### 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.

**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*}

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