I will assume that you understand how to operate with derivatives. For more information on the topic, I recommend visiting Paul's Notes. It's a wonderful resource on all college level calculus related topics.
Problem: How do we describe how quickly the value of a three variable function changes as we move in 3D space?
We can define temperature in a room as \(T(x,y,z)\) where this function tells the temperature as the function of position.
This is more complicated, because we need to account for direction of motion. This can be resolved by taking a partial derivative in each
direction of motion.
\(dT=(\frac{\partial{T}}{\partial{x}}dx) + (\frac{\partial{T}}{\partial{y}}dy) + (\frac{\partial{T}}{\partial{z}}dz) \Rightarrow \underbrace{(\frac{\partial{T}}{\partial{x}}\hat{x} + \frac{\partial{T}}{\partial{y}}\hat{y} + \frac{\partial{T}}{\partial{z}}\hat{z})}_{\nabla{T}(gradient\ of\ T)} \cdot \underbrace{(dx\hat{x}+dy\hat{y}+dz\hat{z})}_{d\overrightarrow{\boldsymbol{l}}} = \nabla{T} \cdot d\overrightarrow{\boldsymbol{l}} \\ dT = \nabla{T} \cdot d\overrightarrow{\boldsymbol{l}} = |\nabla{T}||d\overrightarrow{\boldsymbol{l}}|\cos{\theta}\)
Gradient \(\nabla{T}\) points in the direction of maximum increase of the function T.
The magnitude of \(|\nabla{T}|\) gives the slope (rate of increase) along this maximal direction.
Take partial derivative with respect to each coordiante.
\(\nabla{f}=(\frac{\partial{f}}{\partial{x}}\hat{x} + \frac{\partial{f}}{\partial{y}}\hat{y} + \frac{\partial{f}}{\partial{z}}\hat{z})= e^x\sin(y)\ln(z)\hat{x} + e^x\cos(y)\ln(z)\hat{y} + e^x\sin(y)\frac{1}{z}\hat{z}\)
Take partial derivative with respect to each coordiante.
\( a)\nabla{((\overrightarrow{r_{sep}})^2)}=\nabla{(x-x')^2 + (y-y')^2 + (z-z')^2}=2(x-x')\hat{x} + 2(y-y')\hat{y} + 2(z-z')\hat{z}=2\overrightarrow{\boldsymbol{r}}_{sep}\)
\( b) \nabla{\frac{1}{\overrightarrow{r_{sep}}}}=\nabla{\frac{1}{\sqrt{(x-x')^2 + (y-y')^2 + (z-z')^2}}}=\frac{ -\frac{1}{2}*(\overrightarrow{r_{sep}})^{-1/2}*2(x-x')\hat{x} + -\frac{1}{2}*(\overrightarrow{r_{sep}})^{-1/2}*2(y-y')\hat{y} + -\frac{1}{2}*(\overrightarrow{r_{sep}})^{-1/2}*2(z-z')\hat{z}}{(x-x')^2 + (y-y')^2 + (z-z')^2} = \\
=\frac{ -(\overrightarrow{r_{sep}})^{-1/2}*(x-x')\hat{x} + -(\overrightarrow{r_{sep}})^{-1/2}*(y-y')\hat{y} + -(\overrightarrow{r_{sep}})^{-1/2}*(z-z')\hat{z}}{(x-x')^2 + (y-y')^2 + (z-z')^2}=\frac{ -(\overrightarrow{r_{sep}})^{-1/2}*[(x-x')\hat{x} + (y-y')\hat{y} + (z-z')\hat{z}]}{(x-x')^2 + (y-y')^2 + (z-z')^2} = \frac{ -(\overrightarrow{r_{sep}})^{-1/2}*\overrightarrow{\boldsymbol{r}}_{sep}}{(x-x')^2 + (y-y')^2 + (z-z')^2} = -\frac{\overrightarrow{\boldsymbol{r}}_{sep}}{((x-x')^2 + (y-y')^2 + (z-z')^2)^{3/2}}=-\frac{\hat{\boldsymbol{r}}_{sep}}{(\overrightarrow{r_{sep}})^2}\)
