text:
User:Egm6321.f10.team03/Hwk1

=Homework One Summary= Problems assigned in the first two weeks of class. = Problem 1 - Maglev First and Second Derivative Derivations =

Reading convention
A function $$\,f^1$$ or $$\,f^1(.)$$ has to be read as: "$$f$$ superscript 1". A function $$\,(f^1)^2$$ or $$\,(f^1(.))^2$$has to be read as: "$$f$$ superscript 1, power 2".

Given
Based on the definitions and equations in Maglev Train EOM the axial deformation of the guidance for a High-Speed Train is represented by a function $$\,u^1(s,t)$$. The argument $$\,s$$ itself is also a function of the time variable $$\,t$$, so that for a general function $$\,f$$ the following is valid:
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$$\,{{\left. f(s,t) \right|}_{s={{y}^{1}}(t)}}=f({{y}^{1}}(t),t)$$
 * $$\,(Eq.1.1)$$
 * }

Find
The first and second derivation of $$\,f({{y}^{1}}(t),t)$$ has to be calculated with respect to $$\,t$$.

Solution
First derivation $$\frac{df}{dt}$$: Let's set
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$$\,z:=f({{y}^{1}}(t),t)$$ $$ and
 * $$\,(Eq. 1.2)
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$$\,s:={{y}^{1}}(t)$$ $$ thus we get for $$z$$ :
 * $$\,(Eq. 1.3)
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$$\,z=f(s,t)$$ $$ The absolute derivation of $$z$$ is:
 * $$\,(Eq. 1.4)
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$$\,dz={{f}_{s}}\ ds+{{f}_{t}}\ dt$$ $$ Following the suggestion in Papula, we devide $$(Eq.1.5)$$ by $$\,dt$$ and get:
 * $$\,(Eq. 1.5)
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$$\frac{dz}{dt}={{f}_{s}}\ \frac{ds}{dt}+{{f}_{t}}$$ $$ substituting $$z$$ $$(Eq. 1.2)$$ gives:
 * $$\,(Eq. 1.6)
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$$\frac{df(s,t)}{dt}=\frac{\partial f(s,t)}{\partial s}\frac{\partial s}{\partial t}+\frac{\partial f(s,t)}{\partial t}$$ $$ substituting $$s$$ this time, gives:
 * $$\,(Eq. 1.7)
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$$\frac{df({{y}^{1}}(t),t)}{dt}=\frac{\partial f({{y}^{1}}(t),t)}{\partial s}\frac{\partial {{y}^{1}}(t)}{\partial t}+\frac{\partial f({{y}^{1}}(t),t)}{\partial t}$$ $$ So that we can finally say:
 * $$\,(Eq. 1.8)
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$$\frac{df}{dt}={{f}_{s}}\cdot \dot{y}^{1}+{{f}_{t}}$$ $$
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 * $$\,(Eq. 1.9)
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 * }

Second derivation $$\frac{{{d}^{2}}f}{d{{t}^{2}}}$$: The procedure to calculate the second derivative of $$f$$ is the same used to calculate the first derivative. For a better overview let's assign $$(Eq. 1.9)$$ to a new variable and also parts of the equation to new variables like follows:
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$$\,q:= {{f}_{s}}\cdot \dot{y}^{1}+{{f}_{t}}$$ $$\,r:= {{f}_{s}}$$ $$\,m:= \dot{y}^{1}$$ $$\,h:= {{f}_{t}}$$ Equation $$(Eq. 1.9)$$ can then be written as:
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 * }
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$$\,q=r\cdot m+h$$ $$ Similarly to the calculation of the first derivative, we calculate the total derivative of $$q$$, paying attention to the fact that now we have to derive the prduct $$(r\cdot m)$$ with $$r$$ being a function of $$s$$ and $$t$$ :
 * $$\,(Eq. 1.10)
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 * }
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$$dq=({{r}_{s}}\cdot ds+{{r}_{s}}\cdot dt)\cdot m+r\cdot {{m}_{t}}\cdot dt+{{h}_{t}}\cdot dt+{{h}_{s}}\cdot ds$$ $$ multiplying $$m$$ into the braket expression gives:
 * $$\,(Eq. 1.11)
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$$dq={{r}_{s}}\cdot ds\cdot m+{{r}_{s}}\cdot dt\cdot m+r\cdot {{m}_{t}}\cdot dt+{{h}_{t}}\cdot dt+{{h}_{s}}\cdot ds$$ $$ Dividing also here $$dq$$ by $$dt$$ and substituiting $$r, m, h$$ with their original terms we get:
 * $$\,(Eq. 1.12)
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 * }
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$$\frac{dq}{dt}=\frac{\partial {{f}_{s}}}{\partial s}\ \frac{ds}{dt}\ {{{\dot{y}}}^{1}}+\frac{\partial {{f}_{s}}}{\partial t}\ {{{\dot{y}}}^{1}}+{{f}_{s}}\frac{\partial {{{\dot{y}}}^{1}}}{\partial t}+\frac{\partial {{f}_{t}}}{\partial t}+\frac{\partial {{f}_{t}}}{\partial s}\ \frac{ds}{dt}$$ $$ That in fact after deriving becomes:
 * $$\,(Eq. 1.13)
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 * }
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$$\frac{dq}{dt}={{f}_{ss}}\cdot {{({{{\dot{y}}}^{1}})}^{2}}+{{f}_{st}}\cdot {{{\dot{y}}}^{1}}+{{f}_{s}}\cdot {{{\ddot{y}}}^{1}}+{{f}_{tt}}+{{f}_{ts}}\cdot {{{\dot{y}}}^{1}}$$ $$ Since the derivation sequence doesn't matter for partial derivations, we finally after ordering the terms in $$(Eq. 1.14) $$ get for the second derivative:
 * $$\,(Eq. 1.14)
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 * }
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$$\frac{{{d}^{2}}f}{d{{t}^{2}}}={{f}_{ss}}\cdot {{({{{\dot{y}}}^{1}})}^{2}}+2{{f}_{st}}\cdot {{{\dot{y}}}^{1}}+{{f}_{s}}\cdot {{{\ddot{y}}}^{1}}+{{f}_{tt}}$$ $$ Mike SigilloEgm6321.f10.team03 16:46, 14 September 2010 (UTC)
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 * $$\,(Eq. 1.15)
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 * }

