Section 12.6 : Vector Functions
1. Find the domain for the vector function : \(\displaystyle \vec r\left( t \right) = \left\langle {{t^2} + 1,\frac{1}{{t + 2}},\sqrt {t + 4} } \right\rangle \)
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Start SolutionThe domain of the vector function is simply the largest possible set of \(t\)’s that we can use in all the components of the vector function.
The first component will exist for all values of \(t\) and so we won’t exclude any values of \(t\) from that component.
The second component clearly requires us to avoid \(t = - 2\) so we don’t have division by zero in that component.
We’ll also need to require that \(t \ge - 4\) so avoid taking the square root of negative numbers in the third component.
Show Step 2Putting all of the information from the first step together we can see that the domain of this function is,
\[\require{bbox} \bbox[2pt,border:1px solid black]{{t \ge - 4,\,\,\,\,t \ne - 2}}\]Note that we can’t forget to add the \(t \ne - 2\) onto this since -2 is larger than -4 and would be included otherwise!