Math Insight

Parametrized curve arc length examples

Example 1

Write a parameterization for the straight-line path from the point (1,2,3) to the point (3,1,2). Find the arc length.

Line segment path

Solution: The vector from (1,2,3) to (3,1,2) is $\vc{d} = (3,1,2)-(1,2,3) = (2,-1,-1)$. We can parametrize the line segment by \begin{align*} \dllp(t) = (1,2,3) + t (2,-1,-1), \qquad 0 \le t \le 1 \end{align*}

To find arc length, we calculate \begin{align*} \dllp'(t) &= (2,-1,-1)\\ \| \dllp'(t)\| &= \sqrt{2^2+(-1)^2 + (-1)^2} = \sqrt{6}\\ \end{align*} Therefore, the length of the line segment is \begin{align*} \int_a^b \| \dllp'(t)\| dt = \int_0^1 \sqrt{6} dt = \sqrt{6} \end{align*}

Clearly, it was silly to calculate the length this way. We knew the length of the line segment must be $\| \vc{d} \| = \sqrt{6}$. But, this simply illustrates the method of calculating arc length of parametrized curves.

Example 2

Another parameterization for the line segment of example 1 is \begin{align*} \adllp(t) = (1,2,3) + (e^t-1) (2,-1,-1), \quad 0 \le t \le \log 2. \end{align*} Find the length of the line segment using this parametrization.

It might not be obvious that $\adllp$ parametrizes the same line segment as $\dllp$ from example 1. To see this fact, notice that $(e^t-1)$ is zero when $t=0$, and it is 1 when $t=\log 2$. (As mathematicians often do, we are using $\log t$ to represent the natural logarithm, which is often written as $\ln t$. Hence, $e^{\log 2} = 2$, as required for this example.) Indeed, a particle with position $\adllp(t)$ at time $t$ does move along the straight line from (1,2,3) to (3,1,2) as $t$ goes from 0 to $\log 2$. It just doesn't move at a constant speed. You can read about another example where particles move along the same curve but at different speeds.

Solution: We simply use the definition of arc length to find the length of the line segment using using this parameterization. We calculate \begin{align*} \adllp'(t) &= e^t (2,-1,-1)\\ \|\adllp'(t)\| &= e^t \|(2,-1,-1)\| = e^t \sqrt{6}, \end{align*} so the arc length is \begin{align*} \int_a^b \| \adllp'(t)\| dt &= \int_0^{\log 2} e^t \sqrt{6} dt\\ &= \sqrt{6} (e^{\log 2}- e^0)\\ &= \sqrt{6} (2-1) = \sqrt{6}, \end{align*} which agrees with example 1.

Examples 1 and 2 illustrate an important principle. The length of a curve does not depend on its parametrization. Of course, this makes sense, as the distance a particle travels along a particular route doesn't depend on its speed.

Example 3

Find the arc length of the helix parametrized by $\dllp(t) = (\cos t, \sin t, t)$ for $0 \le t \le 6\pi$. (This was the example used in the introduction to arc length.)

Solution: We calculate \begin{align*} \dllp'(t) &= (-\sin t, \cos t, 1)\\ \|\dllp'(t)\| &= \sqrt{\sin^2 t + \cos^2 t + 1^2} = \sqrt{2}. \end{align*} The length is \begin{align*} \int_0^{6\pi} \sqrt{2} dt = \left.\left.\sqrt{2} t\right|_0^{6\pi}\right. = 6\sqrt{2} \pi \approx 26.7. \end{align*}