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User:Egm6321.f09.team4/HW6

= Homework #6: Team 4 =

Problem #1:
Lecture 31-1:

 * Given :

 * A circular cylindrical coordinate system (see lecture slide for graphic):

 * $$ \begin{align}

\xi_{1} &= r \\ \xi_{2} &= \theta \\ \xi_{3} &= z \end{align} $$

 * This system is related to cartesian coordinates as follows:

 * $$ \begin{align}

x &= r \text{ cos}\left(\theta\right) &= \xi_{1} \text{ cos}\left(\xi_{2}\right) \\ y &= r \text{ sin}\left(\theta\right) &= \xi_{1} \text{ sin}\left(\xi_{2}\right) \\ z &= z                               &= \xi_{3} \end{align} $$

 * Find :

  Obtain $$\left\{\text{d}x_{1},\text{d}x_{2},\text{d}x_{3}\right\}$$ in terms of     $$\left\{\xi_{1},\xi_{2},\xi_{3}\right\}$$.   Find $$\text{ d}s^{2} = \sum_{i}\left(\text{d}x_{i}\right)^{2} = \sum_{k}\left(h_{k}\right)^{2}\left(\text{d}\xi_{k}\right)^{2}$$. Identify $$\left\{h_{i}\right\}$$ in terms of $$\left\{\xi_{j}\right\}$$.   Find Laplacian in this coordinate system.

 * Solution :

 * First we solve for the components of $$\left\{\text{d}x_{1},\text{d}x_{2},\text{d}x_{3}\right\}$$ individually:

 * $$\text{d}x_1$$ is a projection of changes in $$\xi_1$$ and $$\xi_2$$ onto the x axis. Changes in $$\xi_3$$ have no effect on $$\text{d}x_1$$.

 * $$\text{d}x_1=\text{cos }\xi_2 \text{ d}\xi_1 + \lim_{\text{d}\xi_2 \to 0}\left [ \xi_1 \text{cos}\left ( \xi_2 + \text{d}\xi_2 \right )-\xi_1 \text{cos }\xi_2 \right ]$$

 * $$\text{d}x_1=\text{cos }\xi_2 \text{ d}\xi_1 + \frac{\text{d}}{\text{d}\xi_2}\left (\xi_1\text{cos }\xi_2 \right )$$

 * $$\text{d}x_1=\text{cos }\xi_2 \text{ d}\xi_1 - \xi_1\text{sin }\xi_2 \text{ d}\xi_2$$

 * Similarly, for $$\text{d}x_2$$ we project onto the y axis instead of the x axis:

 * $$\text{d}x_2=\text{sin }\xi_2 \text{ d}\xi_1 + \lim_{\text{d}\xi_2\to 0}\left [ \xi_1 \text{sin}\left ( \xi_2+\text{d}\xi_2 \right )-\xi_1\text{sin }\xi_2 \right ]$$

 * $$\text{d}x_2=\text{sin }\xi_2 \text{ d}\xi_1 + \xi_1\text{cos }\xi_2\text{ d}\xi_2$$

 * For $$\text{d}x_3$$, only changes in $$\xi_3$$ have an influence, since changes in $$\xi_1$$ and $$\xi_2$$ are constrained to a single plane perpendicular to z:

 * $$\text{d}x_3=\text{d}\xi_3$$

 * Therefore $$\left\{\text{d}x_{1},\text{d}x_{2},\text{d}x_{3}\right\}=\left\{(\text{cos }\xi_2 \text{ d}\xi_1 - \xi_1\text{sin }\xi_2 \text{ d}\xi_2),(\text{sin }\xi_2 \text{ d}\xi_1 + \xi_1\text{cos }\xi_2\text{ d}\xi_2),\text{d}\xi_3 \right\}$$

 * Next we solve for $$\text{ d}s^{2} = \sum_{i}\left(\text{d}x_{i}\right)^{2}$$ using the solution for $$\left\{\text{d}x_{1},\text{d}x_{2},\text{d}x_{3}\right\}$$above:

 * $$\text{ d}s^{2} = \sum_{i}\left(\text{d}x_{i}\right)^{2}=\left ( \text{cos }\xi_2\text{ d}\xi_1-\xi_1\text{sin }\xi_2 \text{d}\xi_2 \right )^2+\left ( \text{sin }\xi_2\text{ d}\xi_1+\xi_1\text{cos }\xi_2 \text{ d}\xi_2 \right )^2+\left ( \text{ d}\xi_3 \right )^2$$

 * $$=\text{cos}^2\xi_2\left (\text{ d}\xi_1 \right )^2-\cancel{2\xi_1 \text{cos }\xi_2 \text{ sin }\xi_2 \text{ d}\xi_1 \text{d}\xi_2}+\xi_1^2\text{sin}^2\xi_2 (\text{d}\xi_2)^2+\text{sin}^2\xi_2(\text{d}\xi_1)^2+\cancel{2\xi_1 \text{cos }\xi_2 \text{ sin }\xi_2 \text{ d}\xi_1 \text{d}\xi_2}+\xi_1^2\text{cos}^2\xi_2(\text{d}\xi_2)^2+(\text{d}\xi_3)^2$$

 * $$=\text{cos}^2\xi_2\left (\text{ d}\xi_1 \right )^2+\xi_1^2\text{sin}^2\xi_2 (\text{d}\xi_2)^2+\text{sin}^2\xi_2(\text{d}\xi_1)^2+\xi_1^2\text{cos}^2\xi_2(\text{d}\xi_2)^2+(\text{d}\xi_3)^2$$

 * $$=(\text{cos}^2\xi_2+\text{sin}^2\xi_2)\left (\text{ d}\xi_1 \right )^2+(\xi_1^2\text{sin}^2\xi_2 +\xi_1^2\text{cos}^2\xi_2)(\text{d}\xi_2)^2+(\text{d}\xi_3)^2$$

 * Using the trigonometric relation $$\text{sin}^2+\text{cos}^2=1$$, we obtain:

 * $$\text{ d}s^{2} =\left (\text{ d}\xi_1 \right )^2+\xi_1^2(\text{d}\xi_2)^2+(\text{d}\xi_3)^2$$

 * Where:

 * $$h_1=1$$

 * $$h_2=\xi_1$$

 * $$h_3=1$$

 * Next we find the Laplacian in these coordinates for a function $$\Psi$$:

 * $$\Delta \Psi =\frac{1}{h_1 h_2 h_3}\sum_{i=1}^{3}\frac{\partial}{\partial \xi_i}\left [ \frac{h_1 h_2 h_3}{(h_i)^2}\frac{\partial \Psi}{\partial \xi_i} \right ]$$

 * For i=1 we get:

 * $$\frac{1}{\xi_1}\frac{\partial}{\partial \xi_1}\left [ \xi_1\frac{\partial \Psi}{\partial \xi_1} \right ]$$

 * For i=2 we get:

 * $$\frac{1}{\xi_1}\frac{\partial}{\partial \xi_2}\left [ \frac{1}{\xi_1}\frac{\partial \Psi}{\partial \xi_2} \right ]$$

 * For i=3 we get:

 * $$\frac{1}{\xi_1}\frac{\partial}{\partial \xi_3}\left [ \xi_1 \frac{\partial \Psi}{\partial \xi_3} \right ]$$

 * Therefore the overall Laplacian is as follows:

 * $$\Delta \Psi =\frac{1}{\xi_1}\frac{\partial}{\partial \xi_1}\left [ \xi_1\frac{\partial \Psi}{\partial \xi_1} \right ]+\frac{1}{\xi_1}\frac{\partial}{\partial \xi_2}\left [ \frac{1}{\xi_1}\frac{\partial \Psi}{\partial \xi_2} \right ]+\frac{1}{\xi_1}\frac{\partial}{\partial \xi_3}\left [ \xi_1 \frac{\partial \Psi}{\partial \xi_3} \right ]$$

Contributors:

Egm6321.f09.team4.palubin.d 17:52, 16 November 2009 (UTC)

Problem #2:
Lecture 31-1:

 * Given :

 * A spherical coordinate system (see Lecture 29-3):

 * $$ \begin{align}

\xi_{1} &= r \\ \xi_{2} &= \phi \\ \xi_{3} &= \theta \end{align} $$

 * This system is related to cartesian coordinates as follows:

 * $$ \begin{align}

x &= r \text{ cos}\left(\theta\right) \text{ cos}\left(\phi\right) &= \xi_{1} \text{ cos}\left(\xi_{3}\right) \text{ cos}\left(\xi_{2}\right) \\ y &= r \text{ cos}\left(\theta\right) \text{ sin}\left(\phi\right) &= \xi_{1} \text{ cos}\left(\xi_{3}\right) \text{ sin}\left(\xi_{2}\right)\\ z &= r \text{ sin}\left(\theta\right) &= \xi_{1} \text{ sin}\left(\xi_{3}\right) \end{align} $$

 * Find :

  Obtain $$\left\{\text{d}x_{1},\text{d}x_{2},\text{d}x_{3}\right\}$$ in terms of      $$\left\{\xi_{1},\xi_{2},\xi_{3}\right\}$$.

 Find $$\text{ d}s^{2} = \sum_{i}\left(\text{d}x_{i}\right)^{2} = \sum_{k}\left(h_{k}\right)^{2}\left(\text{d}\xi_{k}\right)^{2}$$. Identify $$\left\{h_{i}\right\}$$ in terms of $$\left\{\xi_{j}\right\}$$.

 Find Laplacian in this coordinate system.  </ol></ul>

 * Solution :

 * First we solve for the the individual components of $$\left\{\text{d}x_{1},\text{d}x_{2},\text{d}x_{3}\right\}$$ in terms of $$\left\{\xi_{1},\xi_{2},\xi_{3}\right\}$$

 * To start, we assume that $$\text{d}x_1$$ is equal to the projection of an infinitesimal distance s onto the x-axis in spherical coordinates:

 * $$\text{d}x_1=\text{d}\xi_1\text{cos }\xi_2 \text{cos }\xi_3+ \text{cos }\xi_2\lim_{\text{d}\xi_3\to 0}\left [ r\text{cos}(\xi_3+\text{d}\xi_3)-r\text{cos }\xi_3 \right ]+\text{cos }\xi_3\lim_{\text{d}\xi_2\to 0}\left [ r\text{cos}(\xi_2+\text{d}\xi_2)-r\text{cos }\xi_2 \right ]$$

 * $$=\text{cos }\xi_2 \text{cos }\xi_3\text{ d}\xi_1- r\text{cos }\xi_2\text{sin }\xi_3\text{ d}\xi_3-r\text{cos }\xi_3 \text{sin }\xi_2 \text{ d}\xi_2$$

 * Similarly, for $$\text{d}x_2$$ as projected on the y-axis:

 * $$\text{d}x_2=\text{d}\xi_1\text{sin }\xi_2 \text{cos }\xi_3+ \text{sin }\xi_2\lim_{\text{d}\xi_3\to 0}\left [ r\text{cos}(\xi_3+\text{d}\xi_3)-r\text{cos }\xi_3 \right ]+\text{cos }\xi_3\lim_{\text{d}\xi_2\to 0}\left [ r\text{sin}(\xi_2+\text{d}\xi_2)-r\text{sin }\xi_2 \right ]$$

 * $$=\text{sin }\xi_2 \text{cos }\xi_3\text{ d}\xi_1- r\text{sin }\xi_2\text{sin }\xi_3\text{ d}\xi_3-r\text{cos }\xi_3 \text{cos }\xi_2 \text{ d}\xi_2$$

 * For $$\text{d}x_3$$, the differentials are projected onto the z-axis, yielding:

 * $$\text{d}x_3=\text{d}\xi_1\text{sin }\xi_3+\lim_{\text{d}\xi_3\to 0}\left [ r\text{sin }(\xi_3 + \text{d}\xi_3)-r\text{sin }\xi_3\right ]$$

 * $$=\text{sin }\xi_3\text{ d}\xi_1+r\text{cos }\xi_3\text{ d}\xi_3$$

 * Therefore:

 * $$\left\{\text{d}x_{1},\text{d}x_{2},\text{d}x_{3}\right\}=\left\{(\text{cos }\xi_2 \text{cos }\xi_3\text{ d}\xi_1- r\text{cos }\xi_2\text{sin }\xi_3\text{ d}\xi_3-r\text{cos }\xi_3 \text{sin }\xi_2 \text{ d}\xi_2),(\text{sin }\xi_2 \text{cos }\xi_3\text{ d}\xi_1- r\text{sin }\xi_2\text{sin }\xi_3\text{ d}\xi_3-r\text{cos }\xi_3 \text{cos }\xi_2 \text{ d}\xi_2),(\text{sin }\xi_3\text{ d}\xi_1+r\text{cos }\xi_3\text{ d}\xi_3)\right\}$$

 * Next we solve for $$\text{ d}s^{2}$$:

 * $$\text{ d}s^{2} = \sum_{i}\left(\text{d}x_{i}\right)^{2}=(\text{cos }\xi_2 \text{cos }\xi_3\text{ d}\xi_1- r\text{cos }\xi_2\text{sin }\xi_3\text{ d}\xi_3-r\text{cos }\xi_3 \text{sin }\xi_2 \text{ d}\xi_2)^2+(\text{sin }\xi_2 \text{cos }\xi_3\text{ d}\xi_1- r\text{sin }\xi_2\text{sin }\xi_3\text{ d}\xi_3-r\text{cos }\xi_3 \text{cos }\xi_2 \text{ d}\xi_2)^2+(\text{sin }\xi_3\text{ d}\xi_1+r\text{cos }\xi_3\text{ d}\xi_3)^2$$

 * $$=\text{cos}^2\xi_3\text{cos}^2\xi_2(dr)^2+\xi_1^2\text{cos}^2\xi_2\text{sin}^2\xi_3(d\xi_3)^2+\xi_1^2\text{cos}^2\xi_3\text{sin}^2\xi_2(d\xi_2)^2-2\xi_1\text{cos}^2\xi_2\text{cos}\xi_3\text{sin}\xi_3 (d\xi_1d\xi_3)-\cancel{2\xi_1\text{cos}^2\xi_3\text{sin}\xi_2\text{cos}\xi_2(d\xi_1d\xi_2)}+\cancel{2\xi_1^2\text{cos}\xi_2\text{sin}\xi_3\text{cos}\xi_3\text{sin}\xi_2(d\xi_3d\xi_2)}$$

 * $$+\text{sin}^2\xi_2\text{cos}^2\xi_3(d\xi_1)^2+\xi_1^2\text{sin}^2\xi_2\text{sin}^2\xi_3(d\xi_3)^2+\xi_1^2\text{cos}^2\xi_2\text{cos}^2\xi_3(d\xi_2)^2-2\xi_1\text{sin}^2\xi_2\text{sin}\xi_3\text{cos}\xi_3(d\xi_3d\xi_1)+\cancel{2\xi_1\text{sin}\xi_2\text{cos}\xi_2\text{cos}^2\xi_3(d\xi_2d\xi_1)} -\cancel{2\xi_1^2\text{sin}\xi_2\text{sin}\xi_3\text{cos}\xi_2\text{cos}\xi_3(d\xi_3d\xi_2)}$$

