8. Divergence, Curl and Potentials

Exercises

a. The Del Operator

Compute the gradient of each scalar field.

\(f=xyz\)

\[ \vec\nabla f =\left\langle \partial_x f,\partial_y f,\partial_z f\right\rangle \]

\(\vec\nabla f =\left\langle yz,xz,xy\right\rangle\)

\[\begin {aligned} \vec\nabla f &=\left\langle \partial_x(xyz),\partial_y(xyz),\partial_z(xyz)\right\rangle \\ &=\left\langle yz,xz,xy\right\rangle \\ \end{aligned}\]
\(f=x^2y^3z^4\)

\(\vec\nabla f =\left\langle 2xy^3z^4,3x^2y^2z^4,4x^2y^3z^3\right\rangle\)

\[\begin {aligned} \vec\nabla f &=\left\langle \partial_x(x^2y^3z^4),\partial_y(x^2y^3z^4),\partial_z(x^2y^3z^4)\right\rangle \\ &=\left\langle 2xy^3z^4,3x^2y^2z^4,4x^2y^3z^3\right\rangle \\ \end{aligned}\]
\(f=x^2+y^2+z^2\)

\(\vec\nabla f =\left\langle 2x,2y,2z\right\rangle\)

\[\begin {aligned} \vec\nabla f &=\left\langle \partial_x(x^2+y^2+z^2), \partial_y(x^2+y^2+z^2), \partial_z(x^2+y^2+z^2)\right\rangle \\ &=\left\langle 2x,2y,2z\right\rangle \\ \end{aligned}\]
\(f=z\sin x\cos y\)

\(\vec\nabla f =\left\langle z\cos x\cos y,-z\sin x\sin y,\sin x\cos y\right\rangle\)

\[\begin {aligned} \vec\nabla f &=\left\langle \partial_x(z\sin x\cos y), \partial_y(z\sin x\cos y), \partial_z(z\sin x\cos y)\right\rangle \\ &=\left\langle z\cos x\cos y,-z\sin x\sin y,\sin x\cos y\right\rangle \\ \end{aligned}\]

b. The Divergence Operator

Compute the divergence of each vector field.

\(\vec F=\left\langle xy,yz,zx\right\rangle\)

\[ \vec\nabla\cdot\vec F =\partial_x F_1+\partial_y F_2+\partial_z F_3 \]

\(\vec\nabla\cdot\vec F=y+z+x\)

\[\begin{aligned} \vec\nabla\cdot\vec F &=\partial_x(xy)+\partial_y(yz)+\partial_z(zx) \\ &=y+z+x \end{aligned}\]
\(\vec F=\left\langle yz,xz,xy\right\rangle\)

\(\vec\nabla\cdot\vec F=0\)

\[\begin{aligned} \vec\nabla\cdot\vec F &=\partial_x(yz)+\partial_y(xz)+\partial_z(xy) \\ &=0+0+0=0 \end{aligned}\]
\(\vec F =\left\langle \dfrac{1}{2}z^2x,\dfrac{1}{2}x^2y,\dfrac{1}{2}y^2z\right\rangle\)

\(\vec\nabla\cdot\vec F =\dfrac{1}{2}(x^2+y^2+z^2)\)

\[\begin{aligned} \vec\nabla\cdot\vec F &=\dfrac{\partial}{\partial x}\left(\dfrac{1}{2}z^2x\right) +\dfrac{\partial}{\partial y}\left(\dfrac{1}{2}x^2y\right) +\dfrac{\partial}{\partial z}\left(\dfrac{1}{2}y^2z\right) \\ &=\dfrac{1}{2}(z^2+x^2+y^2) \end{aligned}\]
\(\vec F=\left\langle x\cos z,y\cos z,\sin z\right\rangle\)

\(\vec\nabla\cdot\vec F=3\cos z\)

\[\begin{aligned} \vec\nabla\cdot\vec F &=\partial_x(x\cos z)+\partial_y(y\cos z)+\partial_z(\sin z) \\ &=\cos z+\cos z+\cos z=3\cos z \end{aligned}\]
\(\vec F =\left\langle z\cos x\cos y,-z\sin x\sin y, \sin x\cos y\right\rangle\)

\(\vec\nabla\cdot\vec F=-2z\sin x\cos y\)

\[\begin{aligned} \vec\nabla\cdot\vec F &=\partial_x(z\cos x\cos y) +\partial_y(-z\sin x\sin y) \\ &\quad+\partial_z(\sin x\cos y) \\ &=-z\sin x\cos y-z\sin x\cos y+0 \\ &=-2z\sin x\cos y \end{aligned}\]
\(\vec F=\left\langle x^3,y^3,z^3\right\rangle\)

\(\vec\nabla\cdot\vec F=3x^2+3y^2+3z^2\)

\[\begin{aligned} \vec\nabla\cdot\vec F &=\partial_x(x^3)+\partial_y(y^3)+\partial_z(z^3) \\ &=3x^2+3y^2+3z^2 \end{aligned}\]
\(\vec F=\left\langle e^x,e^y,e^z\right\rangle\)

\(\vec\nabla\cdot\vec F=e^x+e^y+e^z\)

\[\begin{aligned} \vec\nabla\cdot\vec F &=\partial_x(e^x)+\partial_y(e^y)+\partial_z(e^z) \\ &=e^x+e^y+e^z \end{aligned}\]
\(\vec F=\left\langle \ln\left(\dfrac{x}{yz}\right), \ln\left(\dfrac{y}{xz}\right), \ln\left(\dfrac{z}{xy}\right)\right\rangle\)

\(\vec\nabla\cdot\vec F=\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\)

\[\begin{aligned} \vec\nabla\cdot\vec F &=\partial_x\ln\left(\dfrac{x}{yz}\right) +\partial_y\ln\left(\dfrac{y}{xz}\right) +\partial_z\ln\left(\dfrac{z}{xy}\right) \\ &=\dfrac{yz}{x}\dfrac{1}{yz} +\dfrac{xz}{y}\dfrac{1}{xz} +\dfrac{xy}{z}\dfrac{1}{xy} \\ &=\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z} \end{aligned}\]

\[\begin{aligned} \vec\nabla\cdot\vec F &=\partial_x(\ln x-\ln y-\ln z) +\partial_y(\ln y-\ln x-\ln z) \\ &\quad+\partial_z(\ln z-\ln x-\ln y) \\ &=\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z} \end{aligned}\]

Compute the divergence of the gradient of \(f=x^2y^3z^4\), i.e. \(\vec\nabla\cdot\vec\nabla f\).

