To find the line integral of a vector field $f(x, y) = y e^{x^2}$ along the curve $\mathbf{r}(t) = 3t \, \mathbf{i} - 4t \, \mathbf{j}$ for $-1 \leq t \leq 2$, we'll follow these steps:

Step 1: Parameterize $f(x, y)$ along the Curve

We have: $x = 3t, \quad y = -4t$ Substitute these into $f(x, y)$: $f(x, y) = y e^{x^2} = (-4t) e^{(3t)^2} = -4t e^{9t^2}$

Step 2: Calculate $\frac{d\mathbf{r}}{dt}$

To proceed with the line integral, we need $\frac{d\mathbf{r}}{dt}$: $\mathbf{r}(t) = 3t \, \mathbf{i} - 4t \, \mathbf{j}$ Then, $\frac{d\mathbf{r}}{dt} = 3 \, \mathbf{i} - 4 \, \mathbf{j}$ The magnitude $\left\|\frac{d\mathbf{r}}{dt}\right\|$ is: $\left\|\frac{d\mathbf{r}}{dt}\right\| = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$

Step 3: Set Up the Integral

The line integral of $f(x, y)$ along $\mathbf{r}(t)$ is given by: $\int_C f(x, y) \, ds = \int_{-1}^{2} f(x(t), y(t)) \left\|\frac{d\mathbf{r}}{dt}\right\| \, dt$ Substitute $f(x(t), y(t)) = -4t e^{9t^2}$ and $\left\|\frac{d\mathbf{r}}{dt}\right\| = 5$: $\int_{-1}^{2} f(x(t), y(t)) \left\|\frac{d\mathbf{r}}{dt}\right\| \, dt = \int_{-1}^{2} (-4t e^{9t^2}) \cdot 5 \, dt$ $= \int_{-1}^{2} -20t e^{9t^2} \, dt$

Step 4: Solve the Integral

Now we have: $\int_{-1}^{2} -20t e^{9t^2} \, dt$ To integrate this, notice that it is of the form $\int u' e^{u} \, dt$ with $u = 9t^2$ and $du = 18t \, dt$. Thus, we make the substitution $u = 9t^2$, $du = 18t \, dt$, or $\frac{du}{18} = t \, dt$.

Rewrite the integral: $= -20 \int_{-1}^{2} t e^{9t^2} \, dt = -20 \cdot \frac{1}{18} \int_{u = 9(-1)^2}^{u = 9(2)^2} e^u \, du$ $= -\frac{20}{18} \left[ e^u \right]_{9}^{36} = -\frac{10}{9} \left( e^{36} - e^{9} \right)$

Final Answer

$\int_C f(x, y) \, ds = -\frac{10}{9} \left( e^{36} - e^{9} \right)$

Would you like further details on any step or explanation? Here are some related questions you might explore:

1. How is line integral different from a surface integral?
2. What are practical applications of line integrals in physics?
3. How do you interpret the result of a line integral geometrically?
4. Can line integrals be evaluated without parameterization?
5. How would the answer change if the curve was in a 3D space?

Tip: For line integrals over vector fields, parameterizing the curve and computing $\frac{d\mathbf{r}}{dt}$ are key steps for simplification.