Let's go through the step-by-step solution for part (b): ## **Given:** - \( \vec{F}(x, y, z) = x\vec{i} - z\vec{j} + y\vec{k} \) - \( \vec{r}(t) = 2t\vec{i} + 9t\vec{j} - t^2\vec{k} \) for \( -1 \leq t \leq 1 \) We need to **find the value of \( \vec{F}(\vec{r}(t)) \) at \( t = -1, -\frac{1}{2}, \frac{1}{2}, 1 \)**, and **draw the vectors starting at the point \( \vec{r}(t) \) and ending at \( \vec{r}(t) + \vec{F}(\vec{r}(t)) \)**. --- ### **Step 1: Compute \( \vec{r}(t) \) for each value of \( t \)** #### **For \( t = -1 \):** \[ \vec{r}(-1) = 2(-1)\vec{i} + 9(-1)\vec{j} - (-1)^2\vec{k} = -2\vec{i} - 9\vec{j} - 1\vec{k} \] Coordinates: \( (-2, -9, -1) \) #### **For \( t = -\frac{1}{2} \):** \[ \vec{r}\left(-\frac{1}{2}\right) = 2\left(-\frac{1}{2}\right)\vec{i} + 9\left(-\frac{1}{2}\right)\vec{j} - \left(-\frac{1}{2}\right)^2\vec{k} \] \[ = -1\vec{i} - 4.5\vec{j} - 0.25\vec{k} \] Coordinates: \( (-1, -4.5, -0.25) \) #### **For \( t = \frac{1}{2} \):** \[ \vec{r}\left(\frac{1}{2}\right) = 2\left(\frac{1}{2}\right)\vec{i} + 9\left(\frac{1}{2}\right)\vec{j} - \left(\frac{1}{2}\right)^2\vec{k} \] \[ = 1\vec{i} + 4.5\vec{j} - 0.25\vec{k} \] Coordinates: \( (1, 4.5, -0.25) \) #### **For \( t = 1 \):** \[ \vec{r}(1) = 2(1)\vec{i} + 9(1)\vec{j} - (1)^2\vec{k} = 2\vec{i} + 9\vec{j} - 1\vec{k} \] Coordinates: \( (2, 9, -1) \) --- ### **Step 2: Compute \( \vec{F}(\vec{r}(t)) \) for each value of \( t \)** Recall: \( \vec{F}(x, y, z) = x\vec{i} - z\vec{j} + y\vec{k} \) #### **For \( t = -1 \):** \[ x = -2, \quad y = -9, \quad z = -1 \] \[ \vec{F}(-2, -9, -1) = -2\vec{i} - (-1)\vec{j} + (-9)\vec{k} = -2\vec{i} + 1\vec{j} - 9\vec{k} \] End point: \( (-2+(-2), -9+1, -1+(-9)) = (-4, -8, -10) \) #### **For \( t = -\frac{1}{2} \):** \[ x = -1, \quad y = -4.5, \quad z = -0.25 \] \[ \vec{F}(-1, -4.5, -0.25) = -1\vec{i} - (-0.25)\vec{j} + (-4.5)\vec{k} = -1\vec{i} + 0.25\vec{j} - 4.5\vec{k} \] End point: \( (-1+(-1), -4.5+0.25, -0.25+(-4.5)) = (-2, -4.25, -4.75) \) #### **For \( t = \frac{1}{2} \):** \[ x = 1, \quad y = 4.5, \quad z = -0.25 \] \[ \vec{F}(1, 4.5, -0.25) = 1\vec{i} - (-0.25)\vec{j} + 4.5\vec{k} = 1\vec{i} + 0.25\vec{j} + 4.5\vec{k} \] End point: \( (1+1, 4.5+0.25, -0.25+4.5) = (2, 4.75, 4.25) \) #### **For \( t = 1 \):** \[ x = 2, \quad y = 9, \quad z = -1 \] \[ \vec{F}(2, 9, -1) = 2\vec{i} - (-1)\vec{j} + 9\vec{k} = 2\vec{i} + 1\vec{j} + 9\vec{k} \] End point: \( (2+2, 9+1, -1+9) = (4, 10, 8) \) --- ### **Step 3: Final Answers** #### **Summary Table** | \( t \) | Start Point \( \vec{r}(t) \) | \( \vec{F}(\vec{r}(t)) \) | End Point \( \vec{r}(t) + \vec{F}(\vec{r}(t)) \) | |--------------|----------------------------------------|---------------------------------------|------------------------------------------------------| | \( -1 \) | \( (-2, -9, -1) \) | \( (-2, 1, -9) \) | \( (-4, -8, -10) \) | | \( -\frac{1}{2} \) | \( (-1, -4.5, -0.25) \) | \( (-1, 0.25, -4.5) \) | \( (-2, -4.25, -4.75) \) | | \( \frac{1}{2} \) | \( (1, 4.5, -0.25) \) | \( (1, 0.25, 4.5) \) | \( (2, 4.75, 4.25) \) | | \( 1 \) | \( (2, 9, -1) \) | \( (2, 1, 9) \) | \( (4, 10, 8) \) | --- ### **Step 4: Explanation for the Plot** On your graph, **draw a vector starting at each "Start Point" and ending at the corresponding "End Point"** as calculated above. The vector itself is the value of \( \vec{F}(\vec{r}(t)) \) at that point. --- ## **Final Answers** - For \( t = -1 \): starts at \( (-2, -9, -1) \), ends at \( (-4, -8, -10) \) - For \( t = -\frac{1}{2} \): starts at \( (-1, -4.5, -0.25) \), ends at \( (-2, -4.25, -4.75) \) - For \( t = \frac{1}{2} \): starts at \( (1, 4.5, -0.25) \), ends at \( (2, 4.75, 4.25) \) - For \( t = 1 \): starts at \( (2, 9, -1) \), ends at \( (4, 10, 8) \) You can now enter these endpoints in your system to illustrate the vectors correctly!

Certainly! Let's carefully walk through each step of the problem, providing detailed calculations and explanations for each part. --- ## **Given Data:** - Vector field: \[ \vec{F}(x, y, z) = x \mathbf{i} - z \mathbf{j} + y \mathbf{k} \] - Parametric curve: \[ \vec{r}(t) = 2t \mathbf{i} + 9t \mathbf{j} - t^2 \mathbf{k} \] for \( t \in [-1, 1] \). --- ## **Part (b): Find \( \vec{F}(\vec{r}(t)) \) at specific \( t \)-values and draw corresponding vectors** ### **Step 1: Find the points \( \vec{r}(t) \) at \( t = -1, -\frac{1}{2}, \frac{1}{2}, 1 \).** --- ### **Calculations for \( \vec{r}(t) \):** \[ \boxed{ \vec{r}(t) = (x(t), y(t), z(t)) } \] where: \[ x(t) = 2t, \quad y(t) = 9t, \quad z(t) = -t^2 \] --- ### **At \( t = -1 \):** \[ x(-1) = 2 \times (-1) = -2 \] \[ y(-1) = 9 \times (-1) = -9 \] \[ z(-1) = -(-1)^2 = -1 \] **Point:** \[ \boxed{ \vec{r}(-1) = (-2, -9, -1) } \] --- ### **At \( t = -\frac{1}{2} \):** \[ x\left(-\frac{1}{2}\right) = 2 \times \left(-\frac{1}{2}\right) = -1 \] \[ y\left(-\frac{1}{2}\right) = 9 \times \left(-\frac{1}{2}\right) = -4.5 \] \[ z\left(-\frac{1}{2}\right) = - \left(-\frac{1}{2}\right)^2 = - \frac{1}{4} = -0.25 \] **Point:** \[ \boxed{ \vec{r}\left(-\frac{1}{2}\right) = (-1, -4.5, -0.25) } \] --- ### **At \( t = \frac{1}{2} \):** \[ x\left(\frac{1}{2}\right) = 2 \times \frac{1}{2} = 1 \] \[ y\left(\frac{1}{2}\right) = 9 \times \frac{1}{2} = 4.5 \] \[ z\left(\frac{1}{2}\right) = - \left(\frac{1}{2}\right)^2 = - \frac{1}{4} = -0.25 \] **Point:** \[ \boxed{ \vec{r}\left(\frac{1}{2}\right) = (1, 4.5, -0.25) } \] --- ### **At \( t = 1 \):** \[ x(1) = 2 \times 1 = 2 \] \[ y(1) = 9 \times 1 = 9 \] \[ z(1) = -1^2 = -1 \] **Point:** \[ \boxed{ \vec{r}(1) = (2, 9, -1) } \] --- ## **Step 2: Compute \( \vec{F}(\vec{r}(t)) \) at each \( t \).