c) Chain rule: Fox x => \( \frac{\partial}{\partial{x}}\overrightarrow{r_{sep}^n}=n*\overrightarrow{r_{sep}^{n-1}}*\underbrace{\frac{\partial{\overrightarrow{r_{sep}}}}{\partial{x}}}_{\hat{x}}=n*\overrightarrow{r_{sep}^{n-1}}*\hat{x}\)
Chain Rule: \(\require{enclose}
\frac{\partial{f(y,z)}}{\partial{\bar{y}}}=\frac{\partial{f}}{\partial{y}} \enclose{box}[mathcolor="red"]{{\frac{\partial{y}}{\partial{\bar{y}}}}} +\frac{\partial{f}}{\partial{z}} \enclose{box}[mathcolor="red"]{\frac{\partial{z}}{\partial{\bar{y}}}},
\frac{\partial{f(y,z)}}{\partial{\bar{z}}}=\frac{\partial{f}}{\partial{y}} \enclose{box}[mathcolor="red"]{{\frac{\partial{y}}{\partial{\bar{z}}}}} +\frac{\partial{f}}{\partial{z}} \enclose{box}[mathcolor="red"]{\frac{\partial{z}}{\partial{\bar{z}}}},\)
Given system of equations for transformation from y,z to \(\bar{y}, \bar{z}\) from Section 1.5
\(
\left\{
\begin{array}{c}
\bar{y}=y\cos{\phi}+z\sin{\phi} \\
\bar{z}=-y\sin{\phi}+z\cos{\phi}
\end{array}
\right.
\)
Define y and z as functions of \(\bar{y},\bar{z}\) to compute derivatives in red.
Once derivatives in red are computed, the partial derivatives can be expressed as a system of equations that will look just like the system of equations in Section 1.5.
Chain Rule: \(\require{enclose}
\frac{\partial{f(y,z)}}{\partial{\bar{y}}}=\frac{\partial{f}}{\partial{y}} \enclose{box}[mathcolor="red"]{{\frac{\partial{y}}{\partial{\bar{y}}}}} +\frac{\partial{f}}{\partial{z}} \enclose{box}[mathcolor="red"]{\frac{\partial{z}}{\partial{\bar{y}}}},
\frac{\partial{f(y,z)}}{\partial{\bar{z}}}=\frac{\partial{f}}{\partial{y}} \enclose{box}[mathcolor="red"]{{\frac{\partial{y}}{\partial{\bar{z}}}}} +\frac{\partial{f}}{\partial{z}} \enclose{box}[mathcolor="red"]{\frac{\partial{z}}{\partial{\bar{z}}}},\)
Need to Define y and z as functions of \(\bar{y},\bar{z}\). Otherwise we cannot compute derivatives highlighted in red.
\(
\left\{
\begin{array}{c}
\bar{y}=y\cos{\phi}+z\sin{\phi} \\
\bar{z}=-y\sin{\phi}+z\cos{\phi}
\end{array}
\right. \Longrightarrow
\left\{
\begin{array}{c}
\frac{\bar{y}}{\cos{\phi}}=y+z\tan{\phi} \\
\frac{\bar{z}}{\cos{\phi}}=-y\tan{\phi}+z
\end{array}
\right. \Longrightarrow
\left\{
\begin{array}{c}
\bar{y}\sec{\phi}=y+z\tan{\phi} \\
\bar{z}\sec{\phi}=-y\tan{\phi}+z
\end{array}
\right. \Longrightarrow
\left\{
\begin{array}{c}
y = \bar{y}\sec{\phi} - z\tan{\phi} \\
z = \bar{z}\sec{\phi} + y\tan{\phi}
\end{array}
\right.
