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### Section 12.9 : Arc Length with Vector Functions

3. Find the arc length function for $$\vec r\left( t \right) = \left\langle {{t^2},2{t^3},1 - {t^3}} \right\rangle$$.

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

$\vec r'\left( t \right) = \left\langle {2t,6{t^2}, - 3{t^2}} \right\rangle$ $\left\| {\vec r'\left( t \right)} \right\| = \sqrt {4{t^2} + 36{t^4} + 9{t^4}} = \sqrt {{t^2}\left( {4 + 45{t^2}} \right)} = \sqrt {{t^2}} \sqrt {4 + 45{t^2}} = \left| t \right|\sqrt {4 + 45{t^2}} = t\sqrt {4 + 45{t^2}}$

For these kinds of problems make sure to simplify the magnitude as much as you can. It can mean the difference between a really simple problem and an incredibly difficult problem.

Note as well that because we are assuming that we are starting at $$t = 0$$ for this kind of problem it is safe to assume that $$t \ge 0$$ and so $$\sqrt {{t^2}} = \left| t \right| = t$$.

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The arc length function is then,

$s\left( t \right) = \int_{0}^{t}{{u\sqrt {4 + 45{u^2}} \,du}} = \frac{1}{{135}}\left. {{{\left( {4 + 45{u^2}} \right)}^{\frac{3}{2}}}} \right|_0^t = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{1}{{135}}\left[ {{{\left( {4 + 45{t^2}} \right)}^{\frac{3}{2}}} - 8} \right]}}$