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User talk:Babysabre

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User talk:Michaelbrabec

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User talk:InfantGorilla

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User talk:LUGALzu

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Interest in Georgian subjects
We could use some more geographic diversity in the sorts of subjects covered here. If you have particular materials from your university that might be worth posting, this could be reason to set up a Wikiversity incubator where materials of all languages are welcome. –SJ + > 07:40, 11 November 2010 (UTC)

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User talk:Sainsf

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User talk:SpeakEzee

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User talk:Doolder

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User talk:Mehulthakor1987

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User talk:Survivor2Sucess!

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User talk:Quantuk

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text:
Methane

Discovery
Methane was first synthesized from carbon and hydrogen in 1904 by a Russian-American chemist, Vladimir Ipatiev (1867-1952).

Properties
Methane is the major component of natural gas, about 87% by volume. At room temperature and standard pressure, methane is a colorless, odorless gas; the smell characteristic of natural gas as used in homes is an artificial safety measure caused by the addition of an odorant, often methanethiol]] or ethanethiol. Methane has a boiling point of −161 °C (−257.8 °F) at a pressure of one Atmosphere (unit)|atmosphere. As a gas it is flammable only over a narrow range of concentrations (5–15%) in air. Liquid methane does not burn unless subjected to high pressure (normally 4–5 atmospheres).

Quick Facts
Name: Methane

Formula: CH4

Melting Point:

Boiling Point: -161.6 celsius

Appearance:

Molar Mass:

Shape: tetrahedryl

title: Methane


text:
User:Egm6321.f10.team3.franklin/Hwk6

= Problem 3 =

Given
For $$q(x)=\sum_{i=0}^4c_ix^i$$ where $$c_0=2,\ c_1=-5,\ c_2=-3,\ c_3=11,\ c_4=7,\ c_5=6$$

Find
1. Find $$\displaystyle a_i$$ such that $$q=\sum_{i=0}^4a_iP_i .$$

2. Plot $$\displaystyle q=\sum_ic_ix^i$$ and $$\displaystyle q=\sum_ia_iP_i .$$

Solution
$$q(x)=\sum_{i=0}^5c_ix^i$$ can be shown as a column matrix

$$\begin{bmatrix} c_0x^0\\c_1x^1\\c_2x^2\\c_3x^3\\c_4x^4\\c_5x^5 \end{bmatrix}=\begin{bmatrix} 2x^0\\-5x^1\\-3x^2\\11x^3\\7x^4\\6x^5 \end{bmatrix}$$

Next, evaluate $$q(x)=\sum_{i=0}^5a_iP_i$$

$$q(x)=\sum_{i=0}^5a_iP_i=\begin{bmatrix} a_0&a_1&a_2&a_3&a_4&a_5 \end{bmatrix}\begin{bmatrix} P_0\\P_1\\P_2\\P_3\\P_4\\P_5 \end{bmatrix}$$

Setting $$q(x)=\sum_{i=0}^5a_iP_i(x)=\sum_{i=0}^5c_ix^i $$ yields

$$\begin{bmatrix} a_0&a_1&a_2&a_3&a_4&a_5 \end{bmatrix}\begin{bmatrix} P_0\\P_1\\P_2\\P_3\\P_4\\P_5 \end{bmatrix}=\begin{bmatrix} 2x^0\\-5x^1\\-3x^2\\11x^3\\7x^4\\6x^5 \end{bmatrix}$$

From above, we see that

$$a_0P_0=2x^0 \Rightarrow a_0={2/P_0}$$

$$a_1P_1=-5x^1 \Rightarrow a_1={-5x/P_1}$$

$$a_2P_1=-3x^2 \Rightarrow a_2={-3x^2/P_2}$$

$$a_3P_3=11x^3 \Rightarrow a_3={11x^3/P_3}$$

$$a_4P_4=7x^4 \Rightarrow a_4={7x^4/P_4}$$

$$a_5P_5=6x^5 \Rightarrow a_4={6x^4/P_5}$$

Where,

$$\displaystyle P_0(x)=1$$

$$\displaystyle P_1(x)=x$$

$$\displaystyle P_2(x)=1/2(3x^2-1)$$

$$\displaystyle P_3(x)=1/2(5x^3-3x)$$

$$\displaystyle P_4(x)=1/8(35x^4-30x^2+3)$$

$$\displaystyle P_5(x)=1/8(63x^5-70x^3+15)$$

Which gives

$$\displaystyle a_0=2$$

$$\displaystyle a_1={-5x\over x}$$

$$\displaystyle a_2={-3x^2\over 1/2(3x^2-1)}$$

$$\displaystyle a_3={11x^3\over 1/2(5x^3-3x)}$$

$$\displaystyle a_4={7x^4\over 1/8(35x^4-30x^2+3)}$$

$$\displaystyle a_5={6x^4\over 1/8(63x^5-70x^3+15)}$$

Which reduces to

 $$\displaystyle a_0=2$$

$$\displaystyle a_1=-5$$

$$\displaystyle a_2={-2\over (3x^2-1)}-2$$

$$\displaystyle a_3={66\over 5(5x^3-3x)}+22/5$$

$$\displaystyle a_4={24(10x^2-1)\over 5(35x^4-30x^2+3)}+8/5$$

$$\displaystyle a_5={80(14x^3-3)\over 21(63x^5-70x^3+15)}+16/21$$

The plots of $$\displaystyle q=\sum_ic_ix^i$$ and $$\displaystyle q=\sum_ia_iP_i$$ are showed below.

Contributing Members

 * Solved and posted by Egm6321.f10.team3.franklin 16:57, 5 December 2010 (UTC)

= Problem 4 Legendre Polynomials' Orthogonality =

Given
Consider the boundary condition
 * {| style="width:100%" border="0" align="left"

a)f(\theta) = T_0 \cos^6 \theta $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 4.1.a)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

b)f(\theta) = T_0 \exp{\left(-\left(\frac{2\theta}{\pi}\right)^2\right)} $$ $$ where
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 4.1.b)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

a)f(\mu) = T_0 \left(1 - \mu^2\right)^3 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 4.2.a)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

b)f(\mu) = T_0 \exp{\left(-\left(\frac{2\arcsin{\mu}}{\pi}\right)^2\right)} $$ $$ is even, and
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 4.2.b)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\mu = \sin \theta. $$ $$ $$\displaystyle A_n $$ is
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 4.3)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

A_n = \frac{2n + 1}{2}  $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 4.4)
 * }
 * }

1.
Without calculation, find property of $$\displaystyle A_n$$, i.e.
 * {| style="width:100%" border="0" align="left"

A_{2k} \stackrel{?}{=} 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 4.5.a)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

A_{2k+1} \stackrel{?}{=} 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 4.5.b)
 * }
 * }

2.
Compute 3 non-zero coefficients of $$\displaystyle A_n .$$

Solution for 1.
The scalar product of $$\displaystyle f $$ and $$\displaystyle P_n $$ is
 * {| style="width:100%" border="0" align="left"