** Recall: \[ \vec{F}(x, y, z) = x \mathbf{i} - z \mathbf{j} + y \mathbf{k} \] which in component form is: \[ \vec{F} = (x, -z, y) \] --- ### **At \( t = -1 \):** \[ x = -2, \quad y = -9, \quad z = -1 \] \[ \boxed{ \vec{F}(-2, -9, -1) = (-2, -(-1), -9) = (-2, 1, -9) } \] --- ### **At \( t = -\frac{1}{2} \):** \[ x = -1, \quad y = -4.5, \quad z = -0.25 \] \[ \vec{F}(-1, -4.5, -0.25) = (-1, -(-0.25), -4.5) = (-1, 0.25, -4.5) \] --- ### **At \( t = \frac{1}{2} \):** \[ x = 1, \quad y = 4.5, \quad z = -0.25 \] \[ \vec{F}(1, 4.5, -0.25) = (1, -(-0.25), 4.5) = (1, 0.25, 4.5) \] --- ### **At \( t = 1 \):** \[ x = 2, \quad y = 9, \quad z = -1 \] \[ \vec{F}(2, 9, -1) = (2, -(-1), 9) = (2, 1, 9) \] --- ## **Step 3: Determine the "end point" of each vector \( \vec{F}(\vec{r}(t)) \) starting from \( \vec{r}(t) \).** This is simply: \[ \text{End point} = \vec{r}(t) + \vec{F}(\vec{r}(t)) \] --- ### **At \( t = -1 \):** \[ \vec{r}(-1) = (-2, -9, -1) \] \[ \vec{F} = (-2, 1, -9) \] \[ \text{End point} = (-2 + (-2), -9 + 1, -1 + (-9)) = (-4, -8, -10) \] --- ### **At \( t = -\frac{1}{2} \):** \[ \vec{r} = (-1, -4.5, -0.25) \] \[ \vec{F} = (-1, 0.25, -4.5) \] \[ \text{End point} = (-1 + (-1), -4.5 + 0.25, -0.25 + (-4.5)) = (-2, -4.25, -4.75) \] --- ### **At \( t = \frac{1}{2} \):** \[ \vec{r} = (1, 4.5, -0.25) \] \[ \vec{F} = (1, 0.25, 4.5) \] \[ \text{End point} = (1 + 1, 4.5 + 0.25, -0.25 + 4.5) = (2, 4.75, 4.25) \] --- ### **At \( t = 1 \):** \[ \vec{r} = (2, 9, -1) \] \[ \vec{F} = (2, 1, 9) \] \[ \text{End point} = (2 + 2, 9 + 1, -1 + 9) = (4, 10, 8) \] --- ## **## **Final Summary:** | \( t \) | Starting point \( \vec{r}(t) \) | \( \vec{F}(\vec{r}(t)) \) | End point \( \vec{r}(t) + \vec{F}(\vec{r}(t)) \) | |---------|----------------------------------|---------------------------|--------------------------------------------| | -1 | \((-2, -9, -1)\) | \((-2, 1, -9)\) | \((-4, -8, -10)\) | | -½ | \((-1, -4.5, -0.25)\) | \((-1, 0.25, -4.5)\) | \((-2, -4.25, -4.75)\) | | ½ | \((1, 4.5, -0.25)\) | \((1, 0.25, 4.5)\) | \((2, 4.75, 4.25)\) | | 1 | \((2, 9, -1)\) | \((2, 1, 9)\) | \((4, 10, 8)\) | --- ## **How to visualize:** - Draw the initial point \( \vec{r}(t) \). - From that point, draw a vector to the end point \( \vec{r}(t) + \vec{F}(\vec{r}(t)) \). - Repeat for each \( t \). --- **This completes the detailed step-by-step solution for part (b).**

