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Section 1-9 : Arc Length with Vector Functions

4. Find the arc length function for \(\vec r\left( t \right) = \left\langle {4t, - 2t,\sqrt 5 \,\,{t^2}} \right\rangle \).

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We first need the magnitude of the derivative of the vector function. This is,

\[\vec r'\left( t \right) = \left\langle {4, - 2,2\sqrt 5 \,\,t} \right\rangle \] \[\left\| {\vec r'\left( t \right)} \right\| = \sqrt {16 + 4 + 20{t^2}} = \sqrt {20 + 20{t^2}} = \sqrt {20} \sqrt {1 + {t^2}} = 2\sqrt 5 \sqrt {1 + {t^2}} \] Show Step 2

The arc length function is then,

\[s\left( t \right) = \int_{0}^{t}{{2\sqrt 5 \sqrt {1 + {u^2}} \,du}}\]

Do not always expect these integrals to be “simple” integrals. They will often require techniques more involved than just a standard Calculus I substitution. In this case we need the following trig substitution.

\[u = \tan \theta \hspace{0.5in}du = {\sec ^2}\theta \,d\theta \hspace{0.5in}\sqrt {1 + {u^2}} = \sqrt {1 + {{\tan }^2}\theta } = \sqrt {{{\sec }^2}\theta } = \left| {\sec \theta } \right|\]

The limits of the integral become,

\[u = 0:0 = \tan \theta \hspace{0.25in} \to \hspace{0.25in} \theta = 0\hspace{0.25in}u = t > 0:t = \tan \theta \hspace{0.25in} \to \hspace{0.25in} \theta = \tan^{ - 1}\left( t \right)\]

Now, as noted we know that \(t > 0\) and so we can safely assume that from the \(u = t\) limit we will get \(0 < \theta < \frac{\pi }{2}\). This in turn means that we will always be in the first quadrant and we know that secant is positive in the first quadrant. Therefore, we can remove the absolute values bars on the secant above.

The arc length function is now,

\[\begin{align*}s\left( t \right) & = \int_{0}^{{{{\tan }^{ - 1}}\left( t \right)}}{{2\sqrt 5 {{\sec }^3}\theta \,d\theta }} = \sqrt 5 \left. {\left( {\sec \theta \tan \theta + \ln \left| {\sec \theta + \tan \theta } \right|} \right)} \right|_0^{{{\tan }^{ - 1}}\left( t \right)}\\ & = \sqrt 5 \left[ {\sec \left( {{{\tan }^{ - 1}}\left( t \right)} \right)\tan \left( {{{\tan }^{ - 1}}\left( t \right)} \right) + \ln \left| {\sec \left( {{{\tan }^{ - 1}}\left( t \right)} \right) + \tan \left( {{{\tan }^{ - 1}}\left( t \right)} \right)} \right|} \right]\end{align*}\]

Now we know that \(\tan \left( {{{\tan }^{ - 1}}\left( t \right)} \right) = t\) so that will simplify our answer a little. Let’s take a look at the secant term to see if we can simplify that as well. First, from our limit work recall that \(\theta = {\tan ^{ - 1}}\left( t \right)\). Or with a little rewrite we have,

\[\tan \theta = t = \frac{t}{1} = \frac{{{\rm{opposite}}}}{{{\rm{adjacent}}}}\]

Construct a right triangle with opposite side being \(t\) and the adjacent side being 1. The hypotenuse is then \(\sqrt {{t^2} + 1} \). This in turn means that \(\sec \theta = \sqrt {{t^2} + 1} \). So,

\[\sec \left( {{{\tan }^{ - 1}}\left( t \right)} \right) = \sec \left( \theta \right) = \sqrt {{t^2} + 1} \]

With this simplification our arc length function is then,

\[s\left( t \right) = \sqrt 5 \left[ {t\sqrt {{t^2} + 1} + \ln \left| {\sqrt {{t^2} + 1} + t} \right|} \right]\]

There was some slightly unpleasant simplification here but once we did that we got a much nicer arc length function.