\(\vec\nabla\cdot\vec\nabla f=2y^3z^4+6x^2yz^4+12x^2y^3z^2\)

The gradient of \(f\) is: \[ \vec\nabla f=\left\langle 2xy^3z^4, 3x^2y^2z^4, 4x^2y^3z^3\right\rangle \] Then the divergence of the gradient is: \[\begin{aligned} \vec\nabla\cdot\vec\nabla f &=\partial_x(2xy^3z^4)+\partial_y(3x^2y^2z^4)+\partial_z(4x^2y^3z^3) \\ &=2y^3z^4+6x^2yz^4+12x^2y^3z^2 \end{aligned}\]

The divergence of the gradient is an operation called the Laplacian, \(\nabla^2=\vec\nabla\cdot\vec\nabla\), which we will study on a later page. It can be computed directly using the formula: \[ \nabla^2f=\vec\nabla\cdot\vec\nabla f =\dfrac{d^2f}{dx^2}+\dfrac{d^2f}{dy^2}+\dfrac{d^2f}{dz^2} \]
Compute the divergence of the gradient of \(f=xyz\), i.e. \(\vec\nabla\cdot\vec\nabla f\).

The divergence of the gradient is an operation called the Laplacian, \(\nabla^2=\vec\nabla\cdot\vec\nabla\), which we will study on a later page. It can be computed directly using the formula: \[\begin{aligned} \nabla^2f&=\vec\nabla\cdot\vec\nabla f \\ &=\dfrac{d^2f}{dx^2}+\dfrac{d^2f}{dy^2}+\dfrac{d^2f}{dz^2} \\ &=\partial_x^2f+\partial_y^2f+\partial_z^2f \end{aligned}\]

\(\vec\nabla\cdot\vec\nabla(xyz)=0\)

The Laplacian is: \[\begin{aligned} \nabla^2f &=\vec\nabla\cdot\vec\nabla f \\ &=\partial_x^2f+\partial_y^2f+\partial_z^2f \end{aligned}\] We compute the Laplacian directly: \[\begin{aligned} \nabla^2(xyz) &=\vec\nabla\cdot\vec\nabla(xyz) \\ &=\partial_x^2(xyz)+\partial_y^2(xyz)+\partial_z^2(xyz) \\ &=0+0+0=0 \end{aligned}\]

c. The Curl Operator

Compute the curl of each vector field.

\(\vec F=\left\langle xy,yz,zx\right\rangle\)

\[ \vec\nabla\times\vec F =\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ \partial_x & \partial_y & \partial_z \\ F_1 & F_2 & F_3 \end{vmatrix} \]

\(\vec\nabla\times\vec F =\left\langle -y,-z,-x\right\rangle\)

\[\begin{aligned} \vec\nabla\times\vec F &=\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ \partial_x & \partial_y & \partial_z \\ xy & yz & zx \end{vmatrix} \\ &=\hat\imath(0-y)-\hat\jmath(z-0)+\hat k(0-x) \\ &=\left\langle -y,-z,-x\right\rangle \end{aligned}\]
\(\vec F=\left\langle yz,xz,xy\right\rangle\)

\(\vec\nabla\times\vec F =\vec0\equiv\left\langle 0,0,0\right\rangle\)

\[\begin{aligned} \vec\nabla\times\vec F &=\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ \partial_x & \partial_y & \partial_z \\ yz & xz & xy \end{vmatrix} \\ &=\hat\imath(x-x)-\hat\jmath(y-y)+\hat k(z-z) \\ &=\left\langle 0,0,0\right\rangle=\vec0 \end{aligned}\]
\(\vec F =\left\langle \dfrac{1}{2}z^2x,\dfrac{1}{2}x^2y, \dfrac{1}{2}y^2z\right\rangle\)

\(\vec\nabla\times\vec F =\left\langle yz,xz,xy\right\rangle\)

\[\begin{aligned} \vec\nabla\times\vec F &=\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ \partial_x & \partial_y & \partial_z \\ \dfrac{1}{2}z^2x & \dfrac{1}{2}x^2y & \dfrac{1}{2}y^2z \end{vmatrix} \\ &=\hat\imath(yz-0)-\hat\jmath(0-zx)+\hat k(xy-0) \\ &=\left\langle yz,xz,xy\right\rangle \end{aligned}\]
\(\vec F=\left\langle x\cos z,y\cos z,\sin z\right\rangle\)

\(\vec\nabla\times\vec F =\left\langle y\sin z,-x\sin z,0\right\rangle\)

\[\begin{aligned} \vec\nabla\times\vec F &=\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ \partial_x & \partial_y & \partial_z \\ x\cos z & y\cos z & \sin z \end{vmatrix} \\ &=\hat\imath(y\sin z)-\hat\jmath(0--x\sin z)+\hat k(0-0) \\ &=\left\langle y\sin z,-x\sin z,0\right\rangle \end{aligned}\]
\(\vec F =\left\langle z\cos x\cos y,-z\sin x\sin y, \sin x\cos y\right\rangle\)

\(\vec\nabla\times\vec F =\vec0\equiv\left\langle 0,0,0\right\rangle\)

\[\begin{aligned} \vec\nabla\times\vec F &=\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ \partial_x & \partial_y & \partial_z \\ z\cos x\cos y & -z\sin x\sin y & \sin x\cos y \end{vmatrix} \\ &=\hat\imath(-\sin x\sin y--\sin x\sin y) \\ &\quad-\hat\jmath(\cos x\cos y-\cos x\cos y) \\ &\quad+\hat k(-z\cos x\sin y--z\cos x\sin y) \\ &=\left\langle 0,0,0\right\rangle=\vec0 \end{aligned}\]
\(\vec F=\left\langle -y,x,0\right\rangle\)

