text:
User:Egm6321.f10.team2.reserve/HW6

= Problem 6.2 =

From the lecture slide Mtg 33-1

Given

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$$ \displaystyle

(x-1)y''-xy'+y=f(x)$$

(Eq 2.1)
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 * }

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$$ \displaystyle

xy''+2y'+xy=f(x)$$

(Eq 2.2)
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 * }

Find
Solve the equations above for $$\displaystyle f(x)=0$$ and $$\displaystyle f(x)=\sin x$$; so

 * a)	Eq 2.1 with $$\displaystyle f(x)=0$$

 * b)	Eq 2.1 with $$\displaystyle f(x)=\sin x$$

 * c)	Eq 2.2 with $$\displaystyle f(x)=0$$

 * d)	Eq 2.2 with $$\displaystyle f(x)=\sin x$$

Solution
First of all applying the generic form of $$\displaystyle y={{x}^{c}}{{e}^{rx}}$$ and then taking the first and the second derivatives will result in the following:

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$$ \displaystyle

\begin{align} & y'=c{{x}^{c-1}}{{e}^{rx}}+{{x}^{c}}r{{e}^{rx}} \\ & y''=c(c-1){{x}^{c-2}}{{e}^{rx}}+c{{x}^{c-1}}r{{e}^{rx}}+c{{x}^{c-1}}r{{e}^{rx}}+{{x}^{c}}rr{{e}^{rx}} \\ & y''=c(c-1){{x}^{c-2}}{{e}^{rx}}+2c{{x}^{c-1}}r{{e}^{rx}}+{{x}^{c}}{{r}^{2}}{{e}^{rx}} \\ \end{align}$$

(Eqs 2.3)
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 * }

The procedure above will be the same for both part(a) and part(c) parts of the problem.

Part a
Now that we found the derivatives, the solutions can be plugged into the Eq 2.1 that is set $$\displaystyle f(x)=0$$:

 * $$\displaystyle (x-1)y''-xy'+y=0$$

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$$ \displaystyle

(x-1)*\underbrace{[c(c-1){{x}^{c-2}}{{e}^{rx}}+2c{{x}^{c-1}}r{{e}^{rx}}+{{x}^{c}}{{r}^{2}}{{e}^{rx}}]}_{y''}-x*\underbrace{[c{{x}^{c-1}}{{e}^{rx}}+{{x}^{c}}r{{e}^{rx}}]}_{y'}+\underbrace{[{{x}^{c}}{{e}^{rx}}]}_{y}=0$$

(Eq 2.4)
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 * }

Since every term in the equation has $$\displaystyle {{e}^{rx}}$$, it can be taken out of the parenthesis to be eliminated:

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$$ \displaystyle

(x-1)*{{e}^{rx}}[c(c-1){{x}^{c-2}}+2c{{x}^{c-1}}r+{{x}^{c}}{{r}^{2}}]-x*{{e}^{rx}}[c{{x}^{c-1}}+{{x}^{c}}r]+{{e}^{rx}}[{{x}^{c}}]=0$$

(Eq 2.5)
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 * }

To simplify the equation we will go through the following procedure:

 * $$\displaystyle \begin{align}

& (x-1)[c(c-1){{x}^{c-2}}+2c{{x}^{c-1}}r+{{x}^{c}}{{r}^{2}}]-x[c{{x}^{c-1}}+{{x}^{c}}r]+[{{x}^{c}}]=0 \\ & ({{c}^{2}}-c){{x}^{c-1}}+2c{{x}^{c}}r+{{x}^{c+1}}{{r}^{2}}-({{c}^{2}}-c){{x}^{c-2}}-2c{{x}^{c-1}}r-{{x}^{c}}{{r}^{2}}-c{{x}^{c}}-{{x}^{c+1}}r+{{x}^{c}}=0 \\ & {{x}^{c+1}}({{r}^{2}}-r)+{{x}^{c}}(2cr-{{r}^{2}}-c+1)+{{x}^{c-1}}({{c}^{2}}-c-2cr)+{{x}^{c-2}}(c-{{c}^{2}})=0 \\ \end{align}$$

When $$\displaystyle c=1\Rightarrow r=0$$ or $$\displaystyle c=0\Rightarrow r=1$$, which will yield us two homogenous solutions plugging into $$\displaystyle y={{x}^{c}}{{e}^{rx}}$$.

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$$ \displaystyle

\begin{align} & {{y}_{H1}}={{x}^{0}}{{e}^{1x}}={{e}^{x}} \\ & {{y}_{H2}}={{x}^{1}}{{e}^{0x}}=x \\ \end{align}$$

(Eqs 2.6)
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Then,

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$$ \displaystyle
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y={{C}_{1}}{{e}^{x}}+{{C}_{2}}x$$

(Eq 2.7)
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 * }

Part b
When we set $$\displaystyle f(x)=\sin x$$ for Eq 2.1, we get:

 * $$\displaystyle (x-1)y''-xy'+y=\sin x$$

We will rewrite the equation to put in the proper form and find the integrating factor:

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$$ \displaystyle

y''-\frac{x}{(x-1)}y'+\frac{1}{(x-1)}y=\frac{\sin x}{(x-1)}$$

(Eq 2.8) Plugging in all the values into the integrating factor formula, we get:
 *
 * }

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$$ \displaystyle

h(x)=\underbrace_{y_{H1}^{2}}{{e}^{-\int{\frac{x}{(x-1)}dx}}}={{e}^{2x}}{{e}^{-\ln (x-1)-x}}={{e}^{x-\ln (x-1)}}={{e}^{x}}{{e}^{\ln \frac{1}{(x-1)}}}=\frac{(x-1)}$$

(Eq 2.9) Now to find the particular solution we will apply the following formula:
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 * }

 * $$\displaystyle {{y}_{P}}(x)={{u}_{1}}(x)\int{\frac{1}{h(x)}}\left[ \int{h(x)F(x)dx} \right]dx$$

$$ \displaystyle

{{y}_{P}}(x)={{e}^{x}}\int{\frac{(x-1)}}\left[ \int{\frac{{{e}^{x}}\sin x}{(x-1)(x-1)}dx} \right]dx={{e}^{x}}\int{\frac{(x-1)}}\left[ \int{\frac{{{e}^{x}}\sin x}dx} \right]dx$$

(Eq 2.10) Finally combining homogenous solution with the particular solution we get:
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 * }
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$$ \displaystyle
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y={{C}_{1}}{{e}^{x}}+{{C}_{2}}x+{{e}^{x}}\int{\frac{(x-1)}}\left[ \int{\frac{{{e}^{x}}\sin x}dx} \right]dx$$

(Eq 2.11)
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 * }

Part c

 * $$\displaystyle xy''+2y'+xy=0$$

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$$ \displaystyle

x*\underbrace{[c(c-1){{x}^{c-2}}{{e}^{rx}}+2c{{x}^{c-1}}r{{e}^{rx}}+{{x}^{c}}{{r}^{2}}{{e}^{rx}}]}_{y''}+2*\underbrace{[c{{x}^{c-1}}{{e}^{rx}}+{{x}^{c}}r{{e}^{rx}}]}_{y'}+x*\underbrace{[{{x}^{c}}{{e}^{rx}}]}_{y}=0$$

(Eq 2.12) Just like the way it was done in part a, the $$\displaystyle {{e}^{rx}}$$ term will be taken out and the equation will be simplified:
 *
 * }
 * $$\begin{align}

& x*[c(c-1){{x}^{c-2}}+2c{{x}^{c-1}}r+{{x}^{c}}{{r}^{2}}]+2*[c{{x}^{c-1}}+{{x}^{c}}r]+x*[{{x}^{c}}]=0 \\ & ({{c}^{2}}-c){{x}^{c-1}}+2cr{{x}^{c}}+{{r}^{2}}{{x}^{c+1}}+2c{{x}^{c-1}}+2{{x}^{c}}r+{{x}^{c+1}}=0 \\ & {{x}^{c+1}}({{r}^{2}}+1)+{{x}^{c}}(2cr+2r)+{{x}^{c-1}}({{c}^{2}}+c)=0 \\ \end{align}$$

By solving the above equation we get $$\displaystyle c=-1\Rightarrow r=\pm i$$

Plugging these values into $$\displaystyle y={{x}^{c}}{{e}^{rx}}$$, will give us the following equations:

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$$ \displaystyle

\begin{align} & {{y}_{H1}}={{x}^{-1}}{{e}^{ix}}=\frac{x} \\ & {{y}_{H2}}={{x}^{-1}}{{e}^{-ix}}=\frac{1}{x{{e}^{ix}}} \\ \end{align}$$

(Eqs 2.13)
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 * }

Then,

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$$ \displaystyle
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y={{C}_{1}}\frac{x}+{{C}_{2}}\frac{1}{x{{e}^{ix}}}$$

(Eq 2.14)
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 * }

Part d
At last but not least Eq 2.2 will be solved with $$\displaystyle f(x)=\sin x$$.

 * $$\displaystyle xy''+2y'+xy=\sin x$$

Rearranging the equation to put in the proper form and find the integrating factor:

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$$ \displaystyle

y''+\frac{2}{x}y'+y=\frac{\sin x}{x}$$

(Eq 2.15) Plugging in all the values into the integrating factor formula, we get:
 *
 * }

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$$ \displaystyle

h(x)=\frac{{e}^{\int{\frac{2}{x}dx}}}=\frac{{e}^{\ln {{x}^{2}}}}=\frac{{x}^{2}}={{e}^{2ix}}$$

(Eq 2.16) Now to find the particular solution we will apply the following formula:
 *
 * }

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$$ \displaystyle

{{y}_{P}}(x)={{u}_{1}}(x)\int{\frac{1}{h(x)}}\left[ \int{h(x)F(x)dx} \right]dx$$

(Eq 2.17)
 *

 * }

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$$ \displaystyle

{{y}_{P}}(x)=\frac{x}\int{\frac{1}}\left[ \int{\frac{{{e}^{2ix}}\sin x}{x}dx} \right]dx$$

(Eq 2.18) Finally combining homogenous solution with the particular solution we get:
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 * }

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$$ \displaystyle
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y={{C}_{1}}\frac{x}+{{C}_{2}}\frac{1}{x{{e}^{ix}}}+\frac{x}\int{\frac{1}}\left[ \int{\frac{{{e}^{2ix}}\sin x}{x}dx} \right]dx$$

(Eq 2.19)
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 * }

= Problem 6.3 =

From the lecture slide Mtg 33-1

Given

 * $$\displaystyle (1-{{x}^{2}})y''-2xy'+2y=\frac{1}{(1-{{x}^{2}})}$$

Find
Solve equation using direct method.

