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If you are looking for some problems with solutions you can find some by clicking on the "Practice Problems" link above.
If you are looking for some problems with solutions you can find some by clicking on the "Practice Problems" link above.
Section 12.6 : Vector Functions
For problems 1 – 3 find the domain of the given vector function.
- \(\displaystyle \vec r\left( t \right) = \left\langle {\frac{1}{{{t^2} - 1}},\frac{1}{{t + 3}},\frac{1}{{t - 6}}} \right\rangle \)
- \(\vec r\left( t \right) = \left\langle {\sqrt t ,\sqrt {t + 1} ,\sqrt {t + 2} } \right\rangle \)
- \(\vec r\left( t \right) = \left\langle {\ln \left( {t + 7} \right),\ln \left( {t - 3} \right)} \right\rangle \)
For problems 4 – 8 sketch the graph of the given vector function.
- \(\vec r\left( t \right) = \left\langle { - 4,t + 1} \right\rangle \)
- \(\vec r\left( t \right) = \left\langle { - 2\cos \left( t \right),5sin\left( t \right)} \right\rangle \)
- \(\vec r\left( t \right) = \left\langle {\sqrt {t + 2} ,1 - t} \right\rangle \)
- \(\vec r\left( t \right) = \left\langle {2t + 1,{t^2} - 1} \right\rangle \)
- \(\vec r\left( t \right) = \left\langle {{t^2} + 4,6 - {t^2}} \right\rangle \)
For problems 9 – 12 identify the graph of the vector function without sketching the graph.
- \(\vec r\left( t \right) = \left\langle {6,2 + 8t, - 1 + 10t} \right\rangle \)
- \(\vec r\left( t \right) = \left\langle {12t,6 - 8t,4 + 7t} \right\rangle \)
- \(\vec r\left( t \right) = \left\langle {2,6\cos \left( t \right),6\sin \left( t \right)} \right\rangle \)
- \(\vec r\left( t \right) = \left\langle { - 2t,6\cos \left( t \right),6\sin \left( t \right)} \right\rangle \)
For problems 13 – 16 write down the equation of the line segment between the two points.
- The line segment starting at \(\left( {4, - 7} \right)\) and ending at\(\left( {2,0} \right)\).
- The line segment starting at \(\left( { - 1,2} \right)\) and ending at\(\left( {7, - 2} \right)\).
- The line segment starting at \(\left( {4,1, - 3} \right)\) and ending at\(\left( { - 1,2,6} \right)\).
- The line segment starting at \(\left( {1, - 1,9} \right)\) and ending at\(\left( {4, - 7,10} \right)\).