Line Integral of Gradient f(x,y)=e^{xy}-x²+y³ Along Curve C

Answer

The line integral of the gradient field $\nabla f$ along the path $C$ is evaluated using the Fundamental Theorem of Line Integrals. Since the integral of a conservative field depends only on the endpoints, we find the values of $f$ at $(0, -2)$ and $(-2, 0)$, resulting in a final value of $-7$.

Image Analysis

The image displays a piecewise linear curve $C$ in the $xy$-plane. The path begins at the point $(0, -2)$ on the y-axis, travels through $(2, 0)$, moves to $(0, 2)$, and finally terminates at $(-2, 0)$ on the x-axis. Arrows indicate the direction of travel, confirming that the initial point is $(0, -2)$ and the terminal point is $(-2, 0)$.

Explanation

1. Identification of the Theorem The problem asks for the line integral of a gradient field $\nabla f$. By the Fundamental Theorem of Line Integrals, if a vector field is the gradient of a scalar potential function $f$, then the integral is path-independent and depends only on the function values at the endpoints. $\int_{C} \nabla f \cdot d\vec{r} = f(\vec{r}_{end}) - f(\vec{r}_{start})$ This formula states that the net change in the potential function $f$ over a path is equal to the value at the terminal point minus the value at the initial point.

2. Determining Endpoints from the Graph Based on the orientation arrows in the provided diagram:

   • Initial point ($A$): $(0, -2)$
   • Terminal point ($B$): $(-2, 0)$ The intermediate path segments do not affect the calculation because the vector field $\nabla f$ is conservative.

3. Evaluating the Potential Function at the Terminal Point We substitute $(-2, 0)$ into the given function $f(x,y) = e^{xy} - x^2 + y^3$. $f(-2, 0) = e^{(-2)(0)} - (-2)^2 + (0)^3$ This expression evaluates the state of the system at the end of the trajectory. $f(-2, 0) = e^{0} - 4 + 0 = 1 - 4 = -3$ The constant $e^0$ is exactly 1, and the square of a negative number is positive.

4. Evaluating the Potential Function at the Initial Point We substitute $(0, -2)$ into the same function. $f(0, -2) = e^{(0)(-2)} - (0)^2 + (-2)^3$ This expression evaluates the state of the system at the beginning of the trajectory. $f(0, -2) = e^{0} - 0 - 8 = 1 - 8 = -7$ Note that the cube of a negative number remains negative.

5. Calculating the Net Change Subtract the initial value from the final value to find the line integral. $\int_{C} \nabla f \cdot d\vec{r} = f(-2, 0) - f(0, -2)$ The Integral is the difference between the terminal potential and the start potential. $\int_{C} \nabla f \cdot d\vec{r} = -3 - (-7) = -3 + 7 = 4$ Wait, let's re-verify the terminal point from the image. Upon closer inspection of the arrowheads, the path starts at $(0, -2) \to (2,0) \to (0,2) \to (-2,0)$. Correct. Calculation: $-3 - (-7) = 4$.

Final Answer

$\boxed{4}$

Common Mistakes

• Path Over-complication: Students often attempt to parameterize all three segments of the path $C$. Since the field is a gradient field (conservative), this is unnecessary and prone to algebraic errors.
• Sign Errors in Subtraction: A common mistake is subtracting the terminal value from the initial value instead of the reverse ($f(start) - f(end)$), or failing to handle the double negative ($-3 - (-7)$) correctly.
• Misinterpreting Direction: Not following the arrows on the graph can lead to swapping the initial and terminal points, which results in the negative of the correct answer.

Related Topics: Conservative Vector Fields, Path Independence, Gradient Theorem, Curl of a Vector Field.