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\[ \qquad \; \; \text{velocity} \qquad = \qquad {d{\bf x} \over dt} \qquad = \qquad \left( {dx_1 \over dt} , {dx_2 \over dt} , {dx_3 \over dt} \right) \qquad \ = \qquad \dot{\bf x} \qquad = \qquad \dot{x}_i \qquad = \qquad x_{i,t} \]

\[ \text{acceleration} \qquad = \qquad {d{\bf v} \over dt} \qquad = \qquad \left( {dv_1 \over dt} , {dv_2 \over dt} , {dv_3 \over dt} \right) \qquad \ = \qquad \dot{\bf v} \qquad = \qquad \dot{v}_i \qquad = \qquad v_{i,t} \]

\[ \text{acceleration} \qquad = \qquad {d^2{\bf x} \over dt^2} \qquad = \qquad \left( {d^2x_1 \over dt^2} , {d^2x_2 \over dt^2} , {d^2x_3 \over dt^2} \right) \qquad \ = \qquad \ddot{\bf x} \qquad = \qquad \ddot{x}_i \qquad = \qquad x_{i,tt} \]

One can use the derivative with respect to \(\;t\), or the dot, which is probably the most popular, or the comma notation, which is a popular subset of tensor notation. Note that the notation \(x_{i,tt}\) somewhat violates the tensor notation rule of double-indices automatically summing from 1 to 3. This is because time does not have 3 dimensions as space does, so it is understood that no summation is performed.

\[ \dot{\bf v} = (10t, \cos t, 3e^{3t}) \]

\[ \dot{\bf x} = {\bf v} = (-2 \sin t, 2 \cos t, 5) \]

and the acceleration is

\[ \ddot{\bf x} = \dot{\bf v} = {\bf a} = (-2 \cos t, -2 \sin t, 0) \]

which always points toward the center of the pipe.

\[ {\partial f \over \partial y} \quad \qquad \text{or} \qquad \quad {\partial f \over \partial x_2} \quad \qquad \text{or} \qquad \quad f_{,2} \]

where the comma is common tensor notation for a derivative.

In the more general case, differentiation with respect to \(x_j\) is (yes, this is a gradient)

\[ {\partial f \over \partial x_j} \qquad \qquad \text{or} \qquad \qquad f,_j \]

Differentiation of a vector, \({\bf v}\), is

\[ {\partial {\bf v} \over \partial x_j} \qquad \qquad \text{or} \qquad \qquad \left( {\partial \, v_x \over \partial x_j} , {\partial \, v_y \over \partial x_j} , {\partial \, v_z \over \partial x_j} \right) \qquad \qquad \text{or} \qquad \qquad v_{i,j} \]

Differentiation of a tensor, \(\boldsymbol{\sigma}\), is

\[ {\partial \boldsymbol{\sigma} \over \partial x_k} \qquad \qquad \text{or} \qquad \qquad \sigma_{ij,k} \]

As with vectors, every component of a tensor is differentiated.

\[ {\partial {\bf v} \over \partial y} = (-2, 0, 3y^2) \]

\[ \nabla f({\bf x}) = \left( {\partial f \over \partial x_1} , {\partial f \over \partial x_2} , {\partial f \over \partial x_3} \right) \]

\[ \nabla f({\bf x}) = ( 6x, -2 z^2, -4 y z ) \]

Coincidentally, the gradient also gives the direction, or orientation, in space that corresponds to the greatest rate of increase. The following example, in 2D space, demonstrates this.

\[ \nabla f(x,y) \quad = \quad ( 4x, 2y ) \quad = \quad (20, 6) \]

Therefore, at \( {\bf x} = (5,3) \), \(f\) is increasing at the rate of 20 along the \(x\) axis, and at the rate of 6 along the \(y\) axis. \( 20{\bf i} + 6 {\bf j} \) also corresponds to the direction in the \(x,y\) plane along which \(f\) will increase the most quickly.

\[ \begin{eqnarray} \nabla \cdot {\bf v} & = & ( {\partial \over \partial x} {\bf i} + {\partial \over \partial y} {\bf j} + {\partial \over \partial z} {\bf k} ) \cdot( v_x {\bf i} + v_y {\bf j} + v_z {\bf k} ) \\ \\ & = & {\partial v_x \over \partial x} + {\partial v_y \over \partial y} + {\partial v_z \over \partial z} \end{eqnarray} \]

\[ \nabla \cdot {\bf v} \quad = \quad {\partial \over \partial x}(3x^2 - 2y) + {\partial \over \partial y}(z^2 + x) + {\partial \over \partial z}(y^3 - z) \quad = \quad 6x - 1 \]

Curls are calculated as follows.

\[ \begin{eqnarray} \nabla \times {\bf v} & = & ( {\partial \over \partial x} {\bf i} + {\partial \over \partial y} {\bf j} + {\partial \over \partial z} {\bf k} ) \times ( v_x {\bf i} + v_y {\bf j} + v_z {\bf k} ) \\ \\ & = & \left| \matrix { {\bf i\;} & {\bf j\;} & {\bf k\;} \\ {\partial \over \partial x} & {\partial \over \partial y} & {\partial \over \partial z} \\ v_x & v_y & v_z } \right| \\ \\ & = & ({\partial \, v_z \over \partial y} - {\partial \, v_y \over \partial z}) {\bf i} + ({\partial \, v_x \over \partial z} - {\partial \, v_z \over \partial x}) {\bf j} + ({\partial \, v_y \over \partial x} - {\partial \, v_x \over \partial y}) {\bf k} \end{eqnarray} \]

An easy way to get the tensor notation right is to think of \( \nabla \times {\bf v} \) as \( \epsilon_{ijk} \nabla_j v_k \) and note the order of the subscripts. Of course, this reduces to the correct result: \( \epsilon_{ijk} v_{k,j} \).

