Arc Length of Curves in Three-Dimensional Space Examples 2

Recall from the Arc Length of Curves in Three-Dimensional Space that the arc length of a curve given by the vector equation $\vec{r}(t) = (x(t), y(t), z(t))$ that traces out a continuous and bounded curve $C$ once for $a ≤ t ≤ b$ can be calculated 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 will now look at some examples of computing arc lengths of curves. For examples can be found on the Arc Length of Curves in Three-Dimensional Space Examples 1 page.

Example 1

Find the length of the curve defined by the vector-valued function $\vec{r}(t) = t^3 \vec{i} + \frac{\sqrt{6}}{2}t^2 \vec{j} + t \vec{k}$ for $0 ≤ t ≤ 1$.

If we differentiate component by component we get that $\frac{dx}{dt} = 3t^2$, $\frac{dy}{dt} = \sqrt{6}t$, and $\frac{dz}{dt} = 1$. Applying the formula above for arc length and we get that:

(2)

\begin{align} \quad \mathrm{arc \: length} = \int_0^1 \sqrt{ (3t^2)^2 + (\sqrt{6}t)^2 + (1)^2} = \int_0^1 \sqrt{9t^4 + 6t^2 + 1} \: dt = \int_0^1 \sqrt{(3t^2 + 1)^2} \: dt = \int_0^1 3t^2 + 1 \: dt = (t^3 + t) \biggr \rvert _0^1 = 2 \end{align}

Example 2

Compute the arc length of the intersection between the surfaces $x^2 = 2y$ and $3z = xy$ from the origin to $(6, 18, 36)$.

We first need to find parametric equations to describe the intersection. Let $x = t$. Then from the equations above, we have that $y = \frac{t^2}{2}$ and $z = \frac{t^3}{6}$, and so let $\vec{r}(t) = \left ( t, \frac{t^2}{2}, \frac{t^3}{6} \right )$. This vector-valued function represents the intersection between the surfaces $x^2 = 2y$ and $3z = xy$.

Now notice that the point $(0, 0, 0)$ corresponds to $t = 0$, and the point $(6, 18, 36)$ corresponds to $t = 6$.

Now we need to differentiate each component of our vector-valued function to get $\frac{dx}{dt} = 1$, $\frac{dy}{dt} = t$, and $\frac{dz}{dt} = \frac{t^2}{2}$. Applying the formula for arc length and we get that:

(3)

\begin{align} \quad \mathrm{arc \: length} = \int_0^6 \sqrt{ \left ( 1 \right )^2 + \left ( t\right )^2 + \left ( \frac{t^2}{2}\right )^2} \: dt = \int_0^6 \sqrt{1 + t^2 + \frac{t^4}{4}} = \int_0^6 \frac{1}{2} \sqrt{t^4 + 4t^2 + 4} \: dt = \frac{1}{2} \int_0^6 \sqrt{(t^2 + 2)^2} \: dt = \frac{1}{2} \int_0^6 t^2 + 2 \: dt \\ = \frac{1}{2} \left [ \frac{t^3}{3} + 2t \right ] \biggr \rvert_0^6 = \left [ \frac{t^3}{6} + t \right ] \biggr \rvert_0^6= 42 \end{align}