 * $$+\text{sin}^2\xi_3(d\xi_1)^2+\xi_1^2\text{cos}^2\xi_3(d\xi_3)^2+2\xi_1\text{cos}\xi_3\text{sin}\xi_3(d\xi_1d\xi_3)$$

 * $$=[(\text{cos}^2\xi_3\text{cos}^2\xi_2+\text{sin}^2\xi_2\text{cos}^2\xi_3)+\text{sin}^2\xi_3](d\xi_1)^2+[\xi_1^2\text{cos}^2\xi_2\text{sin}^2\xi_3+\xi_1^2\text{sin}^2\xi_2\text{sin}^2\xi_3+\xi_1^2\text{cos}^2\xi_3](d\xi_3)^2$$

 * $$+[\xi_1^2\text{cos}^2\xi_3\text{sin}^2\xi_2+\xi_1^2\text{cos}^2\xi_2\text{cos}^2\xi_3](d\xi_2)^2+[-2\xi_1\text{cos}^2\xi_2\text{cos}\xi_3\text{sin}\xi_3-2\xi_1\text{sin}^2\xi_2\text{sin}\xi_3\cos\xi_3+2\xi_1\cos\xi_3\sin\xi_3](d\xi_1d\xi_3)$$

 * $$=[\text{cos}^2\xi_3(\text{cos}^2\xi_2+\text{sin}^2\xi_2)+\text{sin}^2\xi_3](d\xi_1)^2+\xi_1^2[\text{sin}^2\xi_3(\text{cos}^2\xi_2+\text{sin}^2\xi_2)+\text{cos}^2\xi_3](d\xi_3)^2$$

 * $$+\xi_1^2\text{cos}^2\xi_3[\text{sin}^2\xi_2+\text{cos}^2\xi_2](d\xi_2)^2+2\xi_1\cos\xi_3\sin\xi_3[-\text{cos}^2\xi_2-\text{sin}^2\xi_2+1](d\xi_1d\xi_3)$$

 * Applying the trigonometric relation $$\sin^2 + \cos^2=1$$:

 * $$=[\text{cos}^2\xi_3(1)+\text{sin}^2\xi_3](d\xi_1)^2+\xi_1^2[\text{sin}^2\xi_3(1)+\text{cos}^2\xi_3](d\xi_3)^2+\xi_1^2\text{cos}^2\xi_3[1](d\xi_2)^2+\cancel{2\xi_1\cos\xi_3\sin\xi_3[-1+1](d\xi_1d\xi_3)}$$

 * $$=[1](d\xi_1)^2+\xi_1^2[1](d\xi_3)^2+\xi_1^2\text{cos}^2\xi_3[1](d\xi_2)^2$$

 * Therefore, $$\text{ d}s^{2}=(d\xi_1)^2+\xi_1^2\text{cos}^2\xi_3(d\xi_2)^2+\xi_1^2(d\xi_3)^2$$

 * Where:

 * $$h_1=1$$

 * $$h_2=\xi_1\cos\xi_3$$

 * $$h_3=\xi_1$$

 * The Laplacian for any function $$\Psi$$ in these coordinates can be determined from:

 * $$\Delta \Psi =\frac{1}{h_1 h_2 h_3}\sum_{i=1}^{3}\frac{\partial}{\partial \xi_i}\left [ \frac{h_1 h_2 h_3}{(h_i)^2}\frac{\partial \Psi}{\partial \xi_i} \right ]$$

 * For i=1,

 * $$ \begin{align}

\frac{1}{h_{1} h_{2} h_{3}} \frac{\partial}{\partial \xi_{1}}\left[ \frac{h_{1} h_{2} h_{3}}{\left( h_{1} \right)^{2}}\ \frac{ \partial \Psi }{\partial \xi_{1}} \right] &= \frac{1}{\xi_1^{2}\text{ cos } \xi_3}\ \frac{\partial}{\partial \xi_1}\left[ \frac{\xi_1^{2}\text{ cos } \xi_3}{1^{2}} \frac{\partial \Psi}{\partial \xi_1}\right] \\

&= \frac{1}{\xi_1^{2}\text{ cos } \xi_3}\ \frac{\partial}{\partial \xi_1}\left[ \xi_1^{2}\text{ cos } \xi_3\ \frac{\partial \Psi}{\partial \xi_1}\right] \\

&= \frac{1}{\xi_1^{2}} \frac{\partial}{\partial \xi_1}\left[ \xi_1^2\frac{\partial \Psi}{\partial \xi_1}\right] \end{align} $$

 * For i=2:

 * $$ \begin{align}

\frac{1}{h_{1} h_{2} h_{3}} \frac{\partial}{\partial \xi_{2}}\left[ \frac{h_{1} h_{2} h_{3}}{\left( h_{2} \right)^{2}}\ \frac{ \partial \Psi }{\partial \xi_{2}} \right] &=\frac{1}{\xi_1^2\text{ cos }\xi_3 }\frac{d}{d\xi_2 }\left [ \frac{\xi_1^2 \text{ cos }\xi_3 }{\xi_1^2 \text{ cos}^2 \xi_3 }\cdot \frac{d\Psi }{d\xi_2} \right ] \\

&=\frac{1}{\xi_1^2\text{ cos }\xi_3 }\frac{d}{d\xi_2 }\left [ \frac{1}{\text{ cos }\xi_3} \cdot \frac{d\Psi }{d\xi_2} \right ] \\

&=\frac{1}{\xi_1^2\text{ cos}^2\xi_3 }\frac{d}{d\xi_2 }\left [ \frac{d\Psi }{d\xi_2} \right ] \end{align} $$

 * for i=3:

 * $$ \begin{align}

\frac{1}{h_{1} h_{2} h_{3}} \frac{\partial}{\partial \xi_{3}}\left[ \frac{h_{1} h_{2} h_{3}}{\left( h_{3} \right)^{2}}\ \frac{ \partial \Psi }{\partial \xi_{3}} \right] &= \frac{1}{\xi_1^2 \text{ cos }\xi_3} \frac{\partial}{\partial \xi_3}\left[ \frac{\xi_1^2 \text{ cos }\xi_3}{ \xi_1^{2}}\ \frac{ \partial \Psi }{\partial \xi_3} \right] \\

&=\frac{1}{\xi_1^2 \text{ cos }\xi_3} \frac{\partial}{\partial \xi_3}\left[ \text{ cos }\xi_3 \cdot \frac{ \partial \Psi }{\partial \xi_3} \right] \end{align} $$

 * Therefore the overall Laplacian is as follows:

 * $$\Delta \Psi =\frac{1}{\xi_1^{2}} \frac{\partial}{\partial \xi_1}\left[ \xi_1^2\frac{\partial \Psi}{\partial \xi_1}\right]+\frac{1}{\xi_1^2\text{ cos}^2\xi_3 }\frac{d}{d\xi_2 }\left [ \frac{d\Psi }{d\xi_2} \right ]+\frac{1}{\xi_1^2 \text{ cos }\xi_3} \frac{\partial}{\partial \xi_3}\left[ \text{ cos }\xi_3 \cdot \frac{ \partial \Psi }{\partial \xi_3} \right]$$

Contributors: Egm6321.f09.team4.palubin.d 00:22, 18 November 2009 (UTC)

Problem #3:
Lecture 31-2:

 * Given :

 * Spherical coordinates in standard convention; uses $$\bar{\theta}$$ instead of $$\theta$$ (see Lecture 29-3). The only difference is that $$\bar{\theta}$$ is measured from the positive z-axis instead of the x-y plane.  In other words, $$\bar{\theta} = \frac{\pi}{2}$$ and $$\theta = 0$$ at the equator.

 * This introduces the following changes:

 * $$ \begin{align}

\bar{\theta} &= \frac{\pi}{2} - \theta \\ \\ x &= r \text{ sin}\left(\bar{\theta}\right) \text{ cos}\left(\phi\right) \\ y &= r \text{ sin}\left(\bar{\theta}\right) \text{ sin}\left(\phi\right) \\ z &= r \text{ cos}\left(\bar{\theta}\right) \end{align} $$

 * Find :

 * Obtain an expression for the Laplacian in this coordinate system.

 * Solution :

 * It has been previously determined that for spherical coordinates using $$\theta$$ rather than $$\bar{\theta}$$,

 * $$\text{d}s^2=\text{d}r^2+r^2\cos\theta(\text{d}\phi)^2+r^2(\text{d}\theta)^2$$

 * And since $$\bar{\theta} = \frac{\pi}{2} - \theta $$:

 * $$\text{d}\bar{\theta}=\text{d}(\frac{\pi}{2})-\text{d}\theta=0-\text{d}\theta=-\text{d}\theta$$

 * $$\cos\theta=\sin\bar{\theta}$$

 * Converting to the new coordinates,

 * $$\text{d}s^2=\text{d}r^2+r^2\sin\bar{\theta}(\text{d}\phi)^2+r^2(\text{d}\bar{\theta})^2$$

 * And:

 * $$h_1=1$$

 * $$h_2=r\sin\bar{\theta}$$

 * $$h_3=r$$

 * The Laplacian for any function $$\Psi$$ in these coordinates can be determined from:

 * $$\Delta \Psi =\frac{1}{h_1 h_2 h_3}\sum_{i=1}^{3}\frac{\partial}{\partial \xi_i}\left [ \frac{h_1 h_2 h_3}{(h_i)^2}\frac{\partial \Psi}{\partial \xi_i} \right ]$$

 * For i=1:

 * $$ \begin{align}

\frac{1}{h_{1} h_{2} h_{3}} \frac{\partial}{\partial \xi_{1}}\left[ \frac{h_{1} h_{2} h_{3}}{\left( h_{1} \right)^{2}}\ \frac{ \partial \Psi }{\partial \xi_{1}} \right] &= \frac{1}{r^{2}\text{ sin } \bar{\theta}}\ \frac{\partial}{\partial r}\left[ \frac{r^{2}\text{ sin } \bar{\theta}}{1^{2}} \frac{\partial \Psi}{\partial r}\right] \\

&= \frac{1}{r^{2}\text{ sin } \bar{\theta}}\ \frac{\partial}{\partial r}\left[ r^{2}\text{ sin } \bar{\theta}\ \frac{\partial \Psi}{\partial r}\right] \\

&= \frac{1}{r^{2}} \frac{\partial}{\partial r}\left[ r^2\frac{\partial \Psi}{\partial r}\right] \end{align} $$

 * For i=2:

 * Using a similar procedure as above, we find that:

 * $$ \begin{align}

\frac{1}{h_{1} h_{2} h_{3}} \frac{\partial}{\partial \xi_{2}}\left[ \frac{h_{1} h_{2} h_{3}}{\left( h_{2} \right)^{2}}\ \frac{ \partial \Psi }{\partial \xi_{2}} \right] &=\frac{1}{r^2\text{ sin }\bar{\theta} }\frac{d}{d\phi }\left [ \frac{r^2 \text{ sin }\bar{\theta} }{r^2 \text{ sin}^2 \bar{\theta} }\cdot \frac{d\Psi }{d\phi} \right ] \\

&=\frac{1}{r^2\text{ sin }\bar{\theta} }\frac{d}{d\phi }\left [ \frac{1}{\text{ sin }\bar{\theta}} \cdot \frac{d\Psi }{d\phi} \right ] \\

&=\frac{1}{r^2\text{ sin}^2\bar{\theta} }\frac{d}{d\phi }\left [ \frac{d\Psi }{d\phi} \right ] \end{align} $$

 * For i=3:

 * As above, we find that:

 * $$ \begin{align}

\frac{1}{h_{1} h_{2} h_{3}} \frac{\partial}{\partial \xi_{3}}\left[ \frac{h_{1} h_{2} h_{3}}{\left( h_{3} \right)^{2}}\ \frac{ \partial \Psi }{\partial \xi_{3}} \right] &= \frac{1}{r^2 \text{ sin }\bar{\theta}} \frac{\partial}{\partial \theta}\left[ \frac{r^2 \text{ sin }\bar{\theta}}{ r^{2}}\ \frac{ \partial \Psi }{\partial \bar{\theta}} \right] \\

&=\frac{1}{r^2 \text{ sin }\bar{\theta}} \frac{\partial}{\partial \bar{\theta}}\left[ \text{ sin }\bar{\theta} \cdot \frac{ \partial \Psi }{\partial \bar{\theta}} \right] \end{align} $$

 * The overall Laplacian is then:

 * $$\Delta \Psi =\frac{1}{r^{2}} \frac{\partial}{\partial r}\left[ r^2\frac{\partial \Psi}{\partial r}\right]+\frac{1}{r^2\text{ sin}^2\bar{\theta} }\frac{d}{d\phi }\left [ \frac{d\Psi }{d\phi} \right ]+\frac{1}{r^2 \text{ sin }\bar{\theta}} \frac{\partial}{\partial \bar{\theta}}\left[ \text{ sin }\bar{\theta} \cdot \frac{ \partial \Psi }{\partial \bar{\theta}} \right]$$

Contributors: Egm6321.f09.team4.palubin.d 01:55, 18 November 2009 (UTC)

Problem #4:
Lecture 31-2:

 * Given :

 * $$ \begin{align}

\lambda\left( \lambda + 1 \right) = n\left( n+1\right) \end{align} $$

 * Find :

 * Two solutions for $$\lambda$$ in terms of $$n$$ (use the quadratic equation).