 = \int_{-1}^1 f(\mu) P_n(\mu) \, {\rm d}\mu $$ $$ We can split the integral up in two intervals
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 4a.6)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

 = \int_{-1}^0 f(\mu) P_n(\mu) \, {\rm d}\mu + \int_{0}^1 f(\mu) P_n(\mu) \, {\rm d}\mu $$ $$ When $$\displaystyle n$$ is odd, so is $$\displaystyle P_n$$, and $$\displaystyle f(\mu)$$ is $$\displaystyle even$$. Hence, when $$\displaystyle n$$ is odd, we have that $$\displaystyle n = 2k + 1$$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 4a.7)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

 = \int_{-1}^0 f(-\mu) (-P_{2k+1}(-\mu)) \, {\rm d}\mu + \int_{0}^1 f(\mu) P_{2k+1}(\mu) \, {\rm d}\mu $$ $$ We can substitute $$\displaystyle \xi = -\mu$$ in the first integral so that
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 4a.8)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

<f, P_{2k+1}> = \int_1^0 f(\xi) P_{2k+1}(\xi) \, {\rm d}\xi + \int_0^1 f(\mu) P_{2k+1}(\mu) \, {\rm d}\mu $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 4a.9)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

<f, P_{2k+1}> = - \int_0^1 f(\xi) P_{2k+1}(\xi) \, {\rm d}\xi + \int_0^1 f(\mu) P_{2k+1}(\mu) \, {\rm d}\mu = 0 $$ $$ Hence
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 4a.10)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle A_{2k+1} = 0 $$ $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 *
$$\displaystyle (Eq. 4a.11)
 * }
 * }

Part a.
The expression for $$\displaystyle A_n $$ is given by,

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

\begin{align} A_n & = \frac{2n+1}{2} \int_{-\pi/2}^{\pi/2} f(\theta)P_n(\sin\theta)d(\sin\theta) \\ & = \frac{2n+1}{2} \int_{-1}^{1} f(\mu)P_n(\mu)d\mu \\ \end{align} $$
 * $$\displaystyle
 * $$\displaystyle

 * }.
 * }.

where $$\displaystyle \mu = \sin\theta $$ and $$\displaystyle f(\mu) = T_0 \left(1-\mu^2\right)^3 $$.

$$\displaystyle A_n = 0 $$ for odd values of $$\displaystyle n $$.

n=0
When $$\displaystyle n = 0 $$, we know $$\displaystyle P_n(\mu) = 1 $$ and so,

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

 * $$\displaystyle
 * $$\displaystyle

A_0 = \frac{2(0)+1}{2} \int_{-1}^{1} T_0 \left(1-\mu^2\right)^3 d\mu

$$ $$
 *
$$\displaystyle (Eq. 4a.12)

 * }.
 * }.

Using definite integrals in Matlab, we get,

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

 * $$\displaystyle
 * $$\displaystyle

\int_{-1}^{1}\left(1-\mu^2\right)^3 d\mu = \left[ \frac{5\pi}{16} \right]

$$ $$
 *
$$\displaystyle (Eq. 4a.13)

 * }.
 * }.

Substituting equation Eq(4a.13) in Eq(4a.12),

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

 * $$\displaystyle
 * $$\displaystyle

\Rightarrow A_0 = \frac{T_0}{2}\left[ \frac{5\pi}{16} \right]

$$
 * }.
 * }.

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

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

\Rightarrow A_0 = T_0\left(\frac{5\pi}{32}\right)

$$ $$
 * style= |
 *
$$\displaystyle (Eq. 4a.14)
 * }.
 * }.

n=2
When $$\displaystyle n = 2 $$, we know $$\displaystyle P_n(\mu) = \frac{1}{2} (3

\mu^2 - 1) $$ and so,

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

 * $$\displaystyle
 * $$\displaystyle

A_2 = \frac{2(2)+1}{2} \int_{-1}^{1} T_0 \left(1-\mu^2\right)^3 \left[\frac{1}{2} (3\mu^2 - 1) \right]d\mu

$$ $$
 *
$$\displaystyle (Eq. 4a.15)

 * }.
 * }.

Using definite integrals in Matlab, we get,

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

 * $$\displaystyle
 * $$\displaystyle

\int_{-1}^{1} \left(1-\mu^2\right)^3\left[\frac{1}{2} (3\mu^2 - 1) \right] d\mu = \left[ \frac{-25\pi}{256} \right]

$$ $$
 *
$$\displaystyle (Eq. 4a.16)

 * }.
 * }.

Substituting equation Eq(4a.16) into Eq(4a.15),

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

 * $$\displaystyle
 * $$\displaystyle

\Rightarrow A_2 = \frac{5T_0}{2} \left[ \frac{-25\pi}{256} \right]

$$
 * }.
 * }.

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

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

\Rightarrow A_2 = T_0\left(\frac{-125\pi}{512}\right)

$$ $$
 * style= |
 *
$$\displaystyle (Eq. 4a.17)
 * }.
 * }.

n=4
When $$\displaystyle n = 4 $$, we know $$\displaystyle P_n(\mu) = \frac{1}{8}\left(35\mu^4 - 30\mu^2 + 3\right) $$ and so,

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

 * $$\displaystyle
 * $$\displaystyle

A_4 = \frac{2(2)+1}{2} \int_{-1}^{1} T_0 \left(1-\mu^2\right)^3 \left(\frac{1}{8}\left(35\mu^4 - 30\mu^2 + 3\right)\right)d\mu

$$ $$
 *
$$\displaystyle (Eq. 4a.18)

 * }.
 * }.

Using definite integrals in Matlab, we get

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

 * $$\displaystyle
 * $$\displaystyle

\int_{-1}^{1} \left(1-\mu^2\right)^3 \left(\frac{1}{8}\left(35\mu^4 - 30\mu^2 + 3\right)\right)d\mu = \left[ \frac{45\pi}{2048} \right]

$$ $$
 *
$$\displaystyle (Eq. 4a.19)

 * }.
 * }.

Substituting equation Eq(4a.19) in Eq(4a.18),

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

 * $$\displaystyle
 * $$\displaystyle

\Rightarrow A_4 = \frac{9T_0}{2} \left[ \frac{45\pi}{2048} \right]

$$
 * }.
 * }.

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

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

\Rightarrow A_4 = T_0 \left(\frac{405\pi}{4096}\right)

$$ $$
 * style= |
 *
$$\displaystyle (Eq. 4a.20)
 * }.
 * }.

