Homework #3 Solutions
- Figure out the transformation matrix that corresponds to:
(a) all three transformed axes are the same angle from the Z axis, and
(b) the new x' axis is directly above the X axis (no Y component).
The z-components satisfy \(z_1^2 + z_2^2 + z_3^3 = 1\). But since
\(z_1 = z_2 = z_3\), this becomes \(3 z^2 = 1\). So \(z_1 = z_2 = z_3 = \sqrt{1/3} = 0.5774\).
The \({\bf i'}\) vector has no y-component, so \(x_1^2 + z_1^2 = 1\). This gives
\(x_1 = \sqrt{2/3} = 0.8165\).
For \({\bf j'}\), \(x_2^2 + y_2^2 + z_2^2 = 1\). And since \({\bf k'}\)
can be written in terms of \({\bf j'}\) components as \((x_2, -y_2, z_2)\).
So \({\bf j'} \cdot {\bf k'} = x_2^2 - y_2^2 + z_2^2 = 0\).
Subtracting the two equations gives \(2 y_2^2 = 1\), so \(y_2 = \sqrt{1/2} = 0.7071\),
and that leaves \(x_2 = \sqrt{1/6} = 0.4082\). So
\[
{\bf Q} =
\left[ \matrix{
\;\;\;0.8165 & \;\;\;0.0000 & 0.5774 \\
-0.4082 & \;\;\;0.7071 & 0.5774 \\
-0.4082 & -0.7071 & 0.5774
} \right]
\]
-
What set of Roe angles will produce this transformation?
\[
\psi = \text{Tan}^{-1}(q_{32}/q_{31}) = \text{Tan}^{-1}(-0.7071 / -0.4082) = 240^\circ
\]
\[
\theta = \text{Cos}^{-1}(q_{33}) = \text{Cos}^{-1}(0.5774) = 54.73^\circ
\]
\[
\phi = \text{Tan}^{-1}(q_{23}/-q_{13}) = \text{Tan}^{-1}(0.5774 / -0.5774) = 135^\circ
\]
-
What single rotation axis and angle will produce the transformation?
\[
\alpha = \text{Cos}^{-1} \left\{ {1\over 2} \Big[ \text{tr}({\bf Q}) - 1 \Big] \right\} = 56.6^\circ
\]
\[
p_1 = { q_{23} - q_{32} \over 2 \sin \alpha } = {0.5774 - (-0.7071) \over 2 \sin(56.6^\circ)} = 0.7693
\]
\[
p_2 = { q_{31} - q_{13} \over 2 \sin \alpha } = {-0.4082 - 0.5774 \over 2 \sin(56.6^\circ)} = -0.5903
\]
\[
p_3 = { q_{12} - q_{21} \over 2 \sin \alpha } = {0.0000 - (-0.4082) \over 2 \sin(56.6^\circ)} = 0.2445
\]
-
A 2-D problem:
Given \( {\bf v} = (5,9) \) and
\({\bf A} = \left[
\matrix {
5 & 2 \\
2 & 3
} \right]
\),
apply a 180° coordinate rotation to both and show that the signs
of all the components change on one but not the other. Any insight
on this?
After 180° rotation, \( {\bf v'} = (-5,-9) \) and
\({\bf A'} = \left[
\matrix {
5 & 2 \\
2 & 3
} \right]
\).
So \({\bf v'} = -{\bf v}\) and \({\bf A'} = {\bf A}\).
-
Show that
\(
\nabla \cdot \nabla || {\bf x} || = 2 / || {\bf x} ||
\)
\[
\begin{eqnarray}
\nabla \cdot \nabla || {\bf x} || & = & \nabla^2 || {\bf x} ||
\\
\\
& = & \left[ (x_i x_i)^{1 \over 2} \right]_{,jj}
\\
\\
& = & \left[ {1 \over 2 } (x_i x_i)^{-{1 \over 2}} (x_i x_{i,j} + x_{i,j} x_i) \right]_{,j}
\\
\\
& = & \left[ {1 \over 2 } (x_i x_i)^{-{1 \over 2}} (x_k x_{k,j} + x_{k,j} x_k) \right]_{,j}
\\
\\
& = & \left[ {1 \over 2 } (x_i x_i)^{-{1 \over 2}} (x_k \delta_{kj} + \delta_{kj} x_k) \right]_{,j}
\\
\\
& = & \left[ {1 \over 2 } (x_i x_i)^{-{1 \over 2}} (x_j + x_j) \right]_{,j}
\\
\\
& = & \left[ (x_i x_i)^{-{1 \over 2}} x_j \right]_{,j}
\\
\\
& = & (x_i x_i)^{-{1 \over 2}} x_{j,j} - {1 \over 2 } x_j (x_i x_i)^{-{3 \over 2}} (x_i x_{i,j} + x_{i,j} x_i)
\\
\\
& = & (x_i x_i)^{-{1 \over 2}} x_{j,j} - {1 \over 2 } x_j (x_i x_i)^{-{3 \over 2}} (x_k x_{k,j} + x_{k,j} x_k)
\\
\\
& = & (x_i x_i)^{-{1 \over 2}} \delta_{jj} - {1 \over 2 } x_j (x_i x_i)^{-{3 \over 2}} (x_k \delta_{kj} + \delta_{kj} x_k)
\\
\\
& = & 3 (x_i x_i)^{-{1 \over 2}} - {1 \over 2 } x_j (x_i x_i)^{-{3 \over 2}} (2 x_j )
\\
\\
& = & 3 (x_i x_i)^{-{1 \over 2}} - x_j (x_i x_i)^{-{3 \over 2}} x_j
\\
\\
& = & 3 (x_i x_i)^{-{1 \over 2}} - (x_i x_i)^{-{3 \over 2}} (x_j x_j)
\\
\\
& = & 3 (x_i x_i)^{-{1 \over 2}} - (x_i x_i)^{-{1 \over 2}}
\\
\\
& = & 2 (x_i x_i)^{-{1 \over 2}}
\\
\\
& = & 2 / || {\bf x} ||
\end{eqnarray}
\]