Solution
The two homogeneous were given in class lecture 30-2 as the following:

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$$ \displaystyle

\begin{align} & {{u}_{1}}(x)=x \\ & {{u}_{2}}(x)=\frac{x}{2}\log \left( \frac{1+x}{1-x} \right)-1 \\ \end{align}$$

(Eqs 3.1) $$\displaystyle {{y}_{H}}$$ can be expressed as
 *

 * }
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$$ \displaystyle

{{y}_{H}}={{C}_{1}}x+{{C}_{2}}\left[ \frac{x}{2}\log \left( \frac{1+x}{1-x} \right)-1 \right]$$

(Eq 3.2)
 *

 * }

Using the example from the textbook on page 34 we can use the variation of parameters formula that is

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$$ \displaystyle

y(x)=\int_ – ^{x}{f(s)\left\{ \frac{{{u}_{1}}(s){{u}_{2}}(x)-{{u}_{1}}(x){{u}_{2}}(s)}{W(s)} \right\}}ds$$

(Eq 3.3)
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 * }

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$$ \displaystyle

{{y}_{P}}=\int_ – ^{x}{\left\{ \frac{{{u}_{1}}(s){{u}_{2}}(x)-{{u}_{1}}(x){{u}_{2}}(s)}{(1-{{s}^{2}})[{{u}_{1}}(s)u_{2}^{'}(s)-u_{1}^{'}(s)u_{2}^ – (s)} \right\}}ds$$

(Eq 3.4)
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 * }

 * $$\displaystyle W=\frac{1-{{s}^{2}}}$$

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$$ \displaystyle

\begin{align} {{y}_{P}}(x) & ={{u}_{2}}(x)\int{{{u}_{1}}(x)dx-}{{u}_{1}}(x)\int{{{u}_{2}}(x)dx} \\ & =\left( \frac{x}{2}\log \left( \frac{1+x}{1-x} \right)-1 \right)\int{xdx-}x\int{\left( \frac{x}{2}\log \left( \frac{1+x}{1-x} \right)-1 \right)dx} \\ & =\left( \frac{x}{2}\log \left( \frac{1+x}{1-x} \right)-1 \right)\frac{2}-\left( \frac{{{x}^{3}}-x}{4}\log \left( \frac{1+x}{1-x} \right)-\frac{2} \right) \\ & =\frac{x}{4}\log \left( \frac{1+x}{1-x} \right) \\ \end{align}$$

(Eq 3.5) So we can express the general solution as the following:
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 * }

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$$ \displaystyle
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y={{C}_{1}}x+{{C}_{2}}\left[ \frac{x}{2}\log \left( \frac{1+x}{1-x} \right)-1 \right]+\frac{x}{4}\log \left( \frac{1+x}{1-x} \right)$$

(Eq 3.6)
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 * }

= Problem 6.6 Finding Laplacian in spherical coordinates using Math/Physics convention =

From the lecture slide Mtg 35-4

Given
The Laplacian in spherical coordinates using astronomy convention is ,

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$$ \displaystyle

\Delta \Psi =\frac{1}\frac{\partial }{\partial r}\left( {{r}^{2}}\frac{\partial \Psi }{\partial r} \right)+\frac{1}{{{r}^{2}}{{\cos }^{2}}\theta }\frac{{{\partial }^{2}}\psi }{\partial {{\varphi }^{2}}}+\frac{1}{{{r}^{2}}\cos \theta }\frac{\partial }{\partial \theta }\left( \cos \theta \frac{\partial \Psi }{\partial \theta } \right)$$

(Eq 6.1)
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 * }

where $$\displaystyle \theta \in [-\pi /2,+\pi /2]$$

Find
Laplacian in spherical coordinates using Math/Physics convention: $$\displaystyle r,\bar{\theta },\varphi $$ where,

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$$ \displaystyle

\bar{\theta }=\frac{\pi }{2}-\theta $$

(Eq 6.2)
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Solution
Sustitute Eq 6.2 into Eq 6.1 we get,

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$$ \displaystyle
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\begin{align} \Delta \Psi & =\frac{1}\frac{\partial }{\partial r}\left( {{r}^{2}}\frac{\partial \Psi }{\partial r} \right)+\frac{1}{{{r}^{2}}{{\cos }^{2}}(\pi /2-\bar{\theta })}\frac{{{\partial }^{2}}\psi }{\partial {{\varphi }^{2}}}+\frac{1}{{{r}^{2}}\cos (\pi /2-\bar{\theta })}\frac{\partial }{\partial (\pi /2-\bar{\theta })}\left( \cos (\pi /2-\bar{\theta })\frac{\partial \Psi }{\partial (\pi /2-\bar{\theta })} \right) \\ & =\frac{1}\frac{\partial }{\partial r}\left( {{r}^{2}}\frac{\partial \Psi }{\partial r} \right)+\frac{1}{{{r}^{2}}{{\sin }^{2}}\bar{\theta }}\frac{{{\partial }^{2}}\psi }{\partial {{\varphi }^{2}}}+\frac{1}{{{r}^{2}}\sin \bar{\theta }}\frac{\partial }{\partial \bar{\theta }}\left( \sin \bar{\theta }\frac{\partial \Psi }{\partial \bar{\theta }} \right) \\ \end{align}$$

(Eq6.3)
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 * }

If we don't know Laplacian in astronomy convention at first
Then the transmation between Cartesian coodinates and spherical coordinates(Math/Physics convention) is,

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$$ \displaystyle

\begin{align} & x=r\sin \bar{\theta }\cos \varphi \\ & y=r\cos \bar{\theta }\sin \varphi \\ & z=r\cos \bar{\theta } \\ \end{align}$$

(Eqs 6.4)
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 * }

Hence,

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$$ \displaystyle

\begin{align} & dx=\sin \bar{\theta }\cos \varphi dr+r\cos \bar{\theta }\cos \varphi d\bar{\theta }-r\sin \bar{\theta }\sin \varphi d\varphi \\ & dy=\sin \bar{\theta }\sin \varphi dr+r\cos \bar{\theta }\sin \varphi d\bar{\theta }+r\sin \bar{\theta }\cos \varphi d\varphi \\ & dz=\cos \bar{\theta }dr-r\sin \bar{\theta }d\bar{\theta } \\ \end{align}$$

(Eqs 6.5)
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 * }

Consider the magnitude of an infinitesimal segment $$\displaystyle ds$$,

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$$ \displaystyle

d{{s}^{2}}=d{{r}^{2}}+{{r}^{2}}d{^{2}}+{{r}^{2}}{{\sin }^{2}}\bar{\theta }d{{\varphi }^{2}}$$

(Eq 6.6)
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 * }

Then we have scale factors,

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$$ \displaystyle

\begin{align} & {{h}_{r}}=1 \\ & {{h}_}=r \\ & {{h}_{\varphi }}=r\sin \bar{\theta } \\ \end{align}$$

(Eqs 6.7)
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 * }

From lecture slide 34-2 we know the Laplacian for curvilinear coordinates is:

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$$ \displaystyle

\Delta \Psi =\frac{1}\sum\limits_{i=1}^{3}{\frac{\partial }{\partial {{\xi }_{i}}}\left[ \frac{h_{i}^{2}}\frac{\partial \Phi }{\partial {{\xi }_{i}}} \right]}$$

(Eq 6.8)
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 * }

Substitute Eqs 6.7 into Eq 6.8 we have,

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$$ \displaystyle
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\Delta \Psi =\frac{1}\frac{\partial }{\partial r}\left( {{r}^{2}}\frac{\partial \Psi }{\partial r} \right)+\frac{1}{{{r}^{2}}{{\sin }^{2}}\bar{\theta }}\frac{{{\partial }^{2}}\psi }{\partial {{\varphi }^{2}}}+\frac{1}{{{r}^{2}}\sin \bar{\theta }}\frac{\partial }{\partial \bar{\theta }}\left( \sin \bar{\theta }\frac{\partial \Psi }{\partial \bar{\theta }} \right)$$

(Eq 6.9)
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 * }

= Problem 6.7 Verify expression for $$\displaystyle \Delta \Psi $$ in elliptic coordinates from wikipedia article =

From the lecture slide Mtg 36-1

Given
The most common two-dimension definition of elliptic coordinates $$\displaystyle (\mu, \nu)$$ is

x = a \ \cosh \mu \ \cos \nu $$

y = a \ \sinh \mu \ \sin \nu $$

where $$\displaystyle \mu$$ is a nonnegative real number and $$\displaystyle \nu \in [0, 2\pi].$$

Find
The Laplacian $$\displaystyle \Delta \Psi $$ expressed in elliptic coordinates.

Solution
From the information given we have:

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$$ \displaystyle \begin{align} & dx=a\sinh \mu \cos \nu d\mu -a\cosh \mu \sin \nu dv \\ & dy=a\sinh \mu \cos \nu d\mu +a\cosh \mu \sin \nu dv \\ \end{align}$$

(Eqs 7.1)
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 * }

Consider magnitude of an infinitesimal segment $$\displaystyle ds$$. Before that we should aware that since ellipic coordinates system is among ORTHOGONAL curvilinear coordinates which never have off-diagonal terms in their metric tensor, meaning terms like $$\displaystyle dxdy$$ will never show up in the magnitude of $$\displaystyle ds$$ thus no need to be calculated.

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$$ \displaystyle

\begin{align} d{{s}^{2}} & =d{{x}^{2}}+d{{y}^{2}} \\ & =({{a}^{2}}{{\sinh }^{2}}\mu {{\cos }^{2}}\nu +a{{\cosh }^{2}}\mu {{\sin }^{2}}\nu )(d{{\mu }^{2}}+d{{v}^{2}}) \\ & =\left[ ({{a}^{2}}{{\sinh }^{2}}\mu {{\cos }^{2}}\nu +{{a}^{2}}{{\sinh }^{2}}\mu {{\sin }^{2}}\nu )+(a{{\cosh }^{2}}\mu {{\sin }^{2}}\nu +{{a}^{2}}{{\sinh }^{2}}\mu {{\sin }^{2}}\nu ) \right](d{{\mu }^{2}}+d{{v}^{2}}) \\ & ={{a}^{2}}({{\sinh }^{2}}\mu +{{\sin }^{2}}\nu )d{{\mu }^{2}}+{{a}^{2}}({{\sinh }^{2}}\mu +{{\sin }^{2}}\nu )d{{v}^{2}} \\ \end{align}$$

(Eq 7.2)
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 * }

We can derive scale factors from Eq 7.2,

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$$ \displaystyle

{{h}_{\mu }}={{h}_{\nu }}=a\sqrt{({{\sinh }^{2}}\mu +{{\sin }^{2}}\nu )}$$

(Eq 7.3)
 *

 * }

From lecture slide 34-2 we know the Laplacian for curvilinear coordinates is:

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$$ \displaystyle

\Delta \Psi =\frac{1}\sum\limits_{i=1}^{3}{\frac{\partial }{\partial {{\xi }_{i}}}\left[ \frac{h_{i}^{2}}\frac{\partial \Psi }{\partial {{\xi }_{i}}} \right]}$$

(Eq 7.4)
 *

 * }

and in two-dimensional case:

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$$ \displaystyle

\Delta \Psi =\frac{1}\left[ \frac{\partial }{\partial u}\left( \frac \right)\frac{\partial \Psi }{\partial u}+\frac{\partial }{\partial u}\left( \frac \right)\frac{\partial \Psi }{\partial \nu }+\frac\frac{{{\partial }^{2}}\Psi }{\partial {{u}^{2}}}+\frac\frac{{{\partial }^{2}}\Psi }{\partial {{\nu }^{2}}} \right]$$

(Eq 7.5)
 *

 * }

from which we can see that the sign before scale factors won't matter since the they will be cancelled.

Then we derive $$\displaystyle \Delta \Psi $$ in two dimensional elliptic coordinates by substituting Eq7.3 into Eq7.5,

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$$ \displaystyle
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 * style="width:90%; padding:10px; border:2px solid #8888aa" |

\Delta \Psi =\frac{1}{{{a}^{2}}({{\sinh }^{2}}\mu +{{\sin }^{2}}\nu )}\left( \frac{{{\partial }^{2}}\Psi }{\partial {{u}^{2}}}+\frac{{{\partial }^{2}}\Psi }{\partial {{\nu }^{2}}} \right)$$

(Eq 7.6)
 *

 * }

Compare Eq 7.6 with the $$\displaystyle \Delta \Psi $$ in wikipedia article, we can see they are identical thus it is verified.

