91Ó°ÊÓ

Question: Calculate the circulation of the vector field F(x, y) = (2x - y)i + (x + 3y)j around the closed curve given by x(t) = cos(t) and y(t) = sin(t), where t ranges from 0 to 2π. Answer: To solve this problem, we will follow the steps outlined in the solution above. Step 1: We are given the vector field F(x, y) = (2x - y)i + (x + 3y)j and a closed curve defined by x(t) = cos(t) and y(t) = sin(t), with t ranging from 0 to 2π. We need to calculate the circulation of the vector field around this closed curve. Step 2: The vector field is already expressed in a suitable form: F(x, y) = (2x - y)i + (x + 3y)j. Step 3: The closed curve is parameterized as C(t) = (x(t), y(t)) = (cos(t), sin(t)), with t ranging from 0 to 2π. Step 4: Now we will compute the line integral on the parameterized curve. First, find the derivative of the parameterized curve: C'(t) = (x'(t), y'(t)) = (-sin(t), cos(t)) Next, calculate F(C(t)): F(C(t)) = (2cos(t) - sin(t), cos(t) + 3sin(t)) Now, evaluate the dot product F(C(t)) · C'(t): F(C(t)) · C'(t) = (2cos(t) - sin(t))(-sin(t)) + (cos(t) + 3sin(t))(cos(t)) Now, we will compute the line integral: Circulation = \(\int_0^{2\pi} F(C(t)) \cdot C'(t) dt = \int_0^{2\pi} (2cos^2(t) - cos(t)sin(t) +3sin^2(t)) dt\) To evaluate this integral, use the double angle formulas for cos(2t) and sin(2t): Circulation = \(\int_0^{2\pi} (2(\frac{1 + cos(2t)}{2}) - \frac{1}{2}sin(2t) + \frac{3(1 - cos(2t))}{2}) dt\) Simplify and integrate: Circulation = \(\int_0^{2\pi} (-\frac{1}{2}cos(2t) + \frac{1}{2}sin(2t) + \frac{5}{2}) dt = [-\frac{1}{4}sin(2t) - \frac{1}{4}cos(2t) + \frac{5}{2}t]_0^{2\pi}\) Evaluate the integral at the limits: Circulation = \((-\frac{1}{4}sin(4\pi) - \frac{1}{4}cos(4\pi) + 5\pi) - (-\frac{1}{4}sin(0) - \frac{1}{4}cos(0) + 0) = 5\pi\) Thus, the circulation of the vector field F(x, y) = (2x - y)i + (x + 3y)j around the closed curve given by x(t) = cos(t) and y(t) = sin(t) is 5π.