Questions: Problem 10. (10 points) Show F(x, y)=(2 x y^3+5) i+(3 x^2 y^2+2 e^(2 y)) j is conservative by finding a potential function f for F, and use f to compute ∫C F · d s, where C is the curve given by r(t)=2 sin^3 t i+(2 t/π) sin^12 5t j for 0 ≤ t ≤ π / 2. f(x, y)= ∫C F · d s=

Problem 10. (10 points)

Show
F(x, y)=(2 x y^3+5) i+(3 x^2 y^2+2 e^(2 y)) j
is conservative by finding a potential function f for F, and use f to compute ∫C F · d s, where C is the curve given by
r(t)=2 sin^3 t i+(2 t/π) sin^12 5t j
for 0 ≤ t ≤ π / 2.
f(x, y)=
∫C F · d s=

Transcript text: Problem 10. (10 points) Show \[ \mathbf{F}(x, y)=\left(2 x y^{3}+5\right) \mathbf{i}+\left(3 x^{2} y^{2}+2 e^{2 y}\right) \mathbf{j} \] is conservative by finding a potential function $f$ for $\mathbf{F}$, and use $f$ to compute $\int_{C} \mathbf{F} \cdot d \mathbf{s}$, where $C$ is the curve given by \[ \mathbf{r}(t)=2 \sin ^{3} t \mathbf{i}+\frac{2 t}{\pi} \sin ^{12} 5 t \mathbf{j} \] for $0 \leq t \leq \pi / 2$. \[ \begin{array}{l} f(x, y)=\square \\ \int_{C} \mathbf{F} \cdot d \mathbf{s}=\square \end{array} \] Note: You can earn partial credit on this problem.

Solution

Solution Steps

To show that the vector field $\mathbf{F}(x, y)$ is conservative, we need to find a potential function $f(x, y)$ such that $\nabla f = \mathbf{F}$. This involves integrating the components of $\mathbf{F}$ with respect to their respective variables and ensuring consistency. Once the potential function is found, the line integral $\int_{C} \mathbf{F} \cdot d \mathbf{s}$ can be computed as the difference $f(\mathbf{r}(\text{end})) - f(\mathbf{r}(\text{start}))$, where $\mathbf{r}(t)$ is the parameterization of the curve $C$.

Step 1: Finding the Components of the Vector Field

The vector field is given by

\[ \mathbf{F}(x, y) = (2xy^3 + 5) \mathbf{i} + (3x^2y^2 + 2e^{2y}) \mathbf{j} \]

We identify the components as

\[ F_x = 2xy^3 + 5 \quad \text{and} \quad F_y = 3x^2y^2 + 2e^{2y} \]

Step 2: Finding the Potential Function

To find the potential function $f(x, y)$, we integrate $F_x$ with respect to $x$:

\[ f_x = \int (2xy^3 + 5) \, dx = x^2y^3 + 5x + g(y) \]

Next, we differentiate $f(x, y)$ with respect to $y$ and set it equal to $F_y$:

\[ f_y = \frac{\partial}{\partial y}(x^2y^3 + 5x + g(y)) = 3x^2y^2 + g'(y) \]

Setting $f_y = F_y$:

\[ 3x^2y^2 + g'(y) = 3x^2y^2 + 2e^{2y} \]