Question

Consider the vector field f and the curve c below. f(x, y) = (6 6xy²)i + 6x²yj, c is the arc of the hyperbola y = 1/x from (1, 1) to (5, 1). (a) Find a potential function f such that f = ˆ‡f. (b) Use part (a) to evaluate ∫c ˆ‡f · dr along the given curve c.

Answer 1

The potential function for the given vector field is found by integrating the components, resulting in a function of x and y. To evaluate the line integral along curve c, we calculate the difference in potential function values at the curve's endpoints.

Step-by-step explanation:

The student's question involves finding a potential function for a given vector field and then using it to evaluate a line integral over a specific curve. We can find the potential function f by integrating the components of the vector field, ensuring that the mixed partial derivatives are equal (since the vector field is conservative).

To find the potential function f such that f = ∇f, we integrate the vector field's i component with respect to x and the j component with respect to y. The two should agree on the mixed partials, which they do since f(x, y) = (6 6xy²)i + 6x²yj. Integrating the first component 6 with respect to x gives us 6x, and integrating 6xy² with respect to y gives 2x×y³. Similarly, integrating the second component 6x² with respect to y gives us 3x³y. The potential function is therefore f(x, y) = 6xy + 2x×y³ + 3x³y + C, where C is a constant.

For part (b), since a potential function exists, the line integral of ∇f over curve c can be evaluated by finding the difference in the potential function values at the endpoints of the curve. This simplifies to f(5, 1/5) - f(1, 1), which involves substituting these points into the potential function and subtracting.