The Laplacian in three dimensinal elliptic coordinate systems
The wikipedia article required for verification also provided some outer links to some 3-D elliptic coordinate systems, hence one may want to verify Laplacians in different 3-D elliptic coordinate systems.

Verification of Laplacian in Elliptic cylindrical coordinates
The most common definition of elliptic cylindrical coordinates $$\displaystyle (\mu, \nu, z)$$ is ,

x = a \ \cosh \mu \ \cos \nu $$

y = a \ \sinh \mu \ \sin \nu $$

z = z \!$$

where $$\displaystyle \mu$$ is a nonnegative real number and $$\displaystyle \nu \in [0, 2\pi)$$.

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$$ \displaystyle

\begin{align} & dx=a\sinh \mu \cos \nu d\mu -a\cosh \mu \sin \nu dv \\ & dy=a\sinh \mu \cos \nu d\mu +a\cosh \mu \sin \nu dv \\ & dz=dz \\ \end{align}$$

(Eq 7.7)
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 * }

Then the corresponding magnitude of $$\displaystyle ds$$ is

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$$ \displaystyle

\begin{align} d{{s}^{2}} & =d{{x}^{2}}+d{{y}^{2}}+d{{z}^{2}} \\ & ={{a}^{2}}({{\sinh }^{2}}\mu +{{\sin }^{2}}\nu )d{{\mu }^{2}}+{{a}^{2}}({{\sinh }^{2}}\mu +{{\sin }^{2}}\nu )d{{v}^{2}}+d{{z}^{2}} \\ \end{align}$$

(Eq 7.8)
 *

 * }

Then we have scale factors,

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$$ \displaystyle

\begin{align} & {{h}_{\mu }}={{h}_{\nu }}=a\sqrt{({{\sinh }^{2}}\mu +{{\sin }^{2}}\nu )} \\ & {{h}_{z}}=1 \\ \end{align}$$

(Eq 7.9)
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 * }

From Eq 7.4 we have

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$$ \displaystyle

\Delta \Psi =\frac{1}\left[ \frac{\partial }{\partial u}\left( \frac \right)\frac{\partial \Psi }{\partial u}+\frac{\partial }{\partial u}\left( \frac \right)\frac{\partial \Psi }{\partial \nu }+\frac{\partial }{\partial u}\left( \frac \right)\frac{\partial \Psi }{\partial z}+\frac\frac{{{\partial }^{2}}\Psi }{\partial {{u}^{2}}}+\frac\frac{{{\partial }^{2}}\Psi }{\partial {{\nu }^{2}}}+\frac\frac{{{\partial }^{2}}\Psi }{\partial {{z}^{2}}} \right]$$

(Eq 7.10)
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 * }

Substitute Eq 7.9 into Eq 7.10 we have,

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$$ \displaystyle
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 * style="width:90%; padding:10px; border:2px solid #8888aa" |

\Delta \Psi =\frac{1}{{{a}^{2}}({{\sinh }^{2}}\mu +{{\sin }^{2}}\nu )}\left( \frac{{{\partial }^{2}}\Psi }{\partial {{u}^{2}}}+\frac{{{\partial }^{2}}\Psi }{\partial {{\nu }^{2}}}+{{a}^{2}}({{\sinh }^{2}}\mu +{{\sin }^{2}}\nu )\frac{{{\partial }^{2}}\Psi }{\partial {{z}^{2}}} \right)$$

(Eq 7.11)
 *

 * }

Compare Eq 7.11 with the $$\displaystyle \Delta \Psi $$ in wikipedia article, we can see they are identical thus it is verified.

Verification of Laplacian in Prolate spheroidal coordinates
The most common definition of prolate spheroidal coordinates $$\displaystyle (\mu, \nu, \phi)$$ is

x = a \ \sinh \mu \ \sin \nu \ \cos \phi $$

y = a \ \sinh \mu \ \sin \nu \ \sin \phi $$

z = a \ \cosh \mu \ \cos \nu $$

where $$\displaystyle \mu$$ is a nonnegative real number and $$\displaystyle \nu \in [0, \pi]$$. The azimuthal angle $$\displaystyle \phi$$ belongs to the interval $$\displaystyle [0, 2\pi)$$.

Then we have,

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$$ \displaystyle

\begin{align} & dx=-a\sin \phi \sinh \mu \cos \nu d\phi +a\cos \phi \cosh \mu \sin \nu d\mu +a\cos \phi \sinh \mu \cos \nu dv \\ & dy=a\cos \phi \sinh \mu \sin \nu d\phi +a\sin \phi \cosh \mu \sin \nu d\mu +a\sin \phi \sinh \mu \cos \nu dv \\ & dz=a\sinh \mu \cos \nu d\mu -a\cosh \mu \sin \nu dv \\ \end{align}$$

(Eq 7.12)
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 * }

Then the corresponding magnitude of $$\displaystyle ds$$ is

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$$ \displaystyle

\begin{align} d{{s}^{2}} & =d{{x}^{2}}+d{{y}^{2}}+d{{z}^{2}} \\ & =({{a}^{2}}{{\cosh }^{2}}\mu {{\sin }^{2}}\nu +{{a}^{2}}{{\sinh }^{2}}\mu {{\cos }^{2}}\nu )(d{{\mu }^{2}}+d{{v}^{2}})+{{a}^{2}}{{\sinh }^{2}}\mu {{\sin }^{2}}\nu d{{\phi }^{2}} \\ & ={{a}^{2}}({{\sinh }^{2}}\mu +{{\sin }^{2}}\nu )d{{\mu }^{2}}+{{a}^{2}}({{\sinh }^{2}}\mu +{{\sin }^{2}}\nu )d{{v}^{2}}+{{a}^{2}}{{\sinh }^{2}}\mu {{\sin }^{2}}\nu d{{\phi }^{2}} \\ \end{align}$$

(Eq 7.13)
 *

 * }

Then we have scale factors,

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

$$ \displaystyle

\begin{align} & {{h}_{\mu }}={{h}_{\nu }}=a\sqrt{({{\sinh }^{2}}\mu +{{\sin }^{2}}\nu )} \\ & {{h}_{\phi }}=a\sinh \mu \sin \nu \\ \end{align}$$

(Eq 7.14)
 *

 * }

Substitute Eq 7.14 into Eq 7.10 we have,

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

$$ \displaystyle
 * style="width:90%; padding:10px; border:2px solid #8888aa" |
 * style="width:90%; padding:10px; border:2px solid #8888aa" |

\Delta \Psi =\frac{1}{{{a}^{3}}({{\sinh }^{2}}\mu +{{\sin }^{2}}\nu )\sinh \mu \sin \nu }\left( \begin{align} & a\cosh \mu \sin \nu \frac{\partial \Psi }{\partial \mu }+a\sinh \mu \cos \nu \frac{\partial \Psi }{\partial \nu }+a\sinh \mu \sin \nu \frac{{{\partial }^{2}}\Psi }{\partial {{u}^{2}}} \\  & +a\sinh \mu \sin \nu \frac{{{\partial }^{2}}\Psi }{\partial {{\nu }^{2}}}+\frac{a({{\sinh }^{2}}\mu +{{\sin }^{2}}\nu )}{\sinh \mu \sin \nu }\frac{{{\partial }^{2}}\Psi }{\partial {{\phi }^{2}}} \\ \end{align} \right)$$

(Eq 7.15)
 *

 * }

Compare Eq 7.15 with the $$\displaystyle \Delta \Psi $$ in wikipedia article, we can see they are identical thus it is verified.

= Problem 6.8 =

From the lecture slide Mtg 36-2

Given

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

$$ \displaystyle

{{P}_{0}}(x)=1$$

(Eq 8.1)
 *

 * }

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

$$ \displaystyle

{{P}_{1}}(x)=x$$

(Eq 8.2)
 *

 * }

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

$$ \displaystyle

{{P}_{2}}(x)=\frac{1}{2}(3{{x}^{2}}-1)$$

(Eq 8.3)
 *

 * }

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

$$ \displaystyle

{{P}_{3}}(x)=\frac{1}{2}(5{{x}^{3}}-3x)$$

(Eq 8.4)
 *

 * }

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

$$ \displaystyle

{{P}_{4}}(x)=\frac{35}{8}{{x}^{4}}-\frac{15}{4}{{x}^{2}}+\frac{3}{8}$$

(Eq 8.5)
 *

 * }

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

$$ \displaystyle

{{P}_{n}}(x)=\sum\limits_{i=0}^{[n/2]}{{{(-1)}^{i}}\frac{(2n-2i)!{{x}^{n-2i}}}{{{2}^{n}}i!(n-i)!(n-2i)!}}$$

(Eq 8.6)
 *

 * }

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

$$ \displaystyle

{{P}_{n}}(x)=\sum\limits_{i=0}^{[n/2]}{\frac{1.3.....(2n-2i-1)}{{{2}^{i}}i!(n-2i)!}{{(-1)}^{i}}}{{x}^{n-2i}}$$

(Eq 8.7)
 *

 * }

 * $$\displaystyle [n/2]\equiv $$ integer part of n/2

Find
Verify that Eq8.1 through Eq8.5 can be written as Eq8.6 or Eq8.7

Solution
If we expand Eq8.1 through Eq8.5 by using Eq8.6 and Eq8.7 for each “n” according to 0,1,2,3,4, we obtain:

For n=0, and $$\displaystyle [n/2]=0\equiv 0$$
into Eq8.6,

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

$$ \displaystyle

\begin{align} {{P}_{0}}(x) & =\sum\limits_{i=0}^{[0/2]}{{{(-1)}^{i}}\frac{(2\times 0-2i)!{{x}^{0-(2i)}}}{{{2}^{0}}i!(0-i)!(0-2i)!}}={{(-1)}^{0}}\frac{(0-2\times 0)!{{x}^{0-(2\times 0)}}}{1\times 0!(0)!(0-2\times 0)!}=1\times \frac{1\times 1}{1\times 1\times 1\times 1} \\ & =1 \\ \end{align}$$

(Eq 8.8)
 *

 * }

into Eq8.7,

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

$$ \displaystyle

\begin{align} {{P}_{0}}(x) & =\sum\limits_{i=0}^{[0/2]}{\frac{1.3....(2\times 0-2i-1)}{{{2}^{i}}i!(0-2i)!}{{(-1)}^{i}}}{{x}^{0-(2i)}}=\frac{1....(0-2\times 0-1)}{{{2}^{0}}0!(0-2\times 0)!}{{(-1)}^{0}}{{x}^{0-(2\times 0)}}=\frac{1}{1\times 1\times 1}\times 1\times 1 \\ & =1 \\ \end{align}$$

(Eq 8.9)
 *

 * }

As seen above, Eq8.8, and Eq8.9 are equal to Eq8.1 so we can say that Eq8.6 and Eq8.7 are equal to each other for n=0.