\)
Substitute in order to express y and z in terms of \(\bar{y}, \bar{z}\)
\(
\left\{
\begin{array}{c}
y = \bar{y}\sec{\phi} - z\tan{\phi} \\
z = \bar{z}\sec{\phi} + \tan{\phi}(\bar{y}\sec{\phi} - z\tan{\phi})
\end{array}
\right. \Longrightarrow
\left\{
\begin{array}{c}
y = \bar{y}\sec{\phi} - z\tan{\phi} \\
z = \bar{z}\sec{\phi} + \bar{y}\tan{\phi}\sec{\phi} - z\tan^2{\phi}
\end{array}
\right. \Longrightarrow \
\left\{
\begin{array}{c}
y = \bar{y}\sec{\phi} - z\tan{\phi} \\
z(1 + \tan^2{\phi}) = \bar{z}\sec{\phi} + \bar{y}\tan{\phi}\sec{\phi}
\end{array}
\right. \Longrightarrow
\\
\left\{
\begin{array}{c}
y = \bar{y}\sec{\phi} - z\tan{\phi} \\
z(\frac{\sin^2{\phi} + \cos^2{\phi}}{\cos^2{\phi}}) = \bar{z}\sec{\phi} + \bar{y}\frac{\sin{\phi}}{\cos^2{\phi}}
\end{array}
\right. \Longrightarrow
\left\{
\begin{array}{c}
y = \bar{y}\sec{\phi} - z\tan{\phi} \\
z(\frac{1}{\cos^2{\phi}}) = \bar{z}\sec{\phi} + \bar{y}\frac{\sin{\phi}}{\cos^2{\phi}}
\end{array}
\right. \Longrightarrow
\left\{
\begin{array}{c}
y = \bar{y}\sec{\phi} - z\tan{\phi} \\
z = \bar{z}\cos{\phi} + \bar{y}\sin{\phi}
\end{array}
\right.
\)
Substitute resulting z into function for y
\(
\left\{
\begin{array}{c}
y = \bar{y}\sec{\phi} - \tan{\phi}(\bar{z}\cos{\phi} + \bar{y}\sin{\phi}) \\
z = \bar{z}\cos{\phi} + \bar{y}\sin{\phi}
\end{array}
\right. \Longrightarrow
\left\{
\begin{array}{c}
y = \bar{y}(\sec{\phi} - \sin{\phi}\tan{\phi}) - \bar{z}\sin{\phi} \\
z = \bar{z}\cos{\phi} + \bar{y}\sin{\phi}
\end{array}
\right. \Longrightarrow
\left\{
\begin{array}{c}
y = \bar{y}\cos{\phi} - \bar{z}\sin{\phi} \\
z = \bar{z}\cos{\phi} + \bar{y}\sin{\phi}
\end{array}
\right.
\)
Thus apply chain rule, calculating known partial derivatives. We can express the transformation in the same form as was shown in Section 1.5 of the book.
\(
\left\{
\begin{array}{c}
\frac{\partial{f(y,z)}}{\partial{\bar{y}}}=\frac{\partial{f}}{\partial{y}} \enclose{box}[mathcolor="red"]{\cos{\phi}} +\frac{\partial{f}}{\partial{z}} \enclose{box}[mathcolor="red"]{\sin{\phi}} \\
\frac{\partial{f(y,z)}}{\partial{\bar{z}}}=\frac{\partial{f}}{\partial{y}} \enclose{box}[mathcolor="red"]{-\sin{\phi}} +\frac{\partial{f}}{\partial{z}} \enclose{box}[mathcolor="red"]{\cos{\phi}}
\end{array}
\right. \Longrightarrow
\begin{pmatrix}
\frac{\partial{f(y,z)}}{\partial{\bar{y}}} \\
\frac{\partial{f(y,z)}}{\partial{\bar{z}}} \\
\end{pmatrix} =
\begin{pmatrix}
\cos\phi & \sin\phi \\
-\sin\phi & \cos\phi \\
\end{pmatrix}
\begin{pmatrix}
\frac{\partial{f(y,z)}}{\partial{y}} \\
\frac{\partial{f(y,z)}}{\partial{z}} \\
\end{pmatrix}
\)
Del is a vector operator that acts upon T.
