Question

Find the domain of the vector function.

\({\bf{r}}(t) = \left\langle {\sqrt {4 - {t^2}} ,{e^{ - 3t}},\ln (t + 1)} \right\rangle \)

Short answer

The domain of the vector function is\(\underline {( - 1,2)} \).

Step-by-step solution

Domain of the vector function.

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

Consider the given vector function is \({\rm{r}}(t) = \left\langle {\sqrt {4 - {t^2}} ,{e^{ - 3t}},\ln (t + 1)} \right\rangle \).

Here, the vector component \(\sqrt {4 - {t^2}} \) is well defined for \(4 - {t^2} \ge 0\).

Find the domain of the vector function.

Simplify the inequality \(4 - {t^2} \ge 0\) to obtain its domain as follows.

\(\begin{array}{c}4 - {t^2} \ge 0\\4 \ge {t^2}\\\sqrt {{t^2}} \le \sqrt 4 \\|t| \le 2\\ - 2 \le t \le 2\end{array}\)

Hence, the interval notation for the domain of \(\sqrt {4 - {t^2}} \) is\(( - 2,2)\).

The domain of the component \({e^{ - 3t}}\) is the set of all real numbers.

The vector component \(\ln (t + 1)\) is well defined for\((t + 1) > 0\).

Simplify the inequality \((t + 1) > 0\) to obtain its domain as follows.

\(\begin{array}{c}(t + 1) > 0\\t > - 1\end{array}\)

Hence, the interval notation for the domain of \(\ln (t + 1)\) is\(( - 1,\infty )\).

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

Domain

\(\begin{array}{c}( - 2,2) \cap ( - \infty ,\infty ) \cap ( - 1,\infty )\\ = \left( { - 1,\left. 2 \right)} \right.\end{array}\)

Thus, the domain of the vector function \({\rm{r}}(t) = \left\langle {\sqrt {4 - {t^2}} ,{e^{ - 3t}},\ln (t + 1)} \right\rangle \) is\(\left( { - 1,\left. 2 \right)} \right.\).