Calculus III - Arc Length with Vector Functions (Practice Problems)

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Section 12.9 : Arc Length with Vector Functions

For problems 1 & 2 determine the length of the vector function on the given interval.

$\vec r\left( t \right) = \left( {3 - 4t} \right)\vec i + 6t\,\vec j - \left( {9 + 2t} \right)\vec k$ from $- 6 \le t \le 8$ . Solution
$\vec r\left( t \right) = \left\langle {\frac{1}{3}{t^3},4t,\sqrt 2 {t^2}} \right\rangle$ from $0 \le t \le 2$ . Solution

For problems 3 & 4 find the arc length function for the given vector function.

$\vec r\left( t \right) = \left\langle {{t^2},2{t^3},1 - {t^3}} \right\rangle$ Solution
$\vec r\left( t \right) = \left\langle {4t, - 2t,\sqrt 5 \,\,{t^2}} \right\rangle$ Solution
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. Solution