\(\vec\nabla\times\vec F =\left\langle 0,0,2\right\rangle\)

\[\begin{aligned} \vec\nabla\times\vec F &=\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ \partial_x & \partial_y & \partial_z \\ -y & x & 0 \end{vmatrix} \\ &=\hat\imath(0-0)-\hat\jmath(0-0)+\hat k(1--1) \\ &=\left\langle 0,0,2\right\rangle \end{aligned}\]
\(\vec F=\left\langle e^{yz},e^{xz},e^{xy}\right\rangle\)

\(\vec\nabla\times\vec F =\left\langle x(e^{xy}-e^{xz}), y(e^{yz}-e^{xy}), z(e^{xz}-e^{yz})\right\rangle\)

\[\begin{aligned} \vec\nabla\times\vec F &=\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ \partial_x & \partial_y & \partial_z \\ e^{yz} & e^{xz} & e^{xy} \end{vmatrix} \\ &=\hat\imath(xe^{xy}-xe^{xz}) -\hat\jmath(ye^{xy}-ye^{yz}) +\hat k(ze^{xz}-ze^{yz}) \\ &=\left\langle x(e^{xy}-e^{xz}), y(e^{yz}-e^{xy}), z(e^{xz}-e^{yz})\right\rangle \end{aligned}\]
\(\vec F=\left\langle \ln\left(\dfrac{x}{yz}\right), \ln\left(\dfrac{y}{xz}\right), \ln\left(\dfrac{z}{xy}\right)\right\rangle\)

\(\vec\nabla\times\vec F= =\left\langle \dfrac{1}{z}-\,\dfrac{1}{y}, \dfrac{1}{x}-\,\dfrac{1}{z}, \dfrac{1}{y}-\,\dfrac{1}{x}\right\rangle\)

\[\begin{aligned} \vec\nabla\times\vec F &=\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ \partial_x & \partial_y & \partial_z \\ \ln x-\ln y-\ln z & \ln y-\ln x-\ln z & \ln z-\ln x-\ln y \end{vmatrix} \\ &=\hat\imath\left(-\,\dfrac{1}{y}--\,\dfrac{1}{z}\right) -\hat\jmath\left(-\,\dfrac{1}{x}--\,\dfrac{1}{z}\right) +\hat k \left(-\,\dfrac{1}{x}--\,\dfrac{1}{y}\right) \\ &=\left\langle \dfrac{1}{z}-\,\dfrac{1}{y}, \dfrac{1}{x}-\,\dfrac{1}{z}, \dfrac{1}{y}-\,\dfrac{1}{x}\right\rangle \end{aligned}\]

Compute the curl of the gradient of \(f=x^2y^3z^4\), i.e. \(\vec\nabla\times\vec\nabla f\).

\(\vec\nabla\times\vec\nabla(x^2y^3z^4)=\vec0\equiv\left\langle 0,0,0\right\rangle\)

The gradient of \(f\) is: \[ \vec\nabla f=\left\langle 2xy^3z^4, 3x^2y^2z^4, 4x^2y^3z^3\right\rangle \] Then the curl of the gradient is: \[\begin{aligned} \vec\nabla\times\vec\nabla f &=\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ \partial_x & \partial_y & \partial_z \\ 2xy^3z^4 & 3x^2y^2z^4 & 4x^2y^3z^3 \end{vmatrix} \\ &=\hat\imath(12x^2y^2z^3-12x^2y^2z^3) \\ &\quad-\hat\jmath(8xy^3z^3-8xy^3z^3) \\ &\quad+\hat k (6xy^2z^4-6xy^2z^4) \\ &=\left\langle 0,0,0\right\rangle=\vec0 \end{aligned}\]
Compute the curl of the gradient of \(f=xyz\), i.e. \(\vec\nabla\times\vec\nabla f\).

\(\vec\nabla\times\vec\nabla(xyz)=\vec0\equiv\left\langle 0,0,0\right\rangle\)

The gradient of \(f\) is: \[ \vec\nabla f=\left\langle yz, xz, xy\right\rangle \] Then the curl of the gradient is: \[\begin{aligned} \vec\nabla\times\vec\nabla f &=\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ \partial_x & \partial_y & \partial_z \\ yz & xz & xy \end{vmatrix} \\ &=\hat\imath(x-x)-\hat\jmath(y-y)+\hat k(z-z) \\ &=\left\langle 0,0,0\right\rangle=\vec0 \end{aligned}\]

It is no coincidence that the curl of the gradient is \(\vec0\) in both this and the previous problem. We will see on a later page that \(\vec\nabla\times\vec\nabla f=\vec0\) for any function \(f\).

Compute the divergence of the curl of \(\vec F=\left\langle xy,yz,zx\right\rangle\), i.e. \(\vec\nabla\cdot\vec\nabla\times\vec F\).

\(\vec\nabla\cdot\vec\nabla\times\left\langle xy,yz,zx\right\rangle =0\)

The curl of \(\vec F\) is: \[ \vec\nabla\times\vec F =\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ \partial_x & \partial_y & \partial_z \\ xy & yz & zx \end{vmatrix} =\left\langle -y,-z,-x\right\rangle \] Then the divergence of the curl is: \[\begin{aligned} \vec\nabla\cdot\vec\nabla\times\vec F &=\partial_x(-y)+\partial_y(-z)+\partial_z(-x) \\ &=0+0+0=0 \end{aligned}\]
Compute the divergence of the curl of \(\vec F=\left\langle e^{yz},e^{xz},e^{xy}\right\rangle\), i.e. \(\vec\nabla\cdot\vec\nabla\times\vec F\).