Part b.
The expression for $$\displaystyle A_n $$ is given by,

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

\begin{align} A_n & = \frac{2n+1}{2} \int_{-\pi/2}^{\pi/2} f(\theta)P_n(\sin\theta)d(\sin\theta) \\ & = \frac{2n+1}{2} \int_{-1}^{1} f(\mu)P_n(\mu)d\mu \\ \end{align} $$
 * $$\displaystyle
 * $$\displaystyle

 * }.
 * }.

where $$\displaystyle \mu = \sin\theta $$ and $$\displaystyle f(\mu) = T_0 \exp{\left(-\left(\frac{2\arcsin{\mu}}{\pi}\right)^2\right)}$$.

$$\displaystyle A_n = 0 $$ for odd values of $$\displaystyle n $$.

n=0
When $$\displaystyle n = 0 $$, we know $$\displaystyle P_n(\mu) = 1 $$ and so,

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

 * $$\displaystyle
 * $$\displaystyle

A_0 = \frac{2(0)+1}{2} \int_{-1}^{1} T_0 \exp{\left(-\left(\frac{2\arcsin{\mu}}{\pi}\right)^2\right)} d\mu

$$ $$
 *
$$\displaystyle (Eq. 4b.12)

 * }.
 * }.

Using definite integrals in Matlab, we get,

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

 * $$\displaystyle
 * $$\displaystyle

\int_{-1}^{1}\exp{\left(-\left(\frac{2\arcsin{\mu}}{\pi}\right)^2\right)} d\mu = \left[ \frac{\pi^{\frac{3}{2}} erf\left(1\right)}{2} \right] , $$ $$
 *
$$\displaystyle (Eq. 4b.13)
 * }
 * }

where $$\displaystyle erf\left(x\right) $$ is
 * {| style="width:100%" border="0" align="left"

erf\left(x\right) = \frac{2}{\sqrt{\pi}}\int_{-1}^{x} \left(\exp{-t^2}\right) dt $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 4b.13.a)
 * }
 * }

Substituting equation Eq(4b.13) in Eq(4b.12),
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle
 * $$\displaystyle

\Rightarrow A_0 = \frac{T_0}{2}\left[ \frac{\pi^{\frac{3}{2}} erf\left(1\right)}{2} \right]

$$
 * }.
 * }.

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

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

\Rightarrow A_0 = T_0\left( \frac{\pi^{\frac{3}{2}} erf\left(1\right)}{4} \right)

$$ $$
 * style= |
 *
$$\displaystyle (Eq. 4b.14)
 * }.
 * }.

n=2
When $$\displaystyle n = 2 $$, we know $$\displaystyle P_n(\mu) = \frac{1}{2} (3

\mu^2 - 1) $$ and so,

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

 * $$\displaystyle
 * $$\displaystyle

A_2 = \frac{2(2)+1}{2} \int_{-1}^{1} T_0 \exp{\left(-\left(\frac{2\arcsin{\mu}}{\pi}\right)^2\right)} \left(\frac{1}{2} (3\mu^2 - 1) \right)d\mu

$$ $$
 *
$$\displaystyle (Eq. 15)

 * }.
 * }.

Matlab does not return and analytical solution for the above definite integral. Evaluating the integral in Wolfram Alpha returns

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

 * $$\displaystyle
 * $$\displaystyle

\int_{-1}^{1} \exp{\left(-\left(\frac{2\arcsin{\mu}}{\pi}\right)^2\right)}\left(\frac{1}{2} (3\mu^2 - 1) \right) d\mu = \left[ \right]

$$ $$
 *
$$\displaystyle (Eq. 16)

 * }.
 * }.

Substituting equation (16) in (15),

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

 * $$\displaystyle
 * $$\displaystyle

\Rightarrow A_2 = \frac{5T_0}{2} \left[  \right]

$$
 * }.
 * }.

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

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

\Rightarrow A_2 = T_0\left( \right)

$$ $$
 * style= |
 *
$$\displaystyle (Eq. 17)
 * }.
 * }.

n=4
When $$\displaystyle n = 4 $$, we know $$\displaystyle P_n(\mu) = \frac{1}{8}\left(35\mu^4 - 30\mu^2 + 3\right) $$ and so,

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

 * $$\displaystyle
 * $$\displaystyle

A_4 = \frac{2(2)+1}{2} \int_{-1}^{1} T_0 \exp{\left(-\left(\frac{2\arcsin{\mu}}{\pi}\right)^2\right)} \left(\frac{1}{8}\left(35\mu^4 - 30\mu^2 + 3\right)\right)d\mu

$$ $$
 *
$$\displaystyle (Eq. 18)

 * }.
 * }.

Using definite integrals in Matlab, we get

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

 * $$\displaystyle
 * $$\displaystyle

\int_{-1}^{1} \exp{\left(-\left(\frac{2\arcsin{\mu}}{\pi}\right)^2\right)}\left(\frac{1}{8}\left(35\mu^4 - 30\mu^2 + 3\right)\right)d\mu = \left[ \right]

$$ $$
 *
$$\displaystyle (Eq. 19)

 * }.
 * }.

Substituting equation (19) in (18),

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

 * $$\displaystyle
 * $$\displaystyle

\Rightarrow A_4 = \frac{9T_0}{2} \left[  \right]

$$
 * }.
 * }.

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

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

\Rightarrow A_4 = T_0 \left( \right)

$$ $$
 * style= |
 *
$$\displaystyle (Eq. 20)
 * }.
 * }.

Contributing Members

 * Solved and posted by Egm6321.f10.team3.franklin 16:57, 5 December 2010 (UTC)

title: User:Egm6321.f10.team3.franklin/Hwk6


text: File:2010 11 11 08 24 37.djvu

title: File:2010 11 11 08 24 37.djvu


text:
User:EGM6321.f10.team6.cook/hw6

=Problem 1 - Legendre's Equation =

Problem Statement
Legendre's Equation is, for n=0,1,2,...

when n=2,

Required
If we were given the 1st homogeneous solution,

Show that the 2nd homogeneous solution is

Solution
For the general homogeneous L2_ODE_VC,

let's put the solution like eqn(1.6) to find out the 2nd homogeneous solution.

Then,

Plug in to the eqn(1.5),

So, we can get L2_ODE_VC with missing dep. var. U. To reduce order, let

then,

integrating factor for this ODE would be,

multiply eqn(1.11) to eqn(1.10) on both side.

Therefore, from eqn(1.6),

So we can figure out what the 2nd homogeneous solution is if we have the 1st homogeneous solution.

Let's rewrite the given problem eqn(1.2).

From the problem statement, we have the 1st homogeneous solution, eqn(1.2).

Let's find the integrating factor first. from eqn(1.11),

Hence,

This is the same to the eqn(1.4).

Problem Statement
The Legendre equation that arises as a result of separation of variables may be written as :

One set of solutions to this equation are the Legendre polynomials of the first kind which can be expressed generally as :

Required
Show that for n=0 to n=4 in equation 9.2 are solutions to the Legendre equation in 9.1.

Verification of n=0
By substituting n=0 into equation 9.2 we obtain the following solution to the Legendre equation:

In order to verify that this is a solution we must back substitute 9.3 into 9.1. If we do so we obtain the following:

Clearly for n=0 equation 9.2 provides a solution to the Legendre Equation.

Verification of n=1
By substituting n=1 into equation 9.2 we obtain the following solution to the Legendre equation:

In order to verify that this is a solution we must back substitute 9.6 into 9.1. If we do so we obtain the following:

For n=1 equation 9.2 provides a solution to the Legendre Equation.