Certainly! Here's a detailed step-by-step solution for both parts (a) and (b), with all calculations shown explicitly and without using tables. --- ## **Given Data:** - Vector field: \[ \vec{F}(x, y, z) = x \mathbf{i} - z \mathbf{j} + y \mathbf{k} \] - Parametric curve: \[ \vec{r}(t) = 2t \mathbf{i} + 9t \mathbf{j} - t^2 \mathbf{k} \] for \( t \in [-1, 1] \). --- ## **Part (a): Evaluate the line integral \( \int_C \vec{F} \cdot d\vec{r} \) along the curve \( C \).** --- ### **Step 1: Parameterize the curve** \[ x(t) = 2t, \quad y(t) = 9t, \quad z(t) = -t^2 \] for \( t \in [-1, 1] \). ### **Step 2: Compute derivatives \( dx/dt, dy/dt, dz/dt \)** \[ \frac{dx}{dt} = 2 \] \[ \frac{dy}{dt} = 9 \] \[ \frac{dz}{dt} = -2t \] ### **Step 3: Express \( d\vec{r} \) in terms of \( dt \)** \[ d\vec{r} = \frac{d\vec{r}}{dt} dt = (2, 9, -2t) dt \] ### **Step 4: Express \( \vec{F} \) along the curve** At point \( \vec{r}(t) \): \[ x = 2t, \quad y = 9t, \quad z = -t^2 \] \[ \vec{F}(\vec{r}(t)) = (x, -z, y) = (2t, -(-t^2), 9t) = (2t, t^2, 9t) \] ### **Step 5: Compute the dot product \( \vec{F}(\vec{r}(t)) \cdot \frac{d\vec{r}}{dt} \)** \[ (2t, t^2, 9t) \cdot (2, 9, -2t) = (2t)(2) + (t^2)(9) + (9t)(-2t) \] Calculating each term: - \( (2t)(2) = 4t \) - \( (t^2)(9) = 9t^2 \) - \( (9t)(-2t) = -18 t^2 \) Adding these: \[ 4t + 9t^2 - 18t^2 = 4t - 9t^2 \] --- ### **Step 6: Set up the integral** \[ \int_{t=-1}^{1} [4t - 9t^2] dt \] ### **Step 7: Integrate** \[ \int_{-1}^{1} 4t dt - \int_{-1}^{1} 9t^2 dt \] Calculate each separately: - \( \int_{-1}^{1} 4t dt = 2t^2 \big|_{-1}^{1} \) At \( t=1 \): \( 2(1)^2 = 2 \) At \( t=-1 \): \( 2(-1)^2 = 2 \) Difference: \( 2 - 2 = 0 \) - \( \int_{-1}^{1} 9t^2 dt = 3t^3 \big|_{-1}^{1} \) At \( t=1 \): \( 3(1)^3 = 3 \) At \( t=-1 \): \( 3(-1)^3 = -3 \) Difference: \( 3 - (-3) = 6 \) ### **Step 8: Final value of the line integral** \[ 0 - 6 = -6 \] --- ## **Answer for Part (a):** \[ \boxed{ \int_C \vec{F} \cdot d\vec{r} = -6 } \] --- ## **Part (b): Find \( \vec{F} \) at specific points \( \vec{r}(t) \) for \( t = -1, -\frac{1}{2}, \frac{1}{2}, 1 \), and determine the vectors starting at \( \vec{r}(t) \) and ending at \( \vec{r}(t) + \vec{F}(\vec{r}(t)) \).** --- ### **Step 1: Compute \( \vec{r}(t) \) at each specified \( t \).