\(\vec\nabla\cdot\vec\nabla\times\vec F=0\)

The curl of \(\vec F\) is: \[\begin{aligned} \vec\nabla\times\vec F &=\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ \partial_x & \partial_y & \partial_z \\ e^{yz} & e^{xz} & e^{xy} \end{vmatrix} \\ &=\left\langle x(e^{xy}-e^{xz}), y(e^{yz}-e^{xy}), z(e^{xz}-e^{yz})\right\rangle \end{aligned}\] Then the divergence of the curl is: \[\begin{aligned} \vec\nabla\cdot\vec\nabla\times\vec F &=\partial_x[x(e^{xy}-e^{xz})] \\ &\quad+\partial_y[y(e^{yz}-e^{xy})] \\ &\quad+\partial_z[z(e^{xz}-e^{yz})] \\ &=[(e^{xy}-e^{xz})+x(ye^{xy}-ze^{xz})] \\ &\quad+[(e^{yz}-e^{xy})+y(ze^{yz}-xe^{xy})] \\ &\quad+[(e^{xz}-e^{yz})+z(xe^{xz}-ye^{yz})] \\ &=0 \end{aligned}\]

It is no coincidence that the divergence of the curl is \(0\) in both this and the previous problem. We will see on a later page that \(\vec\nabla\cdot\vec\nabla\times\vec F=0\) for any vector field \(\vec F\).

Compute the curl of the curl of \(\vec F=\left\langle xy,yz,zx\right\rangle\), i.e. \(\vec\nabla\times\vec\nabla\times\vec F\).

\(\vec\nabla\times\vec\nabla\times\left\langle xy,yz,zx\right\rangle =\left\langle 1,1,1\right\rangle\)

The curl of \(\vec F\) is: \[ \vec\nabla\times\vec F =\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ \partial_x & \partial_y & \partial_z \\ xy & yz & zx \end{vmatrix} =\left\langle -y,-z,-x\right\rangle \] Then the curl of the curl is: \[\begin{aligned} \vec\nabla\times\vec\nabla\times\vec F =\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ \partial_x & \partial_y & \partial_z \\ -y & -z & -x \end{vmatrix} =\left\langle 1,1,1\right\rangle \end{aligned}\]

Although there is an identity for the curl of the curl which we will learn on a later page, it is not very useful in computations.

d. Differential Identities

If \(f=2x\cos z\) and \(g=2y\sin z\), compute \(\vec\nabla(fg)\) in two ways.
1. Find \(f\vec\nabla g\) and \(g\vec\nabla f\) and add them to get \(\vec\nabla(fg)\).
  
  \(\begin{aligned} \vec\nabla(fg) &=\left\langle 2y\sin(2z),2x\sin(2z),4xy\cos(2z)\right\rangle \\ &=\left\langle 4y\sin z\cos z,4x\sin z\cos z,4xy\cos^2 z-4xy\sin^2 z\right\rangle \end{aligned}\)
  
  \[\begin{aligned} f\vec\nabla g &=2x\cos z\left\langle 0,2\sin z,2y\cos z\right\rangle \\ &=\left\langle 0,4x\sin z\cos z,4xy\cos^2 z\right\rangle \\[6pt] g\vec\nabla f &=2y \sin z\left\langle 2\cos z,0,-2x\sin z\right\rangle \\ &=\left\langle 4y\sin z\cos z,0,-4xy\sin^2 z\right\rangle \\[6pt] \vec\nabla(fg) &=f\vec\nabla g+g\vec\nabla f \\ &=\left\langle 4y\sin z\cos z,4x\sin z\cos z,4xy\cos^2 z-4xy\sin^2 z\right\rangle \end{aligned}\]
2. Find \(fg\) and then \(\vec\nabla(fg)\).
  
  \(\begin{aligned} \vec\nabla(fg) &=\left\langle 2y\sin(2z),2x\sin(2z),4xy\cos(2z)\right\rangle \\ &=\left\langle 4y\sin z\cos z,4x\sin z\cos z,4xy\cos^2 z-4xy\sin^2 z\right\rangle \end{aligned}\)
  
  \[\begin{aligned} fg&=4xy\cos z\sin z=2xy\sin(2z) \\[6pt] \vec\nabla(fg) &=\left\langle 2y\sin(2z),2x\sin(2z),4xy\cos(2z)\right\rangle \\ &=\left\langle 4y\sin z\cos z,4x\sin z\cos z,4xy\cos^2 z-4xy\sin^2 z\right\rangle \end{aligned}\]
At a point \(P\), we know: \[\begin{aligned} f&=3\qquad& \vec\nabla f&=\left\langle -2,1,3\right\rangle \\ g&=2\qquad& \vec\nabla g&=\left\langle 2,-1,4\right\rangle \end{aligned}\] Find \(\vec\nabla(fg)\) at \(P\).

\(\vec\nabla(fg) =\left\langle 2,-1,18\right\rangle\)

Using the formula, we have: \[\begin{aligned} \vec\nabla(fg) &=f\vec\nabla g+g\vec\nabla f \\ &=3\left\langle 2,-1,4\right\rangle+2\left\langle -2,1,3\right\rangle \\ &=\left\langle 6,-3,12\right\rangle+\left\langle -4,2,6\right\rangle \\ &=\left\langle 2,-1,18\right\rangle \end{aligned}\]
If \(f=xyz\) and \(\vec G=\left\langle xz,yz,z^2\right\rangle\), compute \(\vec\nabla\cdot (f\vec G)\) in two ways.
1. Find \((\vec\nabla f)\cdot\vec G\) and \(f\vec\nabla\cdot\vec G\) and add them to get \(\vec\nabla\cdot(f\vec G)\).
  
  \(\vec\nabla\cdot(f\vec G) =7xyz^2\)
  
  \[\begin{aligned} \vec\nabla f &=\left\langle yz,xz,xy\right\rangle \\ (\vec\nabla f)\cdot\vec G &=xyz^2+xyz^2+xyz^2=3xyz^2 \\[6pt] \vec\nabla\cdot\vec G &=z+z+2z=4z \\ f\vec\nabla\cdot\vec G &=4xyz^2 \\[6pt] \vec\nabla\cdot(f\vec G) &=(\vec\nabla f)\cdot\vec G +f\vec\nabla\cdot\vec G \\ &=3xyz^2+4xyz^2=7xyz^2 \end{aligned}\]
2. Find \(f\vec G\) and then \(\vec\nabla\cdot(f\vec G)\).
  
