Calculus II - Arc Length with Vector Functions

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Section 12.9 : Arc Length with Vector Functions

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

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

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

We need to solve this for $t$ . Doing this gives,

\begin{matrix} {(4 + 45 t^{2})}^{\frac{3}{2}} - 8 & = 135 s {(4 + 45 t^{2})}^{\frac{3}{2}} & = 135 s + 8 4 + 45 t^{2} & = {(135 s + 8)}^{\frac{2}{3}} t^{2} & = \frac{1}{45} [{(135 s + 8)}^{\frac{2}{3}} - 4] \to t = \sqrt{\frac{1}{45} [{(135 s + 8)}^{\frac{2}{3}} - 4]} \end{matrix}

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

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.

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

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

Our position after traveling a distance of 20 is then,

\to r (t (20)) = \to r (2.05633) = ⟨ 4.22849, 17.39035, - 7.69518 ⟩

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