For n=1, and $$\displaystyle [n/2]=0.5\equiv 0$$
into Eq8.6,

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

$$ \displaystyle

\begin{align} {{P}_{1}}(x) & =\sum\limits_{i=0}^{[1/2]}{{{(-1)}^{i}}\frac{(2\times 1-2i)!{{x}^{1-(2i)}}}{{{2}^{1}}i!(1-i)!(1-2i)!}}={{(-1)}^{0}}\frac{(2-2\times 0)!{{x}^{1-(2\times 0)}}}{2\times 0!(1-0)!(1-2\times 0)!}=1\times \frac{2x}{2\times 1\times 1\times 1} \\ & =x \\ \end{align}$$

(Eq 8.10)
 *

 * }

into Eq8.7,

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

$$ \displaystyle

\begin{align} {{P}_{1}}(x) & =\sum\limits_{i=0}^{[1/2]}{\frac{1.3...(2\times 1-2i-1)}{{{2}^{i}}i!(1-2i)!}{{(-1)}^{i}}}{{x}^{1-(2i)}}=\frac{1}{{{2}^{0}}0!(1-2\times 0)!}{{(-1)}^{0}}{{x}^{1-(2\times 0)}}=\frac{1}{1\times 1\times 1}x \\ & =x \\ \end{align}$$

(Eq 8.11)
 *

 * }

As seen above, Eq8.10 and Eq8.11 are equal to Eq8.2 so we can say that Eq8.6 and Eq8.7 are equal to each other for n=1.

For n=2, and $$\displaystyle [n/2]=1\equiv 1$$
into Eq8.6,

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

$$ \displaystyle

\begin{align} {{P}_{2}}(x) & =\sum\limits_{i=0}^{[2/2]}{{{(-1)}^{i}}\frac{(2\times 2-2i)!{{x}^{2-(2i)}}}{{{2}^{2}}i!(2-i)!(2-2i)!}} \\ & =\left( {{(-1)}^{0}}\frac{(4-2\times 0)!{{x}^{2-(2\times 0)}}}{4\times 0!(2-0)!(2-2\times 0)!} \right)+\left( {{(-1)}^{1}}\frac{(4-2\times 1)!{{x}^{2-(2\times 1)}}}{4\times 1!(2-1)!(2-2\times 1)!} \right) \\ & =\left( 1\times \frac{4!{{x}^{2}}}{4\times 0!2!2!} \right)+\left( (-1)\frac{2!{{x}^{0}}}{4\times 1!1!0!} \right) \\ & =\left( \frac{24{{x}^{2}}}{4\times 1\times 2\times 2} \right)-\left( \frac{2\times 1}{4\times 1\times 1\times 1} \right) \\ & =\frac{24{{x}^{2}}}{16}-\frac{2}{4} \\ & =\frac{1}{2}(3{{x}^{2}}-1) \\ \end{align}$$

(Eq 8.12)
 *

 * }

into Eq8.7,

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

$$ \displaystyle

\begin{align} {{P}_{2}}(x) & =\sum\limits_{i=0}^{[2/2]}{\frac{1.3.....(2\times 2-2i-1)}{{{2}^{i}}i!(2-2i)!}{{(-1)}^{i}}}{{x}^{2-(2i)}} \\ & =\frac{1....(4-2\times 0-1)}{{{2}^{0}}.0!(2-2\times 0)!}{{(-1)}^{0}}{{x}^{2-(2\times 0)}}+\frac{1....(4-2\times 1-1)}{{{2}^{1}}.1!(2-2\times 1)!}{{(-1)}^{1}}{{x}^{2-(2\times 1)}} \\ & =\frac{1\times 3}{1\times 1\times 2}{{x}^{2}}-\frac{1}{2\times 1\times 1} \\ & =\frac{1}{2}(3{{x}^{2}}-1) \\ \end{align}$$

(Eq 8.13)
 *

 * }

As seen above, Eq8.12 and Eq8.13 are equal to Eq8.3 so we can say that Eq8.6 and Eq8.7 are equal to each other for n=2.

For n=3, and $$\displaystyle [n/2]=1.5\equiv 1$$
into Eq8.6,

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

$$ \displaystyle

\begin{align} {{P}_{3}}(x) & =\sum\limits_{i=0}^{[3/2]}{{{(-1)}^{i}}\frac{(2\times 3-2i)!{{x}^{3-(2i)}}}{{{2}^{3}}i!(3-i)!(3-2i)!}} \\ & =\left( {{(-1)}^{0}}\frac{(6-2\times 0)!{{x}^{3-(2\times 0)}}}{8\times 0!(3-0)!(3-2\times 0)!} \right)+\left( {{(-1)}^{1}}\frac{(6-2\times 1)!{{x}^{3-(2\times 1)}}}{8\times 1!(3-1)!(3-2\times 1)!} \right) \\ & =\left( 1\times \frac{6!{{x}^{3}}}{8\times 0!3!3!} \right)+\left( (-1)\frac{4!{{x}^{1}}}{8\times 1!2!1!} \right)=\left( \frac{720{{x}^{3}}}{8\times 1\times 6\times 6} \right)-\left( \frac{24x}{8\times 1\times 2\times 1} \right) \\ & =\frac{720{{x}^{3}}}{288}-\frac{24x}{16} \\ & =\frac{1}{2}(5{{x}^{3}}-3x) \\ \end{align}$$

(Eq 8.14)
 *

 * }

into Eq8.7,

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

$$ \displaystyle

\begin{align} {{P}_{3}}(x) & =\sum\limits_{i=0}^{[3/2]}{\frac{1.3.....(2\times 3-2i-1)}{{{2}^{i}}i!(3-2i)!}{{(-1)}^{i}}}{{x}^{3-(2i)}} \\ & =\frac{1.3...(6-2\times 0-1)}{{{2}^{0}}.0!(3-2\times 0)!}{{(-1)}^{0}}{{x}^{3-(2\times 0)}}+\frac{1....(6-2\times 1-1)}{{{2}^{1}}.1!(3-2\times 1)!}{{(-1)}^{1}}{{x}^{3-(2\times 1)}} \\ & =\frac{1\times 3\times 5}{1\times 1\times 6}{{x}^{3}}-\frac{1\times 3}{2\times 1\times 1}x \\ & =\frac{1}{2}(5{{x}^{3}}-3x) \\ \end{align}$$

(Eq 8.15)
 *

 * }

As seen above, Eq8.14 and Eq8.15 are equal to Eq8.4 so we can say that Eq8.6 and Eq8.7 are equal to each other for n=3.

For n=4, and $$\displaystyle [n/2]=2\equiv 2$$
into Eq8.6,

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

$$ \displaystyle

\begin{align} {{P}_{4}}(x) & =\sum\limits_{i=0}^{[4/2]}{{{(-1)}^{i}}\frac{(2\times 4-2i)!{{x}^{4-(2i)}}}{{{2}^{4}}i!(4-i)!(4-2i)!}} \\ & ={{(-1)}^{0}}\frac{(8-2\times 0)!{{x}^{4-(2\times 0)}}}{16\times 0!(4-0)!(4-2\times 0)!}+{{(-1)}^{1}}\frac{(8-2\times 1)!{{x}^{4-(2\times 1)}}}{16\times 1!(4-1)!(4-2\times 1)!}+{{(-1)}^{2}}\frac{(8-2\times 2)!{{x}^{4-(2\times 2)}}}{16\times 2!(4-2)!(4-2\times 2)!} \\ & =\left( 1\times \frac{8!{{x}^{4}}}{16\times 0!\times 4!\times 4!} \right)-\left( 1\times \frac{6!{{x}^{2}}}{16\times 1!\times 3!\times 2!} \right)+\left( 1\times \frac{4!{{x}^{0}}}{16\times 2!\times 2!\times 0!} \right) \\ & =\left( \frac{40320{{x}^{4}}}{16\times 1\times 24\times 24} \right)-\left( \frac{720{{x}^{2}}}{16\times 1\times 6\times 2} \right)+\left( \frac{24}{16\times 2\times 2\times 1} \right) \\ & =\frac{40320{{x}^{4}}}{9216}-\frac{720{{x}^{2}}}{192}+\frac{24}{64} \\ & =\frac{35}{8}{{x}^{4}}-\frac{15}{4}{{x}^{2}}+\frac{3}{8} \\ \end{align}$$

(Eq 8.16)
 *

 * }

into Eq8.7,

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

$$ \displaystyle

\begin{align} {{P}_{4}}(x) & =\sum\limits_{i=0}^{[4/2]}{\frac{1.3.....(2\times 4-2i-1)}{{{2}^{i}}i!(4-2i)!}{{(-1)}^{i}}}{{x}^{4-(2i)}} \\ & =\frac{1.3....(8-2\times 0-1)}{{{2}^{0}}.0!(4-2\times 0)!}{{(-1)}^{0}}{{x}^{4-(2\times 0)}}+\frac{1.3....(8-2\times 1-1)}{{{2}^{1}}.1!(4-2\times 1)!}{{(-1)}^{1}}{{x}^{4-(2\times 1)}}+\frac{1....(8-2\times 2-1)}{{{2}^{2}}.2!(4-2\times 2)!}{{(-1)}^{2}}{{x}^{4-(2\times 2)}} \\ & =\frac{1\times 3\times 5\times 7}{1\times 1\times 24}{{x}^{4}}-\frac{1\times 3\times 5}{2\times 1\times 2}{{x}^{2}}+\frac{1\times 3}{4\times 4\times 1}{{x}^{0}} \\ & =\frac{35}{8}{{x}^{4}}-\frac{15}{4}{{x}^{2}}+\frac{3}{8} \\ \end{align}$$

(Eq 8.17)
 *

 * }

As seen above, Eq8.16 and Eq8.17 are equal to Eq8.5 so we can say that Eq8.6 and Eq8.7 are equal to each other for n=4.

As a result, Eq8.8 through Eq8.17 for n=0,1,2,3,4 are equal to Eq8.1 through Eq8.5. Therefore, it is verified that Eq8.1 through Eq8.5 can be written as Eq8.6 or Eq8.7.

= References =

title: User:Egm6321.f10.team2.reserve/HW6


text:
Category:Resources by need

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Resources need volunteers to help them. Resources are categorized by what type of help is needed.

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

title: Category:Resources by need


text:
Ideas in Geometry/Instructive examples/Lesson 17- Properties of Triangles

3.1.1 There are different ideas about what the center of a triangle might be. Consider the following. Which do you think best describes the "center" of a triangle? :

1. The perpendicular bisectors of the the 3 sides of the triangle that meet at one point= Circumcenter At this point used as the center, a circle can be made to intersect each of the 3 vertices. This line may not always lie inside of the triangle. This site helps us visualize the circumcenter as well as help construct one: http://www.mathopenref.com/constcircumcenter.html

2. The interior bisectors of each angle meet at a point= Incenter At this point used as the center, a circle can be drawn within the triangle that makes the circle tangent to each of the sides. This circle made is called the incircle. This site helps us visualize the incenter as well as help construct one: http://www.mathopenref.com/constincenter.html

3. The meeting point of the 3 altitudes of a triangle= orthocenter What is an altitude? An altitude is a line from the vertex to the opposite line. This line is perpendicular to the opposite line and a right angle is always formed between the line made and the opposite line of the triangle. The orthocenter may not lie inside of the triangle This site helps us visualize the orthocenter as well as help construct one: http://www.mathopenref.com/constorthocenter.html

4. The meeting point of the lines of a triangle that connect the vertex to the midpoint of the opposite side= Centroid The centroid is unique because it is called the "center of mass." This means the triangle can be balanced on a plane at this point. This site helps us visualize the centroid as well as help construct one: http://www.mathopenref.com/constcentroid.html

But how is this different than an orthocenter? The lines used to create the centroid do not necessarily form a right angle with the other side. When constructing the othocenter, the altitudes always form right angles with the opposite sides.

Helpful acronym to remember which constructions always lie inside the triangle : COIN C ircumenter O rthocenter I ncenter N centroid     IN is always inside the triangle!


text:
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text:
Ideas in Geometry/Constructions II

4.2 Trickier Constructions

This lesson focuses on constructing triangles. please note that some of the videos use a protractor to measure angles, we are using a compass to copy the angles.