  \(\vec\nabla\cdot(f\vec G) ==7xyz^2\)
  
  \[\begin{aligned} f\vec G &=xyz\left\langle xz,yz,z^2\right\rangle \\ &=\left\langle x^2yz^2,xy^2z^2,xyz^3\right\rangle \\[6pt] \vec\nabla\cdot(f\vec G) &=2xyz^2+2xyz^2+3xyz^2=7xyz^2 \end{aligned}\]
At a point \(P\), we know: \[\begin{aligned} \vec F&=\left\langle 4,-2,1\right\rangle\qquad& \vec\nabla\times\vec F&=\left\langle 2,3,-2\right\rangle \\ \vec G&=\left\langle 1,-3,2\right\rangle\qquad& \vec\nabla\times\vec G&=\left\langle -2,1,4\right\rangle \end{aligned}\] Find \(\vec\nabla\cdot(\vec F\times\vec G)\) at \(P\).

\(\vec\nabla\cdot(\vec F\times\vec G) =-5\)

Using the formula, we have: \[\begin{aligned} \vec\nabla\cdot(\vec F\times\vec G) &=\vec\nabla\times\vec F\cdot\vec G-\vec F\cdot\vec\nabla\times\vec G \\ &=\left\langle 2,3,-2\right\rangle\cdot\left\langle 1,-3,2\right\rangle -\left\langle 4,-2,1\right\rangle\cdot\left\langle -2,1,4\right\rangle \\ &=(2-9-4)-(-8-2+4)=-5 \end{aligned}\]
If \(f=xyz\) and \(\vec G=\left\langle -yz,xz,z^2\right\rangle\), compute \(\vec\nabla\times (f\vec G)\) in two ways.
1. Find \((\vec\nabla f)\times\vec G\) and \(f\vec\nabla\times\vec G\) and add them to get \(\vec\nabla\times(f\vec G)\).
  
  \(\vec\nabla\times(f\vec G) =\left\langle xz^3-2x^2yz,-yz^3-2xy^2z,4xyz^2\right\rangle\)
  
  \[\begin{aligned} \vec\nabla f &=\left\langle yz,xz,xy\right\rangle \\[6pt] (\vec\nabla f)\times\vec G &=\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ yz & xz & xy \\ -yz & xz & z^2 \end{vmatrix} \\ &=\hat\imath(xz^3-x^2yz)-\hat\jmath(yz^3--xy^2z) \\ &\quad+\hat k(xyz^2--xyz^2) \\ &=\left\langle xz^3-x^2yz,-yz^3-xy^2z,2xyz^2\right\rangle \\[6pt] \vec\nabla\times\vec G &=\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ \partial_x & \partial_y & \partial_z \\ -yz & xz & z^2 \end{vmatrix} \\ &=\hat\imath(0-x)-\hat\jmath(0--y) +\hat k(z--z) \\ &=\left\langle -x,-y,2z\right\rangle \\[6pt] f\vec\nabla\times\vec G &=\left\langle -x^2yz,-xy^2z,2xyz^2\right\rangle \\[6pt] \vec\nabla\times(f\vec G) &=(\vec\nabla f)\times\vec G +f\vec\nabla\times\vec G \\ &=\left\langle xz^3-2x^2yz,-yz^3-2xy^2z,4xyz^2\right\rangle \end{aligned}\]
2. Find \(f\vec G\) and then \(\vec\nabla\times(f\vec G)\).
  
  \(\vec\nabla\times(f\vec G) =\left\langle xz^3-2x^2yz,-yz^3-2xy^2z,4xyz^2\right\rangle\)
  
  \[\begin{aligned} f\vec G &=\left\langle -xy^2z^2,x^2yz^2,xyz^3\right\rangle \\[6pt] \vec\nabla\times(f\vec G) &=\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ \partial_x & \partial_y & \partial_z \\ -xy^2z^2 & x^2yz^2 & xyz^3 \end{vmatrix} \\ &=\hat\imath(xz^3-2x^2yz)-\hat\jmath(yz^3--2xy^2z) +\hat k(2xyz^2--2xyz^2) \\ &=\left\langle xz^3-2x^2yz,-yz^3-2xy^2z,4xyz^2\right\rangle \end{aligned}\]
At a point \(P\), we know: \[\begin{aligned} f&=3\qquad& \vec\nabla f&=\left\langle 3,4,5\right\rangle \\ \vec G&=\left\langle -3,2,4\right\rangle\qquad& \vec\nabla\times\vec G&=\left\langle 2,-1,3\right\rangle \end{aligned}\] Find \(\vec\nabla\times(f\vec G)\) at \(P\).

\(\vec\nabla\times(f\vec G) =\left\langle 12,-30,27\right\rangle\)

\[\begin{aligned} (\vec\nabla f)\times\vec G &=\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ 3 & 4 & 5 \\ -3 & 2 & 4 \end{vmatrix} \\ &=\hat\imath(16-10)-\hat\jmath(12+15)+\hat k(6+12) \\ &=\left\langle 6,-27,18\right\rangle \\[6pt] f\vec\nabla\times\vec G &=3\left\langle 2,-1,3\right\rangle =\left\langle 6,-3,9\right\rangle \\[6pt] \vec\nabla\times(f\vec G) &=(\vec\nabla f)\times\vec G +f\vec\nabla\times\vec G \\ &=\left\langle 12,-30,27\right\rangle \end{aligned}\]
If \(f=\sin x\cos y\), compute the Laplacian \(\nabla^2f\) and show it satisfies \(\nabla^2f=-2f\).

\(\nabla^2f=-2\sin x\cos y=-2f\)

\[\begin{aligned} \nabla^2f &=\partial_x^2(\sin x\cos y)+\partial_y^2(\sin x\cos y) \\ &=\partial_x(\cos x\cos y)+\partial_y(-\sin x\sin y) \\ &=-\sin x\cos y-\sin x\cos y=-2\sin x\cos y=-2f \\ \end{aligned}\]
If \(f=x^2+y^2-2z^2\), compute the Laplacian \(\nabla^2f\) and show it satisfies the Laplace equation \(\nabla^2f=0\).

