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Section 6-6 : 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