 * Solution :

 * $$\lambda\left( \lambda + 1 \right) = n\left( n+1\right)$$
 * $$\lambda^2+\lambda- n\left( n+1\right)=0$$
 * $$\lambda_{1,2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$
 * $$\lambda_{1,2}=\frac{-1\pm \sqrt{1-4(1)(-n(n+1))}}{2}$$
 * $$\lambda_{1,2}=\frac{-1\pm \sqrt{4n^2+4n+1}}{2}$$
 * $$\lambda_{1,2}=\frac{-1\pm \sqrt{(2n+1)^2}}{2}$$
 * $$\lambda_{1,2}=\frac{-1\pm (2n+1)}{2}$$
 * $$\lambda_{1,2}=-\frac{1}{2}\pm \left (n+\frac{1}{2} \right )$$

 * Therefore,
 * $$\lambda_{1}=n$$
 * $$\lambda_{2}=-n-1=-(n+1)$$

Contributors: Egm6321.f09.Team4.meyerat 00:27, 16 November 2009 (UTC)

Problem #5:
Lecture 31-3:

 * Given :

 * $$ \begin{align}

P_{n}\left(x\right) &= \sum_{i=0}^{\left\lfloor n/2\right\rfloor}\ x^{n-2i} \left(-1\right)^{i}\ \frac{\left(2n-2i\right)!}{2^{n}\ i!\ \left(n-i\right)!\ \left(n-2i\right)!} \\ &\text{where } \left\lfloor n/2\right\rfloor = \ \text{floor of } n/2 \end{align} $$

 * Find :

 * Show that the above expression is equivalent to the following:

 * $$ \begin{align}

P_{n}\left(x\right) = \sum_{i=0}^{\left\lfloor n/2 \right\rfloor}\ x^{n-2i} \left(-1\right)^{i}\ \frac{1 \times 3 \times \dots \times \left(2n-2i-1\right)}{2^{i}\ i!\ \left(n-2i\right)!} \end{align} $$

 * Solution :

 * Setting these two equations equal to each other, we can simplify what we're trying to prove:

 * $$ \begin{align}

\sum_{i=0}^{\left\lfloor n/2\right\rfloor}\ x^{n-2i} \left(-1\right)^{i}\ \frac{\left(2n-2i\right)!}{2^{n}\ i!\ \left(n-i\right)!\ \left(n-2i\right)!} &= \sum_{i=0}^{\left\lfloor n/2 \right\rfloor}\ x^{n-2i} \left(-1\right)^{i}\ \frac{1 \times 3 \times \dots \times \left(2n-2i-1\right)}{2^{i}\ i!\ \left(n-2i\right)!} \\

\cancel{x^{n-2i}} \cancel{\left(-1\right)^{i}}\ \frac{\left(2n-2i\right)!}{2^{n}\ \cancel{i!}\ \left(n-i\right)!\ \cancel{\left(n-2i\right)!}} &= \cancel{x^{n-2i}} \cancel{\left(-1\right)^{i}}\ \frac{1 \times 3 \times \dots \times \left(2n-2i-1\right)}{2^{i}\ \cancel{i!}\ \cancel{\left(n-2i\right)!}} \qquad \forall\ i \\

\frac{\left(2n-2i\right)!}{2^{n}\left(n-i\right)!} &= \frac{1\times 3\times \dots \times \left(2n-2i-1\right)}{2^{i}} \qquad \forall\ i \end{align} $$

 * Expanding the left-hand side of this equation:

 * $$ \begin{align}

\frac{\left(2n-2i\right)!}{2^{n}\left(n-i\right)!} &= \frac{\left(2n-2i\right)!}{2^{n-i}2^{i}\left(n-i\right)!} \\ &= \frac{\left(\prod_{k=n-i+1}^{2n-2i} k\right)\left(\prod_{k=1}^{n-i} k\right) \frac{1}{2^{n-i}}}{2^{i}\left(\prod_{k=1}^{n-i} k \right)} \\ &= \frac{\frac{1}{2^{n-i}} \left(\prod_{k=n-1+1}^{2n-2i} k\right) }{2^{i}} \\ &= \frac{\prod_{k=1}^{n-i} \left(2k-1\right)}{2^{i}} \\ &= \frac{1 \times 3 \times \dots \times \left(2n-2i-1\right)}{2^{i}} \quad \text{ Proof.} \end{align} $$

Contributors:
 * Egm6321.f09.team4.sls 19:26, 16 November 2009 (UTC)

Problem #6:
Lecture 31-3:

 * Given :

 * Either of the equations from Problem #5:

 * $$ \begin{align}

P_{n}\left(x\right) &= \sum_{i=0}^{\left\lfloor n/2\right\rfloor}\ x^{n-2i} \left(-1\right)^{i}\ \frac{\left(2n-2i\right)!}{2^{n}\ i!\ \left(n-i\right)!\ \left(n-2i\right)!} \\ &\text{where } \left\lfloor n/2\right\rfloor = \ \text{floor of } n/2 \end{align} $$

 * OR

 * $$ \begin{align}

P_{n}\left(x\right) = \sum_{i=0}^{\left\lfloor n/2 \right\rfloor}\ x^{n-2i} \left(-1\right)^{i}\ \frac{1,3,\dots,\left(2n-2i-1\right)}{2^{i}\ i!\ \left(n-2i\right)!} \end{align} $$

 * Find :

 * $$P_{i}\left(x\right)$$, for $$i=0,1,2,3,4$$.

 * Solution :

 * $$ \begin{align}

P_{0}\left(x\right) &= \sum_{i=0}^{0}\left(\dots\right) = x^{0}\left(-1\right)^{0}\frac{0!}{2^{0}\left(0!\right)\left(0-0\right)!\left(0-0\right)!} \\ &= 1 \\ \\  P_{1}\left(x\right) &= \sum_{i=0}^{0}\left(\dots\right) = x^{1-2\left(0\right)}\left(-1\right)^{0}\frac{\left(2-0\right)!}{2\left(0\right)!\left(1-0\right)!\left(1-0\right)!} \\ &= x \\ \\ P_{2}\left(x\right) &= \sum_{i=0}^{1}\left(\dots\right) = x^{2-2\left(0\right)}\left(-1\right)^{0}\frac{4!}{2^{2}0!\left(2\right)!\left(2\right)!} + x^{2-2}\left(-1\right)^{1}\frac{\left(4-2\right)!}{2^{2}\left(1\right)!\left(1\right)!\left(0\right)!} \\ &=\frac{3}{2}x^{2}-\frac{1}{2} \\ &=\frac{1}{2}\left(3x^{2}-1\right) \\ \\ P_{3}\left(x\right) &= \sum_{i=0}^{1}\left(\dots\right) = x^{3-2\left(0\right)}\left(-1\right)^{0}\frac{6!}{2^{3}\left(0\right)!\left(3\right)!\left(3\right)!} + x^{3-2}\left(-1\right)^{1}\frac{\left(6-2\right)!}{2^{3}\left(1\right)!\left(2\right)!\left(1\right)!} \\ &= x^{3} \frac{\cancel{6}\cdot 5\cdot \cancel{4}\cdot \cancel{3}\cdot \cancel{2}}{2\cdot \cancel{2}\cdot \cancel{2}\cdot \cancel{3}\cdot \cancel{2}\cdot \cancel{3}\cdot \cancel{2}} - x \frac{\cancel{4}\cdot 3\cdot \cancel{2}}{2\cdot \cancel{2}\cdot \cancel{2}\cdot \cancel{2}} \\ &= x^{3}\frac{5}{2} - x\frac{3}{2} \\ &= \frac{1}{2}\left(5x^{3}-3x\right) \\ \\ P_{4}\left(x\right) &= \sum_{i=0}^{2}\left(\dots\right) = x^{4-2\left(0\right)}\left(-1\right)^{0}\frac{8!}{2^{4}\left(0\right)!\left(4\right)!\left(4\right)!} + x^{4-2}\left(-1\right)^{1}\frac{6!}{2^{4}\left(1\right)!\left(3\right)!\left(2\right)!} + x^{4-4}\left(-1\right)^{2}\frac{4!}{2^{4}\left(2\right)!\left(2\right)!\left(0\right)!} \\ &= x^{4}\frac{\cancel{8}\cdot 7\cdot \cancel{6}\cdot 5\cdot \cancel{4}\cdot \cancel{3}\cdot \cancel{2}\cdot}{\cancel{2}\cdot \cancel{2}\cdot \cancel{2}\cdot 2\cdot 4\cdot \cancel{3}\cdot \cancel{2}\cdot \cancel{4}\cdot \cancel{3}\cdot \cancel{2}\cdot} - x^{2}\frac{6\cdot 5\cdot \cancel{4}\cdot \cancel{3}\cdot \cancel{2}\cdot}{\cancel{2}\cdot 2\cdot 2\cdot 2\cdot \cancel{3}\cdot \cancel{2}\cdot \cancel{2}\cdot} + \frac{\cancel{4}\cdot 3\cdot \cancel{2}}{2\cdot 2\cdot 2\cdot \cancel{2}\cdot \cancel{2}\cdot \cancel{2}} \\ &= \frac{35}{8}x^{4} - \frac{15}{4}x^{2} + \frac{3}{8} \end{align} $$

Contributors:
 * Egm6321.f09.team4.sls 18:19, 16 November 2009 (UTC)

Problem #7:
Lecture 32-1:

 * Given :

 * The Legendre polynomials $$P_{0},\dots,P_{4}$$ verified in Problem #6 (see Lecture 31-3):

 * $$ \begin{align}

P_{0}\left(x\right) &= 1 \\ P_{1}\left(x\right) &= x \\ P_{2}\left(x\right) &= \frac{1}{2}\left(3x^{2}-1\right) \\ P_{3}\left(x\right) &= \frac{1}{2}\left(5x^{3}-3x\right) \\ P_{4}\left(x\right) &= \frac{35}{8}x^{4} - \frac{15}{4}x^{2} + \frac{3}{8} \end{align} $$

 * Find :

 * Show that these polynomials are solutions to the Legendre equation (given below; also see Lectures 14-2 and 30-4).

 * $$ \begin{align}

\left(1-x^{2}\right)y'' - 2xy' + n\left(n+1\right)y = 0 \end{align} $$

 * Note: Legendre polynomial $$P_{k}$$ is a solution to the specific Legendre equation given by $$ n = k $$.

 * Solution :

 * k = 0 :

 * Let $$y = P_{0} = 1$$. This results in the following derivative values:

 * $$ \begin{align}

y' &= 0 \\ y'' &= 0 \end{align} $$

 * Plugging these values, along with $$n=0$$, into the Legendre equation:

 * $$ \begin{align}

\left(1-x^{2}\right)\left(0\right) - 2x\left(0\right) + 0\left(0+1\right)\left(1\right) = 0 \\ 0 = 0 \quad \text{Proof.} \end{align} $$

 * k = 1 :

 * Let $$y = P_{1} = x$$. This results in the following derivative values:

 * $$ \begin{align}

y' &= 1 \\ y'' &= 0 \end{align} $$

 * Plugging these values, along with $$n=1$$, into the Legendre equation:

 * $$ \begin{align}

\left(1-x^{2}\right)\left(0\right) - 2x\left(1\right) + 1\left(1+1\right)x &= 0 \\ -2x + 2x &= 0 \\ 0 &= 0 \quad \text{Proof.} \end{align} $$

 * k = 2 :

 * Let $$\begin{align} y = P_{2} = \frac{1}{2}\left(3x^{2}-1\right) \end{align}$$. This results in the following derivative values:

 * $$ \begin{align}

y' &= 3x \\ y'' &= 3 \end{align} $$

 * Plugging these values, along with $$n=2$$, into the Legendre equation:

 * $$ \begin{align}

\left(1-x^{2}\right)\left(3\right) - 2x\left(3x\right) + \left(6\right)\left(\frac{1}{2}\right)\left(3x^{2}-1\right) &= 0 \\ 3 - 3x^{2} - 6x^{2} + 9x^{2} - 3 &= 0 \\ 3 - 3 - 9x^{2} + 9x^{2} &= 0 \\ 0 &= 0 \quad \text{Proof.} \end{align} $$

 * k = 3 :

 * Let $$\begin{align} y = P_{3} = \frac{1}{2}\left(5x^{3}-3x\right) \end{align}$$. This results in the following derivative values:

 * $$ \begin{align}

y' &= \frac{1}{2}\left(15x^{2} - 3\right) \\ y'' &= \frac{1}{2}\left(30x\right) = 15x \end{align} $$

 * Plugging these values, along with $$n=3$$, into the Legendre equation:

 * $$ \begin{align}

\left(1-x^{2}\right)\left(15x\right) - 2x\left(\frac{1}{2}\right)\left(15x^{2}-3\right) + 12\frac{1}{2}\left(5x^{3}-3x\right) &= 0 \\ 15x - 15x^{3} - 15x^{3} + 3x + 30x^{3} - 18x &= 0 \\ \left(15x + 3x - 18x\right) + \left(30x^{3}-15x^{3}-15x^{3}\right) &= 0 \\ 0 &= 0 \quad \text{Proof.} \end{align} $$

 * k = 4 :

 * Let $$\begin{align} \end{align}$$. This results in the following derivative values:

 * $$ \begin{align}

y' &= \frac{35}{2}x^{3} - \frac{15}{2}x \\ y'' &= \frac{105}{2}x^{2} - \frac{15}{2} \end{align} $$

 * Plugging these values, along with $$n=4$$, into the Legendre equation:

 * $$ \begin{align}

\left(1-x^{2}\right)\left(\frac{105}{2}x^{2}-\frac{15}{2}\right) - 2x\left(\frac{35}{2}x^{3} -\frac{15}{2}x\right) + 20 \left(\frac{35}{8}x^{4}-\frac{15}{4}x^{2}+\frac{3}{8}\right) &= 0 \\ \frac{105}{2}x^{2} - \frac{15}{2} - \frac{105}{2}x^{4} + \frac{15}{2}x^{2} - \frac{70}{2}x^{4} + \frac{30}{2}x^{2} + \frac{175}{2}x^{4} - \frac{150}{2}x^{2} + \frac{15}{2} &= 0 \\ \left(-\frac{105}{2}-\frac{70}{2}+\frac{175}{2}\right)x^{4} + \left(\frac{105}{2}+\frac{15}{2}+\frac{30}{2} - \frac{150}{2}\right)x^{2} + \cancel{\left(\frac{15}{2}-\frac{15}{2}\right)} &= 0 \\ \left(-\frac{175}{2}+\frac{175}{2}\right)x^{4} + \left(\frac{150}{2} - \frac{150}{2}\right)x^{2} &= 0 \\ 0 &= 0 \quad \text{Proof.} \end{align} $$

Contributors:
 * Egm6321.f09.team4.sls 20:26, 16 November 2009 (UTC)

Problem #8:
Lecture 33-3:

 * Given :

 * Functions $$f\left(\right)$$ and $$\left\{g_{i}\left(\right)\right\}$$ such that:

 * $$ \begin{align}

f\left(\right) = \sum_{i}\ g_{i}\left(\right) \end{align} $$

 * Find :

<ul> <li> Show that, if $$\left\{g_{i}\right\}$$ is odd, then $$f$$ is odd. </li> <li> Show that, if $$\left\{g_{i}\right\}$$ is even, then $$f$$ is even. </li> </ol></ul>

 * Solution :

 * Start with the definitions of even and odd functions. If a function is even, than

 * $$f(x)=f(-x)$$

 * If the function is odd, than,

 * $$-f(x)=f(-x)$$

 * So, for our problem,

 * $$f(x)=\sum_{i}^{\infty }g_{i}(x)$$
 * $$f(x)=g_{0}(x)+g_{1}(x)+g_{2}(x)+...$$

 * If $$g(x)$$ is odd, than

 * $$g_{0}(-x)+g_{1}(-x)+g_{2}(-x)+...$$
 * $$=-g_{0}(x)-g_{1}(x)-g_{2}(x)-...$$
 * $$=-\left (g_{0}(x)+g_{1}(x)+g_{2}(x)+... \right )$$
 * $$=-f(x)$$

 * Therefore, if $$g_{i}(x)$$ is odd, than $$f(x)$$ is also odd. The same concept applies if $$g(x)$$ is an even function.