\(\nabla^2f=0\)

\[\begin{aligned} \nabla^2f &=\partial_x^2(x^2+y^2-2z^2)+\partial_y^2(x^2+y^2-2z^2)+\partial_z^2(x^2+y^2-2z^2) \\ &=\partial_x(2x)+\partial_y(2y)+\partial_z(-4z) \\ &=2+2-4=0 \\ \end{aligned}\]
If \(f=z\sin x\cos y\), compute the curl of the gradient of \(f\), i.e. \(\vec\nabla\times\vec\nabla f\).

\(\vec\nabla\times\vec\nabla f=\vec0\)

For any function: \[ \vec\nabla\times\vec\nabla f=\vec0 \]
If \(\vec F=\left\langle y\sin z,x\cos z,xy\right\rangle\), compute the divergence of the curl of \(\vec F\), i.e. \(\vec\nabla\cdot\vec\nabla\times\vec F\).

\(\vec\nabla\cdot\vec\nabla\times\vec F=0\)

For any vector field: \[ \vec\nabla\cdot\vec\nabla\times\vec F=0 \]

e. Generalizing Antiderivatives

Verify that \(F=\cos^2 x\) and \(G=-\sin^2 x\) are both antiderivatives of \(f=-2\sin x\cos x\). Why is this possible?

By chain rule: \[F'=2(\cos x)(-\sin x)=-2\sin x\cos x\] Similarly: \[G'=-2(\sin x)(\cos x)=-2\sin x\cos x\] This is possible because \(F\) and \(G\) differ by a constant: \[\cos^2 x=-\sin^2 x +1\]
Verify that \(f=xy^2z^3\) is a scalar potential for \(\vec F=\left\langle y^2z^3,2xyz^3,3xy^2z^2\right\rangle\)

The gradient of \(f=xy^2z^3\) is: \[ \vec\nabla f=\left\langle y^2z^3,2xyz^3,3xy^2z^2\right\rangle=\vec F \]
Verify that \(f=z\sin x\cos y\) is a scalar potential for \(\vec F=\left\langle z\cos x\cos y,-z\sin x\sin y,\sin x\cos y\right\rangle\)

The gradient of \(f=z\sin x\cos y\) is: \[ \vec\nabla f =\left\langle z\cos x\cos y,-z\sin x\sin y,\sin x\cos y\right\rangle =\vec F \]
Verify that \(\vec A=\langle xy,yz,zx\rangle\) is a vector potential for \(\vec F=\left\langle -y,-z,-x\right\rangle\)

The curl of \(\vec A=\left\langle yz,xz,xy\right\rangle\) is: \[\begin{aligned} \vec\nabla\times\vec A &=\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ \partial_x & \partial_y & \partial_z \\ xy & yz & zx \end{vmatrix} \\ &=\left\langle 0-y,0-z,0-x\right\rangle=\vec F \end{aligned}\]
Verify that \(\vec A=\left\langle z\sin y,x\sin z,y\sin x\right\rangle\) and \(\vec A=\left\langle z\sin y-yz\cos x,x\sin z-z\sin x,0\right\rangle\) are both vector potentials for \(\vec F=\left\langle \sin x-x\cos z,\sin y-y\cos x,\sin z-z\cos y\right\rangle\)

The curl of \(\vec A=\left\langle x\sin y,y\sin z,z\sin x\right\rangle\) is: \[\begin{aligned} \vec\nabla&\times\vec A =\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ \partial_x & \partial_y & \partial_z \\ z\sin y & x\sin z & y\sin x \end{vmatrix} \\ &=\left\langle \sin x-x\cos z,\sin y-y\cos x,\sin z-z\cos y\right\rangle \\ &=\vec F \end{aligned}\]

The curl of \(\vec A=\left\langle z\sin y-yz\cos x,x\sin z-z\sin x,0\right\rangle\) is: \[\begin{aligned} \vec\nabla&\times\vec A =\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ \partial_x & \partial_y & \partial_z \\ z\sin y-yz\cos x & x\sin z-z\sin x & 0 \end{vmatrix} \\ &=\left\langle -x\cos z+\sin x,\sin y-y\cos x,\right. \\ &\qquad\qquad\qquad\left.\sin z-z\cos x-z\cos y+z\cos x\right\rangle \\ &=\left\langle \sin x-x\cos z,\sin y-y\cos x,\sin z-z\cos y\right\rangle \\ &=\vec F \end{aligned}\]

f. Scalar Potentials

Find a scalar potential for each vector field or show one does not exist.

\(\vec F=\left\langle y+z,x+z,x+y\right\rangle\)

A scalar potential for \(\vec F\) is any function \(f\) satisfying \(\vec\nabla f=\vec F\). Solve the equations: \[ \partial_x f=F_1 \qquad \partial_y f=F_2 \qquad \partial_z f=F_3 \] for a single function \(f\).

\(f=xy+xz+yz\)

We solve: \[ \partial_x f=y+z \qquad \partial_y f=x+z \qquad \partial_z f=x+y \] The \(x\)-antiderivative of the \(1^\text{st}\) equation is: \[ f=xy+xz+g(y,z) \] The \(y\)-antiderivative of the \(2^\text{nd}\) equation is: \[ f=xy+yz+h(x,z) \] The \(z\)-antiderivative of the \(3^\text{rd}\) equation is: \[ f=xz+yz+k(x,y) \] A scalar potential which satisfies all three conditions is: \[ f=xy+xz+yz \] where \(g(y,z)=yz\), \(h(x,z)=xz\) and \(k(x,y)=xy\).
\(\vec F=\left\langle yz\cos(xyz),xz\cos(xyz),xy\cos(xyz)\right\rangle\)

\(f=\sin(xyz)\)

We solve: \[ \partial_x f=yz\cos(xyz) \qquad \partial_y f=xz\cos(xyz) \qquad \partial_z f=xy\cos(xyz) \] The \(x\)-antiderivative of the \(1^\text{st}\) equation, the \(y\)-antiderivative of the \(2^\text{nd}\) equation and the \(z\)-antiderivative of the \(3^\text{rd}\) equation are all: \[ f=\sin(xyz) \] which is therefore the scalar potential.
\(\vec F=\left\langle y^2+2xz,2xy-3y^2z,x^2-y^3-2z\right\rangle\)

Successively solve the equations: \[ \partial_x f=F_1 \qquad \partial_y f=F_2 \qquad \partial_z f=F_3 \] If you reach a contradiction, there is no solution.

