1. **Sketch the vector-valued function** $\mathbf{r}(t) = \langle 1 + 2t, -1 + 3t \rangle$. - This is a vector function in 2D where the $x$-component is $1 + 2t$ and the $y$-component is $-1 + 3t$. - As $t$ varies, the point moves along a line. - To sketch, find points for some values of $t$: - At $t=0$, $\mathbf{r}(0) = \langle 1, -1 \rangle$. - At $t=1$, $\mathbf{r}(1) = \langle 3, 2 \rangle$. - At $t=-1$, $\mathbf{r}(-1) = \langle -1, -4 \rangle$. - Plot these points and draw a line through them. 2. **Find the limit:** $\lim_{t \to 2} (t\mathbf{i} - 3\mathbf{j} + t^2\mathbf{k})$. - This is a vector function in 3D: $\mathbf{v}(t) = t\mathbf{i} - 3\mathbf{j} + t^2\mathbf{k}$. - Since each component is continuous, the limit is the vector of the limits of each component: - $\lim_{t \to 2} t = 2$ - $\lim_{t \to 2} (-3) = -3$ - $\lim_{t \to 2} t^2 = 4$ - Therefore, the limit is $2\mathbf{i} - 3\mathbf{j} + 4\mathbf{k}$. 3. **Determine whether** $\mathbf{r}(t) = e^{t}\mathbf{i} + \mathbf{j} + \csc(t)\mathbf{k}$ **is continuous at** $t=0$. - The function has three components: - $e^{t}$ is continuous everywhere. - The constant vector $\mathbf{j}$ is continuous. - $\csc(t) = \frac{1}{\sin(t)}$ is undefined at $t=0$ because $\sin(0) = 0$. - Since $\csc(t)$ is not defined at $t=0$, $\mathbf{r}(t)$ is not continuous at $t=0$. **Final answers:** - Sketch: Line through points $(1,-1)$, $(3,2)$, $(-1,-4)$. - Limit: $2\mathbf{i} - 3\mathbf{j} + 4\mathbf{k}$. - Continuity: $\mathbf{r}(t)$ is not continuous at $t=0$ due to $\csc(t)$.