We focus on four main constructions for triangles. This section gives us different constructions to make tirangles when given different parts of the triangle These constructions are done when you are given three different pieces of information. They can be completed using a compass and straight edge.

Side Angle Side (SAS Triangle)

This construction should be used when you are given a side, an angle adjacent to that side, and a side adjacent to that angle. Its like an angle sandwich! 1. Transfer one side that that it starts at the vertex of the angle 2. Transfter the other side so that it starts at the vertex 3. Connect the end points of all moved line segments

SAS video

SIde Side Side - you are given all three sides of the triangle, and no angles.

1. choose a side and select one of its endpoints 2. Draw a circle of radius equal to the length of the seond side aorund the choses end point 3. draw a circle of radius equal o the length of the third side around the other end point 4. connect the end pointd of the first side and the interection of the circles. This is the desired triangle

here is an example of how to do this side side side video

Side Angle Angle

1. start with given side and place adjacent angle at one of its end points 2. move second angle so that it shares a leg with the leg of the first angle- not the leg with the side 3. extend the side past the first angle, forming a new angle with the leg of the second angle 4. move this new angle to the other end point of the slide, extending the legs of this angle and the first angle will produce the desired triangle you are given a side, the angle adjacent, and the angle

here is an example video SAA

Angle Side Side is a construction to be careful of. it can produce more than one triangle, therefore making it an unreliable construction.

title: Ideas in Geometry/Constructions II


text:
Which word? Situation/Position/Condition

All three nouns are similar in meaning, but they have different nuances or shades of meaning and are used in different contexts in different ways. Condition Condition describes the physical state of something or some one. We talk about people or things being in good / bad / terrible / etc condition. Condition can also refer to a health problem: Considering its age, this house is in excellent condition. He was in a terrible condition and had drunk far too much whiskey. She has a severe heart condition and shouldn't be smoking at all. Conditions (plural) refer to the environment in which something occurs. We talk about things happening in or under appalling / terrible conditions. We also talk about people's living or working conditions: The rescue was attempted under extremely difficult conditions and with little chance of success. The refugees were living under appalling conditions with no access to clean water. Are you happy with your working conditions? ~ Yes, they are excellent. I have no complaints. The extremely windy conditions made it difficult for either side to play decent football. Conditions also describe things that must be true or be done before something else can happen. We talk about meeting or satisfying or imposing conditions. In order to qualify for a grant as a postgraduate student, you will need to satisfy certain conditions. The conditions imposed by the university meant that no one was likely to qualify. Terms and conditions describe the business or financial arrangements of an agreement. Make sure you read the terms and conditions carefully before you take out the student loan on condition that Note the expression on condition that for saying that one thing will happen only if another thing happens: You can have the day off tomorrow on condition that you agree to work on Saturday. They spoke to the police about the incident on condition that they would not be called as witnesses. Situation Situation refers to a set of conditions that are in place at a particular time and in a particular place: If the situation had been different, our marriage might have succeeded. I am particularly concerned about the situation in the south of the country where the rules of law and order appear to have broken down. We also talk about an economic or financial situation: The financial situation is dire - the company has failed to make a profit in each of the last four years. situation comedies / situations vacant Note also the compound nouns situation comedies (abbreviated to sitcoms) which describe amusing television drama series revolving around a set of characters in a family or organisation and situations vacant which refers to a column or page in a newspaper where jobs are advertised: 'The Office' is regarded as one of the most original sitcoms the BBC has produced so far this century. If you just want a clerical job, look in the situations vacant column in the local newspaper. position Position is used in a wider variety of contexts than situation or condition. First and foremost, it refers to the way or where somebody or something is placed. I was quite badly injured in the demonstration but managed to drag myself to a sitting position under a tree. This plant loves sunlight and should be placed in an open sunny position in the flower border. This is quite a detailed map showing the position of all the oil refineries. Position can also describe a general situation and in this context can sometimes be replaced by situation: The position / situation is that everyone must be interviewed about the break in by the security services. If I were in your position / situation, I wouldn't dream of sheltering an escaped convict. Position also means opinion, i.e. where you stand or are placed, on an issue: My position on fox hunting is that it is a useful way of keeping the number of foxes down. Position can also refer to a job in a company, a place in a list or where you play in a team sport such as football: Is the assistant manager position still open? ~ No, I'm sorry, it's already been filled. What is their position in the league? ~ They're in forth position at the moment, but if they win today, they'll move up to third. My best position is on the left in midfield, just in front of the back four.

resource

 * http://www.bbc.co.uk/worldservice/learningenglish


text:
Clinical Exercise Biomechanics

Biomechanics explains the mechanical aspects of human movement. Clinical Exercise Biomechanics demonstrates the fundamental mechanical factors associated with human performance in exercise. The description of pathological movement characteristics and the forces associated with such movement will be explored with respect to concepts such as linear and angular motion. Movement screening and postural alignment will be demonstrated. Newton's Laws of motion will also be explored with specific application to a clinical context. The principles of gait will be discussed and include an in-depth examination of pathological gait mechanics and appropriate biomechanical assessment techniques. Variations to walking mechanics when changing directions (e.g. avoiding obstacles) as well as centre of pressure disturbances will also be discussed with reference to falls prevention in older adults. In addition, the changes to movement patterns with various chronic and acute injury are explored. Clinical Exercise Biomechanics provides a sound base of knowledge to apply biomechanical concepts in a clinical exercise setting.

Key Outcomes

 * 1) Demonstrate a thorough understanding of mechanical factors that influence human movement in a clincial context
 * 2) Demonstrated understanding of clinical exercise biomechanics assessment techniques
 * 3) Ability to conduct clinical biomechanical analysis and provide meaningful feedback

Topics

 * Background to clincal exercise biomechanics
 * Biomechanical measurement in a clincial context
 * Muscle and joint forces and moments in exercise
 * Ground reaction forces, impulse and momentum in exercise
 * Motion and joint motion
 * Work, energy and power in exercise
 * Inverse dynamics theory
 * Force and pressure analysis
 * Anatomical models and markers
 * Muscle function and measurement
 * Gait analysis including pathological conditions
 * Clinical assessments including sit-to-stand and stair ascent/decent

Resources

 * Richards, J. (2008), Biomechanics in clinic and research, Philadelphia, USA: Churchil Lvingstone Elsevier.

Assignments

 * Examination assessing application of biomechanical concepts and principles in a clinical context
 * Presentation involving a critical review of the scientific literature surrounding an area of interest in clinical biomechanics
 * Portfolio demonstrating an ability to conduct and present clinical biomechanics research effectively

Coordinator
Keane Wheeler

title: Clinical Exercise Biomechanics


text:
TESOL/Animal idioms

By using concealed cameras, fly on the wall television programmes or films show people doing what they normally do every day. Big Brother is a good example of this type of reality TV programme. We use the expression "fly on the wall" because if a fly settles on your wall at home, you do not normally notice it. We use many animal idioms in English. Here are a selection of those involving both larger and smaller animals. How many of these do you know?

quiz
{The ________ of the Eighties and Nineties is unlikely to return for some time. - lion's share + bull market - stag night - donkey work
 * type="[]"}
 * Incorrect. This idiom refers to the majority of something.
 * Correct! This refers to a situation where the price of shares on the stock market are rising.
 * Incorrect. This idiom refers to a social/drinking evening for the groom's male friends prior to a wedding.
 * Incorrect. This idiom refers to manual labor or routine jobs.

{I decided to ________ and tell him he was upsetting Jane. + take the bull by the horns - flog a dead horse - do the donkey work - could eat a horse
 * type="[]"}
 * Correct! This means to face up to a problem instead of avoiding it.
 * Incorrect. This idiom means to waste time on something unlikely to be successful.
 * Incorrect. This idiom refers to manual labor or routine jobs.
 * Incorrect. This idiom means to be very hungry.

{If you tell him to take it easy, he'll sit back and let you ________. - take the bull by the horns + do the donkey work - flog a dead horse - bull market
 * type="[]"}
 * Incorrect. This means to face up to a problem instead of avoiding it.
 * Correct! This refers to manual labor or routine jobs.
 * Incorrect. This idiom means to waste time on something unlikely to be successful.
 * Incorrect. This refers to a situation where the price of shares on the stock market are rising.

{Invest in shares and the chances are you'll ________. - take the bull by the horns + flog a dead horse - do the donkey work - bull market
 * type="[]"}
 * Incorrect. This means to face up to a problem instead of avoiding it.
 * Correct! This means to waste time on something unlikely to be successful.
 * Incorrect. This idiom refers to manual labor or routine jobs.
 * Incorrect. This refers to a situation where the price of shares on the stock market are rising.

{The ________ of her money - over 80% - went to her nephews and nieces. + lion's share - flog a dead horse - do the donkey work - bull market
 * type="[]"}
 * Correct! This refers to the majority of something.
 * Incorrect. This idiom means to waste time on something unlikely to be successful.
 * Incorrect. This idiom refers to manual labor or routine jobs.
 * Incorrect. This refers to a situation where the price of shares on the stock market are rising.

{I'm having my ________ on the same day as my Sue is having her hen night. - lion's share - flog a dead horse + stag night - bull market
 * type="[]"}
 * Incorrect. This idiom refers to the majority of something.
 * Incorrect. This idiom means to waste time on something unlikely to be successful.
 * Correct! This refers to a social/drinking evening for the groom's male friends prior to a wedding.
 * Incorrect. This refers to a situation where the price of shares on the stock market are rising.

{I skipped breakfast so now I ________. - lion's share - flog a dead horse - stag night + could eat a horse
 * type="[]"}
 * Incorrect. This idiom refers to the majority of something.
 * Incorrect. This idiom means to waste time on something unlikely to be successful.
 * Incorrect. This refers to a social/drinking evening for the groom's male friends prior to a wedding.
 * Correct! This means to be very hungry.

quiz
{Which idiom means "earn enough to buy food and other essentials"? - dog tired + keep the wolf from the door - let sleeping dogs lie - dog days
 * type="[]"}
 * Incorrect. This idiom means "very tired".
 * Correct! This idiom means to earn enough to buy food and other essentials.
 * Incorrect. This idiom means "not interfere; not mention something that could cause trouble".
 * Incorrect. This idiom refers to very hot days.

{Which idiom means "very tired"? + dog tired - let the cat out of the bag - cat nap - raining cats and dogs
 * type="[]"}
 * Correct! This idiom means "very tired".
 * Incorrect. This idiom means "reveal a secret".
 * Incorrect. This idiom refers to a short sleep.
 * Incorrect. This idiom means "raining heavily".

{Which idiom means "not interfere; not mention something that could cause trouble"? - smell a rat - cat nap + let sleeping dogs lie - keep the wolf from the door
 * type="[]"}
 * Incorrect. This idiom means "sense that something is wrong".
 * Incorrect. This idiom refers to a short sleep.
 * Correct! This idiom means "not interfere; not mention something that could cause trouble".
 * Incorrect. This idiom means "earn enough to buy food and other essentials".

{Which idiom means "reveal a secret"? + let the cat out of the bag - dog days - smell a rat - keep the wolf from the door
 * type="[]"}
 * Correct! This idiom means "reveal a secret".
 * Incorrect. This idiom refers to very hot days.
 * Incorrect. This idiom means "sense that something is wrong".
 * Incorrect. This idiom means "earn enough to buy food and other essentials".