Verification of n=2
By substituting n=2 into equation 9.2 we obtain the following solution to the Legendre equation:

In order to verify that this is a solution we must back substitute 9.10 into 9.1. If we do so we obtain the following:

For n=2 equation 9.2 provides a solution to the Legendre Equation.

Verification of n=3
By substituting n=3 into equation 9.2 we obtain the following solution to the Legendre equation:

In order to verify that this is a solution we must back substitute 9.14 into 9.1. If we do so we obtain the following:

For n=3 equation 9.2 provides a solution to the Legendre Equation.

Verification of n=4
By substituting n=4 into equation 9.2 we obtain the following solution to the Legendre equation:

In order to verify that this is a solution we must back substitute 9.18 into 9.1. If we do so we obtain the following:

For n=4 equation 9.2 provides a solution to the Legendre Equation.

title: User:EGM6321.f10.team6.cook/hw6


text:
User:Ottava Rima/102nd Intelligence Wing

On 11 November 2010, I was contacted and asked to perform a review of 102nd Intelligence Wing. Below is my review.

Lead
1. "After the war, the latter was deactivated and eventually the 102nd was formed with a direct lineage." - "the latter" can be confusing. Reuse "318th" for clarity.

2. "In 1946, the 102nd was activated at Logan International Airport where it stayed until 1968, when it moved to Otis. Also beginning in 1946 were regular patrols of the Northeastern United States, which ran in conjunction with the regular duty wings of the Air Force. This practice continued until 1973 when the Air Force removed the last active duty wings at Otis, leaving the 102nd with the duty to patrol the Northeast." - This seems to be a little confused in wording. Try this: "In 1946, he 102nd was activated at Logan International Airport where it began regular patrols of the Northeastern United States, which ran in conjunction with the regular duty wings of the Air Force. In 1968, the 102nd was moved to Otis where it continued its regular patrols until 1973."

3. "In 1961, the wing deployed to France during the Berlin Crisis. In 1989, the wing deployed to Panama during Operation Coronet Nighthawk. " - Two sentences beginning the same way. Perhaps say, "In 1961, the wing deployed to France during the Berlin Crisis and then, in 1989, the wing deployed to Panama during Operation Coronet Nighthawk.

4. "but they arrived too late to help stop the attacks." "help stop" seems inappropriate. Just say "to stop".

5. "the wing came in control of" - "came in control" doesn't make sense. "gained control of" or "gained control over" would make more sense. "Put in charge" would be another phrase to use.
 * Everything is all set here. Kevin Rutherford (talk) 23:09, 21 February 2011 (UTC)

Mission
1. Source used is from here. The site says "To provide world wide precision intelligence and command and control along with trained and experienced Airmen for expeditionary combat support and homeland security." The page copies this verbatum and instead should quote, saying "According to their homepage, the 102nd's mission is '_____'".

2. "This goes along with" - You cannot add that, as that is original research. You can say, "In addition, the website says that their Air Force based mission is '_______'".

3. "The mission of the wing ties into that" - See "2" above.
 * We're good! Kevin Rutherford (talk) 23:47, 21 February 2011 (UTC)

History - Origins
1. Language/Source problems -
 * Page - "After World War I ended, the Army showed interest in organizing aviation assets for the National Guard system. At the time, aircraft were attached to infantry units, similar to other weapons like artillery. Guard units without their own aircraft needed assets from other forces to co-operate with them. The War Department agreed The Guard should organize aviation squadrons as an organic part of the 18 infantry divisions assigned to the National Guard."
 * Source - "After World War I, there was a great interest on the part of the National Guard in military aviation... The War Department agreed that the Guard should organize aviation squadrons as an organic part of the 18 infantry divisions assigned to the National Guard."

You substitute "Army" for "National Guard" and I do not see anything about aircraft being attached to infantry units. The first and last sentence quoted above are taken direct from the source and need to be rewritten.

2. Language/Source problems -
 * Page - "In Massachusetts, the Archie Club, comprising former Army Air Service pilots, lobbied for the formation of an air unit for the Massachusetts National Guard. The state had earlier been allotted the entire 26th Guard Division. On June 27, 1921, the Adjutant General of Massachusetts authorized the organization of the 101st Observation Squadron, and within weeks, 15 veteran World War I pilots were commissioned and assigned to the 101st under the command of Capt. James K. Knowles. The 101st built its own air base on land-filled tidal flats at Jeffries Point, East Boston.[5] The 101st flew its Curtiss JN-4 "Jenny" aircraft throughout New England at air shows, county fairs and other events, and held two-week summer camps to simulate forward deployments.[4] Pilots flew their Curtiss O-11s to temporary fields on Cape Cod while ground crews followed in trucks.[4] One of these fields became Cape Cod Airport"
 * Source - "In Massachusetts, the Archie Club, composed of former Army Air Service pilots, lobbied the adjutant general to request an aviation squadron for the Massachusetts National Guard." "On June 27, 1921, the adjutant general of Massachusetts authorized the organization of the 101st Observation Squadron. Within weeks, 15 veteran World War I pilots were commissioned and assigned to the 101st under the command of Capt. James K. Knowles. The 101st was stationed at the U.S. Army Air Corps base at Jeffries Point, East Boston which later became Logan Airport in honor of Maj. Gen. Edward L. Logan, commanding general of the 26th Division, 1923-1928."

The first lines are stated to be from [5] but are actually directly copied from [4]. The last sentence before the [5] is pulled from the source: "The squadron secured its own training site on the land filled tidal flats at Jeffries Point, East Boston." This, however, is reworded enough to not need to be dramatically changed.

3. "held two-week summer camps to simulate forward deployments" I could not find this part in the source.

4. "Pilots flew their Curtiss O-11s to temporary fields on Cape Cod while ground crews followed in trucks.[4] One of these fields became Cape Cod Airport.[4]" - I could not find either in the sources.

5. "In 1933 Jeffery Field was rebuilt with new hangars and administrative buildings, and renamed Logan Airport in honor of Major General Edward L. Logan, who commanded the 26th Division from 1923 to 1928." - Pulled from two different parts, but deals with enough "fact" and has enough rewording to be acceptable.

6. Language/Source problems -
 * Page - "The 101st was ordered into state service in 1936 and 1938 during a flood and hurricane to fly observation missions and to drop food and equipment to stranded fishermen and the residents of Isle au Haut, Maine"
 * Source - "The 101st was also ordered into state service in 1936 and 1938 during a devastating flood and hurricane to fly observation missions and to drop food and equipment to stranded fisherman and the residents of Isle au Haut, Maine."

This needs to be reworded and separated from the source. Suggestion - "In 1936 and 1938, he 101st was used to aid the residents of Isle au Haut, Maine and stranded fisherman during natural disasters."

