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Section 12.9 : Arc Length with Vector Functions
2. Determine the length of \(\vec r\left( t \right) = \left\langle {\frac{1}{3}{t^3},4t,\sqrt 2 {t^2}} \right\rangle \) from \(0 \le t \le 2\).
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Start SolutionWe first need the magnitude of the derivative of the vector function. This is,
\[\vec r'\left( t \right) = \left\langle {{t^2},4,2\sqrt 2 t} \right\rangle \] \[\left\| {\vec r'\left( t \right)} \right\| = \sqrt {{t^4} + 16 + 8{t^2}} = \sqrt {{t^4} + 8{t^2} + 16} = \sqrt {{{\left( {{t^2} + 4} \right)}^2}} = {t^2} + 4\]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.
Show Step 2The length of the curve is then,
\[L = \int_{0}^{2}{{{t^2} + 4\,dt}} = \left. {\left( {\frac{1}{3}{t^3} + 4t} \right)} \right|_0^2 = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{32}}{3}}}\]Note that if we’d not simplified the magnitude this would have been a very difficult integral to compute!