find the domain of the vector functions, r(t), listed below. using interval notation.\na) r(t) = ⟨ln(4t)…

find the domain of the vector functions, r(t), listed below. using interval notation.\na) r(t) = ⟨ln(4t), √(t + 14), 1/√(15 - t)⟩\nb) r(t) = ⟨√(t - 1), sin(7t), t²⟩\nc) r(t) = ⟨e^(-1t), t/√(t² - 49), t^(1/3)⟩\nsubmit answer next item

find the domain of the vector functions, r(t), listed below. using interval notation.\na) r(t) = ⟨ln(4t), √(t + 14), 1/√(15 - t)⟩\nb) r(t) = ⟨√(t - 1), sin(7t), t²⟩\nc) r(t) = ⟨e^(-1t), t/√(t² - 49), t^(1/3)⟩\nsubmit answer next item

Answer

Explanation:

Step1: Analyze component - by - component for part a

For $\ln(4t)$, we need $4t>0$ (logarithm domain), so $t > 0$. For $\sqrt{t + 14}$, we need $t+14\geq0$, so $t\geq - 14$. For $\frac{1}{\sqrt{15 - t}}$, we need $15 - t>0$, so $t<15$.

Step2: Find the intersection for part a

Taking the intersection of $t > 0$, $t\geq - 14$ and $t<15$, we get $0 < t<15$. In interval notation, it is $(0,15)$.

Step3: Analyze component - by - component for part b

For $\sqrt{t - 1}$, we need $t - 1\geq0$, so $t\geq1$. The functions $\sin(7t)$ and $t^{2}$ have domain all real numbers. So the domain of $\mathbf{r}(t)$ is $[1,\infty)$.

Step4: Analyze component - by - component for part c

The function $e^{-t}$ has domain all real numbers. For $\frac{t}{\sqrt{t^{2}-49}}$, we need $t^{2}-49>0$, which gives $(t - 7)(t + 7)>0$. The solutions are $t>7$ or $t<-7$. The function $t^{\frac{1}{3}}$ has domain all real numbers. So the domain of $\mathbf{r}(t)$ is $(-\infty,-7)\cup(7,\infty)$.

Answer:

a. $(0,15)$ b. $[1,\infty)$ c. $(-\infty,-7)\cup(7,\infty)$