Kurt SchulzeEgm6321.f10.team03 16:46, 14 September 2010 (UTC)

= Problem 2 - Dimensional Analysis=

Given
From the Maglev Train EOM we are given the horizontal force on the wheel/magnet $$\,{{c}_{0}}({{y}^{1}},t)$$ :
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$$\,{{c}_{0}}({{y}^{1}},t)=-{{F}^{1}}[1-\bar{R}\cdot u_{ss}^{2}({{y}^{1}},t)]-{{F}^{2}}\cdot u_{s}^{2}-\frac{T}{R}+M[[1-\bar{R}\cdot u_{ss}^{2}][u_{tt}^{1}-\bar{R}\cdot u_{stt}^{2}]+u_{s}^{2}\cdot u_{tt}^{2}]$$ $$ with :
 * $$\,(Eq. 2.1)
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$$\,[{{F}^{1}}]=F(force)$$: Force acting in the lateral direction of the wheel $$\,[{{F}^{2}}]=F(force)$$: Force acting in the vertical direction of the wheel $$\,[T]=T(torque)$$: Torque at the edge of the wheel $$\,[M]=M(mass)$$: Mass of the wheel $$\,[\bar{R}]=L(length)$$: Distance between the rail-midpoint and the wheel-midpoint $$\,[R]=L(length)$$: Radius of the wheel $$\,[t]=t(time)$$ $$\,[{{u}^{2}}]=L(length)$$: Transverse deformation of the guideway

$$\,\left[ u_{s}^{2} \right]=\left[ \frac{\partial {{u}^{2}}}{\partial s} \right]=\frac{L}{L}= \left( \frac{length}{length} \right)$$: Slope of the deformation of the guideway in the transverse direction

$$\left[ u_{tt}^{1} \right]=\frac{d}{dt}\left[ \frac{\partial {{u}^{1}}}{\partial t} \right]=\left[ \frac{\partial {{t}^{2}}} \right]=\frac=\left( \frac{length}{{time} \cdot {time}} \right)$$ : Acceleration of the axial deformation of the guideway

$$\left[ u_{tt}^{2} \right]=\frac{d}{dt}\left[ \frac{\partial {{u}^{2}}}{\partial t} \right]=\left[ \frac{\partial {{t}^{2}}} \right]=\frac=\left( \frac{length}{{time} \cdot {time}} \right)$$ : Acceleration of the transverse deformation of the guideway $$\left[ u_{ss}^{2} \right]=\frac{d}{ds}\left[ \frac{\partial {{u}^{2}}}{\partial s} \right]=\left[ \frac{\partial {{s}^{2}}} \right]=\frac{L}=\frac{1}{L}\ \ \left( \frac{1}{length} \right)$$ : Curvature of the transverse deformation of the flexible guideway $$\left[ u_{stt}^{2} \right]=\frac=\left( \frac{length}{{length}\cdot{time}\cdot{time}} \right) $$: Acceleration of the slope of the deformation of the guideway in the transverse direction $$\,[1]=1$$
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 * }

Find
Dimensional Analysis of the terms in $$(Eq. 2.1)$$ and physical meaning.

Solution
Analysis of the first term $$\,(-{{F}^{1}}[1-\bar{R}\cdot u_{ss}^{2}({{y}^{1}},t)])$$ :
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$$\,F[1-L\cdot {{L}^{-1}}]=F[1-1]\to$$ Dimension results to be a Force Analysis of the second term $$\,(-{{F}^{2}}\cdot u_{s}^{2}-\frac{T}{R})$$ :
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$$F\cdot \frac{L}{L}-\frac{T}{L}=F-F\to $$ Dimension results to be a Force Analysis of third term $$\,(M[[1-\bar{R}\cdot u_{ss}^{2}][u_{tt}^{1}-\bar{R}\cdot u_{stt}^{2}]+u_{s}^{2}\cdot u_{tt}^{2}])$$ :
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$$\,M\left[ \left[ 1-L\cdot {{L}^{-1}} \right]\left[ \frac{L}-L\cdot \frac{L}{L\cdot {{t}^{2}}} \right]+\frac{L}{L}\cdot \frac{L} \right]=M\cdot \frac{L}+M\cdot \frac{L}=F+F\to $$ Dimension results to be a Force Mike SigilloEgm6321.f10.team03 16:46, 14 September 2010 (UTC)
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Kurt SchulzeEgm6321.f10.team03 16:46, 14 September 2010 (UTC)