Contributors: Egm6321.f09.Team4.meyerat 01:46, 17 November 2009 (UTC)

Problem #9:
Lecture 33-3:

 * Given :

 * Functions $$P_{2k}(x)$$ and $$P_{2k+1}(x)$$

 * Find :

 * Show that $$P_{2k}(x)$$ is even and $$P_{2k+1}(x)$$ is odd.

 * Solution :

 * The equation for $$P_{n}(x)$$ is given as:

 * $$P_{n}(x)=\sum_{i=0}^{[n/2]}(-1)^{i}\frac{(2n-2i)!x^{n-2i}}{2^{n}i!(n-1)!(n-2i)!}$$

 * Substitute in n=2k,

 * $$P_{2k}(x)=\sum_{i=0}^{[2k/2]}(-1)^{i}\frac{(4k-2i)!x^{2k-2i}}{2^{2k}i!(2k-1)!(2k-2i)!}$$

 * This can be thought of as

 * $$P_{2k}(x)=\sum_{i=0}^{[k]}C_kx^{2k-2i}$$

 * where $$C_k$$ is just a coefficient. Therefore, we can rearrange the above as

 * $$P_{2k}(x)=\sum_{i=0}^{[k]}C_k\left (x^{k-i} \right )^2$$

 * Since $$k$$ and $$i$$ are whole, positive integers, and $$k\geq i$$, than $$k-i$$ will always be a positive integer. Therefore, regardless of whether $$x^{k-i}$$ is an odd or even function, it will always be squared, resulting in an even function:

 * Therefore, $$P_{2k}(x)$$ is even.

 * Similarly,

 * $$P_{2k+1}(x)=\sum_{i=0}^{[(2k+1)/2]}A_k\left (x^{2k+1-2i} \right )$$

 * $$P_{2k+1}(x)=\sum_{i=0}^{[(2k+1)/2]}A_k\left (x^{2k-2i} \right )x$$

 * Which is an even function times an odd function $$x$$. And as always, an even function times and odd function is always an odd function.

 * Therefore, $$P_{2k+1}(x)$$ is odd.

Contributors: Egm6321.f09.Team4.meyerat 01:29, 16 November 2009 (UTC)

Problem #10:
Lecture (33-4):

 * Given :

 * $$ \begin{align}

q\left(x\right) &= \sum_{i=0}^{4}\ c_{i} x^{i} \quad \in \mathcal{P}_{4} \\ &\text{where:} \\ &\qquad c_{0} = 3\\ &\qquad c_{1} = 10\\ &\qquad c_{2} = 15\\ &\qquad c_{3} = -1\\ &\qquad c_{4} = 5 \end{align} $$

 * Find :

 * Obtain $$\left\{a_{i}\right\}$$ such that:

 * $$ \begin{align}

\sum_{i=0}^{4}\ a_{i} P_{i} = q\left(x\right) \end{align} $$

 * Note: $$ P_{i} $$ are the Legendre polynomials verified in Problem #6 (see Lecture 31-3):

 * $$ \begin{align}

P_{0}\left(x\right) &= 1 \\ P_{1}\left(x\right) &= x \\ P_{2}\left(x\right) &= \frac{1}{2}\left(3x^{2}-1\right) \\ P_{3}\left(x\right) &= \frac{1}{2}\left(5x^{3}-3x\right) \\ P_{4}\left(x\right) &= \frac{35}{8}x^{4} - \frac{15}{4}x^{2} + \frac{3}{8} \end{align} $$

 * Solution :

 * Plug the Legendre polynomials into the expression containing $$\left\{a_{i}\right\}$$, then group the terms by power of $$x$$:

 * $$ \begin{align}

q\left(x\right) &= \sum_{i=0}^{4}\ a_{i} P_{i} \\ &= a_{0} + a_{1} x + a_{2} \frac{1}{2}\left(3x^{2}-1\right) + a_{3}\frac{1}{2}\left(5x^{3}-3x\right) + a_{4}\left(\frac{35}{8}x^{4}-\frac{15}{4}x^{2}+\frac{3}{8}\right) \\ &= \left(a_{0}-\frac{1}{2}a_{2}+\frac{3}{8}a_{4}\right) + x\left(a_{1}-3a_{3}\right) + x^{2}\left(\frac{3}{2}a_{2}-\frac{15}{4}a_{4}\right) + x^{3}\left(\frac{5}{2}a_{3}\right) + x^{4}\left(\frac{35}{8}a_{4}\right) \end{align} $$

 * Then, we can set each term of this equation equal to the corresponding term in the definition of $$q\left(x\right)$$, obtaining a system of five equations:

 * $$ \begin{align}

\sum_{i=0}^{4}\ a_{i} &P_{i} = \sum_{i=0}^{4}\ c_{i} x^{i}& \\ \\ &\text{(1)  }a_{0} - \frac{1}{2}a_{2} + \frac{3}{8}a_{4} &= 3 \\ &\text{(2) }a_{1} - 3a_{3} &= 10 \\ &\text{(3) }\frac{3}{2}a_{2} - \frac{15}{4}a_{4} &= 15 \\ &\text{(4) }\frac{5}{2}a_{3} &= -1 \\ &\text{(5) }\frac{35}{8}a_{4} &= 5 \end{align} $$

 * The last two equations yield the values of $$a_{3}$$ and $$a_{4}$$ directly:

 * $$ \begin{align}

a_{3} &= \frac{-2}{5} \\ a_{4} &= \frac{8}{7} \end{align} $$

 * $$a_{2}$$ is then obtained from equation (3) by plugging in $$a_{4}$$:

 * $$ \begin{align}

\frac{3}{2}a_{2} &= 15 + \frac{15}{4}a_{4} \\ a_{2} &= 10 + \frac{10}{4}\cdot \frac{8}{7} \\ &= \frac{90}{7} \end{align} $$

 * Similarly, $$a_{1}$$ is obtained from equation (2) by plugging in $$a_{3}$$:

 * $$ \begin{align}

a_{1} &= 10 + 3a_{3} \\ &= 10 + \frac{-6}{5} \\ &= \frac{44}{5} \end{align} $$

 * Finally, $$a_{0}$$ is obtained by inserting $$a_{2}$$ and $$a_{4}$$ into equation (1):

 * $$ \begin{align}

a_{0} &= 3 + \frac{1}{2}a_{2} - \frac{3}{8}a_{4} \\ &= 3 + \frac{45}{7} - \frac{3}{7} \\ &= 3 + \frac{42}{7} \\ &= 9 \end{align} $$

 * In summary, the following expressions for $$\left\{a_{i}\right\}$$ were obtained:

 * $$ \begin{align}

a_{0} &= 9, \quad a_{1} = \frac{44}{5}\\ \\ a_{2} &= \frac{90}{7}, \quad a_{3} = \frac{-2}{5}\\ \\ a_{4} &= \frac{8}{7} \end{align} $$

Contributors:
 * Egm6321.f09.team4.sls 15:04, 16 November 2009 (UTC)

Problem #11:
Lecture (34-2):

 * Given :

 * $$\Psi (1,\theta )=f(\theta )=T_ocos(\theta)$$
 * $$f(\mu )=T_o\sqrt{1-\mu ^2}$$
 * $$A_n=\frac{2n+1}{2}<f,P_n>$$

 * Find :

 * a) Without calculation, find the property of $$A_n$$.
 * b) Complete 3 nonzero coefficients of $$A_n$$ analytically or numerically using Gaussian-Legendre accurately to within 5%.

 * Solution :

 * a)We see that $$f(\mu)=T_o\sqrt{1-\mu ^2}$$ is an even function and we know that $$P_n(\mu)$$ is an even function when $$n$$ is an even integer. Therefore,

 * $$A_n=\frac{2n+1}{2}<f,P_n>=0$$ when n is odd, therefore,
 * $$A_1=A_3=A_5...=0$$.

 * b)The first three coefficients are as follows:

 * $$A_n=\frac{2n+1}{2}\int_{-1}^{1}T_o\sqrt{1-\mu^2}P_n(\mu)d\mu$$

 * $$A_0=\frac{T_o}{2}\int_{-1}^{1}\sqrt{1-\mu^2}d\mu$$

 * From the integration tables, we see that this is,

 * $$A_0=\frac{T_o}{2}\left (\frac{\mu}{2}\sqrt{1-\mu^2}+\frac{1}{2}sin^{-1}(\mu) \right )_{-1}^{1}$$

 * $$A_0=\frac{T_o}{2}\left (\frac{\pi}{4}-\frac{3\pi}{4}\right )$$

 * $$A_0=-\frac{T_o\pi}{4}$$

 * $$A_2=\frac{5T_o}{2}\int_{-1}^{1}\sqrt{1-\mu^2}\left ( \frac{1}{2}(3\mu^2-1) \right )d\mu$$
 * $$A_2=\frac{5T_o}{2}\left [\int_{-1}^{1}\frac{3}{2}\mu^2\sqrt{1-\mu^2}d\mu-\int_{-1}^{1}\frac{1}{2}\sqrt{1-\mu^2}d\mu \right ]$$
 * $$A_2=\frac{5T_o}{2}\left [\frac{3}{2}\int_{-1}^{1}\mu^2\sqrt{1-\mu^2}d\mu-\frac{1}{2}\int_{-1}^{1}\sqrt{1-\mu^2}d\mu \right ]$$
 * $$A_2=\frac{15T_0}{4}\int_{-1}^{1}\mu^2\sqrt{1-\mu^2}d\mu-\frac{5T_0}{4}\int_{-1}^{1}\sqrt{1-\mu^2}d\mu$$

 * From integral tables this becomes,

 * $$A_2=\frac{15T_0}{4}\left [ \frac{\mu^2}{8}(2\mu^2-1)\sqrt{1-\mu^2}+\frac{1}{8}sin^{-1}\mu \right ]_{-1}^{1}-\frac{5T_0}{4}\left [\frac{\mu}{2}\sqrt{1-\mu^2}+\frac{1}{2}sin^{-1}(\mu) \right ]_{-1}^{1}$$

 * $$A_2=\frac{15T_0}{4}\left [\frac{\pi}{16}-\frac{3\pi}{16}\right ]-\frac{5T_0}{4}\left [\frac{\pi}{4}-\frac{3\pi}{4}\right ]$$

 * $$A_2=-\frac{15T_0\pi}{32}+\frac{5T_0\pi}{8}$$

 * $$A_2=\frac{5T_0\pi}{32}$$

 * $$A_4=\frac{9T_o}{2}\int_{-1}^{1}\sqrt{1-\mu^2}\left ( \frac{25}{8}\mu^4-\frac{15}{4}\mu^2+\frac{3}{8} \right )d\mu$$

 * $$A_4=\frac{225T_0}{16}\int_{-1}^{1}\mu^4\sqrt{1-\mu^2}d\mu-\frac{135T_0}{8}\int_{-1}^{1}\mu^2\sqrt{1-\mu^2}d\mu+\frac{27T_0}{16}\int_{-1}^{1}\sqrt{1-\mu^2}d\mu$$

 * $$A_4=\frac{225T_0}{16}\int_{-1}^{1}\mu^4\sqrt{1-\mu^2}d\mu-\frac{135T_0}{8}\left [\frac{\pi}{16}-\frac{3\pi}{16}\right ]+\frac{27T_0}{16}\left [\frac{\pi}{4}-\frac{3\pi}{4}\right ]$$

 * $$A_4=\frac{225T_0}{16}\int_{-1}^{1}\mu^4\sqrt{1-\mu^2}d\mu+\frac{135T_0\pi}{64}-\frac{27T_0\pi}{32}$$

 * Substitute in $$\mu=cos(x)$$ and we obtain

 * $$A_4=-\frac{225T_0}{16}\int_{0}^{\pi}cos^4(x)sin^2(x)dx+\frac{135T_0\pi}{64}-\frac{27T_0\pi}{32}$$

 * $$A_4=-\frac{225T_0}{16}\int_{0}^{\pi}cos^4(x)dx+\frac{225T_0}{16}\int_{0}^{\pi}cos^6(x)dx+\frac{135T_0\pi}{64}-\frac{27T_0\pi}{32}$$

 * From the integration tables

 * $$\int cos^n(x)dx=\frac{1}{n}cos^{n-1}(x)sin(x)-\frac{n-1}{n}\int cos^{n-2}(x)dx$$

 * Therefore, we have,

 * $$A_4=\frac{225T_0}{16}\left [\frac{1}{6}cos^5(x)sin(x) \right ]_{0}^{\pi}+\frac{225T_0}{16}\frac{5}{6}\int_{0}^{\pi}cos^4(x)dx-\frac{225T_0}{16}\int_{0}^{\pi}cos^4(x)dx+\frac{135T_0\pi}{64}-\frac{27T_0\pi}{32}$$

 * Which reduces to,

 * $$A_4=0-\frac{225T_0}{96}\int_{0}^{\pi}cos^4(x)dx+\frac{81T_0\pi}{64}$$

 * $$A_4=0-\frac{225T_0}{96}\left [\left (\frac{1}{4}cos^3(x)sin(x) \right )_{0}^{\pi}+\frac{3}{4}\int_{0}^{\pi} cos^2(x)dx  \right ]+\frac{81T_0\pi}{64}$$

 * $$A_4=0-\frac{225T_0}{96}\left [\left (\frac{1}{4}cos^3(x)sin(x) \right )_{0}^{\pi}+\frac{3}{4}\left (\frac{x}{2}+\frac{ \sin 2x}{4}  \right )_{0}^{\pi} \right ]+\frac{81T_0\pi}{64}$$

 * $$A_4=0-\frac{225T_0}{96}\left [\frac{3\pi}{8}\right ]+\frac{81T_0\pi}{64}$$

 * $$A_4=-\frac{675T_0\pi}{768}+\frac{81T_0\pi}{64}$$

 * $$A_4=\frac{297T_0\pi}{768}=\frac{99T_0\pi}{256}$$

 * In summary,

 * $$A_0=-\frac{T_o\pi}{4}$$
 * $$A_2=\frac{5T_0\pi}{32}$$
 * $$A_4=\frac{99T_0\pi}{256}$$

Contributors: Egm6321.f09.Team4.meyerat 04:20, 18 November 2009 (UTC)

Problem #12:
Lecture (35-3):

 * Given :

 * Wikipedia table for Gaussian Legendre Quadrature

 * Find :

 * Verify the Gaussian Legendre Quadrature table in wikipedia for analytical expressions of $$\left \{ x_{j} \right \}$$ and $$\left \{ w_{j} \right \}$$, $$j=1,2...,n$$ and $$n=1,...5$$ after completing homework on page 31-3. Evaluate numerically $$\left \{ x_{j} \right \}$$ and $$\left \{ w_{j} \right \}$$ and compare results with Abramowitz and Stegun.