{Which idiom means "raining heavily"? - cat nap + raining cats and dogs - dog tired - let sleeping dogs lie
 * type="[]"}
 * Incorrect. This idiom refers to a short sleep.
 * Correct! This idiom means "raining heavily".
 * Incorrect. This idiom means "very tired".
 * Incorrect. This idiom means "not interfere; not mention something that could cause trouble".

{Which idiom means "sense that something is wrong"? + smell a rat - let the cat out of the bag - dog days - cat nap
 * type="[]"}
 * Correct! This idiom means "sense that something is wrong".
 * Incorrect. This idiom means "reveal a secret".
 * Incorrect. This idiom refers to very hot days.
 * Incorrect. This idiom refers to a short sleep.

quiz
{What does the idiom "a wild goose chase" mean? - a search for wild geese + a search that has no chance of success - a search for something valuable - a search that is very easy
 * type="[]"}
 * Incorrect. This is a literal interpretation, not the idiomatic meaning.
 * Correct! This idiom refers to a futile or hopeless search.
 * Incorrect. This idiom doesn't refer to the value of the search.
 * Incorrect. This idiom refers to a difficult or impossible search.

{What does the idiom "up with the lark" mean? - to be with a bird + awake early - to be very active - to be very lazy
 * type="[]"}
 * Incorrect. This is a literal interpretation, not the idiomatic meaning.
 * Correct! This idiom refers to waking up early in the morning.
 * Incorrect. This idiom specifically refers to waking up early.
 * Incorrect. This idiom refers to the opposite of being lazy.

{What does the idiom "chicken out" mean? - to eat chicken + stop participating in something which is too dangerous or difficult - to be very brave - to be very foolish
 * type="[]"}
 * Incorrect. This is a literal interpretation, not the idiomatic meaning.
 * Correct! This idiom refers to backing out of a situation due to fear or apprehension.
 * Incorrect. This idiom refers to the opposite of being brave.
 * Incorrect. This idiom doesn't refer to foolishness.

{What does the idiom "a fish out of water" mean? - a fish that is outside water + uncomfortable - a fish that is very special - a fish that is very common
 * type="[]"}
 * Incorrect. This is a literal interpretation, not the idiomatic meaning.
 * Correct! This idiom refers to feeling out of place or uncomfortable in a particular situation.
 * Incorrect. This idiom doesn't refer to the value or uniqueness of the fish.
 * Incorrect. This idiom doesn't refer to the commonality of the fish.

{What does the idiom "a cold fish" mean? - a fish that is frozen + unemotional - a fish that is very tasty - a fish that is very rare
 * type="[]"}
 * Incorrect. This is a literal interpretation, not the idiomatic meaning.
 * Correct! This idiom refers to someone who is perceived as lacking emotion or warmth.
 * Incorrect. This idiom doesn't refer to the taste of the fish.
 * Incorrect. This idiom doesn't refer to the rarity of the fish.

{What does the idiom "have other fish to fry" mean? - to cook multiple fish + have other things to do or more important matters to attend to - to be very busy with one task - to be very relaxed
 * type="[]"}
 * Incorrect. This is a literal interpretation, not the idiomatic meaning.
 * Correct! This idiom refers to having other priorities or tasks.
 * Incorrect. This idiom refers to having multiple tasks or priorities.
 * Incorrect. This idiom doesn't refer to relaxation.

quiz
{The trip will be amazing, but there's just one ________ that Sue can't drive. - difficulty in the journey + fly in the ointment - bee in the bonnet - bee's knees
 * type="[]"}
 * Incorrect. Think of an idiom that relates to a small issue in an otherwise good situation.
 * Correct! This idiom means a difficulty which prevents total enjoyment.
 * Incorrect. This idiom means being obsessed with something.
 * Incorrect. This idiom refers to someone who is very clever and important.

{Even though he gets angry easily, he wouldn't ________. - hurt a bee + hurt a fly - be a busy bee - have a bee in his bonnet
 * type="[]"}
 * Incorrect. Think of a gentler creature.
 * Correct! This idiom means someone is gentle.
 * Incorrect. This idiom refers to a very busy person.
 * Incorrect. This idiom means being obsessed with something.

{Jacob is very sharp. There are ________. - bees in his bonnet - flies in his ointment + no flies on him - none of the bees' knees
 * type="[]"}
 * Incorrect. This idiom means being obsessed with something.
 * Incorrect. This idiom means a difficulty which prevents total enjoyment.
 * Correct! This idiom means someone is alert and clever, not easy to deceive.
 * Incorrect. This idiom refers to someone who is very clever and important.

{She's always working on something. She's such a ________. + busy bee - fly in the ointment - bee in her bonnet - bee's knees
 * type="[]"}
 * Correct! This idiom refers to a very busy person.
 * Incorrect. This idiom means a difficulty which prevents total enjoyment.
 * Incorrect. This idiom means being obsessed with something.
 * Incorrect. This idiom refers to someone who is very clever and important.

{She can't stop talking about the health benefits of raw fish. She's got a ________ about it. - fly in her ointment + bee in her bonnet - busy bee - bee's knees
 * type="[]"}
 * Incorrect. This idiom means a difficulty which prevents total enjoyment.
 * Correct! This idiom means being obsessed with something.
 * Incorrect. This idiom refers to a very busy person.
 * Incorrect. This idiom refers to someone who is very clever and important.

{With her impressive credentials, she believes she's the ________. - fly in the ointment - bee in her bonnet - busy bee + bee's knees
 * type="[]"}
 * Incorrect. This idiom means a difficulty which prevents total enjoyment.
 * Incorrect. This idiom means being obsessed with something.
 * Incorrect. This idiom refers to a very busy person.
 * Correct! This idiom refers to someone who is very clever and important.

Resources

 * http://www.bbc.co.uk/worldservice/learningenglish

title: TESOL/Animal idioms


text:
Wikiversity:Requests for Deletion/header

__NEWSECTIONLINK__ Deletion guideline | Deletion log | Archives

We welcome and appreciate civil discussion of requests to delete or undelete pages when reasonable objections are made or are likely, the advice in Deletions is followed, and other options have failed. A good attitude is to explain what you have tried, ask for help or advice from fellow Wikiversity participants on what to do now, keep an open mind, accept any community consensus, and focus on how pages can be improved. Finding ways to improve pages is the preferred outcome of any discussion and consensus here. Pages should always be kept when reasonable concerns are adequately addressed. Reasons and responses should be specific and relate to Wikiversity policy or scope in some way, kept brief, and stated in a positive or neutral way. Vague reasons ("out of scope", "disruptive") may be ignored.

A clear consensus should emerge before archiving a request. Often discussion takes a week or more to reach a clear consensus. Remember to add dr to the top of pages nominated for deletion. You can put "keep", "delete", or "neutral" at the beginning of your response, but consensus is established by discussion and reasoning, not mere voting.

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text:
User:EGM6321.f10.team6.yoshioka/HW6

=Problem 3 - Solving the Legendre Equation Using the Direct Method =

Given
The non-homogeneous Legendre Equation with n=1:

And the 1st homogeneous solution is given by:

Find
The solution of the Legendre equation using the direct method.

Solution
From (1) 29-1, let the solution be in the form:

And the derivatives of 3.3:

Substituting equations 3.3, 3.4 and 3.5 into 3.1 we have:

By rearranging the equation 3.6:

Calculating the derivatives of the 1st homogeneous solution (equation 3.2):

Substituting equations 3.2, 3.8 and 3.9 into 3.7:

The equation 3.10 results in:

By applying the reduction of order method as presented in Meeting 5, z(x) can be defined as:

Substituting the equation 3.12 into 3.11:

Rearranging the equation 3.13:

Since the ODE is in the form of:

Calculating the integrating factor by using (1) 10-2

From (6) 10-3 the solution of z(x) is given by:

And the solution of U(x) is given by:

Finally the solution of y(x) is calculated by using (2) 29-3:

The final solution of y(x) is given by:

Given
The spherical coordinates:

Find
The infinitesimal segment ds and the Laplace equation.

Solution
The infinitesimal segment ds in the 3 dimensional space is given by:

Where:

Therefore applying the Einstein summation in the equation 4.5:

The general equation to find $$d{{x}_{i}}$$ is given by:

Calculating $$d{{x}_{1}}$$, $$d{{x}_{2}}$$ and $$d{{x}_{3}}$$ from equation 4.7.

Calculating the squares of equation 4.10, 4.13 and 4.16:

Substituting equations 4.18, 4.20 and 4.22 into 4.6 and grouping them in terms of $$dr$$, $$d\theta $$ and $$d\varphi $$:

Using the identities of $${{\cos }^{2}}(\varphi )$$ and $${{\sin }^{2}}(\varphi )$$ into 4.23:

Finally, the the infinitesimal segment ds in the 3 dimensional space can be defined by:

The Laplace equation in the spherical coordinates is defined by:

Where:

Substituting 4.28, 4.29 and 4.30 into 4.27:

Therefore $$\psi $$ can be calculated by:

Finally the Laplace equation for the spherical coordinates id defined by:

title: User:EGM6321.f10.team6.yoshioka/HW6


text:
User talk:Dejoy

Hi I am Rev. DeJoy Amankwa. A Ghanaian living in Kumasi the garden city of West Africa. I am the founder and thr General Overseer of the Community Church International. An African Independent Church with 24 Asemblies in Ghana. Some of our annual programmes: National Easte Convention;  National Youth Camp
 * National Women's Convention; and National Men's convention. This year Rev Ben. Amankwa of USA was the speaker for the Easter convention., and Evangelist Rebbecca Johnson of USA was the speaker for the Women's convention.

DOORS ARE OPENED for those willing to be speakers for 2011. Programmes. contact links: amankwa@wiki.joy> revdejoy@yahoo.co.uk God bless al readers. Rev.DeJoy--Dejoy 11:14, 15 November 2010 (UTC)

title: User talk:Dejoy


text:
Airplane Flying Handbook/Airport Traffic Patterns and Operations

Airplane Flying Handbook

Airport Traffic patterns and operations
Just as roads and streets are needed in order to utilize automobiles, airports or airstrips are needed to utilize airplanes. Every flight begins and ends at an airport or other suitable landing field. For that reason, it is essential that the pilot learn the traffic rules, traffic procedures, and traffic pattern layouts that may be in use at various airports.

When an automobile is driven on congested city streets, it can be brought to a stop to give way to conflicting traffic; however, an airplane can only be slowed down. Consequently, specific traffic patterns and traffic control procedures have been established at designated airports. The traffic patterns provide specific routes for takeoffs, departures, arrivals, and landings. The exact nature of each airport traffic pattern is dependent on the runway in use, wind conditions, obstructions, and other factors.

Control towers and radar facilities provide a means of adjusting the flow of arriving and departing aircraft, and render assistance to pilots in busy terminal areas. Airport lighting and runway marking systems are used frequently to alert pilots to abnormal conditions and hazards, so arrivals and departures can be made safely.

Airports vary in complexity from small grass or sod strips to major terminals having many paved runways and taxiways. Regardless of the type of airport, the pilot must know and abide by the rules and general operating procedures applicable to the airport being used. These rules and procedures are based not only on logic or common sense, but also on courtesy, and their objective is to keep air traffic moving with maximum safety and efficiency. The use of any traffic pattern, service, or procedure does not alter the responsibility of pilots to see and avoid other aircraft.