\(f=xy^2+x^2z-y^3z-z^2+C\)

We solve: \[ \partial_x f=y^2+2xz \qquad \partial_y f=2xy-3y^2z \qquad \partial_z f=x^2-y^3-2z \] The \(x\)-antiderivative of the \(1^\text{st}\) equation is: \[ f=xy^2+x^2z+g(y,z) \] This needs to satisfy the the \(2^\text{nd}\) equation. So we compute \(\partial_y f\) and plug into the \(2^\text{nd}\) equation: \[ 2xy+\partial_y g=2xy-3y^2z \qquad \text{or} \qquad \partial_y g=-3y^2z \] The solution is \(g(y,z)=-y^3z+h(z)\). So: \[ f=xy^2+x^2z-y^3z+h(z) \] This needs to satisfy the the \(3^\text{rd}\) equation. So we compute \(\partial_z f\) and plug into the \(3^\text{rd}\) equation: \[ x^2-y^3+h'(z)=x^2-y^3-2z \qquad \text{or} \qquad h'(z)=-2z \] A solution is \(h(z)=-z^2+C\). So a scalar potential is: \[ f=xy^2+x^2z-y^3z-z^2+C \]
\(\vec F=\left\langle y^2+2xz,2xy+3y^2z,x^2-y^3-2z\right\rangle\)

There is no scalar potential. The contradiction is shown in the solution and in the remark.

We solve: \[ \partial_x f=y^2+2xz \qquad \partial_y f=2xy+3y^2z \qquad \partial_z f=x^2-y^3-2z \] The \(x\)-antiderivative of the \(1^\text{st}\) equation is: \[ f=xy^2+x^2z+g(y,z) \] This needs to satisfy the the \(2^\text{nd}\) equation. So we compute \(\partial_y f\) and plug into the \(2^\text{nd}\) equation: \[ 2xy+\partial_y g=2xy+3y^2z \qquad \text{or} \qquad \partial_y g=3y^2z \] The solution is \(g(y,z)=y^3z+h(z)\). So: \[ f=xy^2+x^2z+y^3z+h(z) \] This needs to satisfy the the \(3^\text{rd}\) equation. So we compute \(\partial_z f\) and plug into the \(3^\text{rd}\) equation: \[ x^2+y^3+h'(z)=x^2-y^3-2z \qquad \text{or} \qquad h'(z)=-2y^3-2z \] This is a contradiction because \(h\) cannot be a function of \(y\). So there is no scalar potential.

To verify there is no scalar potential, we compute the curl: \[\begin{aligned} \vec\nabla\times\vec F &=\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ \partial_x & \partial_y & \partial_z \\ y^2+2xz & 2xy+3y^2z & x^2-y^3-2z \end{vmatrix} \\ &=\hat\imath(-3y^2-3y^2) -\hat\jmath(2x-2x) +\hat k (2y-2y) \\ &=\left\langle -6y^2,0,0\right\rangle \end{aligned}\] Since this is not \(\vec0\), there cannot be a scalar potential.
\(\vec F=\left\langle \cos x\cos y,\sin x\sin y,\sec^2 z\right\rangle\)

There is no scalar potential, since the curl of \(\vec F\) is not \(\vec0\).

We first compute the curl: \[\begin{aligned} \vec\nabla\times\vec F &=\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ \partial_x & \partial_y & \partial_z \\ \cos x\cos y & \sin x\sin y & \sec^2 z \end{vmatrix} \\ &=\hat\imath(0-0) -\hat\jmath(0-0) +\hat k (\cos x\sin y+\cos x\sin y)\\ &=\left\langle 0,0,2\cos x\sin y\right\rangle \end{aligned}\] Since this is not \(\vec0\), there is no scalar potential.
\(\vec F=\left\langle \cos x\cos y,-\sin x\sin y,\sec^2 z\right\rangle\)

\(f=\sin x\cos y+\tan z\)

We first compute the curl: \[\begin{aligned} \vec\nabla\times\vec F &=\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ \partial_x & \partial_y & \partial_z \\ \cos x\cos y & -\sin x\sin y & \sec^2 z \end{vmatrix} \\ &=\hat\imath(0-0) -\hat\jmath(0-0) +\hat k (-\cos x\sin y+\cos x\sin y) =\vec0 \end{aligned}\] Since this is \(\vec0\), we expect there is a potential and solve: \[ \partial_x f=\cos x\cos y \qquad \partial_y f=-\sin x\sin y \qquad \partial_z f=\sec^2 z \] The \(x\)-antiderivative of the \(1^\text{st}\) equation is: \[ f=\sin x\cos y+g(y,z) \] This needs to satisfy the the \(2^\text{nd}\) equation. So we compute \(\partial_y f\) and plug into the \(2^\text{nd}\) equation: \[ -\sin x\sin y+\partial_y g=-\sin x\sin y \qquad \text{or} \qquad \partial_y g=0 \] The solution is \(g(y,z)=h(z)\). (Don't say \(g=0\) or you will not get a solution!) So: \[ f=\sin x\cos y+h(z) \] This needs to satisfy the the \(3^\text{rd}\) equation. So we compute \(\partial_z f\) and plug into the \(3^\text{rd}\) equation: \[ h'(z)=\sec^2 z \qquad \text{or} \qquad h(z)=\tan z \] Therefore, a scalar potential is: \[ f=\sin x\cos y+\tan z \]

In this case, computing the curl was a waste of time. But we could not know this in advance!

g. Vector Potentials

Find a vector potential for each vector field or show one does not exist.