Jimmy RoarkEgm6321.f10.team3.roark 18:20, 14 September 2010 (UTC)

= Problem 3 - Linearity Test of Maglev Train EOM Coefficient =

Given
$$\displaystyle C_3$$ from the Maglev Train EOM is defined as:

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C_3(Y^1,t) := M\left [\left( 1-\bar Ru^2,_{ss}\left(Y^1,t\right)\right)\right]^2 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq.3.1)
 * }
 * }

Find
Using the definition of linearity of a function, show that $$\displaystyle C_3\ddot{Y^1}$$ is nonlinear, with respect to $$\displaystyle Y^1$$.

Solution
A linear function is a function which obeys two laws, shown below. $$\displaystyle H$$ is an arbitrary function while $$\displaystyle c$$ is an arbitrary constant.

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H(cx) = cH(x) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq.3.2)
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 * }

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H(x+y) = H(x) + H(y) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq.3.3)
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 * }

In order for $$\displaystyle C_3(Y^1,t)\ddot{Y^1}$$ to be linear, the following equation must be true.

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$$
 * $$\displaystyle C_3(cx,t)\ddot{Y^1} = cC_3(x,t)\ddot{Y^1}$$
 *
$$\displaystyle (Eq.3.4)
 *
$$\displaystyle (Eq.3.4)
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 * }

Applying the cumulative property to (Eq.3.1) gives us

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$$
 * $$\displaystyle C_3(cx,t)\ddot{Y^1} = M\left [\left( 1-\bar Ru^2,_{ss}\left(cx,t\right)\right)\right]^2\ddot{Y^1}.$$
 *
$$\displaystyle (Eq.3.5)
 *
$$\displaystyle (Eq.3.5)
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 * }

Let us assume the function $$\displaystyle u^2,_{ss}$$ is linear, (Eq.3.5) becomes

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$$
 * $$\displaystyle C_3(cx,t)\ddot{Y^1} = M\left [\left( 1-c\bar Ru^2,_{ss}\left(x,t\right)\right)\right]^2\ddot{Y^1}.$$
 *
$$\displaystyle (Eq.3.6)
 *
$$\displaystyle (Eq.3.6)
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 * }

Now we evaluate the right half of (Eq.3.4)

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$$
 * $$\displaystyle cC_3(x,t)\ddot{Y^1} = cM\left [\left( 1-\bar Ru^2,_{ss}\left(x,t\right)\right)\right]^2\ddot{Y^1}.$$
 *
$$\displaystyle (Eq.3.7)
 *
$$\displaystyle (Eq.3.7)
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Since

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$$
 * $$\displaystyle C_3(cx,t)\ddot{Y^1} \not= cC_3(x,t)\ddot{Y^1},$$
 *
$$\displaystyle (Eq.3.8)
 *
$$\displaystyle (Eq.3.8)
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 * }

$$\displaystyle C_3(Y^1,t)\ddot{Y^1}$$ is nonlinear because it did not meet the linearity test of a function.

Adam Franklin 18:56, 12 September 2010 (UTC)

= Problem 4 - Coefficients of Linear Combination =

Given
Given a linear, second order ordinary differential equation with varying coefficients (L2-ODE-VC) and with boundary conditions:

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$$ y \left( a\right) =  \alpha  ;  y \left( b\right)  =   \beta $$
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 * }
 * }

From eqn. [4], pg [[media:2010_09_02_13_55_50.djvu|5-3]] of the notes, the ODE has a solution of the form:

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$$ y \left( x\right) =  c {y}^{1}_{H}\left( x\right) + d {y}^{2}_{H} \left( x\right) + {y}_{P} \left( x\right) $$ $$
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 *
$$\displaystyle (Eq.4, Pg. 5-3)
 * }

Find
Find $$c,d$$ in terms of $$\alpha$$ and $$\beta$$

Solution
Boundary conditions
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$$ y \left( a\right) =  \alpha  ;  y \left( b\right)  =   \beta $$ $$
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 *
$$\displaystyle (Eq.4.1)
 * }

Now

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$$ \alpha =  c {y}^{1}_{H}\left( a\right) + d {y}^{2}_{H} \left( a\right) + {y}_{P} \left( a\right) $$ $$
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 *
$$\displaystyle (Eq.4.2)
 * }

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$$ \beta =  c {y}^{1}_{H}\left( b\right) + d {y}^{2}_{H} \left( b\right) + {y}_{P} \left( b\right) $$ $$
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 *
$$\displaystyle (Eq.4.3)
 * }

Solve for $$ c $$ in $$\displaystyle (Eq.4.2)$$

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$$ \Rightarrow c = \frac{\alpha - {y}_{P} \left( a\right) - d {y}^{2}_{H} \left( a\right)}{{y}^{1}_{H}\left( a\right)} $$ $$
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 *
$$\displaystyle (Eq.4.4)
 * }