 * Forward to Standard airport traffic patterns
 * Return to Basic flight maneuvers
 * Return to Airplane Flying Handbook menu


text:
Airplane Flying Handbook/Standard Airport Traffic Patterns

Airplane Flying Handbook

Standard airport traffic patterns
To assure that air traffic flows into and out of an airport in an orderly manner, an airport traffic pattern is established appropriate to the local conditions, including the direction and placement of the pattern, the altitude to be flown, and the procedures for entering and leaving the pattern. Unless the airport displays approved visual markings indicating that turns should be made to the right, the pilot should make all turns in the pattern to the left.

When operating at an airport with an operating control tower, the pilot receives, by radio, a clearance to approach or depart, as well as pertinent information about the traffic pattern. If there is not a control tower, it is the pilot’s responsibility to determine the direction of the traffic pattern, to comply with the appropriate traffic rules, and to display common courtesy toward other pilots operating in the area.

The pilot is not expected to have extensive knowledge of all traffic patterns at all airports, but if the pilot is familiar with the basic rectangular pattern, it will be easy to make proper approaches and departures from most airports, regardless of whether they have control towers. At airports with operating control towers, the tower operator may instruct pilots to enter the traffic pattern at any point or to make a straight-in approach without flying the usual rectangular pattern. Many other deviations are possible if the tower operator and the pilot work together in an effort to keep traffic moving smoothly. Jets or heavy airplanes will frequently be flying wider and/or higher patterns than lighter airplanes, and in many cases will make a straight-in approach for landing.

Compliance with the basic rectangular traffic pattern reduces the possibility of conflicts at airports without an operating control tower. It is imperative that the pilot form the habit of exercising constant vigilance in the vicinity of airports even though the air traffic appears to be light.

The standard rectangular traffic pattern is illustrated in figure 7-1. The traffic pattern altitude is usually 1,000 feet above the elevation of the airport surface. The use of a common altitude at a given airport is the key factor in minimizing the risk of collisions at airports without operating control towers.

It is recommended that while operating in the traffic pattern at an airport without an operating control tower the pilot maintain an airspeed that conforms with the limits established by Title 14 of the Code of Federal Regulations (14 CFR) part 91 for such an airport: no more than 200 knots (230 miles per hour (m.p.h.)). In any case, the speed should be adjusted, when practicable, so that it is compatible with the speed of other airplanes in the pattern.

When entering the traffic pattern at an airport without an operating control tower, inbound pilots are expected to observe other aircraft already in the pattern and to conform to the traffic pattern in use. If other aircraft are not in the pattern, then traffic indicators on the ground and wind indicators must be checked to determine which runway and traffic pattern direction should be used. [Figure 7-2] Many airports have L-shaped traffic pattern indicators displayed with a segmented circle adjacent to the runway. The short member of the L shows the direction in which the traffic pattern turns should be made when using the runway parallel to the long member. These indicators should be checked while at a distance well away from any pattern that might be in use, or while at a safe height well above generally used pattern altitudes. When the proper traffic pattern direction has been determined, the pilot should then proceed to a point well clear of the pattern before descending to the pattern altitude.

When approaching an airport for landing, the traffic pattern should be entered at a 45° angle to the downwind leg, headed toward a point abeam of the midpoint of the runway to be used for landing. Arriving airplanes should be at the proper traffic pattern altitude before entering the pattern, and should stay clear of the traffic flow until established on the entry leg. Entries into traffic patterns while descending create specific collision hazards and should always be avoided.

The entry leg should be of sufficient length to provide a clear view of the entire traffic pattern, and to allow the pilot adequate time for planning the intended path in the pattern and the landing approach.

The downwind leg is a course flown parallel to the landing runway, but in a direction opposite to the intended landing direction. This leg should be approximately 1/2 to 1 mile out from the landing runway, and at the specified traffic pattern altitude. During this leg, the before landing check should be completed and the landing gear extended if retractable. Pattern altitude should be maintained until abeam the approach end of the landing runway. At this point, power should be reduced and a descent begun. The downwind leg continues past a point abeam the approach end of the runway to a point approximately 45° from the approach end of the runway, and a medium bank turn is made onto the base leg.

The base leg is the transitional part of the traffic pattern between the downwind leg and the final approach leg. Depending on the wind condition, it is established at a sufficient distance from the approach end of the landing runway to permit a gradual descent to the intended touchdown point. The ground track of the airplane while on the base leg should be perpendicular to the extended centerline of the landing runway, although the longitudinal axis of the airplane may not be aligned with the ground track when it is necessary to turn into the wind to counteract drift. While on the base leg, the pilot must ensure, before turning onto the final approach, that there is no danger of colliding with another aircraft that may be already on the final approach.

The final approach leg is a descending flightpath starting from the completion of the base-to-final turn and extending to the point of touchdown. This is probably the most important leg of the entire pattern, because here the pilot’s judgment and procedures must be the sharpest to accurately control the airspeed and descent angle while approaching the intended touchdown point. As stipulated in 14 CFR part 91, aircraft while on final approach to land or while landing, have the right-of-way over other aircraft in flight or operating on the surface. When two or more aircraft are approaching an airport for the purpose of landing, the aircraft at the lower altitude has the right-of-way. Pilots should not take advantage of this rule to cut in front of another aircraft that is on final approach to land, or to overtake that aircraft.

The upwind leg is a course flown parallel to the landing runway, but in the same direction to the intended landing direction. The upwind leg continues past a point abeam of the departure end of the runway to where a medium bank 90° turn is made onto the crosswind leg.

The upwind leg is also the transitional part of the traffic pattern when on the final approach and a go-around is initiated and climb attitude is established. When a safe altitude is attained, the pilot should commence a shallow bank turn to the upwind side of the airport. This will allow better visibility of the runway for departing aircraft.

The departure leg of the rectangular pattern is a straight course aligned with, and leading from, the takeoff runway. This leg begins at the point the airplane leaves the ground and continues until the 90° turn onto the crosswind leg is started. On the departure leg after takeoff, the pilot should continue climbing straight ahead, and, if remaining in the traffic pattern, commence a turn to the crosswind leg beyond the departure end of the runway within 300 feet of pattern altitude. If departing the traffic pattern, continue straight out or exit with a 45° turn (to the left when in a left-hand traffic pattern; to the right when in a right-hand traffic pattern) beyond the departure end of the runway after reaching pattern altitude. The crosswind leg is the part of the rectangular pattern that is horizontally perpendicular to the extended centerline of the takeoff runway and is entered by making approximately a 90° turn from the upwind leg. On the crosswind leg, the airplane proceeds to the downwind leg position.

Since in most cases the takeoff is made into the wind, the wind will now be approximately perpendicular to the airplane’s flightpath. As a result, the airplane will have to be turned or headed slightly into the wind while on the crosswind leg to maintain a ground track that is perpendicular to the runway centerline extension. Additional information on airport operations can be found in the Aeronautical Information Manual (AIM).

 * Forward to Normal approach and landing
 * Return to Airport traffic patterns and operations
 * Return to Airplane Flying Handbook menu


text:
Numerical Analysis/stability of RK methods/quizzes

{ the A-stability is characterized by: - all the points in the left half of the complex plane. - it a special case of the absolute stability. - inludes all $$\{ z \in \mathbb{C} | \mathrm{Re}(z) < 0 \}$$, + all of the above
 * type=""}

{Euler method is: - A-stable + not A-stable
 * type=""}

{the absolute stability region for RK4 is greater than the A-stability region for the same method:
 * type=""}

+ false - True - True for certain values of $$ b_j$$.

{A numerical method is stable if : + small change in the initial condition will produce small change in subsequent steps. - small change in the initial condition will produce huge change in subsequent steps. - big change in the initial conditions produce oscillatory solution at the end.
 * type=""}

{$$ \lambda =1$$ is always a root of the characteristic polynomial of the multistep method:
 * type=""}

+ True - False - usually not the case

{the root of the C.P. can be real or complex,and the method still be stable: + True - False
 * type=""}

{A multi- step method is strongly stable if : - $$ \lambda =1$$ is the only root of magnitude 1. - all other roots has magnitude <1 + the first and the second sentence - just one root inside the unit circle if it has more than one.
 * type=""}

{if more than one root has magnitude equal to 1, and the others are less than one, the method is - strongly stable. - A-stable + weakly stable. - Unstable.
 * type=""}

{The numerical method is unstable if - $$ \lambda > 1$$ for at least one root - $$ Re(\lambda) > 1$$ for at least one root - $$ Re(\lambda) \leqslant 1$$ for at least one root + the firs and the second sentences.
 * type=""}

{the absolute stabity region for the explicit Euler's method is + unit cicle in the complex plane, its center shifted to the left, by one unit. - unit cicle in the complex plane, its center shifted to the rigt, by one unit. - the whole left side of the complex plane. - the method is unstable.
 * type=""}

{ None of the RK methods is A-Stable: + True - False
 * type=""}

{All explicit methods are : - A-Stable + Not A-Stable. - Unstable. - None of the above
 * type=""}

{implicit multistep methods are A-stable if the have order at most -1 +2 -3 -they are always A-stable.
 * type=""}


text:
Numerical Analysis/stability of RK methods/exercises

Ex:1
find the stability function for RK2 which is given by: $$ y_{n+1}=y_{n}+hf(t+\frac {h}{2},y_{n}+\frac {hf(t_n,y_n)}{2})$$

Solution: applying this method to the test equation $$ y'=\lambda y$$
 * we get $$ \begin{align}

y_{n+1} &=y_n+h \lambda y_n+\frac {(h \lambda)^2}{2}y_n \\ \Rightarrow \end{align}$$ the stability polynomial $$ \phi(z)=1+h \lambda +\frac {(h \lambda)^2}{2}$$

Ex:2
find the absolute stability region for RK2. Solution: by setting $$ z = h \lambda $$
 * $$ \Rightarrow $$ the abs.stability region is given by $$\{z\in \mathbb{C}||1+ h \lambda + \frac {(h \lambda)^2}{2}| < 1 \}$$

Ex:3
find the characteristic polynomial for RK2. Solution: it is $$ r^{i+1}=(1+z+\frac {z^2}{2})r^i$$ divide both sides of the equation by $$ r^i;r \ne 0$$ you get $$ r^{i+1}=(1+z+\frac {z^2}{2})r^i$$
 * $$ \Rightarrow \phi(r,z)=r-1+z+\frac {z^2}{2}$$

Ex:4
is RK2 stable, if it is what type of stability. Solution: you get $$ r^{i+1}=(1+z+\frac {z^2}{2})r^i$$ by setting z=0,
 * $$ \Rightarrow r=1$$
 * so the method is strongly stable since r=1, is the only root, and has a value of 1.