7. "The 101st gained attention when it played a part in the United States Army Air Service's flight around the world.[4] The flight was the crowning achievement in United States aviation because it succeeded where five other nations had failed. The 101st serviced the aircraft at Boston which was on the flight leg from Canada to Washington D.C..[4] It also serviced the Spirit of St. Louis when Charles Lindbergh visited the state.[6]" - I could no find this in the source. The word "Spirit", "Louis", "Lindbergh", etc, do not appear.
 * Everything is done. Kevin Rutherford (talk) 23:47, 21 February 2011 (UTC)

History - Cold War
Continued on November 16, 2010.

Source 4 currently not working. Will have to recheck later.

1. Only line that can be checked in "Cold War": "From 1956 to 1976, the 102nd was commanded by Brigadier General Charles W. Sweeney, pilot of the B-29 Superfortress that dropped the Fat Man nuclear bomb on Nagasaki, Japan."

The time "1956" does not appear in the source. "Sweeney" doesn't appear in the source.

2. "Berlin Wall Crisis" begins with: "On August 16, 1961, when the Berlin Wall crisis was unfolding, several United States Air Force Reserve units were notified of their pending recall to active duty. On October 1 the Massachusetts Air National Guard's 102nd Tactical Fighter Wing and its three squadrons, the 101st, 131st and 138th assumed active duty at Otis Air Force Base."

The words "August 16" do not appear in the source. First specific date mentioned was November 1st. Only 102, 104, and 174 are mentioned, not the squadrons above. These are stated as being the subordinate unites in source 7 used elsewhere (see below at 3 for it being used).

3. "The wing had 82 Sabres, plus two C-47 Skytrains and six T-33 Shooting Stars for support and training purposes. During the crisis, the wing controlled the 102nd Tactical Fighter Group, the 104th Tactical Fighter Group, and the 174th Tactical Fighter Group from New York"

The word "82" does not appear in the source nor do any of the other numbers. The plane names do appear.

4. Unsourced sentences: "Starting on December 5, 1961 the 102nd began deploying to Wheelus Air Base, Libya for gunnery training.During its time in Europe, the 102nd participated in several United States Air Force and North Atlantic Treaty Organization exercises, including a deployment to Leck Air Base, West Germany near the Danish border. At Leck, ground and support crews from both countries exchanged duties, learning how to perform aircraft maintenance and operational support tasks."

5. "Regular Air Force personnel and a group of Air National Guard personnel who volunteered to remain on active duty formed the 480th Tactical Fighter Squadron of the newly activated 366th Tactical Fighter Wing."

The word "480" doesn't appear in the source.

6. "In 1968, the 102nd Tactical Fighter Wing moved to Otis Air Force Base, and was reassigned from Air Defense Command to Tactical Air Command the next year. The wing flew the F-84F Thunderstreak from 1964 until June 1971, when a squadron of F-100D Super Sabres was transferred directly from units fighting the Vietnam War.[7] These were superseded soon after by the Mach 2 F-106 Delta Darts and on June 10, 1972, the unit became the 102nd Air Defense Wing.[7] On December 30, 1973, Otis Air Force Base was inactivated and transferred to the Massachusetts Air National Guard as Otis Air National Guard Base.[7]"

Second sentence is directly copied from the source: "The Wing flew the F-84F Thunderstreak until June 1971, when a squadron of F-100D Supersabres was transferred directly from the conflict in Vietnam." The word "1973" does no appear in the source.

7. "The wing intercepted Soviet Tupolev Tu-95 Bear bombers on many occasions, the first of which occurred off Long Island on April 25, 1975."

Fine.

8. Don't have full article to check sourcing of "Many of these incidents involved escorting the Bears to Cuba. The wing occasionally shadowed drug smuggling aircraft, and on one occasion was scrambled to escort an unidentified object, which later turned out to be a weather balloon."

9. "In 1976, the 102nd Fighter Interceptor Group was deactivated and re-formed as the 102nd Fighter Interceptor Wing, assuming authority for the 177th and 125th Fighter Interceptor Groups in Atlantic City, New Jersey, and Jacksonville, Florida, and for the 107th and 147th Fighter Interceptor Groups, flying F-4C Phantom at Niagara Falls, New York, and Ellington Field, Texas.[7]"

Part of the first line is copied from source: "In 1976, the 102nd Fighter Interceptor Group was deactivated with the 102nd Fighter Interceptor Wing assuming working command authority. " Parts of the rest of the sentence copied from source: "authority for the 177th and 125th Fighter Interceptor Groups ... located at Atlantic City, NJ, and Jacksonville, Fla. respectively, and also for the 107th and 147th Fighter Interceptor Groups (flying F-4C aircraft) at Niagara, N.Y., and Ellington ANG Base, Texas. "

10. "The 102nd Fighter Interceptor Wing deactivated its F-106s on January 5, 1988. Between January and April 1988, the wing converted to the F-15A Eagle, which it received from a unit deactivating at Minot Air Force Base. It then resumed its alert commitment at Otis, and also established a new Detachment 1 at Loring Air Force Base. The 102nd was the first Air National Guard unit to be equipped with the F-15."

The word "1988" does not appear in the source. The word "Minot" does not appear in the source. The word "Loring" does not appear in the source.

11. "On January 24, 1989, airmen monitoring the radar at the Northeast Air Defense Sector at Griffiss Air Force Base spotted a plane which was not following any known flight plan. The order was then given to "scramble the Eagles," after repeated attempts to contact the pilot failed. Two jets then took off from Loring to search for the "unknown rider." The pilots later came across a plane that was blacked out, with no lights on on the inside or outside. The pilot, Diego Jose Ganuza was a narcotics smuggler from Colombia's Medellin drug cartel. The 500 kilograms of cocaine that he was carrying had a street value of two hundred million dollars."

The last two sentences are similar to the source: "the pilot was Diego Jose Ganuza, a narcotics smuggler working for Colombia's Medellin drug cartel. He was ferrying 500 kilograms of cocaine with a street value of about $200 million."
 * All is well! Kevin Rutherford (talk) 23:09, 21 February 2011 (UTC)

History - Post Cold War
1. A lot of information from a tiny source, which is a bad sign. Article says: "The wing continued its air defense mission after the fall of the Soviet Union. In 1992, the wing deployed eight pilots, five F-15 Eagles, and 48 maintenance and security personnel, for five days training at Canadian Forces Base Goose Bay, Labrador, Canada." 1992 is not mentioned at all nor any of the planes.

2. Same source cites this: "The same year, with the reorganization of the Air Force, the wing was reassigned from the disbanding Tactical Air Command to the new Air Combat Command". No 1992 means no "same year". The rest of the paragraph has the same problem.

3. This isn't the best source but it is used (for the most part) properly in the beginning second paragraph.

4. "The planes were joined by F-16s from the 177th Fighter Wing in Atlantic City, New Jersey. " This is derived from "Later, F-16s are scrambled from Atlantic City, New Jersey. ", which is almost original research but is okay as there was only one wing located in Atlantic City.

