APEX Calculus

1. The arrow sizes are constant along any horizontal line, so if one were to draw a small box anywhere on the graph, it would seem that the same amount of fluid would enter the box as exit. Therefore it seems the divergence is zero; it is, as

   \begin{equation*} \divv\vec F = \nabla \cdot \vec F = M_x + N_y = \frac{\partial}{\partial x}(y) + \frac{\partial}{\partial y}(0) = 0\text{.} \end{equation*}

   At any point on the \(x\)-axis, arrows above it move to the right and arrows below it move to the left, indicating that a cork placed on the axis would spin clockwise. A cork placed anywhere above the \(x\)-axis would have water above it moving to the right faster than the water below it, also creating a clockwise spin. A clockwise spin also appears to be created at points below the \(x\)-axis. Thus it seems the curl should be negative (and not zero). Indeed, it is:

   \begin{equation*} \curl \vec F = \nabla\times\vec F = N_x-M_y = \frac{\partial}{\partial x}(0) - \frac{\partial}{\partial y}(y) = -1\text{.} \end{equation*}

2. It appears that all vectors that lie on a circle of radius \(r\text{,}\) centered at the origin, have the same length (and indeed this is true). That implies that the divergence should be zero: draw any box on the graph, and any fluid coming in will lie along a circle that takes the same amount of fluid out. Indeed, the divergence is zero, as

   \begin{equation*} \divv\vec F = \nabla \cdot \vec F = M_x + N_y = \frac{\partial}{\partial x}(-y) + \frac{\partial}{\partial y}(x) = 0\text{.} \end{equation*}

   Clearly this field moves objects in a circle, but would it induce a cork to spin? It appears that yes, it would: place a cork anywhere in the flow, and the point of the cork closest to the origin would feel less flow than the point on the cork farthest from the origin, which would induce a counterclockwise flow. Indeed, the curl is positive:

   \begin{equation*} \curl \vec F = \nabla\times\vec F = N_x-M_y = \frac{\partial}{\partial x}(x) - \frac{\partial}{\partial y}(-y) = 1-(-1) = 2\text{.} \end{equation*}

   Since the curl is constant, we conclude the induced spin is the same no matter where one is in this field.

3. At the origin, there are many arrows pointing out but no arrows pointing in. We conclude that at the origin, the divergence must be positive (and not zero). If one were to draw a box anywhere in the field, the edges farther from the origin would have larger arrows passing through them than the edges close to the origin, indicating that more is going from a point than going in. This indicates a positive (and not zero) divergence. This is correct:

   \begin{equation*} \divv\vec F = \nabla \cdot \vec F = M_x + N_y = \frac{\partial}{\partial x}(x) + \frac{\partial}{\partial y}(y) = 1+1=2\text{.} \end{equation*}

   One may find this curl to be harder to determine visually than previous examples. One might note that any arrow that induces a clockwise spin on a cork will have an equally sized arrow inducing a counterclockwise spin on the other side, indicating no spin and no curl. This is correct, as

   \begin{equation*} \curl \vec F = \nabla\times\vec F = N_x-M_y = \frac{\partial}{\partial x}(y) - \frac{\partial}{\partial y}(x) = 0\text{.} \end{equation*}

4. One might find this divergence hard to determine visually as large arrows appear in close proximity to small arrows, each pointing in different directions. Instead of trying to rationalize a guess, we compute the divergence:

   \begin{equation*} \divv\vec F = \nabla \cdot \vec F = M_x + N_y = \frac{\partial}{\partial x}(\cos y) + \frac{\partial}{\partial y}(\sin x) = 0\text{.} \end{equation*}

   Perhaps surprisingly, the divergence is 0. Will all the loops of different directions in the field, one is apt to reason the curl is variable. Indeed, it is:

   \begin{equation*} \curl \vec F = \nabla\times\vec F = N_x-M_y = \frac{\partial}{\partial x}(\sin x) - \frac{\partial}{\partial y}(\cos y) = \cos x + \sin y\text{.} \end{equation*}

   Depending on the values of \(x\) and \(y\text{,}\) the curl may be positive, negative, or zero.

We compute the divergence and curl of each field following the definitions.

1. \begin{align*} \divv \vec F \amp = \nabla \cdot \vec F = M_x+N_y+P_z = 2x+0+0= 2x\\ \curl\vec F \amp = \nabla \times \vec F = \langle P_y-N_z,M_z-P_x,N_x - M_y\rangle\\ \amp = \langle 1 - (-1), 1-1,-1-(1)\rangle = \langle 2,0,-2\rangle\text{.} \end{align*}

   For this particular field, no matter the location in space, a spin is induced with axis parallel to \(\langle 2,0,-2\rangle\text{.}\)

2. \begin{align*} \divv \vec F \amp = \nabla \cdot \vec F = M_x+N_y+P_z = ye^{xy}+0+0= ye^{xy}\\ \curl\vec F \amp = \nabla \times \vec F = \langle P_y-N_z,M_z-P_x,N_x - M_y\rangle\\ \amp = \langle 1-\cos(x+z), -2x, \cos(x+z) - xe^{xy}\rangle \end{align*}

The point mass pulls toward the origin, so at \((x,y,z)\text{,}\) the force will pull in the direction of \(\langle -x, -y, -z\rangle\text{.}\) To get the proper magnitude, it will be useful to find the unit vector in this direction. Dividing by its magnitude, we have

The magnitude of the force is inversely proportional to the square of the distance between the two points. Letting \(k\) be the constant of proportionality, we have the magnitude as \(\ds\frac{k}{x^2+y^2+z^2}\text{.}\) Multiplying this magnitude by the unit vector above, we have the desired vector field:

We leave it to the reader to confirm that \(\divv \vec F = 0\) and \(\curl \vec F = \vec 0\text{.}\)

The analogous planar vector field is given in Figure 14.2.19. Note how all arrows point to the origin, and the magnitude gets very small when “far” from the origin.

Given \(f\text{,}\) we find \(\vec F = \nabla f = \langle -2x,-4y\rangle\text{.}\) A graph of \(\vec F\) is given in Figure 14.2.21.(a). In Figure 14.2.21.(b), the vector field is given along with a graph of the surface itself; one can see how each vector is pointing in the direction of “steepest uphill”, which, in this case, is not simply just “toward the origin.”

We leave it to the reader to confirm that \(\divv \vec F = -6\) and \(\curl \vec F = 0\text{.}\)