1. **Problem:** Match the vector equation $\mathbf{r}(t)$ with its corresponding graph. 2. **Understanding vector equations:** Each vector equation describes a curve or line in 3D space parameterized by $t$. The components of $\mathbf{r}(t)$ give the $x$, $y$, and $z$ coordinates as functions of $t$. 3. **Analyze each vector equation:** - $\mathbf{r}(t) = t\mathbf{i} + (1 - t)\mathbf{j}$, $0 \leq t \leq 1$: - $x = t$, $y = 1 - t$, $z = 0$ (since no $k$ component). - This is a line segment in the $xy$-plane from $(0,1,0)$ to $(1,0,0)$. - $\mathbf{r}(t) = \mathbf{i} + \mathbf{j} + t\mathbf{k}$, $-1 \leq t \leq 1$: - $x=1$, $y=1$, $z=t$. - A vertical line segment along $z$ from $(1,1,-1)$ to $(1,1,1)$. - $\mathbf{r}(t) = 2\cos t\mathbf{i} + 2\sin t\mathbf{j}$, $0 \leq t \leq 2\pi$: - $x=2\cos t$, $y=2\sin t$, $z=0$. - A circle of radius 2 in the $xy$-plane. - $\mathbf{r}(t) = t\mathbf{i}$, $-1 \leq t \leq 1$: - $x=t$, $y=0$, $z=0$. - A line segment along the $x$-axis from $(-1,0,0)$ to $(1,0,0)$. - $\mathbf{r}(t) = t\mathbf{i} + t\mathbf{j} + t\mathbf{k}$, $0 \leq t \leq 2$: - $x=t$, $y=t$, $z=t$. - A line segment from $(0,0,0)$ to $(2,2,2)$ along the line $x=y=z$. - $\mathbf{r}(t) = t\mathbf{j} + (2 - 2t)\mathbf{k}$, $0 \leq t \leq 1$: - $x=0$, $y=t$, $z=2-2t$. - A line segment from $(0,0,2)$ to $(0,1,0)$ in the $yz$-plane. - $\mathbf{r}(t) = (t^2 - 1)\mathbf{j} + 2t\mathbf{k}$, $-1 \leq t \leq 1$: - $x=0$, $y=t^2 - 1$, $z=2t$. - A parabola in the $yz$-plane. - $\mathbf{r}(t) = 2\cos t\mathbf{i} + 2\sin t\mathbf{k}$, $0 \leq t \leq \pi$: - $x=2\cos t$, $y=0$, $z=2\sin t$. - A half circle of radius 2 in the $xz$-plane from $(2,0,0)$ to $(-2,0,0)$. 4. **Match with graphs:** - Graph a: Points at $x=1$ and $z=-1$ with a diagonal vector arrow suggest a 3D line segment involving $x$ and $z$. - Graph b: Points at $z=2$ and a vector arrow downward along $z$ suggest a vertical segment along $z$. 5. **Conclusion:** - For graph b, the vector equation $\mathbf{r}(t) = t\mathbf{j} + (2 - 2t)\mathbf{k}$, $0 \leq t \leq 1$ fits perfectly because it starts at $z=2$ and moves downward. - For graph a, the vector equation $\mathbf{r}(t) = \mathbf{i} + \mathbf{j} + t\mathbf{k}$, $-1 \leq t \leq 1$ fits because it has fixed $x=1$, $y=1$ and varies $z$ from $-1$ to $1$. **Final answers:** - Graph a matches equation 2. - Graph b matches equation 6.