Ex:5
Determine the stability of Back ward Euler method. Solution: $$ y_{n+1}=y_n+hf(t_{n+1},y_{n+1})$$
 * by applying this method to the test equation
 * then $$ y_{n+1}=y_n+h\lambda y_{n+1}$$
 * and so $$ y_{n+1}=\frac{1}{1+h\lambda }y_{n}$$
 * call $$ z=h\lambda$$ and so:
 * $$ G(z)\longrightarrow 0$$ as $$ Re(z)\longrightarrow \infty$$
 * and so this is L-Stable method when applied to stiff equation.


text:
User talk:Sjcriscione

 Hello Sjcriscione, 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. --La comadreja formerly AFriedman RESEARCH (talk) 17:30, 15 November 2010 (UTC)

 Hi! Our name is Vicky-Verity. We are twins and are Wikiversity mascots. We think Wikiversity is a great place to learn and make friends (when we aren't sleeping). Leave us a message!

title: User talk:Sjcriscione


text:
User talk:Loopie

 Hello Loopie, 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. --La comadreja formerly AFriedman RESEARCH (talk) 17:30, 15 November 2010 (UTC)

 Hi! My name is WikiversityJack. I am a pumpkin and a Wikiversity mascot. My aunt told me that fun + learning = Wikiversity. Let me know if I can help you out.

title: User talk:Loopie


text:
Ideas in Geometry/Hyperbolic Geometry

What is Hyperbolic Geometry? - Hyperbolic geometry allows us to follow Euclid's first four axioms then replacing the last one: Parallel Postulate. What are the axioms that hold true in this geometry then? 1. A line can be drawn from a point to any other point. 2. A finite line can e extended indefinately. 3. A circle can be drawn, given a center and a radius. 4. All right angles are 90 degrees. 5. Given a line and a point not on that line there exists more than one parallel to the given line through the given point. What does this mean? In hyperbolic geometry the Poincaré Half-Plane Model we work in the upper half-plane (x,y) when x > 0. - here points are still points the difference is that lines are actually arcs inside the boundary where the intersection point with the boundary forms a right angle. This causes a small distortion around the boundary causing these distances to look much smaller and closer together than they truely are when near the bourndary. The closer an object is to a boundary the larger it is with more area it can hold inside of it. This also means that the shortest distance between two points on a line lies between the two points the farthest from the boundary line.

title: Ideas in Geometry/Hyperbolic Geometry


text:
United States Law/Constitutional/Outline

Before we had a constitution in this country, we had the Articles of Confederation. They established a system of government where each state had more power than the central government. What the states did to support the central government was voluntary.

The Constitution replaced the Articles of Confederation and established a much more powerful central government that could make the states does certain things, for example, to Raise and support armies. The Constitution did leave the States some power such as the power to provide for the health, safety and welfare of its citizens which is the main reason that most criminal laws are state laws. It is also the reason most domestic and criminal laws are state laws, and why each state has its educational system.

When the constitution was established, there was still a fear of a strong central government because of what had happened with England. The United States Constitution is the oldest working constitution in the world. One of the reasons why it still works is that the words in it are not specific which lets it be reinterpreted by the courts to change with the times. For example, the Constitution did not provide that a person charged with a crime had a right to counsel before the 1950’s this meant that if you could afford a Lawyer you could have one. Then the Supreme Court reinterpreted what the words said. When it held that if a person that has been charged with a crime and is too poor to afford a Lawyer the government will provide one for him. Although the writers of the constitution realized that they had to limit the power of the central government they thought the best way to do this was to divide the power into different pots. So that one branch of government didn’t have too much power. Another way they limited power was to provide ways that one group could check up on another. This whole concept is called checks and balances.

Three pots the power is divided into
 * 1) Congress
 * 2) President
 * 3) Courts
 * 4) States

Article I
Gives Congress the power to enact statutes, Congress can override the president's vote if 2/3 of both House vote for the bill. The Senate must confirm the presidents’ appointments.
 * Congress

Congress holds the pocket book to provide the financial support for these troops. They have the power “to raise and support armies.” and the power of impeachment -to include certain government officials including the President. The House of Representatives decides whether to charge formally a government official and the Senate holds the trial. If the official is convicted, he is removed from office.

 * Two big parts of article
 * 1) The power to regulate interstate commerce.
 * 2) Congress can do anything necessary and proper to carry out its powers.

Requirements to run for the senate and house of reps Senate:
 * Shall choose other officers
 * Must be at least 30 yrs old
 * Must have nine yrs of citizenship in the USA
 * Must be a resident of the state they are running in
 * Two from each state make up Congress
 * 100 members in Congress

House of Representatives
 * Enacts laws
 * 435 members this is based on population
 * Prints money
 * Build post offices
 * Bankruptcy
 * Tax and pay taxes

Interstate- between states Intrastate- within a state

Power to regulate interstate commerce among the states is used by Congress and the courts to expand significantly the power of the federal government. This clause was originally intended to regulate trade between the states and over the years, it has been interpreted to mean that Congress can pass laws affecting interstate commerce. If goods go from one state to another, it can be regulated by the federal government. For example, the Supreme Court said if a person grows vegetables in their garden and eats them then the vegetable can still be regulated by Congress. If the crop does not travel between states, the fact that the farmer’s growing and eating their vegetables means that they are not consuming vegetables produced by others that did travel between the States. Therefore, less is purchased from the farmers, and this affects interstate commerce and gives the Supreme Court jurisdiction over such cases.

Article 1 Section 8

 * Lists the powers of Congress
 * Necessary and proper clause make all laws that are necessary and proper for carrying into execution of preceding powers means = we can make all laws that are necessary and proper to carry out all of those listed powers.
 * Power to regulate interstate commerce. The necessary and proper clause used more than any other clause to expand the power of government.

Once the US Supreme Court has decided that the law meets the requirements of the Constitution the only thing left to do is would be to amend the constitution and the constitution has only been amended 28 times in 200 years.

Article 1 Section 9

 * Puts restriction on Congress

 * 1) Writ of habeas corpus. Courts often call it the great writ. It challenges the lawfulness of confinement is a way that a person who is in prison or held in any way may challenge whether it’s lawful.  A person can’t be confined unless there is probable cause that they have committed a crime or it they are legally insane.
 * 2) Prisoners who are convicted of crimes can file motions for writs of habeas corpus very frequently asking the court to order the custodian of the prison to bring him into court and show the Judge that his confinement is lawful. The writ of habeas corpus is considered to be one of the foundations of our democracy. If we didn’t have writ, the police could take you into custody without any reason and confine you indefinably. The only time the writ can be suspended is in a time of war by Congress.  The last time that this happened was during WWII after the Japanese invaded Pearl Harbor. They then passed a law that all people with Japanese ancestors, could be confined in camps, even if the individual was born in the USA. They were given a certain amount of days to get their affairs in order and then sent off to the camps instill the end of the war. They lost their businesses & homes because they couldn’t work. They were placed into camps because of the fear that they would be more loyal to Japan and become spies.

 * 1) No Ex-post facto laws. The government can't make an act a crime after the act was committed. Nor can they increase the punishment for a crime after the act was committed.

 * 1) Can’t pass bills of attainder. A law that would allow punishment of a person for an offense without a trial.

Article II- Presidents
The President may vote a proposed law (bill) sent to him by Congress. The chief executive has the job to carry out the laws that Congress has passed. The president can appoint ambassadors to foreign countries and appoint federal Judges (even the Supreme Court).

As commander in chief - this means that he is the leader of the military and can send troops all over the world, he can do this without Congress formally declaring war. If Congress doesn’t like where troops are sent, they have the power to raise and support armies.

===== Article II Section II ===== The President can grant reprieves and pardons to people convicted of federal crimes, not state crimes. The last known time of this occurrence was right after Nixon resigned. His Vice President Gerald Ford became President, and he pardoned Nixon for any crimes he had committed with relation to Watergate.

Article III-Courts
This article created the Supreme Court, which is the 3rd branch of government. Because the Constitution establishes the article, neither Congress nor the president can distort it. Once a bill becomes law the courts have the power to interpret the words, or strike the law down.

Article IV- States
Whatever power isn’t given to the central government are reserved to the states. States have the authority to provide for the health, safety and welfare of all of their citizens, the reason most criminal and domestic laws are State laws.

Full Faith and Credit Clause
Only applies to civil judgments of every other State.
 * Ex. Divorce in NY and you moved to North Carolina. North Carolina must recognize the divorce.
 * Ex. A Civil money judgment in one state you can send that decision to the state court of the state where the defendant is living and they are supposed to help you collect the judgment.

Extradition applies only in felony criminal cases. It is the process by which a fleeing felon is returned to the state from which he fled, in a felony crime, the maximum is over two years. The governor of one state must request the return of the felon from the other state (Between governors) the governor of the asylum state (where he fled to) has the choice to send the person back or not.

Article V
This article talks about amending the constitution and is the main reason the Constitution has only been amended 27 times in over 200 years. The words in the constitution are so vague the courts can reinterpret what they mean to change with the times. (Judicial review - power of the court to judge a law by the requirements of the Constitution) It is so hard to get thru the process.

For proposal of an Amendment to pass it must have a majority vote by both houses.

Most Common Way to Amend an Article is a proposal in which 2/3 of both House of Congress vote for the amendment. Last time Congress proposed an amendment was with the ERA= equal rights amendment and allowed Women will have equal rights to men. The amendment got through the proposal stage and was then sent to the state legislatures for them to vote on it.

To allow ratification- ¾ of state legislatures vote to approve an amendment. Rather than formally amending the Constitution it has been changed over the years by how the words in it are interpreted.

Article VI-
Supremacy Clause. If there is ever a conflict between a state law and federal law, the federal law will win.

Article VII-
Explains how the states would use procedures to use the constitution.

The 1st ten amendments are called the Bill of Rights, and they were added to the Constitution shortly after it was written for the final approval of the constitution. The Reason they are called the bill of rights is that they protected the individual from the power of the government. The dividing the power between the different branches of the government and the system of checks and balances that were established in the body of the constitution (the 1st six articles) weren’t enough protection for the individual. We needed something affirmative.
 * AMENDMENTS

1st Amendment
Freedom of religion. This amendment is the basis of a democratic society and grants the right to believe anything you want to religiously. It allows a right to express your beliefs. As far as exercising the right to religion it can't be limited unless it threatens the safety of others. The amendment defines the separation of church and state and says that our government can’t have a state religion even though most countries have a specified religion.

4th Amendment
Protects Individuals against unreasonable search and seizure and keeps law enforcement from being too enthusiastic about fighting crime by putting a natural party between the person and the police. Who with a cool head can decide is there proof that more likely then not evidence of a particular crime is in a given place.

With most searches, the search can’t be made without an independent official deciding that there is probable cause that evidence of a crime is in a particular place. A search warrant must specifically describe the place to be searched and the items to be seized.

5th Amendment
The privilege against self-incrimination. A person can't be forced our government to give information against himself that would lead to his being charged with or convicted of a crime. This rule applies to what the individual knows in his head. Such as where the dead body is, were stolen merchandise. It doesn’t apply to physical characteristics such as fingerprints.

6th Amendment
The right to Counsel. Provides that a person has a right to an attorney when charged with a crime that will, in fact, result in incarceration if he is convicted. Also provides that a person accused of an criminal offence has the right to confront witnesses against him.

Indigent - Too poor to afford a lawyer

8th Amendment
Provides against cruel and unusual punishment and excessive bail. Bail is designed only to ensure that a person will show up for trial. It can not be more than is reasonably necessary to ensure the person will appear for trial.

10th Amendment
The powers not given to the federal government in the Constitution are reserved to the states. When the 1st ten amendments were added to the constitution, they only protected the individual against the power of the federal government. They didn’t protect against the state government. The reason for this was that the people were more afraid of the federal government and not the state. As the years passed, it became apparent that the states were worse than the federal government in not protecting people's rights.

So after the civil war, the following amendments were added the 13th, 14th, 15th.

13th Amendment
Abolishes Slavery

14th Amendment
No state shall deprive any person of life, liberty or property without due process of law. Used to make some of the rights in the bill of rights apply to the states. The courts have decided due process to mean “fundamental fairness in judicial procedures.”

15th Amendment
A person can’t be denied the right to vote because of race.

After the 13th, 14th, and 15th were passed people lost interest in protecting the citizens. It was in this period (1850-1900) that we had the rise of big business. The theory was that the government should keep its hands off business. It wasn't until the 1920’s that the 1st right of the Bill of Rights was applied to the states.


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User talk:Swtbbygrl

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title: User talk:Swtbbygrl