 * Solution :

 * For reference only, the wikipedia table is shown below:

Contributors:

Problem #13:
Lecture (36-4):

 * Given :

 * $$ \begin{align}

\tanh \left(x\right) = \frac{\exp\left(x\right) - \exp\left(-x\right)}{\exp\left(x\right) + \exp\left(-x\right)} \end{align} $$

 * Find :

 * Show that:

 * $$ \begin{align}

\tanh^{-1} \left(x\right) = \frac{1}{2}\ln\left(\frac{1+x}{1-x}\right) \end{align} $$.

 * Solution :

 * Start with the following statement:

 * $$ \begin{align}

\tanh\left( \ \tanh^{-1}\left(x\right)\ \right) &= x \\ \tanh\left(\theta\right) &= x, \quad \text{(where } \theta = \tanh^{-1}\left(x\right) \text{)} \end{align} $$

 * Insert $$\text{ } \theta$$ into the definition of $$tanh(x)$$:

 * $$ \begin{align}

\left(\frac{\exp\left(\theta\right)-\exp\left(-\theta\right)}{\exp\left(\theta\right) + \exp\left(-\theta\right)}\right) &= x \\ \exp\left(\theta\right)-\exp\left(-\theta\right) &= x\exp\left(\theta\right) + x\exp\left(-\theta\right) \\ \exp\left(\theta\right) - x\exp\left(\theta\right) &= \exp\left(-\theta\right) + x\exp\left(-\theta\right) \\ \exp\left(\theta\right)\left(1-x\right) &= \exp\left(-\theta\right)\left(1+x\right)\\ \exp\left(2\theta\right) &= \frac{1+x}{1-x} \\ 2 \theta &= \ln\left(\frac{1+x}{1-x}\right) \\ \theta &= \frac{1}{2}\ln\left(\frac{1+x}{1-x}\right) \end{align} $$

 * Since $$\theta = \tanh^{-1}\left(x\right)$$:

 * $$ \begin{align}

\tanh^{-1}\left(x\right) = \frac{1}{2}\ln\left(\frac{1+x}{1-x}\right) \end{align} $$

Contributors:
 * Egm6321.f09.team4.sls 17:20, 16 November 2009 (UTC)

title: User:Egm6321.f09.team4/HW6


text: File:HW6 Fig1.pdf

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

title: File:HW6 Fig1.pdf


text: File:HW6 Fig2.pdf

title: File:HW6 Fig2.pdf


text:
Somology

Somology refers to the study of the caring of the body, e.g., in nursing.

Notation
Notation: let the symbol Def. indicate that a definition is following.

Notation: let the symbols between [ and ] be replacement for that portion of a quoted text.

Notation: let the symbol ... indicate unneeded portion of a quoted text.

Sometimes these are combined as [...] to indicate that text has been replaced by ....

Theoretical somology
Def. "understanding the body as an integration of the object body (the thing) into experience so that it is simultaneously an object, a means of experience, a means of expression, a manner of presence among other people, and a part of one's personal identity" is called somology.

title: Somology


text:
User:Yawgyimah/Phd dissertation in thermo/resource economics

Yawgyimah/Phd dissertation in thermo / resource economics


text:
Psychological well-being

According to Pavot and Diener (1993), psychological well-being refers to the subjective experience of two aspects of one's psychological experience:
 * 1) Emotional or affective experience (i.e., positive and negative affect)
 * 2) Conceptual or cognitive experience (i.e., satisfaction with life, relationships, work, and leisure)

title: Psychological well-being


text:
User:B9 hummingbird hovering/Blog/Learning Tibetan Weblog/Archive1

[url=][/url]. ..


text:
Data screening

 * 1) Data screening should be conducted prior to data recoding and data analysis, to help ensure the integrity of the data.
 * 2) It is only necessary to screen the data for the variables and cases used for the analyses  presented in the lab report.
 * 3) Data screening means checking data for errors and fixing or removing these errors. The goal is to maximise "signal" and minimise "noise" by identifying and fixing or removing errors.
 * 4) Keep a record of data screening steps undertaken and any changes made to the data. This should be summarised in a one to two paragraph section called "Data screening" at the beginning of the Results.
 * 5) Fixing or removing incorrect data:
 * 6) Erroneous data can be changed to missing data. Alternatively, if a correct value can be presumed, then this can be entered.
 * 7) To change data in a cell, open the data file and using the data view, left-click on the cell. Delete the data to make it missing data (sysmis). Or change the value in the cell by typing in the new value. Repeat for each problematic value.
 * 8) It is probably best to make erroneous data missing unless the correct value is obvious (e.g., if 77 was entered, it might reasonably be deduced that 7 was intended) in which case the incorrect value can be replaced with a best guess correct value.
 * 9) For cases with a lot of erroneous data, it is probably best to remove the entire case (i.e., delete the whole row).
 * 10) Out-of-range values:
 * 11) Out of range values are either below the minimum or above the maximum possible value.
 * 12) To know what the in-range values are, check:
 * 13) the survey
 * 14) download the data and check the SPSS Value Labels in Variable View
 * 15) Identify out-of-range values by obtaining descriptive statistics (in SPSS, use Analyze - Descriptive Statistics - Descriptives) to examine the minimum and maximum values for all variables of interest. Or try this syntax - DESCRIPTIVES VARIABLES=all /STATISTICS=MEAN STDDEV MIN MAX.
 * 16) In the SPSS Data View, sort variables with out-of-range values in ascending or descending order to help identify the case(s) which has(have) the out-of-range values. Alternatively, use search and find to identify the case(s) with out-of-range values.
 * 17) Decide whether to accept, replace, or remove out-of-range values.
 * 18) Unusual cases:
 * 19) Unusual cases occur when a case's responses are very different from the pattern of responses by most other respondents.
 * 20) In SPSS, Data - Identify Unusual cases - Enter several to many variables (e.g., all the Time Management Skill variables).
 * 21) The results will flag the top 10 anomalous cases.
 * 22) Look carefully at each of these cases' responses to the target variables - do they appear to be legitimate (e.g., are their out-of-range values, or consider if the data was fabricated, often the responses to the reverse scored items aren't fabricated in the expected direction). If so, consider removing the case.
 * 23) Duplicate cases:
 * 24) Duplicate cases occur when two or more cases have identical or near-identical data
 * 25) In SPSS, Data - Identify Duplicate cases - Enter several to many variables.
 * 26) Consider whether to remove both cases (e.g., the integrity of the data may be in doubt because it may have been fabricated and then duplicated?) or to retain one copy of each case and delete duplicates.
 * 27) Manual check for other anomalies
 * 28) Check carefully through the data file (case by case and variable by variable) looking for and addressing any oddities.
 * 29) Empty cases: e.g., cases with no or little data could be removed
 * 30) Cases with responses which lack meaningful variation: (e.g., 5 5 5 5 5 5 5 5 5) or which exhibit obvious arbitrary patterns (e.g., zig-zag - 1 2 3 4 5 4 3 2 1 2 3 4 5) - such responses are unlikely to be valid and probably should be deleted.

title: Data screening


text:
User talk:Rosannew~enwikiversity

 Hello Rosannew, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

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And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Mu301Bot 08:18, 13 November 2009 (UTC)

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Hello,

The developer team at Wikimedia is making some changes to how accounts work, as part of our on-going efforts to provide new and better tools for our users like cross-wiki notifications. These changes will mean you have the same account name everywhere. This will let us give you new features that will help you edit and discuss better, and allow more flexible user permissions for tools. One of the side-effects of this is that user accounts will now have to be unique across all 900 Wikimedia wikis. See the announcement for more information.

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Renamed
 This account has been renamed as part of single-user login finalisation. If you own this account you can |log in using your previous username and password for more information. If you do not like this account's new name, you can choose your own using this form after logging in: . -- Keegan (WMF) (talk) 06:20, 19 April 2015 (UTC)

title: User talk:Rosannew~enwikiversity


text:
User talk:AvsB

 Hello AvsB, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random project.
 * Browse Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Mu301Bot 08:18, 13 November 2009 (UTC)

Feel free to leave a message on my talk page if you have any questions or need help. --mikeu talk 08:18, 13 November 2009 (UTC)

title: User talk:AvsB


text:
New Zealand Law/Criminal/Assault

Wounding with Intent
A person is guilty of the offence who: (a) with intent to cause gbh – maims, disfigures or causes gbh (b) with intent to injure – maims, disfigures or causes gbh. maim: to cause a person to be unable to uses one of his members

Injury with Intent
A person is guilty of the offence who: (a) with intent to cause gbh – injures anyone (b) with intent to injure – injures anyone

Injury by Unlawful Act
Injuring where if death had occurred it would have been manslaughter.

Aggravated Wounding/Injury
A person is guilty of the offence who with intent to:
 * (a) commit a crime;
 * (b) avoid arrest;
 * (c) avoid detection; or
 * (d) facilitate flight
 * either:
 * (i) wounds, maims, disfigures, causes gbh, stupefies, renders unconscious, renders incapable of offering resistence; or
 * (ii) injures anyone.

maim: to cause a person to be unable to use one of his members.

Aggravated Assault
A person is guilty of the offence who either: (a) assault with intent to commit a crime (b) obstructing a constable in the execution of his duty.

Assault with Intent to Injure
A person is guilty of the offence who with intent to injure, assaults anyone.

R v Kerr

 * Facts
 * The defendant, a gang member, went looking for a rival gang member.
 * The victim attacked the defendant with a pipe.
 * The defendant pulled out a knife, stabbed and killed the victim; the defendant was subsequently charged with murder.

 * Held
 * 1) Judge should leave defence to jury unless it is impossible for jury to entertain reasonable doubt.
 * 2) Quoting Palmer
 * - a person under attack cannot weigh to a nicety the exact measure of necessary defensive action;
 * - if a person did what he honestly and instinctively considered necessary, it is pretty powerful evidence that it was reasonable defensive action.

R v Nazif

 * Question
 * Can a person consent to assault?

 * Held
 * 1) Common law defence saved by s 20 Crimes Act except where not in the public interest.
 * 2) Honest belief in consent sufficient (unless otherwise provided in statute).
 * 3) Belief need not be reasonable but goes to whether belief actually held.

R v Mwai

 * Facts
 * M had unprotected sex with several women after becoming HIV positive. Charged with causing gbh.

 * Held
 * 1) Not necessary for act to be the sole cause, enough that is a sufficient cause.
 * 2) Virus and women’s immune system were causes, but def was a substantial cause.

Lazarus v Police

 * Facts:
 * Defendant broke window of a car with a golf club. Some of the glass struck an occupant.

 * Held:

 * It is only assault if the defendant;
 * (i) intended to apply force to someone; or
 * (ii) intended someone to feel personally threatened.

cf Kerr – where it was held that there can be a threat even if the victim is unaware.

title: New Zealand Law/Criminal/Assault


text:
User talk:B9 hummingbird hovering/rdo la gser zhun

Dedication & Invocation: Ocean of milk

"In the midst of your daily work, in the tumult and storm of life, pause for a while, again and again." J. P. Vaswani


text:
User:EgyWiz

work in progress...

title: User:EgyWiz


text:
Template talk:Make a portal/Tab1

See also: discussions on Tab2 and Tab3. - I suspect this collection of templates may need refactoring - with themes and other cool features (e.g. something more flexible than Template:Module Intro Page with tabs.

Ideas:

Possible? K 15:44, 13 November 2009 (UTC)

title: Template talk:Make a portal/Tab1


text:
User:Egm6321.f09.Team1.vasquez/HW6

= Problem 1: Laplacian in Circular Cylindrical Coordinates =

From lecture slide 31-1.

Given
The equations for circular cylinder coordinates are given by:

 * {| style="width:100%" border="0" align="left"

\begin{array}{*{20}l} x_1 & = & r \cos \theta & = & \xi_1 \cos \xi_2 \\ x_2 & = & r \sin \theta & = & \xi_1 \sin \xi_2 \\ x_3 & = & z & = & \xi_3 & \end{array} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1)
 * }
 * }

Find
1. Find $$\displaystyle \{{\rm d} x_i\} = \{{\rm d} x_1, {\rm d} x_2, {\rm d} x_3\} $$ in terms of $$\displaystyle \{\xi_j\} = \{\xi_1, \xi_2, \xi_3\} $$ and $$\displaystyle \{{\rm d} \xi_k\} $$

2. Find $$\displaystyle {\rm d} s^2 = \sum_i ({\rm d} x_i)^2 = \sum_k (h_k)^2 ({\rm d} \xi_k)^2 $$. Identify $$\displaystyle \{h_i\}$$ in terms of $$\displaystyle \{ \xi_j \}$$.

