Calculus III - Arc Length with Vector Functions

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Section 12.9 : Arc Length with Vector Functions

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1. Determine the length of $\vec r\left( t \right) = \left( {3 - 4t} \right)\vec i + 6t\,\vec j - \left( {9 + 2t} \right)\vec k$ from $- 6 \le t \le 8$ .

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

$\vec r'\left( t \right) = - 4\vec i + 6\,\vec j - 2\vec k$

$\left\| {\vec r'\left( t \right)} \right\| = \sqrt {16 + 36 + 4} = \sqrt {56} = 2\sqrt {14}$ Show Step 2

The length of the curve is then,

$L = \int_{{ - 6}}^{8}{{2\sqrt {14} \,dt}} = \left. {2\sqrt {14} t} \right|_{ - 6}^8 = \require{bbox} \bbox[2pt,border:1px solid black]{{28\sqrt {14} }}$