5. "The F-15s initially intercepted the aircraft off the coast of eastern Canada, trailing at a distance of 10-mile (16 km), which decreased to 5-mile (8.0 km) as the plane got closer to the airport. The fighters then did a low fly-by as the plane landed. They continued to circle around the airport until they returned back to Otis." Problem. Source says: "... intercept the... aircraft off the coast of eastern Canada,... trail... a distance of... ten miles. As the plane approaches JFK Airport, the fighters move in to a distance of five miles. They circle overhead for a while... and then return to their bases." This needs to be further changed from the source but it does describe facts.

6. Last sentence in "Local defense" is fine.

7. "Deployments" uses the source mentioned above in number 1. This is a problem as dates do not match.

8. "The wing again deployed more than 350 personnel to the Middle East and Europe in 2000 to participate in Operation Southern Watch.[8]" The source isn't the best, and it will need to be altered further from the source: "The wing deployed again more than 350 personnel to the Middle East and Europe in 2000 to participate in Operation Southern Watch."

9. The 9/11 section uses this source in an okay manner but the quote needs to have a citation after it. Some needs to be altered like: "On his radio, he called Major Daniel Nash, who was the other alert pilot who was on duty and told him to get ready for the alert call. They then got suited up and headed out to their jets." which is from "On his... radio he calls Major Daniel Nash, who along with Duffy is an “alert” pilot on duty at this time, and instructs him to suit up ready for any scramble call."

10. Article: "The Northeast Air Defense Sector responded by ordering Major Kevin Nasypany, the commander of the Weapons team which controlled the fighters, to put the jets on "battle stations." " Source: "At NEADS... orders his Weapons Team, which controls the fighters, to put the Otis planes on “battle stations.” The rest appears to be okay.

11. Article: "Major Naspany was then asked what to do with the fighters and he responded by saying" Source: "Major Kevin Nasypany... is asked what to do with the Otis fighters. He responds:" The actual quotes are okay.

12. Article: "From 9:09 to 9:13 the F-15s stayed in the holding pattern[20][21][22] until heading for Manhattan where they arrived at 9.25 to establish a Combat Air Patrol (CAP) over the area.[21" Source: "From 9:09 to 9:13, the Otis fighters stayed in this holding pattern." and "They arrived at 9:25 and established a combat air patrol (CAP) over the city."

13. The whole fourth, fifth, and sixth paragraph in 9/11 needs to be rewritten, as you can see from the first line compared to the paragraph in the source: Article: "the senior battle staff were watching the news when they saw Flight 175 hit the South Tower. One commander immediately shouted out, "We need to go to battle staff!"" Source: "They watch as the live television coverage shows Flight 175 crashing into the South Tower... One commander immediately shouts out, “We need to go to battle staff!” "

Just remove the whole "Conspiracy link" section and here is why:

14. In "Conspiracy link": I can't find the conspiracy in the source.

15. "The questions arising from the response time of the jets come from the practice of Cold War era policies which prohibited the immediate response to an emergency like a hijacking." I don't see the words "cold war" in the source.

16. "First responder and pilot Daniel Nash said that he could not recall being told that the North Tower " Daniel Nash's name is not used in the source. It is in the comments section which is not reliable. This is true for all uses of the source.

17. The last sentence using sources for Supersonic speed and the rest do not deal with 9/11 and would be original research.
 * I didn't remove the last section yet, but everything else is done. Kevin Rutherford (talk) 23:09, 21 February 2011 (UTC)


text:
User talk:Digawm

 Hello Digawm, 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.
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 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
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 * 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. --Ottava Rima (talk) 19:10, 11 November 2010 (UTC)

title: User talk:Digawm


text: File:2010 11 11 15 13 58.djvu

title: File:2010 11 11 15 13 58.djvu


text:
User:Egm6321.f10.team03.sigillo/Hwk6

=Problem 4: Infinitesimal segment ds in spherical coordinates and spherical Laplace equation"=

Given
The spherical coordinates in the cartesian coordinate system are given as
 * {| style="width:100%" border="0"

$$x={{x}_{1}}=r\cos \left( \theta \right)\cos \left( \varphi  \right)$$
 * $$\,(Eq. 4.1) $$
 * style= |
 * }

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

$$y={{x}_{2}}=r\cos \left( \theta \right)\sin \left( \varphi  \right)$$
 * $$\,(Eq. 4.2) $$
 * style= |
 * }

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

$$z={{x}_{3}}=r\sin \left( \theta \right)$$
 * $$\,(Eq. 4.3) $$
 * style= |
 * }

Find
The infinitesimal segment ds and the Laplace equation for spherical coordinates have to be found.

Solution
Segment ds: The infinitesimal segment ds in the three dimensional space can be expressed as
 * {| style="width:100%" border="0"

$$d{{s}^{2}}=\sum\limits_{i=1}^{3}$$ Thus the first step is to derive expressions for $$d{{x}_{1}}$$,$$d{{x}_{2}}$$ and $$d{{x}_{3}}$$
 * $$\,(Eq. 4.4) $$
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$$d{{x}_{1}}=\frac{\partial \left( r\cos \left( \theta \right)\cos \left( \varphi  \right) \right)}{\partial r}dr+\frac{\partial \left( r\cos \left( \theta  \right)\cos \left( \varphi  \right) \right)}{\partial \theta }d\theta +\frac{\partial \left( r\cos \left( \theta  \right)\cos \left( \varphi  \right) \right)}{\partial \varphi }d\varphi =\underbrace{\cos \left( \theta  \right)\cos \left( \varphi  \right)dr}_{a}-\underbrace{r\cos \left( \varphi  \right)\sin \left( \theta  \right)d\theta }_{b}-\underbrace{r\cos \left( \theta  \right)\sin \left( \varphi  \right)d\varphi }_{c}$$

 * $$\,(Eq. 4.5) $$
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$$d{{x}_{2}}=\frac{\partial \left( r\cos \left( \theta \right)\sin \left( \varphi  \right) \right)}{\partial r}dr+\frac{\partial \left( r\cos \left( \theta  \right)\sin \left( \varphi  \right) \right)}{\partial \theta }d\theta +\frac{\partial \left( r\cos \left( \theta  \right)\sin \left( \varphi  \right) \right)}{\partial \varphi }d\varphi =\underbrace{\cos \left( \theta  \right)\sin \left( \varphi  \right)dr}_{d}-\underbrace{r\sin \left( \varphi  \right)\sin \left( \theta  \right)d\theta }_{e}+\underbrace{r\cos \left( \theta  \right)\cos \left( \varphi  \right)d\varphi }_{f}$$

 * $$\,(Eq. 4.6) $$
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$$d{{x}_{3}}=\frac{\partial \left( r\sin \left( \theta \right) \right)}{\partial r}dr+\frac{\partial \left( r\sin \left( \theta  \right) \right)}{\partial \theta }d\theta +\frac{\partial \left( r\sin \left( \theta  \right) \right)}{\partial \varphi }d\varphi =\sin \left( \theta  \right)dr+r\cos \left( \theta  \right)d\theta $$

