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Section 12.9 : Arc Length with Vector Functions

In this section we’ll recast an old formula into terms of vector functions. We want to determine the length of a vector function,

\[\vec r\left( t \right) = \left\langle {f\left( t \right),g\left( t \right),h\left( t \right)} \right\rangle \]

on the interval \(a \le t \le b\).

We actually already know how to do this. Recall that we can write the vector function into the parametric form,

\[x = f\left( t \right)\hspace{0.25in}y = g\left( t \right)\hspace{0.25in}z = h\left( t \right)\]

Also, recall that with two dimensional parametric curves the arc length is given by,

\[L = \int_{{\,a}}^{{\,b}}{{\sqrt {{{\left[ {f'\left( t \right)} \right]}^2} + {{\left[ {g'\left( t \right)} \right]}^2}} \,dt}}\]

There is a natural extension of this to three dimensions. So, the length of the curve \(\vec r\left( t \right)\) on the interval \(a \le t \le b\) is,

\[L = \int_{{\,a}}^{{\,b}}{{\sqrt {{{\left[ {f'\left( t \right)} \right]}^2} + {{\left[ {g'\left( t \right)} \right]}^2} + {{\left[ {h'\left( t \right)} \right]}^2}} \,dt}}\]

There is a nice simplification that we can make for this. Notice that the integrand (the function we’re integrating) is nothing more than the magnitude of the tangent vector,

\[\left\| {\vec r'\left( t \right)} \right\| = \sqrt {{{\left[ {f'\left( t \right)} \right]}^2} + {{\left[ {g'\left( t \right)} \right]}^2} + {{\left[ {h'\left( t \right)} \right]}^2}} \]

Therefore, the arc length can be written as,

\[L = \int_{{\,a}}^{{\,b}}{{\left\| {\vec r'\left( t \right)} \right\|\,dt}}\]
Let’s work a quick example of this.

Example 1 Determine the length of the curve \(\vec r\left( t \right) = \left\langle {2t,3\sin \left( {2t} \right),3\cos \left( {2t} \right)} \right\rangle \) on the interval \(0 \le t \le 2\pi \).
Show Solution

We will first need the tangent vector and its magnitude.

\[\begin{align*}\vec r'\left( t \right) & = \left\langle {2,6\cos \left( {2t} \right), - 6\sin \left( {2t} \right)} \right\rangle \\ \left\| {\vec r'\left( t \right)} \right\| & = \sqrt {4 + 36{{\cos }^2}\left( {2t} \right) + 36{{\sin }^2}\left( {2t} \right)} = \sqrt {4 + 36} = 2\sqrt {10} \end{align*}\]

The length is then,

\[\begin{align*}L & = \int_{{\,a}}^{{\,b}}{{\left\| {\vec r'\left( t \right)} \right\|\,dt}}\\ & = \int_{{\,0}}^{{\,2\pi }}{{2\sqrt {10} \,dt}}\\ & = 4\pi \sqrt {10} \end{align*}\]

We need to take a quick look at another concept here. We define the arc length function as,

\[s\left( t \right) = \int_{{\,0}}^{{\,t}}{{\left\| {\vec r'\left( u \right)} \right\|\,du}}\]

Before we look at why this might be important let’s work a quick example.

Example 2 Determine the arc length function for \(\vec r\left( t \right) = \left\langle {2t,3\sin \left( {2t} \right),3\cos \left( {2t} \right)} \right\rangle \).
Show Solution

From the previous example we know that,

\[\left\| {\vec r'\left( t \right)} \right\| = 2\sqrt {10} \]

The arc length function is then,

\[s\left( t \right) = \int_{{\,0}}^{{\,t}}{{2\sqrt {10} \,du}} = \left( {2\sqrt {10} \,u} \right)_0^t = 2\sqrt {10} \,t\]

Okay, just why would we want to do this? Well let’s take the result of the example above and solve it for \(t\).

\[t = \frac{s}{{2\sqrt {10} }}\]

Now, taking this and plugging it into the original vector function and we can reparametrize the function into the form, \(\vec r\left( {t\left( s \right)} \right)\). For our function this is,

\[\vec r\left( {t\left( s \right)} \right) = \left\langle {\frac{s}{{\sqrt {10} }},3\sin \left( {\frac{s}{{\sqrt {10} }}} \right),3\cos \left( {\frac{s}{{\sqrt {10} }}} \right)} \right\rangle \]

So, why would we want to do this? Well with the reparameterization we can now tell where we are on the curve after we’ve traveled a distance of \(s\) along the curve. Note as well that we will start the measurement of distance from where we are at \(t = 0\).

Example 3 Where on the curve \(\vec r\left( t \right) = \left\langle {2t,3\sin \left( {2t} \right),3\cos \left( {2t} \right)} \right\rangle \) are we after traveling for a distance of \(\displaystyle \frac{{\pi \sqrt {10} }}{3}\)?
Show Solution

To determine this we need the reparameterization, which we have from above.

\[\vec r\left( {t\left( s \right)} \right) = \left\langle {\frac{s}{{\sqrt {10} }},3\sin \left( {\frac{s}{{\sqrt {10} }}} \right),3\cos \left( {\frac{s}{{\sqrt {10} }}} \right)} \right\rangle \]

Then, to determine where we are all that we need to do is plug in \(s = \frac{{\pi \sqrt {10} }}{3}\) into this and we’ll get our location.

\[\vec r\left( {t\left( {\frac{{\pi \sqrt {10} }}{3}} \right)} \right) = \left\langle {\frac{\pi }{3},3\sin \left( {\frac{\pi }{3}} \right),3\cos \left( {\frac{\pi }{3}} \right)} \right\rangle = \left\langle {\frac{\pi }{3},\frac{{3\sqrt 3 }}{2},\frac{3}{2}} \right\rangle \]

So, after traveling a distance of \(\frac{{\pi \sqrt {10} }}{3}\) along the curve we are at the point \(\left( {\frac{\pi }{3},\frac{{3\sqrt 3 }}{2},\frac{3}{2}} \right)\).