Let's solve the problem step by step: --- ### **Given:** Vector field: \[ \vec{F}(x, y) = x \mathbf{i} + \mathbf{j} \] Initial position: At \( t = 0 \), \( (x_0, y_0) = (1, 2) \) --- ## **(A) Sketch the vector field over \([-3, 3]\)** - At each point \((x, y)\), the vector is \((x, 1)\). - This means *horizontal* component depends on \(x\), *vertical* component is always 1. - For example: - At \((0, 0)\): vector = \((0, 1)\) (straight up) - At \((2, 0)\): vector = \((2, 1)\) (right and up) - At \((-2, 0)\): vector = \((-2, 1)\) (left and up) - The field has arrows that slant more to the right as \(x\) increases and more to the left as \(x\) decreases, but always point up. --- ## **(B) Euler’s Method with Step Size \(0.5\)** The system of ODEs for the flow line: \[ \frac{dx}{dt} = x, \quad \frac{dy}{dt} = 1 \] Euler's method formulas: \[ x_{n+1} = x_n + h f(x_n, y_n) \\ y_{n+1} = y_n + h g(x_n, y_n) \] where \(h = 0.5\). Initial: \(x_0 = 1\), \(y_0 = 2\) **Step 1: \(t_0 = 0\), \(t_1 = 0.5\)** - \(x_1 = x_0 + 0.5 \cdot x_0 = 1 + 0.5 \times 1 = 1.5\) - \(y_1 = y_0 + 0.5 \cdot 1 = 2 + 0.5 = 2.5\) **Step 2: \(t_1 = 0.5\), \(t_2 = 1.0\)** - \(x_2 = x_1 + 0.5 \cdot x_1 = 1.5 + 0.5 \times 1.5 = 1.5 + 0.75 = 2.25\) - \(y_2 = y_1 + 0.5 \cdot 1 = 2.5 + 0.5 = 3.0\) **Approximate location at \(t=1\):** \[ \boxed{(2.25,\ 3.0)} \] --- ## **(C) Find a Flow Line for \(\vec{F}\)** Solve: \[ \frac{dx}{dt} = x \implies x(t) = x_0 e^{t} \] \[ \frac{dy}{dt} = 1 \implies y(t) = t + y_0 \] Given \(x_0 = 1, y_0 = 2\): \[ \boxed{x(t) = e^{t},\quad y(t) = t + 2} \] Or eliminate \(t\): \[ x = e^{y-2} \] This is the flow line equation. --- ## **(D) Draw the Flow Line on the Vector Field** - On the vector field, sketch the curve \(x = e^{y-2}\) starting at \((1,2)\). - It is an exponential curve rising as \(y\) increases. --- ### **Summary of Answers** **(A)** Sketch vectors \((x, 1)\) at each point in \([-3, 3]^2\). **(B)** Euler’s method gives approximate position at \(t=1\): \(\boxed{(2.25,\, 3.0)}\) **(C)** Flow line: \(\boxed{x = e^{y-2}}\) **(D)** On the vector field sketch, draw the curve \(x = e^{y-2}\) starting from \((1,2)\). If you'd like, I can draw the vector field and the flow line for you!

Certainly! Here's a visualization to help you understand the vector field and the flow line: ### 1. **Vector Field \(\vec{F}(x, y) = x \mathbf{i} + \mathbf{j}\)** - At each point \((x, y)\), vectors point **upward** with a horizontal component equal to \(x\). - For example: - When \(x=0\), vectors point straight up \((0,1)\). - When \(x>0\), vectors lean **right and up**. - When \(x<0\), vectors lean **left and up**. ### 2. **Flow Line Equation** - The flow line starting at \((1,2)\) is: \[ x = e^{y-2} \] - This is an **exponential curve** passing through \((1, 2)\). ### 3. **Drawing the Flow Line** - At \(y=2\), \(x = e^{0} = 1\), so it starts at \((1,2)\). - As \(y\) increases, \(x\) increases exponentially. - As \(y\) decreases, \(x\) decreases exponentially. --- ### **Visualization Sketch:** Here's a simplified sketch representation: ``` ^ | / (Flow line: x = e^{y-2}) | / | / | / | / | / |-----*----------------> y | (1,2) | | (Other vectors pointing upward with horizontal component depending on x) ``` - The curve starts at \((1,2)\) and rises steeply as \(y\) increases. - The vector arrows around the line point mostly upward, leaning to the right or left depending on \(x\). --- **Note:** To see an accurate and detailed drawing, you might consider plotting this using graphing software like Desmos, GeoGebra, or a graphing calculator. Would you like me to generate a digital plot for you?

Certainly! Here's a detailed explanation and a sketch for **Part (A)** — the vector field \(\vec{F}(x,y) = x \mathbf{i} + \mathbf{j}\) over the domain \([-3, 3]\). --- ### **Part (A): Sketch of the Vector Field \(\vec{F}(x, y) = x \mathbf{i} + \mathbf{j}\)** **Key points:** - At each point \((x, y)\), the vector is \((x, 1)\). - The vertical component is always 1, so all vectors point **upward**. - The horizontal component is equal to \(x\), so: - When \(x > 0\), vectors point **upward and to the right**. - When \(x < 0\), vectors point **upward and to the left**. - When \(x = 0\), vectors point **straight up**. --- ### **Sketch instructions:** 1. **Draw the coordinate axes**: \(x\) from \(-3\) to \(3\), \(y\) from \(-3\) to \(3\). 2. **At selected points**, draw vectors: - \((x, y)\) with arrow from the point pointing in the direction \((x, 1)\). - The length of each arrow is proportional to the magnitude \(\sqrt{x^2 + 1}\) (not necessarily to scale here, just direction). 3. **Sample vectors:** - At \((0, 0)\): vector points straight up, \((0, 1)\). - At \((1, 0)\): vector \((1, 1)\), pointing diagonally up and right. - At \((-1, 0)\): vector \((-1, 1)\), pointing diagonally up and left. - At \((3, 0)\): \((3, 1)\), pointing up and to the right. - At \((-3, 0)\): \((-3, 1)\), pointing up and to the left. Similarly, at \((x, y)\) for various \(x\), the arrows will lean right or left accordingly, always pointing upward. --- ### **Visual Sketch (ASCII approximation):** ``` y ^ | / / / | / / / | / / / | / / / | / / / | / / / | *--------------*--------------*--------------> x | (-3, y) (0, y) (3, y) | ``` - The arrows near \((-3, y)\) point up and to the **left**. - Near \((0, y)\), arrows point straight up. - Near \((3, y)\), arrows point up and to the **right**. --- ### **Digital Plot (Recommendation):** For a precise visual, I recommend plotting with software like GeoGebra or Desmos: - Plot points over \([-3,3]\) in x and y. - At each point, draw an arrow from \((x,y)\) to \((x + \text{scale} \times x, y + \text{scale} \times 1)\) with an appropriate scale for visibility. --- ### **Summary:** - The vectors always point **upward** (vertical component = 1). - The **horizontal** component varies with \(x\): positive to the right, negative to the left. - The overall pattern is a "fan" of arrows pointing upward, leaning right on the right side, and leaning left on the left side. --- If you want, I can generate an actual plot image or provide code to create one!