Plug $$\displaystyle (Eq.4.4)$$ into $$\displaystyle (Eq.4.3)$$ to eliminate $$ c $$

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$$ \beta =  \frac{{y}^{1}_{H}\left( b\right)}{{y}^{1}_{H}\left( a\right)} \left[ \alpha - {y}_{P} \left( a\right) - d {y}^{2}_{H} \left( a\right) \right] + d {y}^{2}_{H} \left( b\right) + {y}_{P} \left( b\right) $$
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 * }
 * }

For ease of notation, Let $$ \frac{{y}^{1}_{H}\left( b\right)}{{y}^{1}_{H}\left( a\right)} = \xi $$

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$$ \Rightarrow \beta = \xi\left[ \alpha - {y}_{P} \left( a\right) - d {y}^{2}_{H} \left( a\right) \right] + d {y}^{2}_{H} \left( b\right) + {y}_{P} \left( b\right) $$ $$
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 *
$$\displaystyle (Eq.4.5)
 * }

Solve $$\displaystyle (Eq.4.5)$$ for $$ d $$

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$$ d = \frac{ \beta - {y}_{P} \left( b\right) + \xi \left[ {y}_{P} \left( a\right) - \alpha\right]}{{y}^{2}_{H} \left( b\right) - \xi {y}^{2}_{H} \left( a\right)} $$ $$
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 *
$$\displaystyle (Eq.4.6)
 * }

Subsitute $$\displaystyle (Eq.4.6)$$ back into $$\displaystyle (Eq.4.4)$$ to obtain $$ c $$

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$$ c = \frac{\alpha - {y}_{P} \left( a\right) - \left[\frac{ \beta - {y}_{P} \left( b\right) + \xi \left[ {y}_{P} \left( a\right) - \alpha\right]}{{y}^{2}_{H} \left( b\right) -  \xi {y}^{2}_{H} \left( a\right)} \right]{y}^{2}_{H} \left( a\right)}{{y}^{1}_{H}\left( a\right)} $$ $$
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 *
$$\displaystyle (Eq.4.7)
 * }

$$ c $$ and $$ d $$ can be re-writen by substituting $$ \frac{{y}^{1}_{H}\left( b\right)}{{y}^{1}_{H}\left( a\right)} = \xi $$

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$$ c = \frac{\alpha - {y}_{P} \left( a\right) - \left[\frac{ \beta - {y}_{P} \left( b\right) + \frac{{y}^{1}_{H}\left( b\right)}{{y}^{1}_{H}\left( a\right)} \left[ {y}_{P} \left( a\right) - \alpha\right]}{{y}^{2}_{H} \left( b\right) - \frac{{y}^{1}_{H}\left( b\right)}{{y}^{1}_{H}\left( a\right)} {y}^{2}_{H} \left( a\right)} \right]{y}^{2}_{H} \left( a\right)}{{y}^{1}_{H}\left( a\right)} $$ $$
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 * style="width:25%; padding:1px; border:2px solid #0000FF" |
 *
$$\displaystyle (Eq.4.8)
 * }

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$$ d = \frac{ \beta - {y}_{P} \left( b\right) + \frac{{y}^{1}_{H}\left( b\right)}{{y}^{1}_{H}\left( a\right)} \left[ {y}_{P} \left( a\right) - \alpha\right]}{{y}^{2}_{H} \left( b\right) - \frac{{y}^{1}_{H}\left( b\right)}{{y}^{1}_{H}\left( a\right)}{y}^{2}_{H} \left( a\right)} $$ $$ Egm6321.f10.team3.roark 11:05, 14 September 2010 (UTC)
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 * style="width:25%; padding:1px; border:2px solid #0000FF" |
 *
$$\displaystyle (Eq.4.9)
 * }

Egm6321.f10.team3.cook 11:12, 14 September 2010 (UTC)

Egm6321.f10.team3.Hong SJ 11:12, 14 September 2010 (UTC)

= Problem 5 - Verification of Homogenous Solutions of Legendre Differential Equation for n = 1=

Given
Legendre differential equation:
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{{L}_{2}}\left( y \right) = (1-x^2)y'' - 2xy' +n(n+1)y = 0 $$ $$ Homogenous solutions of the above equation for $$\displaystyle  n = 1 $$ [[media:2010_09_02_14_58_46.djvu|(Mtg 6-page1)]], are given by:
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq.5.1)
 * }
 * }
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$$ $$
 * $$y_{H}^{1}\left( x \right)=x\equiv P_1(x)
 * $$y_{H}^{1}\left( x \right)=x\equiv P_1(x)
 *
$$\displaystyle (Eq.5.2)
 * }
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 * {| style="width:100%" border="0" align="left"

$$ $$ Where $$y_{H}^{1}\left( x \right)$$ is the first homogenous solution and $$y_{H}^{2}\left( x \right)$$ is the second homogenous solution
 * $$y_{H}^{2}\left( x \right)=\frac{x}{2}.\log \left( \frac{1+x}{1-x} \right)-1\equiv Q_1(x)
 * $$y_{H}^{2}\left( x \right)=\frac{x}{2}.\log \left( \frac{1+x}{1-x} \right)-1\equiv Q_1(x)
 *
$$\displaystyle (Eq.5.3)
 * }
 * }

