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If you are looking for some problems with solutions you can find some by clicking on the "Practice Problems" link above.
If you are looking for some problems with solutions you can find some by clicking on the "Practice Problems" link above.
Section 12.9 : Arc Length with Vector Functions
For problems 1 – 3 determine the length of the vector function on the given interval.
- \(\vec r\left( t \right) = 4\cos \left( {2t} \right)\vec i + 3t\,\vec j - 4\sin \left( {2t} \right)\vec k\) from \(0 \le t \le 3\pi \).
- \(\vec r\left( t \right) = \left\langle {9 - 2t,4 + 2t,\sqrt 2 \,{t^2}} \right\rangle \) from \(0 \le t \le 1\).
- \(\vec r\left( t \right) = 2t\,\vec i + \frac{1}{2}{t^2}\,\vec j + \ln \left( {{t^2}} \right)\vec k\) from \(1 \le t \le 3\).
For problems 4 – 6 find the arc length function for the given vector function.
- \(\vec r\left( t \right) = \left\langle {8t,6 + t, - 7t} \right\rangle \)
- \(\displaystyle \vec r\left( t \right) = \left\langle {8t,4{t^{\frac{3}{2}}},3} \right\rangle \)
- \(\vec r\left( t \right) = \left\langle {{{\bf{e}}^{4t}}\sin \left( t \right),{{\bf{e}}^{4t}}\cos \left( t \right),2} \right\rangle \)
- Determine where on the curve given by \(\vec r\left( t \right) = \left\langle {8t,4{t^{\frac{3}{2}}},3} \right\rangle \) we are after traveling a distance of 4.
- Determine where on the curve given by \(\vec r\left( t \right) = \left\langle {{{\bf{e}}^{4t}}\sin \left( t \right),{{\bf{e}}^{4t}}\cos \left( t \right),2} \right\rangle \) we are after traveling a distance of 15.