Answer
$-\infty < t < \infty$
$\mathbf{r}(\pi) = -\mathbf{i} - 3\pi\mathbf{j}$.
Work Step by Step
Step 1
Given the vector function, the standard form of a vector function can be given as: \[ \mathbf{r}(t) = x(t)\mathbf{i} + y(t)\mathbf{j} \] Now compare the two functions together: \[ \cos(t)\mathbf{i} - 3t\mathbf{j} \] The parametric equations can be represented as: \[ x(t) = \cos(t) \quad \text{(1)} \] \[ y(t) = -3t \quad \text{(2)} \]
Step 2
The natural domains of the component functions, which are (1) and (2), are: \[ (-\infty, \infty) \quad \text{and} \quad (-\infty, \infty) \] So, the natural domain of $\mathbf{r}(t)$ consists of all values of $t$ such that $-\infty < t < \infty$.
Step 3
Now, replace $t$ with $t_0$ to find the value of $\mathbf{r}(t_0)$. We have: \[ \mathbf{r}(t_0) = \cos(t_0)\mathbf{i} - 3t_0\mathbf{j} \] Then, substitute $t_0 = \pi$ in $\mathbf{r}(t_0)$. We have: \[ \mathbf{r}(\pi) = (\cos(\pi))\mathbf{i} - (3\pi)\mathbf{j} = (-1)\mathbf{i} - (3\pi)\mathbf{j} \] Therefore, the domain of $\mathbf{r}(t)$ is $-\infty < t < \infty$, and $\mathbf{r}(\pi) = (-1)\mathbf{i} - (3\pi)\mathbf{j}$.