** - For \( t = -1 \): \[ x = 2(-1) = -2 \] \[ y = 9(-1) = -9 \] \[ z = -(-1)^2 = -1 \] **Point:** \[ \boxed{ \vec{r}(-1) = (-2, -9, -1) } \] - For \( t = -\frac{1}{2} \): \[ x = 2 \times \left(-\frac{1}{2}\right) = -1 \] \[ y = 9 \times \left(-\frac{1}{2}\right) = -4.5 \] \[ z = - \left(-\frac{1}{2}\right)^2 = - \frac{1}{4} = -0.25 \] **Point:** \[ \boxed{ \vec{r}\left(-\frac{1}{2}\right) = (-1, -4.5, -0.25) } \] - For \( t = \frac{1}{2} \): \[ x = 2 \times \frac{1}{2} = 1 \] \[ y = 9 \times \frac{1}{2} = 4.5 \] \[ z = - \left(\frac{1}{2}\right)^2 = - \frac{1}{4} = -0.25 \] **Point:** \[ \boxed{ \vec{r}\left(\frac{1}{2}\right) = (1, 4.5, -0.25) } \] - For \( t=1 \): \[ x=2(1)=2, \quad y=9(1)=9, \quad z=-(1)^2=-1 \] **Point:** \[ \boxed{ \vec{r}(1) = (2, 9, -1) } \] --- ### **Step 2: Compute \( \vec{F}(\vec{r}(t)) \) at each point** Recall: \[ \vec{F}(x, y, z) = (x, -z, y) \] - At \( t = -1 \), where \( (x, y, z) = (-2, -9, -1) \): \[ \vec{F} = (-2, -(-1), -9) = (-2, 1, -9) \] - At \( t = -\frac{1}{2} \), with \( (-1, -4.5, -0.25) \): \[ \vec{F} = (-1, -(-0.25), -4.5) = (-1, 0.25, -4.5) \] - At \( t=\frac{1}{2} \), with \( (1, 4.5, -0.25) \): \[ \vec{F} = (1, -(-0.25), 4.5) = (1, 0.25, 4.5) \] - At \( t=1 \), with \( (2, 9, -1) \): \[ \vec{F} = (2, -(-1), 9) = (2, 1, 9) \] --- ### **Step 3: Determine the end points of the vectors starting at \( \vec{r}(t) \)** Add \( \vec{F}(\vec{r}(t)) \) to \( \vec{r}(t) \): - For \( t=-1 \): \[ (-2, -9, -1) + (-2, 1, -9) = (-2 + (-2), -9 + 1, -1 + (-9)) = (-4, -8, -10) \] - For \( t=-\frac{1}{2} \): \[ (-1, -4.5, -0.25) + (-1, 0.25, -4.5) = (-2, -4.25, -4.75) \] - For \( t=\frac{1}{2} \): \[ (1, 4.5, -0.25) + (1, 0.25, 4.5) = (2, 4.75, 4.25) \] - For \( t=1 \): \[ (2, 9, -1) + (2, 1, 9) = (4, 10, 8) \] --- ## **Final summary:** - At \( t=-1 \): start point \( (-2, -9, -1) \), vector \( (-2, 1, -9) \), end point \( (-4, -8, -10) \). - At \( t=-\frac{1}{2} \): start point \( (-1, -4.5, -0.25) \), vector \( (-1, 0.25, -4.5) \), end point \( (-2, -4.25, -4.75) \). - At \( t=\frac{1}{2} \): start point \( (1, 4.5, -0.25) \), vector \( (1, 0.25, 4.5) \), end point \( (2, 4.75, 4.25) \). - At \( t=1 \): start point \( (2, 9, -1) \), vector \( (2, 1, 9) \), end point \( (4, 10, 8) \). --- This completes the detailed step-by-step solution for both parts.

Certainly! Here are the final answers for both parts: **Part (a):** The value of the line integral along the curve is: \(\boxed{-6}\) **Part (b):** At \( t = -1 \): start at \((-2, -9, -1)\), vector \((-2, 1, -9)\), end at \((-4, -8, -10)\) At \( t = -\frac{1}{2} \): start at \((-1, -4.5, -0.25)\), vector \((-1, 0.25, -4.5)\), end at \((-2, -4.25, -4.75)\) At \( t = \frac{1}{2} \): start at \((1, 4.5, -0.25)\), vector \((1, 0.25, 4.5)\), end at \((2, 4.75, 4.25)\) At \( t = 1 \): start at \((2, 9, -1)\), vector \((2, 1, 9)\), end at \((4, 10, 8)\) These are the final concise answers for both parts.