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Section 12.6 : Vector Functions

2. Find the domain for the vector function : \(\vec r\left( t \right) = \left\langle {\ln \left( {4 - {t^2}} \right),\sqrt {t + 1} } \right\rangle \)

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The 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.

We know that we can’t take logarithms of negative values or zero and so from the first term we know that we’ll need to require that,

\[4 - {t^2} > 0\hspace{0.5in} \to \hspace{0.5in} - 2 < t < 2\]

We’ll also need to require that \(t \ge - 1\) so avoid taking the square root of negative numbers in the second component.

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Putting all of the information from the first step together we can see that the domain of this function is,

\[\require{bbox} \bbox[2pt,border:1px solid black]{{ - 1 \le t < 2}}\]

Remember that we want the largest possible set of \(t\)’s for which all the components will exist. So we can’t take values of \( - 2 < t < - 1\) because even though those are okay in the first component but they aren’t in the second component. Likewise, even though we can include \(t \ge 2\) in the second component we can’t plug them into the first component and so we can’t include them in the domain of the function.