Find
We need to prove that:
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 * $${{L}_{2}}\left( y_{H}^{1} \right)={{L}_{2}}\left( y_{H}^{2} \right)=0$$
 * }
 * }
 * }

Solution
Let us consider first $${{L}_{2}}\left( y_{H}^{1} \right)$$:

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$$ Substitute Eq.2 for $${{L}_{2}}\left( y_{H}^{1} \right)$$ in Eq.5.4 =>
 * $${{L}_{2}}\left( y_{H}^{1} \right)=\left( 1-{{x}^{2}} \right){{\left( y_{H}^{1} \right)}^{''}}-2x{{\left( y_{H}^{1} \right)}^{'}}+2\left( y_{H}^{1} \right)$$
 *
$$\displaystyle (Eq.5.4)
 *
$$\displaystyle (Eq.5.4)
 * }
 * }

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 * $${{L}_{2}}\left( y_{H}^{1} \right)=\left( 1-{{x}^{2}} \right){{\left( x \right)}^{''}}-2x{{\left( x \right)}^{'}}+2\left( x \right)$$
 * }
 * }
 * }

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$$ {{L}_{2}}\left( y_{H}^{1} \right)=\left( 1-{{x}^{2}} \right)\left( 0 \right)-2x\left( 1 \right)+2\left( x \right)=0$$ $$
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 * style="width:20%; padding:1px; border:2px solid #0000FF" |
 *
$$\displaystyle (Eq.5.5)
 * }
 * }

Now consider $${{L}_{2}}\left( y_{H}^{2} \right)$$:

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$$ Determine first derivative of $$y_{H}^{2}$$ (Eq.5.2) using Product rule, Logarithmic derivative rule ($$ (\ln f)'= \frac{f'}{f} \quad$$) and chain rule of differentiation:
 * $${{L}_{2}}\left( y_{H}^{2} \right)=\left( 1-{{x}^{2}} \right){{\left( y_{H}^{2} \right)}^{''}}-2x{{\left( y_{H}^{2} \right)}^{'}}+2\left( y_{H}^{2} \right)$$
 *
$$\displaystyle (Eq.5.6)
 *
$$\displaystyle (Eq.5.6)
 * }
 * }

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 * $${{\left( y_{H}^{2} \right)}^{'}}=\frac{x}{2}.\frac{d}{dx}\left( \log \left( \frac{1+x}{1-x} \right) \right)+\frac{d}{dx}\left( \frac{x}{2} \right).\log \left( \frac{1+x}{1-x} \right)-\frac{d}{dx}\left( 1 \right)$$
 * }
 * }
 * }

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 * $${{\left( y_{H}^{2} \right)}^{'}}=\frac{x}{2}\left( \frac{1}{\left( \frac{1+x}{1-x} \right)}.\frac{d}{dx}\left( \frac{1+x}{1-x} \right) \right)+\frac{1}{2}\log \left( \frac{1+x}{1-x} \right)+0$$
 * }
 * }
 * }

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 * $${{\left( y_{H}^{2} \right)}^{'}}=\frac{x}{2}\left( \frac{1-x}{1+x} \right)\left( \frac{\left( 1-x \right)+\left( 1+x \right)} \right)+\frac{1}{2}\log \left( \frac{1+x}{1-x} \right)$$
 * }
 * }
 * }

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$${{\left( y_{H}^{2} \right)}^{'}}=\frac{x}{1-{{x}^{2}}}+\frac{1}{2}\log \left( \frac{1+x}{1-x} \right)$$ $$
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 * style="width:20%; padding:1px; border:2px solid #8888aa" |
 *
$$\displaystyle (Eq.5.7)
 * }
 * }

Second derivative of $$y_{H}^{2}$$ (Eq.5.2) is:

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 * $${{\left( y_{H}^{2} \right)}^{''}}={{\left( {{\left( y_{H}^{2} \right)}^{'}} \right)}^{'}}=\frac{d}{dx}\left( \frac{x}{1-{{x}^{2}}}+\frac{1}{2}\log \left( \frac{1+x}{1-x} \right) \right)$$
 * }
 * }
 * }

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 * $${{\left( y_{H}^{2} \right)}^{''}}=\frac{d}{dx}\left( \frac{x}{1-{{x}^{2}}} \right)+\frac{d}{dx}\left( \frac{1}{2}\log \left( \frac{1+x}{1-x} \right) \right)$$
 * }
 * }
 * }

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

 * $${{\left( y_{H}^{2} \right)}^{''}}=\left( \frac{\left( 1-{{x}^{2}} \right)-x.\left( -2x \right)} \right)+\frac{1}{2}\left( \frac{1-x}{1+x} \right)\left( \frac{\left( 1-x \right)+\left( 1+x \right)} \right)$$
 * }
 * }
 * }

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

 * $${{\left( y_{H}^{2} \right)}^{''}}=\frac{1+{{x}^{2}}}+\frac{1}{2}\left( \frac{1-x}{1+x} \right)\left( \frac{2} \right)$$
 * }
 * }
 * }

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

 * $${{\left( y_{H}^{2} \right)}^{''}}=\frac{1+{{x}^{2}}}+\frac{1}$$
 * }
 * }
 * }

