Arc Length of Curves in Three-Dimensional Space Examples 1

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 2 page.

Example 1

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

We can rewrite our vector-valued function as $\vec{r}(t) = (\sqrt{2}t, e^t, e^{-t})$. When we differentiate each term, we get that $\frac{dx}{dt} = \sqrt{2}$, $\frac{dy}{dt} = e^t$, and $\frac{dz}{dt} = -e^{-t}$. Applying this to our formula for arc length above, and noticing that $(e^t + e^{-t})^2 = e^2t + 2e^{-t + t} + e^{-2t} = e^{2t} + 2 + e^{-2t}$ and we have that:

(2)

\begin{align} \quad \mathrm{arc \: length} = \int_0^1 \sqrt{ (\sqrt{2})^2 + (e^t)^2 + (-e^{-t})^2} \: dt = \int_0^1 \sqrt{2 + e^{2t} + e^{-2t}} \: dt = \int_0^1 \sqrt{(e^t + e^{-t})^2} \: dt = \int_0^1 e^t + e^{-t} \: dt = [ e^t - e^{-t} ] \biggr \rvert _0^1 = e - \frac{1}{e} \end{align}

Example 2

Verify that the circumference of the unit circle is $2\pi$.

We can describe an arbitrary unit circle by the vector-valued function $\vec{r}(t) = (\cos t, \sin t, 0)$ for $0 ≤ t ≤ 2\pi$. If we differentiate each component of this equation, we get that $\frac{dx}{dt} = -\sin t$, $\frac{dy}{dt} = \cos t$ and $\frac{dz}{dt} = 0$. Applying our formula above and we get that

(3)

\begin{align} \quad \mathrm{arc \: length} = \int_0^{2\pi} \sqrt{(-\sin t)^2 + (\cos t)^2} \: dt = \int_0^{2\pi} \sqrt{\sin ^2 t + \cos ^2 t} \: dt = \int_0^{2\pi} 1 \: dt = t \biggr \rvert_0^{2\pi} = 2\pi \end{align}

The reader should check that the circumference of the unit circle is still $2\pi$ if instead we have $\pi ≤ t ≤ 3\pi$, or any other interval of $t$ of length $2 \pi$.

Example 3

Find the length of the curve defined by the vector-valued function $\vec{r}(t) = (3 \cos t, t, 3 \sin t)$ for $-4 ≤ t ≤ 4$.

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

(4)

\begin{align} \quad \mathrm{arc \: length} = \int_{-4}^4 \sqrt{ \left ( -3 \sin t\right )^2 + \left ( 1\right )^2 + \left ( 3 \cos t\right )^2 } \: dt = \int_{-4}^{4} \sqrt{10} \: dt = \sqrt{10}t \biggr \rvert_{-4}^{4} = 4 \sqrt{10} + 4 \sqrt{10} = 8 \sqrt{10} \end{align}