Calculus II - Arc Length with Vector Functions

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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 $\left( {\frac{\pi }{3},\frac{{3\sqrt 3 }}{2},\frac{3}{2}} \right)$ .