For Questions 1, 2, and 3 we are given a vector field and a closed path. Since we have our Path as a vector, we can use the General Line Integral formula to find the Circulation.- SraCirculation...

For Questions 1, 2, and 3 we are given a vector field and a closed path.

Extracted text: Vector Field: F(x,y) = (y, x) and Closed Path: r(1) = (3 cos(r). - 3 sin(r)> Since we have our Path as a vector, we can use the General Line Integral formula to find the Circulation. - Sra Circulation = F(1). r'(1) dt This will tell us how much of the vector field is in the same direction as the path. In other words, think of it as how much work the vector field is applying to a sailboat traveling that path. Given the vector field and closed path below Vector Field: F(x.y) = (?, x) and Closed Path: r(1) = (3 cos(r). - 3 sin(1)> Use the closed path to convert the vector field into a parameterized vector field. F(1) = (3 cos(f), - 3 sin(r)> B F(1) = (9, - 3) F(1) = (9 cos²(1). – 3 sin(1)> O F(1) = (9 sin²), 3 cos(1)> Given the vector field and closed path below Vector Field: F(x,y) = (?. x) and Closed Path: r(1) = (3 cos(1). - 3 sin(f) > Find the derivative of the closed path. (A r'(1) = (-3 sin (1), – 3 cos(1)> (B r(1) = (3 cos(1), - 3 sin(1) > r'(1) = (-3 sin (1), 3 cos(r)> D r'() = (3 sin (f), - 3 cos(f) > Now that you have the parameterized vector field and the derivative of the closed path, we can use the general line integral formula to find the circulation in this situation. - Srm Ciculation = r'(1) dt and interval: [0, 2x] F) - r'(1) dt = (B) F(1) - r'(1) dt = - 9% © Ji r'(1) dt = 18x F(1) . r'(1) di = 0