Both vector fields have circles as flow lines, but due to their different spatial behaviors, only vec G has a zero curl due to its inherent rotational symmetry.
**(a) vec F(x, y) = < -y, x > and origin (0, 0):**
1. **Origin as a flow line:** To see if the origin is a flow line, we need to check if the vector field at the origin points along the tangent of any potential curve passing through the origin. In this case, any circle centered at the origin is a potential curve. The tangent vector at any point on a circle of radius r centered at the origin is <cos(t), sin(t)>, where t is the angle between the positive x-axis and the point. Evaluating vec F at the origin gives <-0, 0>. This matches the x-component of the tangent vector for any angle (cos(t)). Therefore, the origin is a flow line for any circle centered at the origin.
**(b) vec G(x, y) = < -y/(x^2 + y^2), x/(x^2 + y^2) > and circle as a flow line:**
1. **Circle as a flow line:** We need to find an angular speed (frequency) at which the vector field at any point on the circle points along the tangent vector. Consider a circle of radius r centered at the origin and parameterized by x = r cos(t) and y = r sin(t). The tangent vector at any point (x, y) on this circle is <-r sin(t), r cos(t)>. Evaluating vec G at this point gives <-sin(t)/(r^2), cos(t)/(r^2)>. The key is to find a value of t for which these two vectors are proportional. Setting <-r sin(t), r cos(t)> = k * <-sin(t)/(r^2), cos(t)/(r^2)> for some constant k, we get k = r^2. Solving for t, we find t = nπ, where n is any integer. This means that at angles 0, π, 2π, etc., the vector field points along the tangent vector for any radius r. Therefore, a circle centered at the origin is a flow line for these specific angles.
**(c) Computing curls:**
1. **Curl of vec F:** Recall that curl in 2D is defined as curl(F) = ∂F_x/∂y - ∂F_y/∂x. For vec F, ∂F_x/∂y = 1 and ∂F_y/∂x = -1. Therefore, curl(F) = 1 - (-1) = 2.
2. **Curl of vec G:** Similarly, curl(G) = (∂G_x/∂y - ∂G_y/∂x) = (-2x)/(x^2 + y^2) - (2y)/(x^2 + y^2) = 0.
**(d) Explanation for zero curl in vec G:**
The fact that vec G has a zero curl indicates that it exhibits **rotational symmetry**, meaning its behavior is the same at any point on a circle centered at the origin. The angular speed required for the circle to be a flow line (angles 0, π, 2π, etc.) corresponds to a constant "rotation" around the origin, explaining the zero curl. On the other hand, vec F has a non-zero curl because it does not exhibit rotational symmetry. The vector field points inwards towards the origin for any point inside the circle (except the origin itself) and outwards for any point outside the circle, creating a net rotational effect that manifests in a non-zero curl.
In summary, both vector fields have circles as flow lines, but due to their different spatial behaviors, only vec G has a zero curl due to its inherent rotational symmetry.
The probable question can be: We are going to revisit the basic premise of the last problem, this time with two new vector fields.
(a) For the vector field vec F(x, y) = \langle- y, x\rangle , verify that origin, is a flow line of vec F a circle of arbitrary fixed radius, centered at the
(b) For the vector field vec G(x, y) = \langle(- y)/(x ^ 2 + y ^ 2), x/(x ^ 2 + y ^ 2)\rangle , verify that a circle of arbitrary fixed radius, centered at the origin, is a flow line of vec G . It may help to remember that you may need to choose the frequency, or angular speed, of motion around the circle to be a special value to ensure that your answer is a flow line.
(c) Treating the vector fields from the previous two parts as 3d-vector fields (with z-component iden- tically equal to zero), compute the curls of vec F and G.
(d) One of your answers from the previous part should be the zero vector. Explain why, even though these two vector fields both have circles as flow lines, the curl is zero for one and non-zero for the other.