\(\vec F=\left\langle -x,y,y-x\right\rangle\)

A vector potential for \(\vec F=\left\langle F_1,F_2,F_3\right\rangle\) is any vector field \(\vec A=\left\langle A_1,A_2,A_3\right\rangle\) satisfying \(\vec\nabla\times\vec A=\vec F\). So we need to solve the equations: \[\begin{aligned} \partial_y A_3-\partial_z A_2&=F_1 \\ \partial_z A_1-\partial_x A_3&=F_2 \\ \partial_x A_2-\partial_y A_1&=F_3 \end{aligned}\] for \(A_1\), \(A_2\) and \(A_3\). There is always a solution with \(A_3=0\). So we actually need to solve: \[\begin{aligned} \partial_z A_2&=-F_1 \\ \partial_z A_1&=F_2 \\ \partial_x A_2-\partial_y A_1&=F_3 \end{aligned}\]

\(\vec A=\left\langle yz+xy,xz+xy,0\right\rangle\)
There are many other solutions.

We assume \(A_3=0\). So we need to solve: \[\begin{aligned} \partial_z A_2&=-F_1=x \\ \partial_z A_1&=F_2=y \\ \partial_x A_2-\partial_y A_1&=F_3=y-x \end{aligned}\] The first \(2\) equations say: \[ A_2=xz+f(x,y) \qquad A_1=yz+g(x,y) \] We substitute into the \(3^\text{rd}\) equation: \[ z+\partial_x f-z-\partial_y g=y-x \] One solution is \(f=xy\) and \(g=xy\). So \[ \vec A=\left\langle yz+xy,xz+xy,0\right\rangle \] There are many other solutions.
\(\vec F=\left\langle x\cos z,y\cos z,2x-2\sin z\right\rangle\)

Successively solve the equations: \[\begin{aligned} \partial_z A_2&=-F_1 \\ \partial_z A_1&=F_2 \\ \partial_x A_2-\partial_y A_1&=F_3 \end{aligned}\] If you reach a contradiction, there is no solution.

\(\vec A=\left\langle y\sin z,-x\sin z+x^2,0\right\rangle\)
There are many other solutions.

We assume \(A_3=0\). So we need to solve: \[\begin{aligned} \partial_z A_2&=-F_1=-x\cos z \\ \partial_z A_1&=F_2=y\cos z \\ \partial_x A_2-\partial_y A_1&=F_3=2x-2\sin z \end{aligned}\] The first \(2\) equations say: \[ A_2=-x\sin z+f(x,y) \qquad A_1=y\sin z+g(x,y) \] We substitute into the \(3^\text{rd}\) equation: \[ -\sin z+\partial_x f-\sin z-\partial_y g=2x-2\sin z \qquad \text{or} \qquad \partial_x f-\partial_y g=2x \] One solution is \(f=x^2\) and \(g=0\). So \[ \vec A=\left\langle y\sin z,-x\sin z+x^2,0\right\rangle \] There are many other solutions.
\(\vec F=\left\langle x\cos z,y\cos z,2x-\sin z\right\rangle\)

There is no vector potential. The contradiction is shown in the solution and in the remark.

We assume \(A_3=0\). So we need to solve: \[\begin{aligned} \partial_z A_2&=-F_1=-x\cos z \\ \partial_z A_1&=F_2=y\cos z \\ \partial_x A_2-\partial_y A_1&=F_3=2x-\sin z \end{aligned}\] The first \(2\) equations say: \[ A_2=-x\sin z+f(x,y) \qquad A_1=y\sin z+g(x,y) \] We substitute into the \(3^\text{rd}\) equation: \[ -\sin z+\partial_x f-\sin z-\partial_y g=2x-\sin z \qquad \text{or} \qquad \partial_x f-\partial_y g=2x+\sin z \] This is a contradiction because \(f\) and \(g\) cannot be functions of \(z\). So there is no vector potential.

To verify there is no vector potential, we compute the divergence: \[\begin{aligned} \vec\nabla\cdot\vec F &=\partial_x(x\cos z)+\partial_y(y\cos z)+\partial_z(2x-\sin z) \\ &=\cos z+\cos z-\cos z=\cos z \ne 0 \end{aligned}\] Since this is not \(0\), there cannot be a vector potential.
\(\vec F=\left\langle yz,xz,xy\right\rangle\)

\(\vec A =\left\langle \dfrac{xz^2}{2},-\,\dfrac{yz^2}{2}+\dfrac{x^2y}{2},0\right\rangle\)

We first compute the divergence: \[ \vec\nabla\cdot\vec F =\partial_x(yz)+\partial_y(xz)+\partial_z(xy) =0+0+0=0 \] Since this is 0, we expect there is a potential and solve: \[\begin{aligned} \partial_z A_2&=-F_1=-yz \\ \partial_z A_1&=F_2=xz \\ \partial_x A_2-\partial_y A_1&=F_3=xy \end{aligned}\] The first \(2\) equations say: \[ A_2=-\,\dfrac{yz^2}{2}+f(x,y) \qquad A_1=\dfrac{xz^2}{2}+g(x,y) \] We substitute into the \(3^\text{rd}\) equation: \[ \partial_x f-\partial_y g=xy \] A solution is \(f=\dfrac{x^2y}{2}\) and \(g=0\). So a vector potential is: \[ \vec A =\left\langle \dfrac{xz^2}{2},-\,\dfrac{yz^2}{2}+\dfrac{x^2y}{2},0\right\rangle \]
\(\vec F=\left\langle x,y,z\right\rangle\)

There is no vector potential because \(\vec\nabla\cdot\vec F=3\ne0\).

We first compute the divergence: \[ \vec\nabla\cdot\vec F =\partial_x(x)+\partial_y(y)+\partial_z(z) =1+1+1=3 \] Since this is not \(0\), there is no vector potential.

PY: Checked to here.

Review Exercises

PY: Add the last problem from Edfinity Stokes.>

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