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

Section 12.7 : Calculus with Vector Functions

For problems 1 – 6 evaluate the given limit.

  1. \(\mathop {\lim }\limits_{t \to 0} \left( {\cos \left( {2t} \right)\,\vec i - {{\bf{e}}^{4 - t}}\vec j + \left( {{t^2} + 3t - 9} \right)\vec k} \right)\)
  2. \(\displaystyle \mathop {\lim }\limits_{t \to 4} \left\langle {\frac{{t - 4}}{{{t^2} - 3t - 4}},\frac{{{t^2} - 4t}}{{16 - {t^2}}}} \right\rangle \)
  3. \(\displaystyle \mathop {\lim }\limits_{t \to 0} \left\langle {\frac{{\sin \left( t \right)}}{{2t}},\frac{{1 - \cos \left( t \right)}}{t}, - 3} \right\rangle \)
  4. \(\displaystyle \mathop {\lim }\limits_{t \to - 8} \left( {\frac{{{{\bf{e}}^{t{\,^2} - \,64}} - 1}}{{t + 8}}\,\vec i + \frac{{\sin \left( {t + 8} \right)}}{{t + 8}}\vec j - \vec k} \right)\)
  5. \(\displaystyle \mathop {\lim }\limits_{t \to - \infty } \left\langle {\frac{{5{t^2} - 8t + 1}}{{12 + 5{t^2}}},\frac{{2 + {t^3}}}{{1 + {t^2} + {t^4}}}} \right\rangle \)
  6. \(\displaystyle \mathop {\lim }\limits_{t \to \infty } \left\langle {\ln \left( {1 - \frac{4}{t}} \right),{{\bf{e}}^{\frac{1}{{t{\,^2}}}}},2} \right\rangle \)

For problems 7 – 11 compute the derivative of the given vector function.

  1. \(\displaystyle \vec r\left( t \right) = \left\langle {\sqrt {3t} ,\frac{1}{{{t^4}}},\frac{1}{{2t}}} \right\rangle \)
  2. \(\vec r\left( t \right) = \cos \left( {2t} \right)\vec i - \sin \left( {2t} \right)\vec j + \ln \left( {2t} \right)\vec k\)
  3. \(\vec r\left( t \right) = \left\langle {{{\bf{e}}^{t{\,^2} - 1}},4 - \sec \left( {2t} \right),7} \right\rangle \)
  4. \(\vec r\left( t \right) = \sin \left( t \right)\cos \left( t \right)\,\,\vec i - {t^3}\ln \left( {{t^2}} \right)\vec j\)
  5. \(\displaystyle \vec r\left( t \right) = \left\langle {\frac{1}{{{{\left( {{t^2} - 4} \right)}^2}}},\frac{{{t^3}}}{{{t^2} + 2}},\frac{{{t^2} + 2}}{{{t^3}}}} \right\rangle \)

For problems 12 – 15 evaluate the given integral.

  1. \(\displaystyle \int{{\vec r\left( t \right)\,dt}}\), where \(\displaystyle \vec r\left( t \right) = \left\langle {\frac{1}{{{t^3}}} - {t^3},5t,\frac{1}{{6t}} - \frac{8}{{{t^4}}}} \right\rangle \)
  2. \(\displaystyle \int{{\vec r\left( t \right)\,dt}}\), where \(\vec r\left( t \right) = \left( {{t^2} - 5} \right){{\bf{e}}^{t{\,^3} - 15t}}\,\vec i + 4t\sqrt {{t^2} + 1} \vec j - {\sin ^2}\left( {5t} \right)\vec k\)
  3. \(\displaystyle \int_{0}^{1}{{\vec r\left( t \right)\,dt}}\) where \(\vec r\left( t \right) = \left\langle {t\cos \left( {\pi t} \right),8t - 2,12{t^3} - {{\bf{e}}^{2t}}} \right\rangle \)
  4. \(\displaystyle \int_{1}^{{ - 1}}{{\vec r\left( t \right)\,dt}}\) where \(\vec r\left( t \right) = \tan \left( t \right)\,\,\vec i - {\sin ^3}\left( t \right){\cos ^2}\left( t \right)\vec j - 8t\)