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

$${{\left( y_{H}^{2} \right)}^{''}}=\frac{2}$$ $$ Substituting Eqs.5.3,7,8 into Eq.5.5,6 yields,
 * style="width:20%; padding:1px; border:2px solid #8888aa" |
 * style="width:20%; padding:1px; border:2px solid #8888aa" |
 *
$$\displaystyle (Eq.5.8)
 * }
 * }
 * {| style="width:100%" border="0"

 * $${{L}_{2}}\left( y_{H}^{2} \right)=\left( 1-{{x}^{2}} \right)\left( \frac{2} \right)-2x\left( \frac{x}{1-{{x}^{2}}}+\frac{1}{2}\log \left( \frac{1+x}{1-x} \right) \right)+2\left( \frac{x}{2}\log \left( \frac{1+x}{1-x} \right)-1 \right)$$
 * }
 * }
 * }

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

 * $${{L}_{2}}\left( y_{H}^{2} \right)=\frac{2}{\left( 1-{{x}^{2}} \right)}-\frac{2{{x}^{2}}}{1-{{x}^{2}}}-x\log \left( \frac{1+x}{1-x} \right)+x\log \left( \frac{1+x}{1-x} \right)-2$$
 * }
 * }
 * }

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

$${{L}_{2}}\left( y_{H}^{2} \right)=\frac{2-2{{x}^{2}}-2\left( 1-{{x}^{2}} \right)}{\left( 1-{{x}^{2}} \right)}=0$$ $$
 * style="width:20%; padding:1px; border:2px solid #0000FF" |
 * style="width:20%; padding:1px; border:2px solid #0000FF" |
 *
$$\displaystyle (Eq.5.9)
 * }
 * }

Eqs.5 and 9 verify that :{| style="width:10%" border="0" $${{L}_{2}}\left( y_{H}^{1} \right)={{L}_{2}}\left( y_{H}^{2} \right)=0$$ Egm6321.f10.team3.Sudheesh 18:11, 14 September 2010 (UTC)
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * }
 * }

Egm6321.f10.team3.Hong SJ 11:12, 14 September 2010 (UTC)

=Contributing Team Members=
 * Adam Franklin 18:56, 12 September 2010 (UTC) Author Problem 3
 * Sudheesh 18:11, 14 September 2010 (UTC) Author Problem 5
 * Jimmy Roark 14:24, 15 September 2010 (UTC) Co-Author Problem 4; Proofread and edited Problem 2
 * Michele Sigilló 22:07, 12 September 2010 (UTC) Author Problem 1 and 2
 * Kurt Schulze 01:10, 13 September 2010 (UTC) Proofread Problem Problem 2
 * Chris Cook 11:06, 14 September 2010 (UTC) Co-Author Problem 4
 * Hong SJ 11:12, 14 September 2010 (UTC) Proofread Problem 4 and 5

=Notes and references=

title: User:Egm6321.f10.team03/Hwk1


text: File:Glycogen and glucose.jpg

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

title: File:Glycogen and glucose.jpg


text:
User:Thezero~enwikiversity

BADGES

title: User:Thezero~enwikiversity


text: File:InsulinPYY.jpg

title: File:InsulinPYY.jpg


text:
User talk:Egm6321.f10.team4.petralanda.n

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title: User talk:Egm6321.f10.team6.csquared


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User talk:Thezero~enwikiversity

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title: User talk:Thezero~enwikiversity


text:
User talk:Egm6321.f10.team03

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title: User talk:Egm6321.f10.team03


text:
Conservation Entrepreneurship/Scenarios and the Future

Mental Models

Concepts derived from Forrester who created systems language, Sterman, his collaborator, and applied to business thinking by Peter Senge, The Learning Organization.

Mental models describe how we think, often with a small number of variable, narrow boundaries, and with linear relationships. For example, I have a mental model that says if you increase law enforcement, deforestation will decrease. However, this model is flawed: it only includes 2 variables, but market demands and prices will also have a strong influence on Scenarios


text:
WikipediaOS/Other examples

Should you have more examples that help demonstrate the similarities between Wikipedia and Open Source development processes, please list and comment them here:

 * Example #1 - Name, mail...

title: WikipediaOS/Other examples


text:
WikipediaOS/Task specialization

Hypothesis
Different tasks among wikipedians (wikigardening, adding external resources, proofreading, programming boots, etc) is equivalent to the same type of specialization among contributors in the development of free source software (adding code, testing programs, correcting bugs, translating, improving usability, etc).

