger. Let $\Delta t_i = t_i - t_{i-1}$ length of our subintervals of $I$, and let $\Delta \vec{r_i} = \vec{r_i} - \vec{r_{i-1}}$. Then we can rewrite the equation above as:

(2)Therefore as the number of subdivisions, $n \to \infty$, and as the largest subdivisions, $\Delta t_i \to 0$ (this ensures us that we don't allow one subinterval to stay large), then our sum approximates the actual length better and better, and so:

(3)Theorem 1: Let $\vec{r_1}(t)$ be a vector-valued function that represents a continuous bounded curve $C$ for $t \in [a, b]$, and let $\vec{r_2}(t)$ be a vector-valued function that also represents the same curve $C$ for $t \in [c, d]$, that is $\vec{r_1}(t)$ and $\vec{r_2}(t)$ are two parameterizations of $C$, and suppose that they are both one-to-one on their respective domains and give $C$ the same orientation, that is $\vec{r_1}(a) = \vec{r_2}(c)$ and $\vec{r_1}(b) = \vec{r_2}(d)$. Then for each $t \in [a, b]$, and for the unique differentiable function $u : [a, b] \to [c, d]$, $u = u(t)$ such that $\vec{r_1}(t) = \vec{r_2}(u(t))$, we have that $\int_a^b \biggr \| \frac{d}{dt} \vec{r_1}(t) \biggr \| \: dt = \int_c^d \biggr \| \frac{d}{du} \vec{r_2} (u) \biggr \| \: du$, that is the arc length of $C$ is independent of the parameterization of $C$. |

**Proof:**Since $\vec{r_1}(t) =\vec{r_2}(u(t))$ and $u$ is differentiable, one-to-one, and preserves orientation, then $\frac{du}{dt} ≥ 0$ on the interval $[a, b]$. By the chain rule we have that:

- Substituting this into our arc length formulas, we have that:

Definition: Let $\vec{r}(t) = (x(t), y(t), z(t))$ be a vector-valued function traced by the curve $C$ for $t \in [a, b]$. Then the Arc Length Function for C is given by $s(t) = \int_a^t \| \vec{r'}(u) \| \: du = \int_a^t \sqrt{ \left ( \frac{dx}{du} \right )^2 + \left ( \frac{dy}{du} \right )^2 + \left ( \frac{dz}{du} \right )^2} \: du$, where $s(t)$ is the length of the curve $C$ between $\vec{r}(a)$ to $\vec{r}(t)$, and differentiating both sides of this equation we get $\frac{ds}{dt} = \| \vec{r'}(t) \|$. |