The squared values for Eq. 4.5, 4.6 and 4.7 are
 * $$\,(Eq. 4.7) $$
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$$\begin{align} & d{{x}_{1}}^{2}={{\left( a-b-c \right)}^{2}}=\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2bc-2ac-2ab \right) \\ & \quad \ \ \ ={{\cos }^{2}}\left( \theta \right){{\cos }^{2}}\left( \varphi  \right)d{{r}^{2}}+{{r}^{2}}{{\cos }^{2}}\left( \varphi  \right){{\sin }^{2}}\left( \theta  \right)d{{\theta }^{2}}+{{r}^{2}}{{\cos }^{2}}\left( \theta  \right){{\sin }^{2}}\left( \varphi  \right)d{{\varphi }^{2}} \\ & \quad \quad \ \ +2{{r}^{2}}\cos \left( \theta \right)\cos \left( \varphi  \right)\sin \left( \theta  \right)\sin \left( \varphi  \right)d\theta d\varphi  \\ & \quad \quad \ \ -2r{{\cos }^{2}}\left( \theta \right)\cos \left( \varphi  \right)\sin \left( \varphi  \right)drd\varphi  \\ & \quad \quad \ \ -2r\cos \left( \theta \right){{\cos }^{2}}\left( \varphi  \right)\sin \left( \theta  \right)drd\theta  \\ \end{align}$$

 * $$\,(Eq. 4.8) $$
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$$\begin{align} & d{{x}_{2}}^{2}={{\left( d-e+f \right)}^{2}}=\left( {{d}^{2}}+{{e}^{2}}+{{f}^{2}}+2df-2de-2ef \right) \\ & \quad \quad ={{\cos }^{2}}\left( \theta \right){{\sin }^{2}}\left( \varphi  \right)d{{r}^{2}}+{{r}^{2}}{{\sin }^{2}}\left( \varphi  \right){{\sin }^{2}}\left( \theta  \right)d{{\theta }^{2}}+{{r}^{2}}{{\cos }^{2}}\left( \theta  \right){{\cos }^{2}}\left( \varphi  \right)d{{\varphi }^{2}} \\ & \quad \quad \ \ +2r{{\cos }^{2}}\left( \theta \right)\cos \left( \varphi  \right)\sin \left( \varphi  \right)drd\varphi  \\ & \quad \quad \ \ -2r\cos \left( \theta \right)\sin \left( \theta  \right){{\sin }^{2}}\left( \varphi  \right)drd\theta  \\ & \quad \quad \ \ -2{{r}^{2}}\cos \left( \theta \right)\cos \left( \varphi  \right)\sin \left( \theta  \right)\sin \left( \varphi  \right)d\theta d\varphi  \\ \end{align}$$
 * $$\,(Eq. 4.9) $$
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$$d{{x}_{3}}^{2}={{\sin }^{2}}\left( \theta \right)d{{r}^{2}}+{{r}^{2}}{{\cos }^{2}}\left( \theta  \right)d{{\theta }^{2}}+2r\cos \left( \theta  \right)\sin \left( \theta  \right)drd\theta $$
 * $$\,(Eq. 4.10) $$
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Adding the Eqs. 4.8, 4.9 and 4.10 together will be performed in the following manner: regroup the common terms, factorize and add
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$$d\theta d\varphi \to \left[ 2{{r}^{2}}\cos \left( \theta \right)\cos \left( \varphi  \right)\sin \left( \theta  \right)\sin \left( \varphi  \right)-2{{r}^{2}}\cos \left( \theta  \right)\cos \left( \varphi  \right)\sin \left( \theta  \right)\sin \left( \varphi  \right) \right]d\theta d\varphi =0$$

 * $$\,(Eq. 4.11) $$
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$$drd\varphi \to \left[ -2r{{\cos }^{2}}\left( \theta \right)\cos \left( \varphi  \right)\sin \left( \varphi  \right)+2r{{\cos }^{2}}\left( \theta  \right)\cos \left( \varphi  \right)\sin \left( \varphi  \right) \right]=0$$
 * $$\,(Eq. 4.12) $$
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$$\begin{align} & drd\theta \to \left[ -2r\cos \left( \theta \right){{\cos }^{2}}\left( \varphi  \right)\sin \left( \theta  \right)-2r\cos \left( \theta  \right)\sin \left( \theta  \right){{\sin }^{2}}\left( \varphi  \right)+2r\cos \left( \theta  \right)\sin \left( \theta  \right) \right]drd\theta  \\ & \quad \quad \quad \ =2r\cos \left( \theta \right)\sin \left( \theta  \right)\left[ -{{\cos }^{2}}\left( \theta  \right)-{{\sin }^{2}}\left( \theta  \right)+1 \right]drd\theta =0 \\ \end{align}$$

 * $$\,(Eq. 4.13) $$
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$$d{{r}^{2}}\to \left[ {{\cos }^{2}}\left( \theta \right){{\cos }^{2}}\left( \varphi  \right)+{{\cos }^{2}}\left( \theta  \right)\left( 1-{{\cos }^{2}}\left( \varphi  \right) \right)+\left( 1-{{\cos }^{2}}\left( \theta  \right) \right) \right]d{{r}^{2}}=1d{{r}^{2}}$$
 * $$\,(Eq. 4.14) $$
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$$d{{\theta }^{2}}\to \left[ {{r}^{2}}\left( 1-{{\sin }^{2}}\left( \varphi \right) \right){{\sin }^{2}}\left( \theta  \right)+{{r}^{2}}{{\sin }^{2}}\left( \varphi  \right){{\sin }^{2}}\left( \theta  \right)+{{r}^{2}}\left( 1-{{\sin }^{2}}\left( \theta  \right) \right) \right]d{{\theta }^{2}}={{r}^{2}}d{{\theta }^{2}}$$
 * $$\,(Eq. 4.15) $$
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$$d{{\varphi }^{2}}\to \left[ {{r}^{2}}{{\cos }^{2}}\left( \theta \right)\left( 1-{{\cos }^{2}}\left( \varphi  \right) \right)+{{r}^{2}}{{\cos }^{2}}\left( \theta  \right){{\cos }^{2}}\left( \varphi  \right) \right]d{{\varphi }^{2}}={{r}^{2}}{{\cos }^{2}}\left( \theta  \right)d{{\varphi }^{2}}$$
 * $$\,(Eq. 4.16) $$
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As it can be seen only Eq. 4.14, 4.15 and 4.16 are non-zero. Thus the infinitesimal segment ds is just the summation of these three expressions:
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$$d{{s}^{2}}=\sum\limits_{i=1}^{3}=1d{{r}^{2}}+{{r}^{2}}d{{\theta }^{2}}+{{r}^{2}}{{\cos }^{2}}\left( \theta \right)d{{\varphi }^{2}}$$
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$$\,(Eq. 4.17) $$
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Spherical Laplace equation:

