Divergence & Curl: Computing Operators for Common 2D & 3D Vector Fields

To find the divergence of the vector field F, we need to calculate the sum of the partial derivatives of its components. The vector field is given by P = 3x, Q = 5y, and R = -2z. We start by taking the partial derivative of P with respect to x, which gives us 3. Next, we take the partial derivative of Q with respect to y, which gives us 5. Finally, we take the partial derivative of R with respect to z, which results in -2. Adding these three values together, we get 3 + 5 + (-2), which equals 6. This positive value indicates that the fluid is expanding at every point in the field.

The divergence of a vector field is the dot product of the del operator and the field (∇ · F), which results in a scalar field (a function that assigns a scalar value to each point in space), representing the rate of expansion per unit volume at that point. The curl of a vector field is the cross product of the del operator and the field (∇ × F), which results in a vector field, representing the axis and magnitude of rotation at each point.

To calculate the curl of F, we use the determinant of a matrix involving the unit vectors i, j, k, the partial derivative operators, and the components of F, which are P=y, Q=-x, and R=0. We calculate the i-component as the partial derivative of 0 with respect to y minus the partial derivative of -x with respect to z, which is 0 - 0 = 0. We calculate the j-component as the partial derivative of y with respect to z minus the partial derivative of 0 with respect to x, which is 0 - 0 = 0. Finally, we calculate the k-component as the partial derivative of -x with respect to x minus the partial derivative of y with respect to y. This gives us -1 - 1, which equals -2. Thus, the resulting vector is < 0, 0, -2 >.

In the physics of rotating fluids, the velocity field v is defined by the cross product of the angular velocity vector w and the position vector r. When we calculate the curl of this specific velocity field, the result is a vector that points in the same direction as the angular velocity axis. Through the calculation of the partial derivatives involved in the curl operation, the result comes out to be exactly twice the angular velocity vector. This means the curl vector represents twice the speed of the angular rotation, pointing along the axis of rotation.

A vector field is incompressible, or solenoidal, if its divergence is equal to zero. We first calculate the divergence of F = < ax, y, 3z >. The partial derivative of the first component ax with respect to x is a. The partial derivative of the second component y with respect to y is 1. The partial derivative of the third component 3z with respect to z is 3. Summing these derivatives gives us the divergence expression a + 1 + 3, which simplifies to a + 4. For the field to be incompressible, this sum must equal zero. Therefore, we set a + 4 = 0 and solve for a, obtaining a = -4.

We are asked to find the curl of a gradient field. First, we determine the gradient of the scalar function f(x, y, z) = x² + y² + z², which results in the vector field F = < 2x, 2y, 2z >. Next, we calculate the curl of this vector field. However, there is a fundamental vector calculus identity stating that the curl of the gradient of any scalar function is always the zero vector. If we perform the calculation manually, the cross partial derivatives (like the partial of 2z with respect to y and the partial of 2y with respect to z) are all zero. Thus, the result is the zero vector < 0, 0, 0 >.

To find the divergence, we compute the partial derivative of each component with respect to its corresponding variable and sum them up. The first component is P = ex sin(y). The partial derivative of P with respect to x is ex sin(y). The second component is Q = ex cos(y). The partial derivative of Q with respect to y is -ex sin(y). The third component is R = z. The partial derivative of R with respect to z is 1. Adding these three results together, we get ex sin(y) + (-ex sin(y)) + 1. The first two terms cancel each other out, leaving a final answer of 1.

We first calculate the curl of the velocity field V = < -y, x, 0 >. The k-component of the curl is found by taking the partial derivative of x (the j-component of V) with respect to x, which is 1, and subtracting the partial derivative of -y (the i-component of V) with respect to y, which is -1. This calculation gives us 1 - (-1) = 2. The i and j components of the curl are zero. So, the curl is < 0, 0, 2 > and its magnitude is 2. Since the curl is twice the angular velocity (curl V = 2w), a magnitude of 2 implies the angular velocity w is 1. Therefore, the paddle wheel spins because of the rotation.