3. Find the Laplacian in these curvilinear coordinates.

Solution
1. We can take the derivative of each $$\displaystyle \{x_i\}$$ in Eq. 1 with respect to $$\displaystyle \xi_i$$ as follows

 * {| style="width:100%" border="0" align="left"

{\rm d} x_1 = \frac{\partial x_1}{\partial \xi_1} {\rm d} \xi_1 + \frac{\partial x_1}{\partial \xi_2} {\rm d} \xi_2 + \frac{\partial x_1}{\partial \xi_3} {\rm d} \xi_3 = \cos \xi_2 {\rm d} \xi_1 - \xi_1 \sin \xi_2 {\rm d} \xi_2 + 0 {\rm d} \xi_3 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 2)
 * }
 * }

 * {| style="width:100%" border="0" align="left"

{\rm d} x_2 = \frac{\partial x_2}{\partial \xi_1} {\rm d}\xi_1 + \frac{\partial x_2}{\partial \xi_2} {\rm d}\xi_2 + \frac{\partial x_2}{\partial \xi_3} {\rm d}\xi_3 = \sin \xi_2 {\rm d}\xi_1 + \xi_1 \cos \xi_2 {\rm d} \xi_2 + 0 {\rm d} \xi_3 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 3)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

{\rm d}x_3 = \frac{\partial x_3}{\partial \xi_1} {\rm d}\xi_1 + \frac{\partial x_3}{\partial \xi_2} {\rm d}\xi_2 + \frac{\partial x_3}{\partial \xi_3} {\rm d}\xi_3 = (0) {\rm d}\xi_1 + (0) {\rm d} \xi_2 + {\rm d} \xi_3 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 4)
 * }
 * }

Therefore,

 * {| style="width:100%" border="0" align="left"

$$\displaystyle \begin{array}{*{20}l} {\rm d}x_1 & = & \cos \xi_2 {\rm d}\xi_1 - \xi_1 \sin \xi_2 {\rm d} \xi_2 \\ {\rm d}x_2 & = & \sin \xi_2 {\rm d}\xi_1 + \xi_1 \cos \xi_2 {\rm d} \xi_2 \\ {\rm d}x_3 & = & {\rm d} \xi_3 \end{array} $$ $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * $$\displaystyle (Eq. 5)
 * }
 * }

2. Using the definition for $$\displaystyle {\rm d} s $$ gives

 * {| style="width:100%" border="0" align="left"

{\rm d}s^2 = \cos^2 \xi_2 \left({\rm d}\xi_1 \right)^2 - \cancel{2 \xi_1 \sin \xi_2 \cos \xi_2 {\rm d}\xi_1 {\rm d} \xi_2} + \xi_1^2 \sin^2 \xi_2 \left({\rm d} \xi_2\right)^2 + \sin^2 \xi_2 \left({\rm d}\xi_1 \right)^2 + \cancel{2 \xi_1 \sin \xi_2 \cos \xi_2 {\rm d}\xi_1 {\rm d} \xi_2} + \xi_1^2 \cos^2 \xi_2 \left({\rm d} \xi_2 \right)^2 + \left( {\rm d} \xi_3 \right)^2 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 6)
 * }
 * }

Simplifying Eq. 6 yields

 * {| style="width:100%" border="0" align="left"

$$\displaystyle {\rm d}s^2 = \left({\rm d}\xi_1 \right)^2 + \xi_1^2 \left({\rm d} \xi_2\right)^2 + \left( {\rm d} \xi_3 \right)^2 $$ $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * $$\displaystyle (Eq. 7)
 * }
 * }

Therefore,

 * {| style="width:100%" border="0" align="left"

$$\displaystyle h_1 = 1, \quad h_2 = \xi_1, \quad h_3 = 1 $$ $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * $$\displaystyle (Eq. 8)
 * }
 * }

3. The Laplacian operator over a function $$\displaystyle \psi$$ in 3-D curvilinear coordinates is given by

 * {| style="width:100%" border="0" align="left"

\Delta \psi= \nabla^2 \psi= \frac{1}{h_1 h_2 h_3} \sum_{i=1}^3 \frac{\partial}{\partial \xi_i} \left[ \frac{h_1 h_2 h_3}{h_i^2} \frac{\partial \psi}{\partial \xi_i} \right] $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 9)
 * }
 * }

Substituting the results given in Eq. 8 into Eq. 9 gives

 * {| style="width:100%" border="0" align="left"

\Delta \psi = \frac{1}{\xi_1} \left[ \frac{\partial}{\partial \xi_1} \left( \xi_1 \frac{\partial \psi}{\partial \xi_1} \right) + \frac{\partial}{\partial \xi_2} \left( \frac{\xi_1}{\xi_1^2} \frac{\partial \psi}{\partial \xi_2} \right) + \frac{\partial}{\partial \xi_3} \left(\xi_1 \frac{\partial}{\partial \xi_3} \right) \right] $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 10)
 * }
 * }

Simplifying yields

 * {| style="width:100%" border="0" align="left"

$$\displaystyle \Delta \psi = \frac{1}{\xi_1} \frac{\partial}{\partial \xi_1} \left(\xi_1 \frac{\partial \psi}{\partial \xi_1} \right) + \frac{1}{\xi_1^2} \frac{\partial^2 \psi}{\partial \xi_2^2} + \frac{\partial^2 \psi}{\partial \xi_3^2} $$ $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * $$\displaystyle (Eq. 11)
 * }
 * }

= Problem 2: Laplacian in Spherical Coordinates =

From lecture slide 31-1.

Given
The equations for spherical coordinates are given by:

 * {| style="width:100%" border="0" align="left"

\begin{array}{*{20}l} x_1 & = & r \cos \theta \cos \phi & = & \xi_1 \cos \xi_3 \cos \xi_2 \\ x_2 & = & r \cos \theta \sin \phi & = & \xi_1 \cos \xi_3 \sin \xi_2 \\ x_3 & = & r \sin \theta & = & \xi_1 \sin \xi_3 & \end{array} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1)
 * }
 * }

Find
1. Find $$\displaystyle \{{\rm d} x_i\} = \{{\rm d} x_1, {\rm d} x_2, {\rm d} x_3\} $$ in terms of $$\displaystyle \{\xi_j\} = \{\xi_1, \xi_2, \xi_3\} $$ and $$\displaystyle \{{\rm d} \xi_k\} $$

2. Find $$\displaystyle {\rm d} s^2 = \sum_i ({\rm d} x_i)^2 = \sum_k (h_k)^2 ({\rm d} \xi_k)^2 $$. Identify $$\displaystyle \{h_i\}$$ in terms of $$\displaystyle \{ \xi_j \}$$.

3. Find the Laplacian in these curvilinear coordinates.

Solution
1. We can take the derivative of each $$\displaystyle \{x_i\}$$ in Eq. 1 with respect to $$\displaystyle \xi_i$$ as follows

 * {| style="width:100%" border="0" align="left"

{\rm d} x_1 = \frac{\partial x_1}{\partial \xi_1} {\rm d} \xi_1 + \frac{\partial x_1}{\partial \xi_2} {\rm d} \xi_2 + \frac{\partial x_1}{\partial \xi_3} {\rm d} \xi_3 = \cos \xi _3 \cos \xi _2 {\rm d}\xi _1 - \xi _1 \cos \xi _3 \sin \xi _2 {\rm d}\xi _2  - \xi _1 \sin \xi _3 \cos \xi _2 {\rm d}\xi _3 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 2)
 * }
 * }

 * {| style="width:100%" border="0" align="left"

{\rm d} x_2 = \frac{\partial x_2}{\partial \xi_1} {\rm d}\xi_1 + \frac{\partial x_2}{\partial \xi_2} {\rm d}\xi_2 + \frac{\partial x_2}{\partial \xi_3} {\rm d}\xi_3 = \cos \xi _3 \sin \xi _2 {\rm d}\xi _1  + \xi _1 \cos \xi _3 \cos \xi _2 {\rm d}\xi _2  - \xi _1 \sin \xi _3 \sin \xi _2 {\rm d}\xi _3 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 3)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

{\rm d}x_3 = \frac{\partial x_3}{\partial \xi_1} {\rm d}\xi_1 + \frac{\partial x_3}{\partial \xi_2} {\rm d}\xi_2 + \frac{\partial x_3}{\partial \xi_3} {\rm d}\xi_3 = \sin \xi _3 {\rm d}\xi _1 + (0){\rm d}\xi _2  + \xi _1 \cos \xi _3 {\rm d}\xi _3 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 4)
 * }
 * }

Therefore,

 * {| style="width:100%" border="0" align="left"

$$\displaystyle \begin{array}{*{20}l} {\rm d}x_1 & = & \cos \xi _3 \cos \xi _2 {\rm d}\xi _1 - \xi _1 \cos \xi _3 \sin \xi _2 {\rm d}\xi _2  - \xi _1 \sin \xi _3 \cos \xi _2 {\rm d}\xi _3 \\ {\rm d}x_2 & = & \cos \xi _3 \sin \xi _2 {\rm d}\xi _1 + \xi _1 \cos \xi _3 \cos \xi _2 {\rm d}\xi _2  - \xi _1 \sin \xi _3 \sin \xi _2 {\rm d}\xi _3 \\ {\rm d}x_3 & = & \sin \xi _3 {\rm d}\xi _1 + \xi _1 \cos \xi _3 {\rm d}\xi _3 \end{array} $$ $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * $$\displaystyle (Eq. 5)
 * }
 * }

2. Using the definition for $$\displaystyle {\rm d} s $$ gives

 * {| style="width:100%" border="0" align="left"

\begin{array}{*{20}l} {\rm d}s^2 & = & \quad \left(\cos \xi _3 \cos \xi _2 {\rm d}\xi _1 - \xi _1 \cos \xi _3 \sin \xi _2 {\rm d}\xi _2  - \xi _1 \sin \xi _3 \cos \xi _2 {\rm d}\xi _3 \right)^2\\ {} & {} & + \left(\cos \xi _3 \sin \xi _2 {\rm d}\xi _1 + \xi _1 \cos \xi _3 \cos \xi _2 {\rm d}\xi _2  - \xi _1 \sin \xi _3 \sin \xi _2 {\rm d}\xi _3 \right)^2 \\ {} & {} & + \left(\sin \xi _3 {\rm d}\xi _1 + \xi _1 \cos \xi _3 {\rm d}\xi _3 \right)^2 \end{array} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 6)
 * }
 * }

Simplifying Eq. 6 yields

 * {| style="width:100%" border="0" align="left"

$$\displaystyle {\rm d}s^2 = \left({\rm d}\xi_1 \right)^2 + \xi_1^2 \cos^2 \xi_3 \left({\rm d} \xi_2\right)^2 + \xi_1^2 \left( {\rm d} \xi_3 \right)^2 $$ $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 *
$$\displaystyle (Eq. 7)
 * }
 * }

Therefore,

 * {| style="width:100%" border="0" align="left"

$$\displaystyle h_1 = 1, \quad h_2 = \xi_1 \cos \xi_3, \quad h_3 = \xi_1 $$ $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 *
$$\displaystyle (Eq. 8)
 * }
 * }

3. The Laplacian operator over a function $$\displaystyle \psi$$ in 3-D curvilinear coordinates is given by

 * {| style="width:100%" border="0" align="left"

\Delta \psi= \nabla^2 \psi= \frac{1}{h_1 h_2 h_3} \sum_{i=1}^3 \frac{\partial}{\partial \xi_i} \left[ \frac{h_1 h_2 h_3}{h_i^2} \frac{\partial \psi}{\partial \xi_i} \right] $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 9)
 * }
 * }

Substituting the results given in Eq. 8 into Eq. 9 gives

 * {| style="width:100%" border="0" align="left"

\Delta \psi = \frac{1}{\xi_1^2 \cos \xi_3} \left[ \frac{\partial}{\partial \xi_1} \left( \xi_1^2 \cos \xi_3 \frac{\partial \psi}{\partial \xi_1} \right) + \frac{\partial}{\partial \xi_2} \left( \frac{\xi_1^2 \cos \xi_3}{\xi_1^2 \cos^2 \xi_3} \frac{\partial \psi}{\partial \xi_2} \right) + \frac{\partial}{\partial \xi_3} \left(\frac{\xi_1^2 \cos \xi_3}{\xi_1^2} \frac{\partial}{\partial \xi_3} \right) \right] $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 10)
 * }
 * }

Simplifying yields

 * {| style="width:100%" border="0" align="left"

$$\displaystyle \Delta \psi = \frac{1}{\xi_1^2} \frac{\partial}{\partial \xi_1} \left(\xi_1^2 \frac{\partial \psi}{\partial \xi_1} \right) + \frac{1}{\xi_1^2 \cos^2 \xi_3} \frac{\partial^2 \psi}{\partial \xi_2^2} + \frac{1}{\xi_1^2 \cos \xi_3}\left(\cos \xi_3\frac{\partial^2 \psi}{\partial \xi_3^2} \right) $$ $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 *
$$\displaystyle (Eq. 11)
 * }
 * }

= Problem 3: Laplacian in Spherical Coordinates (Math/Phys. convention) = From lecture slide 31-2.

Given
The equations for spherical coordinates with Math/Phys. convention are given by:

 * {| style="width:100%" border="0" align="left"

\begin{array}{*{20}l} x_1 & = & r \sin {\bar \theta } \cos \phi & = & \xi_1 \sin {\bar \xi_3 } \cos \xi_2 \\ x_2 & = & r \sin {\bar \theta } \sin \phi & = & \xi_1 \sin {\bar \xi_3 } \sin \xi_2 \\ x_3 & = & r \cos {\bar \theta } & = & \xi_1 \cos {\bar \xi_3 } & \end{array} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 1)
 * }
 * }

Find
Find the Laplacian in these coordinates considering that

 * {| style="width:100%" border="0" align="left"

\bar \theta = \frac{\pi}{2} - \theta $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 2)
 * }
 * }

Solution
Eq. 2 can be written as follows

 * {| style="width:100%" border="0" align="left"

\bar \xi_3 = \frac{\pi}{2} - \xi_3 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 3)
 * }
 * }

Therefore,

 * {| style="width:100%" border="0" align="left"

\cos \xi_3 = \sin \bar \xi_3 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 4)
 * }
 * }

 * {| style="width:100%" border="0" align="left"

\sin \xi_3 = \cos \bar \xi_3 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 5)
 * }
 * }

 * {| style="width:100%" border="0" align="left"

\partial \xi_3 = - \partial \bar \xi_3 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 6)
 * }
 * }

Substituting Eqs. 4, 5 and 6 into Eq.11 from problem 2 gives

 * {| style="width:100%" border="0" align="left"

$$\displaystyle \Delta \psi = \frac{1}{\xi_1^2} \frac{\partial}{\partial \xi_1} \left( \xi_1^2 \frac{\partial \psi}{\partial \xi_1} \right) + \frac{1}{\xi_1^2 \sin^2 \bar \xi_3} \frac{\partial^2 \psi}{\partial \xi_2^2} - \frac{1}{\xi_1^2 \sin \bar \xi_3} \frac{\partial}{\partial \bar \xi_3} \left[ \sin \bar \xi_3 \left(- \frac{\partial \psi}{\partial \bar \xi_3} \right) \right] $$ $$
 *
$$\displaystyle (Eq. 7)
 * }
 * }

Simplifying Eq. 7 yields

 * {| style="width:100%" border="0" align="left"

$$\displaystyle \Delta \psi = \frac{1}{\xi_1^2} \frac{\partial}{\partial \xi_1} \left( \xi_1^2 \frac{\partial \psi}{\partial \xi_1} \right) + \frac{1}{\xi_1^2 \sin^2 \bar \xi_3} \frac{\partial^2 \psi}{\partial \xi_2^2} + \frac{1}{\xi_1^2 \sin \bar \xi_3} \frac{\partial}{\partial \bar \xi_3} \left( \sin \bar \xi_3 \frac{\partial \psi}{\partial \bar \xi_3} \right) $$ $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 *
$$\displaystyle (Eq. 8)
 * }
 * }

= Problem 4: Heat problem in a sphere. Separation of variables =

From lecture slide 31-2.

