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Section 1-9 : Arc Length with Vector Functions

5. Determine where on the curve given by \(\vec r\left( t \right) = \left\langle {{t^2},2{t^3},1 - {t^3}} \right\rangle \) we are after traveling a distance of 20.

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Start Solution

From Problem 3 above we know that the arc length function for this vector function is,

\[s\left( t \right) = \frac{1}{{135}}\left[ {{{\left( {4 + 45{t^2}} \right)}^{\frac{3}{2}}} - 8} \right]\]

We need to solve this for \(t\). Doing this gives,

\[\begin{align*}{\left( {4 + 45{t^2}} \right)^{\frac{3}{2}}} - 8 & = 135s\\ {\left( {4 + 45{t^2}} \right)^{\frac{3}{2}}} & = 135s + 8\\ 4 + 45{t^2} & = {\left( {135s + 8} \right)^{\frac{2}{3}}}\\ {t^2} & = \frac{1}{{45}}\left[ {{{\left( {135s + 8} \right)}^{\frac{2}{3}}} - 4} \right]\hspace{0.25in} \to \hspace{0.25in}t = \sqrt {\frac{1}{{45}}\left[ {{{\left( {135s + 8} \right)}^{\frac{2}{3}}} - 4} \right]} \end{align*}\]

Note that we only used the positive \(t\) after taking the root because the implicit assumption from the arc length function is that \(t\) is positive.

Show Step 2

We could use this to reparametrize the vector function however that would lead to a particularly unpleasant function in this case.

The key here is to simply realize that what we are being asked to compute is the value of the reparametrized vector function, \(\vec r\left( {t\left( s \right)} \right)\) when \(s = 20\). Or, in other words, we want to compute \(\vec r\left( {t\left( {20} \right)} \right)\).

So, first,

\[t\left( {20} \right) = \sqrt {\frac{1}{{45}}\left[ {{{\left( {135\left( {20} \right) + 8} \right)}^{\frac{2}{3}}} - 4} \right]} = 2.05633\]

Our position after traveling a distance of 20 is then,

\[\vec r\left( {t\left( {20} \right)} \right) = \vec r\left( {2.05633} \right) = \require{bbox} \bbox[2pt,border:1px solid black]{{\left\langle {4.22849,17.39035, - 7.69518} \right\rangle }}\]