A vector field is defined as irrotational if its curl is the zero vector. We test the options by calculating their curls. For option C, F = < yz, xz, xy >, let's check the components. The i-component of the curl is the partial of xy with respect to y (which is x) minus the partial of xz with respect to z (which is x), resulting in 0. The j-component is the partial of yz with respect to z (which is y) minus the partial of xy with respect to x (which is y), resulting in 0. The k-component is the partial of xz with respect to x (which is z) minus the partial of yz with respect to y (which is z), resulting in 0. Since all components are zero, this field is irrotational.

A vector field F can be the curl of another field G only if the divergence of F is zero. We check the divergence of the options. For option A, div(< x, y, z >) = 1 + 1 + 1 = 3. Since the divergence is not zero, this field cannot be the curl of any vector field. Options B, C, and D all have zero divergence.

We calculate the divergence by summing the partial derivatives of the components. The partial derivative of x with respect to x is 1. The partial derivative of y with respect to y is 1. The partial derivative of z with respect to z is 1. Adding these together, 1 + 1 + 1 = 3. Since the divergence is a positive constant (3) everywhere, including at the point (1, 1, 1), it indicates that there is a net outward flow or expansion of the vector field at that point. The fluid is acting as a source.

This describes an "irrotational vortex." While the air particles are revolving around the center of the tornado (macroscopic circulation), the local curl being zero outside the center means the particles are not spinning about their own individual centers of mass. Imagine a face drawn on a balloon moving around the tornado; if the curl is zero, the face always points in the same compass direction (e.g., always North) as it orbits the center, rather than turning to face the center continuously. This is a classic application of interpreting curl in fluid dynamics.

We can solve this by direct calculation or by using a vector identity. Using the identity div(curl F) = 0, we know the answer immediately. To verify with calculation: First, find the curl of F. The i-component is (partial of zx w.r.t y) - (partial of yz w.r.t z) = 0 - y = -y. The j-component is (partial of xy w.r.t z) - (partial of zx w.r.t x) = 0 - z = -z. The k-component is (partial of yz w.r.t x) - (partial of xy w.r.t y) = 0 - x = -x. So curl F = < -y, -z, -x >. Now, find the divergence of this result. The partial of -y w.r.t x is 0. The partial of -z w.r.t y is 0. The partial of -x w.r.t z is 0. The sum is 0 + 0 + 0 = 0.

The divergence of a vector field F = <P, Q, R> is given by ∇·F = ∂P/∂x + ∂Q/∂y + ∂R/∂z. For the vector field F(x, y, z) = <x², y³, z⁴>, we identify P = x², Q = y³, and R = z⁴. We compute each partial derivative separately. First, ∂P/∂x = ∂/∂x(x²) = 2x. Next, ∂Q/∂y = ∂/∂y(y³) = 3y². Finally, ∂R/∂z = ∂/∂z(z⁴) = 4z³. Adding these three partial derivatives together gives us the divergence: ∇·F = 2x + 3y² + 4z³. This result represents the net rate at which fluid is emerging from a point, with positive values indicating a source and negative values indicating a sink.

The divergence is calculated as ∇·F = ∂P/∂x + ∂Q/∂y + ∂R/∂z where F = <P, Q, R>. For F(x, y, z) = <sin(x), cos(y), tan(z)>, we identify P = sin(x), Q = cos(y), and R = tan(z). We compute each partial derivative. First, ∂P/∂x = ∂/∂x(sin(x)) = cos(x). Next, ∂Q/∂y = ∂/∂y(cos(y)) = -sin(y) because the derivative of cosine is negative sine. Finally, ∂R/∂z = ∂/∂z(tan(z)) = sec²(z) because the derivative of tan(z) is sec²(z). Adding these three results together gives us the divergence: ∇·F = cos(x) + (-sin(y)) + sec²(z) = cos(x) - sin(y) + sec²(z).