Question

Within the cube \(\\{(x, y, z):|x| \leq 1,\) \(|y| \leq 1,|z| \leq 1\\},\) where does div \(\mathbf{F}\) have the greatest magnitude when \(\mathbf{F}=\left\langle x^{2}-y^{2}, x y^{2} z, 2 x z\right\rangle ?\)

Short answer

Based on the given vector field and the cube's boundaries, the divergence (div) of the vector field has the greatest magnitude of 6 within the cube. This occurs at points on the boundary, such as (x,y,z) = (1,1,0).

Step-by-step solution

Compute div \(\mathbf{F}\)

To calculate the divergence of the vector field \(\mathbf{F}\), we'll find the partial derivatives of its components with respect to their corresponding variables and then sum them up: div \(\mathbf{F} = \frac{\partial}{\partial x}(x^2 - y^2) + \frac{\partial}{\partial y}(xy^2z) + \frac{\partial}{\partial z}(2xz)\). Now, let's compute these derivatives: \[ \frac{\partial}{\partial x}(x^2 - y^2) = 2x, \] \[ \frac{\partial}{\partial y}(xy^2z) = 2xyz, \] \[ \frac{\partial}{\partial z}(2xz) = 2x. \] Thus, div \(\mathbf{F} = 2x + 2xyz + 2x = 2x(1+yz+1) = 2x(2+yz)\).

Find critical points

Next, we'll find the critical points of div \(\mathbf{F}\) by setting its partial derivatives equal to zero and solving for the coordinates \((x, y, z)\): \[ \frac{\partial}{\partial x}(2x(2+yz)) = 0, \] \[ \frac{\partial}{\partial y}(2x(2+yz)) = 0, \] \[ \frac{\partial}{\partial z}(2x(2+yz)) = 0. \] Now, let's find these derivatives: \[ \frac{\partial}{\partial x}(2x(2+yz)) = 2(2+yz), \] \[ \frac{\partial}{\partial y}(2x(2+yz)) = 2xz, \] \[ \frac{\partial}{\partial z}(2x(2+yz)) = 2xy. \] Setting these partial derivatives to zero, we have three cases. Case 1: \(2(2+yz)=0,\) implying \(y=-\frac{2}{z}\). But because \(|y|\leq 1\) and \(|z|\leq 1\), there is no solution within the cube. Case 2: \(2xz = 0\). We have two subcases: - When \(x=0\), we have no restriction on \(y\) or \(z\). So this gives us the points on the \(y-z\) plane. - When \(z=0\), we have no restriction on \(x\) or \(y\). So this gives us the points on the \(x-y\) plane. Case 3: \(2xy = 0\). We have two subcases: - When \(x=0\), we again get the points on the \(y-z\) plane, previously considered in Case 2. - When \(y=0\), we have no restriction on \(x\) or \(z\). So this gives us the points on the \(x-z\) plane.

Determine maximum div \(\mathbf{F}\) within boundaries

To find the maximum div \(\mathbf{F}\) within the cube, we need to consider the values of div \(\mathbf{F}\) on the critical points obtained in Step 2, and at the boundaries of the cube, namely when \(x=\pm1\), \(y=\pm1\), and \(z=\pm1\). Div \(\mathbf{F} = 2x(2+yz)\). We know that \(|x|\leq1\), \(|y|\leq1\), and \(|z|\leq1\). 1. At \(x=0\), div \(\mathbf{F}=0\). 2. At \(y=0\), div \(\mathbf{F}=4x\) which is \(4\) for \(x=1\) and \(-4\) for \(x=-1\). 3. At \(z=0\), div \(\mathbf{F}=4x\) which is \(4\) for \(x=1\) and \(-4\) for \(x=-1\). 4. At \(x=\pm1\) and \(y=\pm1\), div \(\mathbf{F}=\pm2(2\pm z)\) which vary between \(-4\) to \(6\). Now maximum positive div \(\mathbf{F}\) is \(6\), and minimum (negative) div \(\mathbf{F}\) is \(-4\). Since \(6 > |-4|\), the greatest magnitude for div \(\mathbf{F}\) within the cube is \(6\), which occurs on the boundary at points such as \((x,y,z) = (1,1,0)\).

Key concepts

Vector Field

A vector field can be thought of as a function that assigns a vector to each point in space. Imagine the wind blowing across a field; each point in the field has a wind vector associated with it, showing both direction and strength. In mathematical terms, a vector field

• has components along the coordinate axes, for example, \(\mathbf{F} = \langle P(x,y,z), Q(x,y,z), R(x,y,z) \rangle \),
• where \(P, Q, R\) are functions that describe how each component of the vector changes with position.

This concept allows us to analyze how a quantity varies in space. For vector fields like \(\mathbf{F} = \langle x^2-y^2, xy^2z, 2xz \rangle\), each vector's components are determined by the expressions given for each dimension. Understanding vector fields is essential as they form the basis for concepts like divergence, which reflects how these vectors spread or converge at a point.

Partial Derivatives

Partial derivatives are a critical tool in calculus used to examine how a function changes as just one of its variables changes. While a normal derivative gives you the rate of change concerning one variable,

• partial derivatives extend this to functions of multiple variables.
• With a notational format like \(\frac{\partial P}{\partial x}\), you aim to see how the function \(P\) changes as \(x\) changes while keeping all other variables constant.

For a vector field like \(\mathbf{F} = \langle x^2-y^2, xy^2z, 2xz \rangle\),

• computing its divergence involves partial derivatives of the field's components \(P, Q, R\).
• You perform partial differentiation with respect to each coordinate variable, which here means changing one while looking at the overall effect on div \(\mathbf{F}\).

Mastering partial derivatives is crucial for deeper understanding and for calculating changes within multi-dimensional functions.

Critical Points

Critical points occur where the derivative (or gradient in higher dimensions) of a function is zero, indicating potential maxima, minima, or saddle points. In a single-variable function, this is simply the point where the function's slope is zero. However,

• for functions of several variables, such as those in vector fields, you look at the partial derivatives.

Consider \(div \mathbf{F} = 2x(2+yz)\):

• finding critical points means setting its partial derivatives to zero.

This gives you equations to solve for the coordinates, \(x, y, z\), that might present maxima or minima. Critical points reveal where significant changes occur in the vector field's divergence and are often found at boundaries or sample points specific to your region of interest.

Cube Boundaries

Cube boundaries refer to the edges and faces of a cube within which the problem is confined. Understanding boundaries is crucial when considering constraints in optimization problems or when assessing certain parameters like divergence. In the context of the cube defined by \(|x| \leq 1, |y| \leq 1, |z| \leq 1\),

• you're working within a space limited by these parameters.
• It's like working within a box and understanding that you cannot exceed its edges in any direction.

Finding where the divergence of a vector field reaches its greatest magnitude within these limits involves testing critical points and examining various surfaces and edges of the cube. Wherever

• \(x = \pm 1\), \(y = \pm 1\), or \(z = \pm 1\),\

you check how factors influence div \(\mathbf{F}\) and identify what happens at these boundaries, such as finding \(6\) as the greatest magnitude in this scenario.