Votes/comments

 * For me is one of the more clear similarities between Wikipedia and OS development, since it represents and its probably the key of wide participation (a wide offer of thing to do, for so many different people and skills) --Esenabre 22:40, 3 September 2010 (UTC)

title: WikipediaOS/Task specialization


text: File:2010 09 02 13 55 50.djvu

title: File:2010 09 02 13 55 50.djvu


text: File:2010 09 02 14 58 46.djvu

title: File:2010 09 02 14 58 46.djvu


text:
User talk:Erkan Yilmaz/observations

monthly view stats
only for the main page, Erkan Yilmaz uses the Chat + Identi.ca 19:29, 2 February 2012 (UTC)

 * There are 3 peaks
 * 2008-06 + 2008-08
 * could be because of Composing free and open online educational resources?
 * I took part (2008: March-May) mainly to help teachers get familiar with WV
 * Checkuser status?
 * nomination (May 2008): Nominations for checkuser/Erkan Yilmaz
 * granted (August 2008)
 * 2009: Oct-Dec
 * this could be interesting...
 * only 23 edits in 2009, last one: 8 August 2009
 * I started my own wiki in 2009 March:, Erkan Yilmaz uses the Chat + Identi.ca 02:05, 3 February 2012 (UTC)

language of blog
Spricht man hier auch deutsch? my english is very bad! i dont speak english correct. --jom, 16 July 2011
 * In general EN is preferred here, for the German WV go here
 * Why was this question relevant for you back then? Erkan Yilmaz uses the Chat + Identi.ca 12:59, 20 October 2011 (UTC)

title: User talk:Erkan Yilmaz/observations


text:
User talk:Kenzy~enwikiversity

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title: User talk:Kenzy~enwikiversity


text:
User talk:Stephmartin

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 Hello , you can call me Jacob Frank. I'd like to be your mascot and welcome you to Wikiversity. Even though I was a plain and untutored man back in 18th-century Europe, that still didn't stop me from contributing my ideas to the world. Neither should it stop you. As a religious leader who developed the doctrine of "purification through transgression," my advice to you is to be bold and to ignore all rules—yes, these are two of Wikiversity's official policies! Happy editing.

Motivation and emotion
Hi. Welcome to the unit and to Wikiversity! I look forward to learning with you and I hope you find the topic and the unit rewarding. Please feel free to let me know if you have any suggestions or would like help at any point along the way. Sincerely, James Neill, -- Jtneill - Talk - c 03:36, 6 September 2010 (UTC)

-- Jtneill - Talk - c 22:30, 14 December 2010 (UTC)

title: User talk:Stephmartin


text:
Teaching EFL Listening via FUN WITH ENGLISH Books/7A/Comic Books

Preface
Started supplementing textbook exercises with outside materials, including an audio recording I made with my colleague Mr. Brown.

Focuses

 * Listening Set Skill = "D" Listen for gist
 * Listening Curriculum Statement = "6" Listen for gist
 * Listening Specific Skill = "b" Recognize varying purposes of discussion.
 * Speaking Set Skill = "A" Awareness of Context
 * Speaking Curriculum Statement = "1" Context & Register

Learning Outcome

 * Students will be able to explain the purpose of a discussion and the difference between formal and informal conversation.

Teaching Materials

 * FUN WITH ENGLISH 7A Textbook
 * Mr. Brown & Mr. Danoff Chapter 3 Audio Clip
 * Comic Strips

0 Before class
Ask if everyone has their textbooks and pens. If they don't tell them to go get them, if they're at home, maybe have them write their names on the board, depending on how they behave.

1 Greeting

 * Teacher Questions - After they answer, continue with follow up questions, e.g., Why?, When?, Where?, How?
 * "How are you?" to specific students
 * "What did you do last night?"
 * "What did you eat for breakfast?"
 * What are you going to do after school today?
 * Do you like comic books?
 * What is your favorite comic book?
 * Try to really engage in a dialogue about this, have them teach you the character's names, the plot and who writes them.
 * Who is your favorite comic book character?
 * "Do you have any questions for me?

2 Page 10 Let's Get Ready A & B

 * Ask them to complete on their own, walk around to make sure everyone is working.
 * When asking questions, do not just ask what are the answers to 1 to 6. Ask, "What is the Hulk?" and change the order, e.g., ask "Who is 5?" before "Who is 3?"
 * After B, ask "What character has ___ power??"

3 Page 11 Let's Listen 1 A

 * Divide the class in half, have the students read.
 * Go over the pronunciation and meaning of "character", "appearance" and "cape" beforehand.
 * Ask, "What are John and Karen talking about?"
 * If no response, "Are they talking about what comic book characters they are afraid of? Are they ..." work your way to the correct answer.

4 Chapter 3 Dialogue
0:03 - 0:11 Mr. Danoff - Mr. Brown and Mr. Danoff just finished watching "The Hulk" movie. 0:14 - 0:17 Mr. Danoff - Wow that was an awful movie. 0:18 - 0:20 Mr. Brown - Why was it awful? I really liked it. 0:20 - 0:27 Mr. Danoff - Ugh. The writing was bad, the acting was bad, and I was bored. 0:28 - 0:32 Mr. Brown - But the Hulk is so scary. Which superhero are you scared of? 0:33 - 0:34 Mr. Danoff - Batman! [Hard to understand] 0:34 - 0:36 Mr. Brown - Batman? But, Batman is good. 0:37 - 0:42 Mr. Danoff - He has a scary mask, OK? Which super hero are you scared of? 0:42 - 0:45 Mr. Brown - I am scared of The Hulk. He is so angry.
 * Explain that now we are going to listen to a conversation between two teachers: Mr. Brown and Mr. Danoff. They are discussing "The Hulk" movie and what comic book characters they are afraid of. Do you know the Hulk? Do you know afraid? Please close your textbooks and listen.
 * Before it begins, go over the vocabulary of: awful, bored and scared.
 * Stop after each person speaks the first time through, making sure they understand.
 * Dialogue
 * Depending on their understanding, play it all the way through again and then ask questions, or stop and ask questions.
 * Possible Questions
 * Did Mr. Danoff like the movie?
 * DId Mr. Brown like the movie?
 * Who is Mr. Danoff scared of?
 * Who is Mr. Brown scared of?
 * Why is Mr. Danoff scared of Batman?
 * Why is Mr. Brown scared of The Hulk?

