To calculate the line integral of the vector field $\mathbf{F} = (2x + y) \mathbf{i} + (x - 3y) \mathbf{j}$ along the line segment from $(1, 2)$ to $(4, 6)$, we follow these steps:

Step 1: Parameterize the line segment

We need to parameterize the line segment from $(1, 2)$ to $(4, 6)$. Let the parameter $t$ vary from $t = 0$ to $t = 1$, and define the parametric equations for the line segment.

The general formula for parameterizing a line segment from point $P_0(x_0, y_0)$ to $P_1(x_1, y_1)$ is: $x(t) = (1 - t)x_0 + t x_1, \quad y(t) = (1 - t)y_0 + t y_1$

For our case, $P_0 = (1, 2)$ and $P_1 = (4, 6)$, so: $x(t) = (1 - t)(1) + t(4) = 1 + 3t$ $y(t) = (1 - t)(2) + t(6) = 2 + 4t$ Thus, the parameterized line is: $x(t) = 1 + 3t, \quad y(t) = 2 + 4t$ for $0 \leq t \leq 1$.

Step 2: Compute the derivatives of $x(t)$ and $y(t)$

We now compute the derivatives of $x(t)$ and $y(t)$ with respect to $t$: $\frac{dx}{dt} = 3, \quad \frac{dy}{dt} = 4$

Step 3: Express the vector field $\mathbf{F}$ in terms of $t$

Substitute $x(t) = 1 + 3t$ and $y(t) = 2 + 4t$ into the vector field $\mathbf{F}$: $\mathbf{F}(x(t), y(t)) = (2x(t) + y(t)) \mathbf{i} + (x(t) - 3y(t)) \mathbf{j}$ Substituting $x(t)$ and $y(t)$: $\mathbf{F}(x(t), y(t)) = (2(1 + 3t) + (2 + 4t)) \mathbf{i} + ((1 + 3t) - 3(2 + 4t)) \mathbf{j}$ Simplify the expressions: $\mathbf{F}(x(t), y(t)) = (2 + 6t + 2 + 4t) \mathbf{i} + (1 + 3t - 6 - 12t) \mathbf{j}$ $\mathbf{F}(x(t), y(t)) = (4 + 10t) \mathbf{i} + (-5 - 9t) \mathbf{j}$

Step 4: Compute the dot product $\mathbf{F} \cdot d\mathbf{r}$

The differential $d\mathbf{r}$ is: $d\mathbf{r} = \frac{dx}{dt} \mathbf{i} + \frac{dy}{dt} \mathbf{j} = 3 \mathbf{i} + 4 \mathbf{j}$ Now compute the dot product $\mathbf{F}(x(t), y(t)) \cdot d\mathbf{r}$: $\mathbf{F}(x(t), y(t)) \cdot d\mathbf{r} = (4 + 10t)(3) + (-5 - 9t)(4)$ Simplify the dot product: $\mathbf{F}(x(t), y(t)) \cdot d\mathbf{r} = 12 + 30t - 20 - 36t$ $\mathbf{F}(x(t), y(t)) \cdot d\mathbf{r} = -8 - 6t$

Step 5: Integrate with respect to $t$

To compute the line integral, we integrate $\mathbf{F}(x(t), y(t)) \cdot d\mathbf{r}$ over $t$ from 0 to 1: $\int_0^1 (-8 - 6t) \, dt$ First, integrate each term: $\int_0^1 (-8) \, dt = -8t \Big|_0^1 = -8(1) + 8(0) = -8$ $\int_0^1 (-6t) \, dt = -3t^2 \Big|_0^1 = -3(1^2) + 3(0^2) = -3$ Adding these together: $-8 - 3 = -11$

Final Answer:

The value of the line integral is $\boxed{-11}$.

Would you like any further clarifications or details on this process?

Here are 5 related questions:

1. How would you parameterize a line segment between two arbitrary points?
2. What is the physical interpretation of a line integral in a vector field?
3. How would you calculate the line integral over a curved path instead of a straight line?
4. What happens if the vector field is conservative? How does this affect the line integral?
5. How does the choice of parameterization affect the calculation of the line integral?

Tip: For line integrals, parameterization of the path is crucial; ensure that the path is parameterized correctly before proceeding with the calculations.