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### Section 12.6 : Vector Functions

For problems 1 – 3 find the domain of the given vector function.

1. $$\displaystyle \vec r\left( t \right) = \left\langle {\frac{1}{{{t^2} - 1}},\frac{1}{{t + 3}},\frac{1}{{t - 6}}} \right\rangle$$
2. $$\vec r\left( t \right) = \left\langle {\sqrt t ,\sqrt {t + 1} ,\sqrt {t + 2} } \right\rangle$$
3. $$\vec r\left( t \right) = \left\langle {\ln \left( {t + 7} \right),\ln \left( {t - 3} \right)} \right\rangle$$

For problems 4 – 8 sketch the graph of the given vector function.

1. $$\vec r\left( t \right) = \left\langle { - 4,t + 1} \right\rangle$$
2. $$\vec r\left( t \right) = \left\langle { - 2\cos \left( t \right),5sin\left( t \right)} \right\rangle$$
3. $$\vec r\left( t \right) = \left\langle {\sqrt {t + 2} ,1 - t} \right\rangle$$
4. $$\vec r\left( t \right) = \left\langle {2t + 1,{t^2} - 1} \right\rangle$$
5. $$\vec r\left( t \right) = \left\langle {{t^2} + 4,6 - {t^2}} \right\rangle$$

For problems 9 – 12 identify the graph of the vector function without sketching the graph.

1. $$\vec r\left( t \right) = \left\langle {6,2 + 8t, - 1 + 10t} \right\rangle$$
2. $$\vec r\left( t \right) = \left\langle {12t,6 - 8t,4 + 7t} \right\rangle$$
3. $$\vec r\left( t \right) = \left\langle {2,6\cos \left( t \right),6\sin \left( t \right)} \right\rangle$$
4. $$\vec r\left( t \right) = \left\langle { - 2t,6\cos \left( t \right),6\sin \left( t \right)} \right\rangle$$

For problems 13 – 16 write down the equation of the line segment between the two points.

1. The line segment starting at $$\left( {4, - 7} \right)$$ and ending at$$\left( {2,0} \right)$$.
2. The line segment starting at $$\left( { - 1,2} \right)$$ and ending at$$\left( {7, - 2} \right)$$.
3. The line segment starting at $$\left( {4,1, - 3} \right)$$ and ending at$$\left( { - 1,2,6} \right)$$.
4. The line segment starting at $$\left( {1, - 1,9} \right)$$ and ending at$$\left( {4, - 7,10} \right)$$.