As with cross products, the fact that \(j\) and \(k\) both occur twice in \( \epsilon_{ijk} v_{k,j} \) dictates that both are automatically summed from 1 to 3. The term expands to

\[ \matrix { \epsilon_{ijk} v_{k,j} & = & \epsilon_{i11} v_{1,1} & + & \epsilon_{i12} v_{2,1} & + & \epsilon_{i13} v_{3,1} & + & \\ & & \epsilon_{i21} v_{1,2} & + & \epsilon_{i22} v_{2,2} & + & \epsilon_{i23} v_{3,2} & + & \\ & & \epsilon_{i31} v_{1,3} & + & \epsilon_{i32} v_{2,3} & + & \epsilon_{i33} v_{3,3} } \]

\[ \matrix { \epsilon_{2jk} v_{k,j} & = & \epsilon_{211} v_{1,1} & + & \epsilon_{212} v_{2,1} & + & \epsilon_{213} v_{3,1} & + & \\ & & \epsilon_{221} v_{1,2} & + & \epsilon_{222} v_{2,2} & + & \epsilon_{223} v_{3,2} & + & \\ & & \epsilon_{231} v_{1,3} & + & \epsilon_{232} v_{2,3} & + & \epsilon_{233} v_{3,3} } \]

All subscripts are now specified, and this permits evaluation of all alternating tensors. All of them will equal zero except two, leaving

\[ \epsilon_{2jk} v_{k,j} \; = \; v_{1,3} - v_{3,1} \; = \; \frac{ \partial \, v_x}{\partial z} - \frac{\partial \, v_z}{\partial x} \]

which is again consistent with the determinant result (as it must be). Results for the x

\[ \begin{eqnarray} x & = & X \cos (\omega \, t) - Y \sin (\omega \, t) \\ y & = & X \sin (\omega \, t) + Y \cos (\omega \, t) \\ z & = & Z \end{eqnarray} \]

where \({\bf X}\) is the vector of original coordinates of each point at \(t = 0\), and \({\bf x}\) is the vector of that point's coordinates at any other time, \(t\).

Note that this is common in Continuum Mechanics to use \({\bf X}\) as the position vector at \(t = 0\), the so-called

The velocity vector is

\[ {\bf v} = {\partial {\bf x} \over \partial t} = \left( \begin{eqnarray} & - \omega X \sin (\omega \, t) - \omega Y \cos (\omega \, t) \\ & + \omega X \cos (\omega \, t) - \omega Y \sin (\omega \, t) \\ & 0 \end{eqnarray} \right) \]

which simplifies to

\[ {\bf v} = ( - \omega \, y, \omega \, x, 0 ) \]

making the curl of the velocity vector relatively simple to compute.

\[ \nabla \times {\bf v} = ( 0, 0, 2 \omega ) \]

As stated above, the curl is related to rotations. It turns out that \( \nabla \times {\bf v} \) gives the axis of rotation, and \( \frac{1}{2} | \nabla \times {\bf v} | \) is the rotational rate. So \( \frac{1}{2} ( \nabla \times {\bf v} ) \) gives

\[ \frac{1}{2} ( \nabla \times {\bf v} ) = ( 0, 0, \omega ) \]

\[ \nabla^2 \! f({\bf x}) \equiv {\partial^{\,2} \! f({\bf x}) \over \partial \, x^2} + {\partial^{\,2} \! f({\bf x}) \over \partial \, y^2} + {\partial^{\,2} \! f({\bf x}) \over \partial \, z^2} \]

Start by calculating the gradient of \(f({\bf x})\).

\[ \nabla f({\bf x}) = \left( 6 x^2 y, 2 x^3 - z \cos(y), - \sin(y) \right) \]

And the divergence of the gradient (which is the Laplacian after all) is

\[ \nabla^2 \! f({\bf x}) = \nabla \cdot \nabla f({\bf x}) = 12 x y + z \sin(y) \]

\[ {d \over dt} ( {\bf a} \cdot {\bf b} ) = {d \, {\bf a} \over dt} \cdot {\bf b} + {\bf a} \cdot {d \, {\bf b} \over dt} \]

while the derivative of a cross product is

\[ {d \over dt} ( {\bf a} \times {\bf b} ) = {d \, {\bf a} \over dt} \times {\bf b} + {\bf a} \times {d \, {\bf b} \over dt} \]

and the derivative of a diadic product is

\[ {d \over dt} ( {\bf a} \otimes {\bf b} ) = {d \, {\bf a} \over dt} \otimes {\bf b} + {\bf a} \otimes {d \, {\bf b} \over dt} \]

\[ {\bf a} \cdot {\bf b} = 5 t^3 + \sin^2 t + 6t e^t \]

and the derivative is

\[ {d \over dt} ( {\bf a} \cdot {\bf b} ) = 15 t^2 + 2 \sin t \cos t + 6 (t + 1) e^t \]

Applying the differentiation product rule gives the same result.

\[ \begin{eqnarray} {d \, {\bf a} \over dt} \cdot {\bf b} + {\bf a} \cdot {d \, {\bf b} \over dt} & = & (5, \cos t, e^t) \cdot (t^2, \sin t, 6t) + (5t, \sin t, e^t) \cdot (2t, \cos t, 6)\\ \\ & = & 15 t^2 + 2 \sin t \cos t + 6 (t + 1) e^t \end{eqnarray} \]

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