Question

Find the domain of the vector function.

\({\bf{r}}(t) = \frac{{t - 2}}{{t + 2}}{\bf{i}} + \sin t{\bf{j}} + \ln \left( {9 - {t^2}} \right){\bf{k}}\)

Short answer

The domain of the vector function is\(( - 2,3)\).

Step-by-step solution

Step 1:Domain of vector component.

The domain of a vector function is the domain of the vector components.

Consider the given vector function is\({\rm{r}}(t) = \frac{{t - 2}}{{t + 2}}{\rm{i}} + \sin t{\rm{j}} + \ln \left( {9 - {t^2}} \right){\bf{k}}\).

The vector component form of the vector function is \(\left\langle {\frac{{t - 2}}{{t + 2}},\sin t,\ln \left( {9 - {t^2}} \right)} \right\rangle \)

Here, the vector component \(\frac{{t - 2}}{{t + 2}}\) is well defined, if\(t + 2 > 0\).

Find the domain of the vector function. 

Simplify the inequality \(t + 2 > 0\) to obtain its domain as follows.

\(\begin{array}{c}t + 2 > 0\\t > - 2\end{array}\)

Hence, the interval notation for the domain of \(\frac{{t - 2}}{{t + 2}}\) is \(( - 2,\infty )\).

The domain of the component \(\sin t\) is the set of all real numbers.

Here, the vector component \(\ln \left( {9 - {t^2}} \right)\) is well defined, if\(\left( {9 - {t^2}} \right) > 0\).

Simplify the inequality \(\left( {9 - {t^2}} \right) > 0\) to obtain its domain as follows.

\(\begin{array}{c}\left( {9 - {t^2}} \right) > 0\\9 > {t^2}\\\sqrt {{t^2}} < \sqrt 9 \\ - 3 < t < 3\end{array}\)

Hence, the interval notation for the domain of \(\ln \left( {9 - {t^2}} \right)\) is\(( - 3, - 3)\).

The domain of the vector function is the intersection of the domain of each component.

Thus, the domain of the vector function \({\rm{r}}(t) = \frac{{t - 2}}{{t + 2}}{\rm{i}} + \sin t{\rm{j}} + \ln \left( {9 - {t^2}} \right){\bf{k}}\) is\(( - 2,3)\).