Given
The following equation

 * {| style="width:100%" border="0" align="left"

\lambda ( \lambda + 1 ) = n(n+1) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 1)
 * }
 * }

Find
Find the two solutions

 * {| style="width:100%" border="0" align="left"

\lambda_1 = n $$ $$ and
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 2)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\lambda_2 = -(n + 1) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 3)
 * }
 * }

Solution
Eq. 1 can be rewritten as follows

 * {| style="width:100%" border="0" align="left"

\lambda ^2 + \lambda - n(n+1) = 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 4)
 * }
 * }

The roots for this quadratic equation are given by

 * {| style="width:100%" border="0" align="left"

\lambda _{1,2} = \frac{2} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 5)
 * }
 * }

 * {| style="width:100%" border="0" align="left"

\lambda _{1,2} =  - \frac{1}{2} \pm \sqrt {\frac{1}{4} + n(n + 1)} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 6)
 * }
 * }

 * {| style="width:100%" border="0" align="left"

\lambda _{1,2} =  - \frac{1}{2} \pm \left( {n + \frac{1}{2}} \right) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 7)
 * }
 * }

Taking the positive and negative signs respectively in Eq. 7 yields

 * {| style="width:100%" border="0" align="left"

$$\displaystyle \lambda_1 = n $$ $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 *
$$\displaystyle (Eq. 8)
 * }
 * }

 * {| style="width:100%" border="0" align="left"

$$\displaystyle \lambda_2 = -\left( n +1 \right) $$ $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 *
$$\displaystyle (Eq. 9)
 * }
 * }

= Problem 5: Legendre Polynomials, equations equivalence =

From lecture slide 31-3.

Given
The Legendre polynomials can generated as follows

 * {| style="width:100%" border="0" align="left"

P_n(x) = \sum_{i=0}^{[n/2]} \frac{(-1)^i (2 n - 2 i)! x^{n - 2i}}{2^n i! (n-i)! (n-2i)!} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 1)
 * }
 * }

where $$\displaystyle [n/2]$$ is the integer part of $$\displaystyle n/2$$.

Find
Show that Eq. 1 is equivalent to

 * {| style="width:100%" border="0" align="left"

P_n(x) = \sum_{i=0}^{[n/2]} \frac{1 \cdot 3 \cdot \cdot \cdot \left(2n - 2i - 1\right)}{2^i i! (n-2i)!} (-1)^i x^{n-2i} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 2)
 * }
 * }

Solution
Eq. 1 can be re-written as follows:

 * {| style="width:100%" border="0" align="left"

P_n(x) = \sum_{i=0}^{[n/2]} \frac{(2 n - 2 i)!}{2^n (n-i)! } \frac{(-1)^i x^{n-2i}}{i!(n-2i)!}
 * $$\displaystyle
 * $$\displaystyle

$$ $$
 *
$$\displaystyle (Eq. 3)
 * }
 * }

The denominator of the left fraction in Eq. 3 can be written as

 * {| style="width:100%" border="0" align="left"

P_n(x) = \sum_{i=0}^{[n/2]} \frac{(2 n - 2 i)!}{2^i 2^{n-i} (n-i)! } \frac{(-1)^i x^{n-2i}}{i!(n-2i)!}
 * $$\displaystyle
 * $$\displaystyle

$$ $$
 *
$$\displaystyle (Eq. 4)
 * }
 * }

Rearrange Eq. 4 and substitute $$\displaystyle {\rm 2}^ = \prod\limits_{k = 1}^{n - i} 2 $$ and $$\displaystyle (n - i)! = \prod\limits_{k = 1}^{n - i} k $$

 * {| style="width:100%" border="0" align="left"

P_n(x) = \sum_{i=0}^{[n/2]} \frac{(2 n - 2 i)!}{ 2^i \prod\limits_{k = 1}^{n - i} 2 \prod\limits_{k = 1}^{n - i} k } \frac{(-1)^i x^{n-2i}}{ i!(n-2i)!}
 * $$\displaystyle
 * $$\displaystyle

$$ $$
 *
$$\displaystyle (Eq. 5)
 * }
 * }

Since the products in the denominator of Eq. 5 have the same limits, we have that

 * {| style="width:100%" border="0" align="left"

P_n(x) = \sum_{i=0}^{[n/2]} \frac{(2 n - 2 i)!}{2^i \prod\limits_{k = 1}^{n - i} 2k } \frac{(-1)^i x^{n-2i}}{ i!(n-2i)!} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 6)
 * }
 * }

Expanding the numerator and denominator in Eq. 6 gives

 * {| style="width:100%" border="0" align="left"

P_n(x) = \sum_{i=0}^{[n/2]} \frac{1 \cdot \cancel{2}\cdot 3 \cdot \cancel{4} \cdot 5 \cdots (2 n - 2 i-1) \cdot \cancel{2( n - i)}}{2^i \quad \cancel{2} \cdot \cancel{4} \cdots  \cancel{2(n-i)} } \frac{(-1)^i  x^{n-2i}}{ i!(n-2i)!} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 7)
 * }
 * }

Simplifying Eq. 7 yields

 * {| style="width:100%" border="0" align="left"

$$\displaystyle P_n(x) = \sum_{i=0}^{[n/2]} \frac{1 \cdot 3 \cdot 5 \cdots  (2 n - 2 i-1) }{ 2^i i!(n-2i)! } (-1)^i x^{n-2i} $$ $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 *
$$\displaystyle (Eq. 8)
 * }
 * }

Since Eqs. 2 and 8 are the same, Eqs. 1 and 2 are equivalent.

= Problem 6: Legendre Polynomials, generating the first five polynomials =

From lecture slide 31-3.

Given
The first five Legendre polynomials are given by

 * {| style="width:100%" border="0" align="left"

P_0(x) = 1 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 1)
 * }
 * }

 * {| style="width:100%" border="0" align="left"

P_1(x) = x $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 2)
 * }
 * }

 * {| style="width:100%" border="0" align="left"

P_2(x) = \frac{1}{2} \left(3 x ^ 2 - 1\right) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 3)
 * }
 * }

 * {| style="width:100%" border="0" align="left"

P_3(x) = \frac{1}{2} \left(5 x ^ 3 - 3 x\right) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 4)
 * }
 * }

 * {| style="width:100%" border="0" align="left"

P_4(x) = \frac{35}{8} x ^ 4 - \frac{15}{4} x ^ 2 + \frac{3}{8} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 5)
 * }
 * }

Find
Verify that Eqs. 1, 2, 3, 4, and 5 can be written as

 * {| style="width:100%" border="0" align="left"

P_n(x) = \sum_{i=0}^{[n/2]} \frac{(-1)^i (2 n - 2 i)! x^{n - 2i}}{2^n i! (n-i)! (n-2i)!} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 6)
 * }
 * }

or

 * {| style="width:100%" border="0" align="left"

P_n(x) = \sum_{i=0}^{[n/2]} \frac{1 \cdot 3 \cdot \cdot \cdot \left(2n - 2i - 1\right)}{2^i i! (n-2i)!} (-1)^i x^{n-2i} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 7)
 * }
 * }

Solution
Since Eqs. 6 and 7 are equivalent we only need to verify that Eqs. 1 to 5 can be generated with one of them.

1. Substituting $$\displaystyle n=0$$ into Eq. 6 gives

 * {| style="width:100%" border="0" align="left"

P_0(x) = \sum_{i=0}^{[0/2]} \frac{(-1)^i (2 (0) - 2 i)! x^{0 - 2i}}{2^0 i! (0-i)! (0-2i)!}= \sum_{i=0}^{0} \frac{(-1)^i ( - 2 i)! x^{- 2i}}{i! (-i)! (-2i)!} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 8)
 * }
 * }

Evaluating the sum gives

 * {| style="width:100%" border="0" align="left"

P_0(x) = \frac{0!}{0!0!0!} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 9)
 * }
 * }

Evaluating Eq. 9 yields

 * {| style="width:100%" border="0" align="left"

$$\displaystyle P_0(x) = 1 $$ $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 *
$$\displaystyle (Eq. 10)
 * }
 * }

2. Substituting $$\displaystyle n=1$$ into Eq. 6 gives

 * {| style="width:100%" border="0" align="left"

P_1(x) = \sum_{i=0}^{[1/2]} \frac{(-1)^i (2 (1) - 2 i)! x^{1 - 2i}}{2^1 i! (1-i)! (1-2i)!}= \sum_{i=0}^{0} \frac{(-1)^i (2 - 2 i)! x^{1 - 2i}}{2 i! (1-i)! (1-2i)!} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 11)
 * }
 * }

Evaluating the sum gives

 * {| style="width:100%" border="0" align="left"

P_1(x) = \frac{(-1)^0 \cancel{2!} x}{\cancel{2}0!1!1!} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 12)
 * }
 * }

Evaluating Eq. 12 yields

 * {| style="width:100%" border="0" align="left"

$$\displaystyle P_1(x) = x $$ $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 *
$$\displaystyle (Eq. 13)
 * }
 * }

3. Substituting $$\displaystyle n=2$$ into Eq. 6 gives

 * {| style="width:100%" border="0" align="left"

P_2(x) = \sum_{i=0}^{[2/2]} \frac{(-1)^i (2 (2) - 2 i)! x^{2 - 2i}}{2^2 i! (2-i)! (2-2i)!}= \sum_{i=0}^{1} \frac{(-1)^i (4 - 2 i)! x^{2 - 2i}}{4 i! (2-i)! (2-2i)!} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 14)
 * }
 * }

Evaluating the sum gives

 * {| style="width:100%" border="0" align="left"

P_2(x) = \frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot x^{2}}{4 \cdot 1 \cdot 2! \cdot 2!} - \frac{2!}{4 \cdot 1 \cdot 1 \cdot 1} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 15)
 * }
 * }

Evaluating Eq. 15 yields

 * {| style="width:100%" border="0" align="left"

$$\displaystyle P_2(x) = \frac{1}{2}(3x^2-1) $$ $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 *
$$\displaystyle (Eq. 16)
 * }
 * }

4. Substituting $$\displaystyle n=3$$ into Eq. 6 gives

 * {| style="width:100%" border="0" align="left"

P_3(x) = \sum_{i=0}^{[3/2]} \frac{(-1)^i (2 (3) - 2 i)! x^{3 - 2i}}{2^3 i! (3-i)! (3-2i)!}= \sum_{i=0}^{1} \frac{(-1)^i (6 - 2 i)! x^{3 - 2i}}{8 i! (3-i)! (3-2i)!} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 17)
 * }
 * }

Evaluating the sum gives

 * {| style="width:100%" border="0" align="left"

P_3(x) = \frac{6! \cdot x^{3}}{8 \cdot 3! \cdot 3!} - \frac{4! \cdot x}{8 \cdot 2!} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 18)
 * }
 * }

Evaluating Eq. 18 yields

 * {| style="width:100%" border="0" align="left"

$$\displaystyle P_3(x) = \frac{1}{2}(5x^3-3x) $$ $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 *
$$\displaystyle (Eq. 19)
 * }
 * }

5. Substituting $$\displaystyle n=4$$ into Eq. 6 gives

 * {| style="width:100%" border="0" align="left"

P_4(x) = \sum_{i=0}^{[4/2]} \frac{(-1)^i (2 (4) - 2 i)! x^{4 - 2i}}{2^4 i! (4-i)! (4-2i)!}= \sum_{i=0}^{2} \frac{(-1)^i (8 - 2 i)! x^{4 - 2i}}{16 i! (4-i)! (4-2i)!} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 20)
 * }
 * }

Evaluating the sum gives

 * {| style="width:100%" border="0" align="left"

P_4(x) = \frac{ 8! \cdot x^{4}}{16 \cdot (4)! \cdot (4)!} -  \frac{6! \cdot x^{2}}{16 \cdot 3! \cdot 2!} +  \frac{ 4! }{16 \cdot 2! \cdot 2!} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 21)
 * }
 * }

Evaluating Eq. 21 yields

 * {| style="width:100%" border="0" align="left"

$$\displaystyle P_4(x) = \frac{35}{8} x ^ 4 - \frac{15}{4} x ^ 2 + \frac{3}{8} $$ $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 *
$$\displaystyle (Eq. 22)
 * }
 * }

title: User:Egm6321.f09.Team1.vasquez/HW6


text:
Talk:Textbook

What kind of textbook we can discuss here?
 * Pretty much any kind - or discuss about textbooks in general - Wikibooks though is the place for creating online textbooks. -- Jtneill - Talk - c 22:49, 13 November 2009 (UTC)

title: Talk:Textbook


text:
UTPA STEM/CBI Courses/Statics

Topics

-

Force Equilibrium

Truss Bridge

title: UTPA STEM/CBI Courses/Statics


text:
UTPA STEM/CBI Courses/CAM

-

Introduction

-

Computer Aided Manufacturing is a very important process when designing and maintaining the automation of manufacturing systems.

Do you have control!!!

Can you open and close a grarge door? Can you make a robot turn? Solar cell works most efficiently when the cells are faced to the sun. Where is your solar cell is facing? All these problems start with very simple DC motor control. If you know how to make motor turn CW (clockwise) and CCW (counter-clockwise) directions, then you have control.

-

Topics

-

Lab 01 Do you have control???

Lab 02 Let's get cars moving!

-

title: UTPA STEM/CBI Courses/CAM


text:
UTPA STEM/CBI Courses/Computer Networks

Topics

-

Socket Programming

-

title: UTPA STEM/CBI Courses/Computer Networks


text:
UTPA STEM/CBI Courses Organized by STEM Category/Biology

Courses
Anatomy and Physiology

Aquatic Entomology

Animal Parasitology

/Botany/

Comparative Embryology

/Life Sciences, Survey of/

Microbiology

Nutrition

Organic Chemistry

Topics
Cancer Biology

The Cell

Surface Area to Volume Ratio

-


text:
UTPA STEM/CBI Courses/Calculus2

Course Description:

Representing Functions As Power Series

title: UTPA STEM/CBI Courses/Calculus2


text:
UTPA STEM/CBI Courses/Organic Chemistry

Topics

-

/Chromatography/

/Molecular Shapes/

/Redox of Organic Molecules/

-

title: UTPA STEM/CBI Courses/Organic Chemistry


text:
UTPA STEM/CBI Courses/Mechatronics

Topics

-

/Basic Electronics/

/Thermocouples, Amplifiers and Comparators/

-

title: UTPA STEM/CBI Courses/Mechatronics


text:
Pi-Conjugated Materials

Course Title: Organic Chemistry I (CHEM 2302)

Lecture Topic: Acid-Bases

Instructor: Jose J. Gutierrez

Institution: The University of Texas Pan American

Backwards Design
Course Objectives

 * Primary Objectives- By the next class period students will be able to:
 * Identify acids and bases
 * Predict the outcome of acid-base reactions
 * Identify the difference between redox reactions on organic molecules

 * Sub Objectives- The objectives will require that students be able to:
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 * Difficulties- Students may have difficulty:
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 * Real-World Contexts- There are many ways that students can use this material in the real-world, such as:
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Model of Knowledge

 * Concept Map
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 * Content Priorities
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 * Important to Do and Know
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 * Worth Being Familiar with
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Assessment of Learning

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Legacy Cycle
OBJECTIVE

By the next class period, students will be able to: The objectives will require that students be able to:
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THE CHALLENGE

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MULTIPLE PERSPECTIVES

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RESEARCH & REVISE

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TEST YOUR METTLE

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GO PUBLIC

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title: Pi-Conjugated Materials


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UTPA STEM/CBI Courses/Biomedical

Emergency Response Robot

title: UTPA STEM/CBI Courses/Biomedical


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UTPA STEM/CBI Courses/CAM/Challenge Name

Course Title: CAM

Lecture Topic: PLC Programming # 1

Instructor: Kevin Lee

Institution: UTPA/ MANE

Backwards Design
Course Objectives

 * Primary Objectives- By the next class period students will be able to:
 * Understand ladder logic and wiring motor for CW and CCW motion.
 * Wiring a motor that opens a garage door up and down.

