Pauls Online Notes
Pauls Online Notes
Home / Calculus III / 3-Dimensional Space / Arc Length with Vector Functions
Show Mobile Notice Show All Notes Hide All Notes
Mobile Notice
You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.
Assignment Problems Notice
Please do not email me to get solutions and/or answers to these problems. I will not give them out under any circumstances nor will I respond to any requests to do so. The intent of these problems is for instructors to use them for assignments and having solutions/answers easily available defeats that purpose.

Section 6-9 : Arc Length with Vector Functions

For problems 1 – 3 determine the length of the vector function on the given interval.

  1. \(\vec r\left( t \right) = 4\cos \left( {2t} \right)\vec i + 3t\,\vec j - 4\sin \left( {2t} \right)\vec k\) from \(0 \le t \le 3\pi \).
  2. \(\vec r\left( t \right) = \left\langle {9 - 2t,4 + 2t,\sqrt 2 \,{t^2}} \right\rangle \) from \(0 \le t \le 1\).
  3. \(\vec r\left( t \right) = 2t\,\vec i + \frac{1}{2}{t^2}\,\vec j + \ln \left( {{t^2}} \right)\vec k\) from \(1 \le t \le 3\).

For problems 4 – 6 find the arc length function for the given vector function.

  1. \(\vec r\left( t \right) = \left\langle {8t,6 + t, - 7t} \right\rangle \)
  2. \(\displaystyle \vec r\left( t \right) = \left\langle {8t,4{t^{\frac{3}{2}}},3} \right\rangle \)
  3. \(\vec r\left( t \right) = \left\langle {{{\bf{e}}^{4t}}\sin \left( t \right),{{\bf{e}}^{4t}}\cos \left( t \right),2} \right\rangle \)
  4. 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.
  5. 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.