The Laplace equation can be written in general as
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$$\Delta \Psi =\frac{1}\sum\limits_{i=1}^{3}{\frac{\partial }{\partial {{\zeta }_{i}}}}\left[ \frac\frac{\partial \Psi }{\partial {{\zeta }_{i}}} \right]$$ To get the spehrical Laplace equation we re-define the following variables
 *
$$\,(Eq. 4.18) $$
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$$\begin{align} & r={{\zeta }_{1}} \\ & \theta ={{\zeta }_{2}} \\ & \varphi ={{\zeta }_{3}} \\ \end{align}$$ Equation 4.17 can now be written as
 *
$$\,(Eq. 4.19) $$
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$$d{{s}^{2}}=1{{\left( d{{\zeta }_{1}} \right)}^{2}}+{{\zeta }_{1}}^{2}{{\left( d{{\zeta }_{2}} \right)}^{2}}+{{\zeta }_{1}}^{2}{{\cos }^{2}}\left( {{\zeta }_{2}} \right){{\left( d{{\zeta }_{3}} \right)}^{2}}={{\left( {{h}_{1}}\left( \zeta \right) \right)}^{2}}{{\left( d{{\zeta }_{1}} \right)}^{2}}+{{\left( {{h}_{2}}\left( \zeta  \right) \right)}^{2}}{{\left( d{{\zeta }_{2}} \right)}^{2}}+{{\left( {{h}_{3}}\left( \zeta  \right) \right)}^{2}}{{\left( d{{\zeta }_{3}} \right)}^{2}}$$ where the coefficients $${{h}_{i}}\left( \zeta \right)$$ are
 *
$$\,(Eq. 4.20) $$
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$$\begin{align} & {{h}_{1}}\left( \zeta \right)=1 \\ & {{h}_{2}}\left( \zeta \right)={{\zeta }_{1}} \\ & {{h}_{3}}\left( \zeta \right)={{\zeta }_{1}}\cos \left( {{\zeta }_{2}} \right) \\ \end{align}$$ Now Eq. 4.18 can be re-written as
 *
$$\,(Eq. 4.21) $$
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$$\Delta \Psi =\frac{1}{{{\zeta }_{1}}^{2}\cos \left( {{\zeta }_{2}} \right)}\sum\limits_{i=1}^{3}{\frac{\partial }{\partial {{\zeta }_{i}}}}\left[ \frac{{{\zeta }_{1}}^{2}\cos \left( {{\zeta }_{2}} \right)}\frac{\partial \Psi }{\partial {{\zeta }_{i}}} \right]$$ for the different cases where the index variable i has values of 1,2 and three, the terms of the summation are:
 *
$$\,(Eq. 4.22) $$
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$$\begin{align} & i=1\to \frac{\partial }{\partial {{\zeta }_{1}}}\left[ \frac{{{\zeta }_{1}}^{2}\cos \left( {{\zeta }_{2}} \right)}\frac{\partial \Psi }{\partial {{\zeta }_{1}}} \right] \\ & i=2\to \frac{\partial }{\partial {{\zeta }_{2}}}\left[ \frac{{{\zeta }_{1}}^{2}\cos \left( {{\zeta }_{2}} \right)}{{{\zeta }_{1}}^{2}}\frac{\partial \Psi }{\partial {{\zeta }_{2}}} \right]=\frac{\partial }{\partial {{\zeta }_{2}}}\left[ \cos \left( {{\zeta }_{2}} \right)\frac{\partial \Psi }{\partial {{\zeta }_{2}}} \right] \\ & i=3\to \frac{\partial }{\partial {{\zeta }_{3}}}\left[ \frac{{{\zeta }_{1}}^{2}\cos \left( {{\zeta }_{2}} \right)}{{{\zeta }_{1}}^{2}{{\cos }^{2}}\left( {{\zeta }_{2}} \right)}\frac{\partial \Psi }{\partial {{\zeta }_{3}}} \right]=\frac{\partial }{\partial {{\zeta }_{3}}}\left[ \frac{1}{\cos \left( {{\zeta }_{2}} \right)}\frac{\partial \Psi }{\partial {{\zeta }_{3}}} \right] \\ \end{align}$$
 *
$$\,(Eq. 4.23) $$
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So that the Laplace equation can be finally written as being:
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$$\Delta \Psi =\frac{1}{{{\zeta }_{1}}^{2}\cos \left( {{\zeta }_{2}} \right)}\left\{ \frac{\partial }{\partial {{\zeta }_{1}}}\left[ \frac{{{\zeta }_{1}}^{2}\cos \left( {{\zeta }_{2}} \right)}\frac{\partial \Psi }{\partial {{\zeta }_{1}}} \right]+\frac{\partial }{\partial {{\zeta }_{2}}}\left[ \cos \left( {{\zeta }_{2}} \right)\frac{\partial \Psi }{\partial {{\zeta }_{2}}} \right]+\frac{\partial }{\partial {{\zeta }_{3}}}\left[ \frac{1}{\cos \left( {{\zeta }_{2}} \right)}\frac{\partial \Psi }{\partial {{\zeta }_{3}}} \right] \right\}$$
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$$\,(Eq. 4.24) $$
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Additoinal links and useful informations on spherical coordinates and Laplace equations
The following page delivers a short description on how to convert cartesian coordinates into spherical or cylindrical coordinates in Matlab with an already existing function:

Matlab-spherical-transformation

This Wolfram page gives a more in depth insight into spherical cooridnates that is worth to take a look at. Note, that the angle $$\theta $$ was selected to be 90° minus the $$\theta $$ used in the problem above.

Wolfram-SphericalCoordinates

In this page the Wolfram-team shows how to solve the Laplace equation with spherical coordinates using the method of separation of vaiables, like discussed in class:

Wolfram-Laplaces Equation SphericalCoordinates

This paper from the University of Texas-State shows also the solution of the Laplace equation, described in a very elegant way:

Texas-State-University:spherical_laplace.pdf

Contributing Members
Solved and posted by Egm6321.f10.team03.sigillo 19:21, 12 November 2010 (UTC)

test
=Problem 7: Laplace equation in elliptic coordinates=

Given
The article for elliptic coordinates discussed in class is given with it's Laplace equation. The cartesian coordinates are expressed as

where a is the distance from the origin to the foci of the ellips and

Find
The expression for the Laplace equation in the mentioned article needs to be verified using the steps learned for cylindrical and spherical coordinates

Solution
Like in probelm 4, we find an expression for the infinitesimal segment ds in the elliptic coordinate system though:

Therefore we find first the expressions for $$d{{x}_{1}}$$ and $$d{{x}_{2}}$$, that are:

Because of the axis symmetry we can exclude the dependency on a and cancel all terms related to $$da$$. The squared term of Eq. 7.5 and 7.6 are:

To find the expression for the segment ds we regroup, factorize and add the common terms:

Thus the segment ds is just a summation over the $$d{{\mu }^{2}}$$ and $$d{{\nu }^{2}}$$ terms:

Now, the infinitesimal segment ds can be expressed in general as where

We recall that the Laplace equation can be written as:

And in our specific case we can re-write the equation as:

Which is the same expression for the Laplace equation as in the article discussed in class.

Contributing Members
Solved and posted by Egm6321.f10.team03.sigillo 21:32, 13 November 2010 (UTC)

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