Paul's Online Notes
Paul's Online Notes
Home / Calculus III / 3-Dimensional Space / 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.

Section 12.6 : Vector Functions

For problems 1 & 2 find the domain of the given vector function.

  1. \(\displaystyle \vec r\left( t \right) = \left\langle {{t^2} + 1,\frac{1}{{t + 2}},\sqrt {t + 4} } \right\rangle \) Solution
  2. \(\vec r\left( t \right) = \left\langle {\ln \left( {4 - {t^2}} \right),\sqrt {t + 1} } \right\rangle \) Solution

For problems 3 – 5 sketch the graph of the given vector function.

  1. \(\vec r\left( t \right) = \left\langle {4t,10 - 2t} \right\rangle \) Solution
  2. \(\displaystyle \vec r\left( t \right) = \left\langle {t + 1,\frac{1}{4}{t^2} + 3} \right\rangle \) Solution
  3. \(\vec r\left( t \right) = \left\langle {4\sin \left( t \right),8\cos \left( t \right)} \right\rangle \) Solution

For problems 6 & 7 identify the graph of the vector function without sketching the graph.

  1. \(\vec r\left( t \right) = \left\langle {3\cos \left( {6t} \right), - 4,\sin \left( {6t} \right)} \right\rangle \) Solution
  2. \(\vec r\left( t \right) = \left\langle {2 - t,4 + 7t, - 1 - 3t} \right\rangle \) Solution

For problems 8 & 9 write down the equation of the line segment between the two points.

  1. The line segment starting at \(\left( {1,3} \right)\) and ending at\(\left( { - 4,6} \right)\). Solution
  2. The line segment starting at \(\left( {0,2, - 1} \right)\) and ending at\(\left( {7, - 9,2} \right)\). Solution