\(\nabla = \hat{x}\frac{\partial{}}{\partial{x}} + \hat{y}\frac{\partial{}}{\partial{y}} +\hat{z}\frac{\partial{}}{\partial{z}}\)
Definition of divergence: \(\nabla \cdot \boldsymbol{v} = (\hat{x}\frac{\partial{}}{\partial{x}} + \hat{y}\frac{\partial{}}{\partial{y}} +\hat{z}\frac{\partial{}}{\partial{z}}) \cdot (v_x\hat{x} + v_y\hat{y} + v_z\hat{z}) = \frac{\partial{v_x}}{\partial{x}} + \frac{\partial{v_y}}{\partial{y}} + \frac{\partial{v_z}}{\partial{z}}\) (scalar)
Geometric definition: How much vector sperads out from the point in question. \(V\) is not a constant vector, it is a vector function, and thus value of divergence will also be a function, whose value can differ for each point in space.
Expand \(\hat{r}\) and \(r^2\) in terms of x, y and z. Then find divergence of the resulting function.
Analogous to problem 1.14, by transformation of coordinates. Refer to Section 1.5.
We have a vector function \(v(y, z)\) transformed into
\( \begin{pmatrix}
\bar{v_y} \\
\bar{v_z} \\
\end{pmatrix} =
\begin{pmatrix}
\cos\phi & \sin\phi \\
-\sin\phi & \cos\phi \\
\end{pmatrix} \begin{pmatrix}
v_y \\
v_z \\
\end{pmatrix}
\).Find partial derivatives with respect to \(\bar{y}\) and \(\bar{z}\).
You will use chain rule in a similar fashion as in problem 1.14.
To find the necessary partial derivatives you will first need to rearrange the system of equations analogous to 1.14.
In short, you mush show that \(\frac{\partial{\bar{v_y}}}{\partial{\bar{y}}} + \frac{\partial{\bar{v_z}}}{\partial{\bar{z}}} =
\frac{\partial{v_y}}{\partial{y}} + \frac{\partial{v_z}}{\partial{z}} \)
Definition of curl:
\(\nabla \times \boldsymbol{v} =
\begin{vmatrix}
\hat{\boldsymbol{\imath}} & \hat{\boldsymbol{\jmath}} & \hat{\boldsymbol{k}} \\
\frac{\partial{}}{\partial{x}} & \frac{\partial{}}{\partial{y}} & \frac{\partial{}}{\partial{z}} \\
v_x & v_y & v_z
\end{vmatrix} =
\boldsymbol{\hat{x}}(\frac{\partial{v_z}}{\partial{y}} - \frac{\partial{v_y}}{\partial{z}})
- \boldsymbol{\hat{y}}(\frac{\partial{v_z}}{\partial{x}} - \frac{\partial{v_x}}{\partial{z}})
+ \boldsymbol{\hat{z}}(\frac{\partial{v_y}}{\partial{x}} - \frac{\partial{v_x}}{\partial{y}}) \)
If you are uncomfortable with the notion of curl, I highly recommend heading over to Paul's Notes for more information.
Geometric definition: How much vector swirls around a given point. If we were to model a hurricane, the eye of a hurricane would be the point of maximum curl.
Simple rules for gradients, divergence and curl:
Product rules:
Find \((\boldsymbol{A} \cdot \boldsymbol{\nabla})\boldsymbol{B}\)
Find \((\boldsymbol{\hat{r}} \cdot \boldsymbol{\nabla})\boldsymbol{\hat{r}}\)
Second derivatives available:
Only two kinds of special second derivatives: the Laplacian and gradient of divergence.