Paul's Online Notes
Home / Calculus III / 3-Dimensional Space / Calculus with Vector Functions
Show All Notes Hide All Notes

Section 12.7 : Calculus with Vector Functions

For problems 1 – 3 evaluate the given limit.

  1. \(\displaystyle \mathop {\lim }\limits_{t \to 1} \left\langle {{{\bf{e}}^{t - 1}},4t,\frac{{t - 1}}{{{t^2} - 1}}} \right\rangle \) Solution
  2. \(\displaystyle \mathop {\lim }\limits_{t \to - 2} \left( {\frac{{1 - {{\bf{e}}^{t + 2}}}}{{{t^2} + t - 2}}\vec i + \vec j + \left( {{t^2} + 6t} \right)\vec k} \right)\) Solution
  3. \(\displaystyle \mathop {\lim }\limits_{t \to \infty } \left\langle {\frac{1}{{{t^2}}},\frac{{2{t^2}}}{{1 - t - {t^2}}},{{\bf{e}}^{ - t}}} \right\rangle \) Solution

For problems 4 – 6 compute the derivative of the given vector function.

  1. \(\vec r\left( t \right) = \left( {{t^3} - 1} \right)\vec i + {{\bf{e}}^{2t}}\,\vec j + \cos \left( t \right)\vec k\) Solution
  2. \(\vec r\left( t \right) = \left\langle {\ln \left( {{t^2} + 1} \right),t{{\bf{e}}^{ - t}},4} \right\rangle \) Solution
  3. \(\displaystyle \vec r\left( t \right) = \left\langle {\frac{{t + 1}}{{t - 1}},\tan \left( {4t} \right),{{\sin }^2}\left( t \right)} \right\rangle \) Solution

For problems 7 – 9 evaluate the given integral.

  1. \(\displaystyle \int{{\vec r\left( t \right)\,dt}}\), where \(\displaystyle \vec r\left( t \right) = {t^3}\,\vec i - \frac{{2t}}{{{t^2} + 1}}\vec j + {\cos ^2}\left( {3t} \right)\vec k\) Solution
  2. \(\displaystyle \int_{{ - 1}}^{2}{{\vec r\left( t \right)\,dt}}\) where \(\vec r\left( t \right) = \left\langle {6,6{t^2} - 4t,t{{\bf{e}}^{2t}}} \right\rangle \) Solution
  3. \(\displaystyle \int{{\vec r\left( t \right)\,dt}}\), where \(\vec r\left( t \right) = \left\langle {\left( {1 - t} \right)\cos \left( {{t^2} - 2t} \right),\cos \left( t \right)\sin \left( t \right),{{\sec }^2}\left( {4t} \right)} \right\rangle \) Solution