 * Sub Objectives- The objectives will require that students be able to:
 * Understand PLC structure
 * Understand Ladder Logic
 * Identify necessary components
 * Wiring a motor that can rotate CW and CCW.

 * Difficulties- Students may have difficulty:
 * Understanding PLC architecture
 * Writing a ladder logic
 * Wiring a motor

 * Real-World Contexts- There are many ways that students can use this material in the real-world, such as:
 * Garage door opener
 * Radar rotating control
 * Solar cell aim to the Sun.

Model of Knowledge

 * Concept Map
 * How to write a ladder logic
 * Identify flow chart
 * Identify input
 * How a motor works?
 * Electric wiring

 * Content Priorities
 * Enduring Understanding
 * How to program a PLC
 * Motor control

 * Important to Do and Know
 * Become familiar programming and wiring for a motor control
 * Worth Being Familiar with
 * DC Motor
 * Basic wiring skills

Assessment of Learning

 * Formative Assessment
 * In Class (groups)
 * text
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 * Homework (individual)
 * text
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 * Summative Assessment
 * text
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Legacy Cycle
OBJECTIVE

By the next class period, students will be able to: The objectives will require that students be able to:
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THE CHALLENGE

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GENERATE IDEAS

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MULTIPLE PERSPECTIVES

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RESEARCH & REVISE

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TEST YOUR METTLE

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GO PUBLIC

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UTPA STEM/CBI Courses/Graduate Seminar (combined with BVP-MATH 4318)

Topics

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How to use Maxwell Model to Reconstruct Radar Image

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UTPA STEM/CBI Courses/Graduate Seminar (combined with BVP-MATH 4318)/How to use Maxwell Model to Reconstruct Radar Image

Course Title: How to use Maxwell Model to Reconstruct Radar Image

Lecture Topic:

Instructor: Zhijun Qiao

Institution: UTPA - Math Dept

Backwards Design
Course Objectives

 * Primary Objectives- By the next class period students will be able to:
 * Understand the model and how to solve the Maxwell equation

 * Sub Objectives- The objectives will require that students be able to:
 * What basic concepts
 * What approach/method used

 * Difficulties- Students may have difficulty:
 * How to apply the boundary values to get solution
 * students may encounter the problems how to solve the ODE reduced from the Maxwell equation

 * Real-World Contexts- There are many ways that students can use this material in the real-world, such as:
 * Radar Wave
 * Water Wave

Model of Knowledge

 * Concept Map
 * Boundary Condition
 * Domain for Solution
 * Equations nature (describe magnetic-wave)

 * Content Priorities
 * Enduring Understanding
 * Derivatives
 * Origin of Maxwell equation

 * Important to Do and Know
 * Separation of Variabls

 * Worth Being Familiar with
 * Calculus III
 * ODEs

Assessment of Learning

 * Formative Assessment
 * In Class (groups)
 * Discussion of Separation of Variabls
 * Discussion of solving ODEs
 * Homework (individual)
 * Get familiar with ODEs and solve them
 * Calculus III
 * Summative Assessment
 * Discussion in group's conclusion

Legacy Cycle
OBJECTIVE

By the next class period, students will be able to: The objectives will require that students be able to:
 * understand the math model -- PDEs (each term in model)
 * understand boundary conditions
 * understand initial conditions and scource signal function
 * explain model by themself
 * try to replated the system to the radar signal process
 * figure out source function to match the initial or boundary conditions.

THE CHALLENGE

How to use Maxwell euqations to reconstruct the radar image for real targets

GENERATE IDEAS

This seminar is to mentor the graduate students participating in my DoD research project on PDE Analysis and Radar Image Reconstruction to enhance achievement level and retention rate here at UTPA. The program students are anticipated to apply mathematical and engeneering methodology to deal with radar imaging problems appearing in military, air force and naval affairs. I deliver PDEs lectures to my DoD program students and direct them to do preliminary work for my DoD research project on Radar Image Reconstruction.

Ideas: solving model using fundamental solutions; point spread functions; delta functions; Solutions for Inverse Problems based on radar data; and numeric simulations.

MULTIPLE PERSPECTIVES

Handle the method of separation of variables and wave front set, and delta function/point spread function skills.

RESEARCH & REVISE

This is my DoD research project on PDE Analysis and Radar Image Reconstruction. The program students are anticipated to apply mathematical and engeneering methodology to deal with radar imaging problems appearing in military, air force and naval affairs. I will give special PDEs lectures to my DoD program students and direct them to do preliminary work for my DoD research project on Radar Image Reconstruction.

TEST YOUR METTLE

Verify every step works fine. We plan to use radar data to test our procedure and solution of inverse problem as well as numeric simulations. GO PUBLIC

Finally, our students will attend national and international conferences to show their work done through the project. Also, we plan to publish our work in some refereed journal or proceedings.

Pre-Lesson Quiz

 * 1) Linear PDEs
 * 2) Initial Conditions
 * 3) Boundary Conditions
 * 4) Separation of Variables
 * 5) Transformation through some variables
 * 6) Integrals formulas
 * 7) Partial Derivatives
 * 8) Electro-magnetic wave model
 * 9) Range Image
 * 10) Crossing Range Image
 * 11) Delta function
 * 12) Point Spread Function
 * 13) Sinc function
 * 14) Time sampling
 * 15) Fourier integrals
 * 16) Fourier series
 * 17) Fourier Integral Operators

Test Your Mettle Quiz

 * 1) Scattering Data
 * 2) Incidental Data
 * 3) Electro-magnetic wave
 * 4) Slow time image
 * 5) Fast time image
 * 6) Fundamental solutions
 * 7) Fourier Integral Operators
 * 8) Time sampling
 * 9) Inverse problem
 * 10) Sinc function and refectivity function
 * 11) Numeric Simulations

129.113.150.41 21:06, 19 January 2010 (UTC)It would be useful to include the actual questions for the pre-lesson quiz and test your mettle quiz


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UTPA STEM/CBI Courses/Surface Area to Volume Ratio

Course Title: General Biology 2

Lecture Topic: Surface Area to Volume Ratio

Instructor: Matthew Terry

Institution: UTPA

Backwards Design
Course Objectives

 * Primary Objectives- By the next class period students will be able to:
 * Know what happens to surface area as volume increases
 * Be able to give real world biological examples of the consequences of this
 * Be aware of ways organisms get around this issue
 * Predict habitat of aquatic insects based on morphology

 * Sub Objectives- The objectives will require that students be able to:
 * Calculate surface area and volume
 * Think of examples from the biological world

 * Difficulties- Students may have difficulty:
 * Graphing

 * Real-World Contexts- There are many ways that students can use this material in the real-world, such as:
 * Determine if a structure functions as an organ for diffusion or temperature regulation based on morphology
 * Be able to explain how specific behaviors compensate for these principles

Model of Knowledge

 * Concept Map
 * What is surface area?
 * What is volume?
 * How are volume and surface area connected?
 * How does the shape of an animal affect these?
 * Real world examples (Emperor penguins, lungs, gills, small intestine, whales, etc.)
 * How are diffusion and heat conservation connected?

 * Content Priorities
 * Enduring Understanding
 * Surface Area to Volume ratio increase exponentially
 * This phenomenon affects the shape and behavior of organisms
 * Important to Do and Know
 * Calculate and graph surface area and volume for cube as it increases in size
 * Use EXCEL to observe the relations of this change
 * Demonstrate diffusion using agar and dye
 * Demonstrate cooling of water in different volumes and shapes
 * Worth Being Familiar with
 * How do insects compensate for this?
 * What organs are shaped to use this phenomenon?
 * Penguin example

Assessment of Learning

 * Formative Assessment
 * In Class (groups)
 * Compare measurements and graphs of different groups
 * Get examples of organs and/or organism behavior that demonstrate this principle
 * Place aquatic insects in their environment based on how they get their oxygen
 * Homework (individual)
 * Complete and turn in a lab report
 * Summative Assessment
 * Weekly Quiz
 * Final Exam

Legacy Cycle
OBJECTIVE

By the next class period, students will be able to: The objectives will require that students be able to: THE CHALLENGE
 * Know what happens to surface area as volume increases
 * Be able to give real world biological examples of the consequences of this
 * Know ways that organisms get around this issue
 * Predict habitat of aquatic insects based on morphology
 * Calculate volume and surface area
 * Use excel
 * Make logical predictions based on the learned concepts and observation

It's Not Easy Being Big: How does the relationship between surface area and volume impact living organisms?

GENERATE IDEAS

In class brainstorming and discussion

MULTIPLE PERSPECTIVES
 * Emperor Penguin example
 * Tissue pictures

RESEARCH & REVISE

Written and oral feedback from students

TEST YOUR METTLE
 * Open oral questions to the class
 * Written quiz at the beginning of the next lab

GO PUBLIC

Groups present their graphical results to the class

Pre-Lesson Quiz

 * 1) How do you calculate the volume of a cube?
 * 2) How do you calculate the surface area of a cube?
 * 3) What happens to the surface area of a cube as you increase its size (volume)?
 * 4) Does this occur at the same rate as you increase volume?
 * 5) Would this process change if you used a different shape, such as a sphere?
 * 6) What cools faster, 100 ml of water or 1000 ml of water?
 * 7) How might you slow the rate at which the water cools?
 * 8) Name an organ for which diffusion is a major part of its function.
 * 9) Name one strategy that organisms use to slow the amount of heat they lose.
 * 10) Why does a kangaroo lick its arms?

Test Your Mettle Quiz

 * 1) Which graph best represents the surface area to volume ratio of an elephant as it grows from a baby to an adult?
 * 2) What is the function of the alveoli in your lungs?
 * 3) How does asthma interfere with this function?
 * 4) Why are there no small marine mammals?
 * 5) Why are there small marine animals that are not mammals?
 * 6) What are three forms of insulation for arctic animals?
 * 7) How do emperor penguins survive the Antarctic winters?
 * 8) Gills in the diagram below are colored red. Which of the insects comes from an oxygen poor environment?
 * 9) Why are there no large insects?
 * 10) Describe the phenomenon of counter exchange and how it helps endotherms maintain their body temperature.


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UTPA STEM/CBI Courses/CAM/Course Description

Computer Aided Design is very important part of automation of manufacturing system.

Do you have control!!!


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UTPA STEM/CBI Courses/Calculus/Solving Right Triangles

Course Title: PreCalculus (Math 2412)

Lecture Topic: Solving Right Triangles

Instructor: Lily Rai

Institution: South Texas College

Backwards Design
Course Objectives

 * Primary Objectives- By the end of the next class period students will be able to:
 * Solve application problems involving angles of elevation, angles of depression and bearing.

 * Sub Objectives- The objectives will require that students be able to:
 * Identify angles in standard position, co-terminal angles and reference angles.
 * Define trigonometric functions using right triangles.
 * Explain angles of elevation, angles of depression and bearing.

 * Difficulties-
 * Some students may face difficulty in distinguishing between angles in standard position, reference angles and co-terminal angles.
 * Many students will have difficulty in understanding the word problems and translating them into mathematical equations.

 * Real-World Contexts- There are many ways that students can use this material in the real-world, such as:
 * There are many real life applications. One of them is the design of highway curves where outside of a curve is often slightly elevated or inclined above the inside of the curve. This inclination is called Super-elevation.
 * Another application is finding the minimum distance of a plant from a fence which requires full sun during winter.
 * Calculating the distance between two cities or two ships, or the distance flown by a plane.
 * Measuring the height of buildings, length of shadows, height of an antenna etc.

Model of Knowledge

 * Concept Map
 * Understand angles in standard position, reference angles and co-terminal angles
 * Define Trigonometric Functions using Right triangles
 * Solve right triangles given: angle and side, two sides
 * Find angle of elevation when lengths are known
 * Solve problems involving bearing angles of elevation and depression

 * Content Priorities
 * Enduring Understanding
 * Solving right triangles given different conditions
 * Important to Do and Know
 * Given an application problem, understand, sketch, analyze, solve, and check for correctness.
 * Worth Being Familiar with
 * Pythagorean Theorem, Bearing: a concept in navigation; angles of elevation and depression
 * Calculating distance between any two points given the coordinates
 * To find the length of an arc; to measure the area of a sector

Assessment of Learning

 * Formative Assessment
 * In Class (groups)
 * Students to work in groups on selected application problems and present it to the entire class
 * Homework (individual)
 * Students will work on assigned problems and bring the difficulties to the next class
 * Summative Assessment
 * In class quiz over the lesson

Legacy Cycle
OBJECTIVE

By the next class period, students will be able to: The objectives will require that students be able to:
 * Differentiate between angles in standard position, co-terminal angles, and reference angles
 * Solve the right triangle given an angle or a side or two sides
 * Solve application problems involving bearing, and angles of elevation and depression
 * Recognize a word problem as an application problem involving right triangle trigonometry
 * Solve application problems involving bearing, angles of elevation and depression, and area of a sector

THE CHALLENGE

The students will be presented with the following problem: "When an automobile travels around a curve on a highway, objects like trees, buildings and fences situated on the curve may obstruct a driver's vision. How far inside the curve of some land must be cleared to provide visibility for a safe stopping distance?"

GENERATE IDEAS

Students will be given a few minutes to individually write down their answers to the questions. Students may respond with answers as: 1) Need to know the arc length of the curve and also the area of the sector 2) Need to know the central angle and radius 3) Speed limit posted, speed of the vehicle etc. 4) Need to come up with the formula for the distance

MULTIPLE PERSPECTIVES

The Faculty will review the concepts of length of an arc, area of a sector; and then will introduce the concepts of trigonometric functions using right triangles.

RESEARCH & REVISE

Students would be provided with the actual questions on modeling highway curves and stopping distance. Students will work in groups.

TEST YOUR METTLE

The groups will make presentations to the class and will receive the feedback. The Faculty will give the correct analysis and steps to solve the problem. Students will be presented with a quiz at the beginning of the following class. GO PUBLIC

Students will be able to answer similar questions presented at the beginning of the lesson and would be asked to turn in a brief write up.