5 Page 12 Let's Listen 2 A

 * Teacher reads, explain to students it will only be read once so they should pay close attention.
 * Ask in reverse chronological order to keep the students on their toes.

6 Page 12 Let's Listen 2 B

 * Students read first time, teacher reads again if necessary.
 * Ask "Why is Tim's cousin asking questions?"
 * If no response, "Because Tim's cousin loves Spiderman? Because .." work your way to the correct answer.

7 Page 13 Let's Listen 3

 * Give students the different parts, with the whole class being "students".
 * First time everyone reads and listens together.
 * Read it again, with new students playing the roles.
 * Third time teacher reads, and students fill in the answers on page 13.
 * Ask, "What are the students talking about?"
 * If no response, "Are they talking about a movie? Are the ..." work your way to the correct answer.

8 Hand around Comic Book Strips

 * Hand around English comics from a newspaper or that you print from the internet for the students to look at. Have enough so each pair of students gets something.
 * Tell the students they can get a new strip if they come to the front.
 * Either walk around engaging with the students helping them, or sit at the front and let them work through them at their own pace.
 * Be sure to collect them when you finish.

Extra Time

 * Competitive Hangman
 * "How high can you clap?" Game
 * Simon Says
 * Telephone

Review

 * Can I create different activities for different levels of students?
 * This particular class changes people almost every week. Difficult for me. Should I do more photos or what?
 * I told one advanced student I would grade her writing outside of class if she did it.

Next Week

 * One female student without internet access asked for ways to improve her spoken English.

Acknowledgments

 * How high can you clap inspired by Clap/Don't Clap from Teach English in Asia.
 * Two Grade 8 students at Anqing Foreign Language School who gave me the idea to play competitive hangman.
 * The comics I used in class were from the Chicago Sun-Times and the Chicago Tribune, kindly sent to China by my father.


text:
User talk:Ambreesh~enwikiversity

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Unfortunately, your account clashes with another account also called Ambreesh. To make sure that both of you can use all Wikimedia projects in future, we have reserved the name Ambreesh~enwikiversity that only you will have. If you like it, you don't have to do anything. If you do not like it, you can pick out a different name.

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Renamed
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title: User talk:Ambreesh~enwikiversity


text:
User talk:Poulamedasgupta

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title: User talk:Poulamedasgupta


text:
User:Pathak.pratik/Books/Spanish:Learning Through English

Learning Through English

 * Spanish 1
 * Spanish 1/Countries
 * Spanish 1/Chapter 1 (The Basics)
 * Spanish 1/Chapter 2 (Activities)
 * Spanish 1/Chapter 3 (Adjectives)
 * Spanish 1/Chapter 4 (School)
 * Spanish 1/Chapter 5 (The Classroom)
 * Spanish 1/Chapter 6 (Food & Drink)
 * Spanish 1/Chapter 7 (Health)
 * Spanish 1/Chapter 8 (Destinations)
 * Spanish 1/Chapter 9 (Recreation & Lifestyle)
 * Spanish 1/Chapter 10 (Family & Celebrations)
 * Spanish 1/Chapter 11 (Adjectives & In a restaurant)
 * Spanish 1/Chapter 12 (The Bedroom)
 * Spanish 1/Chapter 13 (The Household)
 * Spanish 1/Chapter 14 (Shopping)
 * Spanish 1/Chapter 15 (At the Mall)
 * Spanish 1/Chapter 16 (Vacation)
 * Spanish 1/Chapter 18 (Television)
 * Spanish 1/Chapter 19 (Computers)
 * Spanish 1/Linguistic characteristics
 * Spanish 2
 * Spanish 2/Chapter 1 (Classroom Events)
 * Spanish 2/Chapter 2 (Free Time)
 * Spanish 2/Chapter 3 (Daily Activities)
 * Spanish 2/Chapter 4 (Fashion)
 * Spanish 2/Chapter 5 (Errands)
 * Spanish 2/Chapter 6 (On the Road)
 * Spanish 2/Chapter 7 (Childhood)
 * Spanish 2/Chapter 8 (Celebrations)
 * Spanish 2/Chapter 9 (Emergencies)
 * Spanish 2/Chapter 10 (Accidents)
 * Spanish 2/Chapter 11 (Television)
 * Spanish 2/Chapter 12 (Movies)
 * Spanish 2/Chapter 13 (Cooking)
 * Spanish 2/Chapter 14 (Picnics)
 * Spanish 2/Chapter 15 (Planning a trip)
 * Spanish 2/Review


text:
User talk:Pathak.pratik

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User:Egm6321.f10.team3.Sudheesh

Hello World,

Welcome  This my Principles of Engineering Analysis-I (EGM 6321) Class page. You will find all my course related work here!!

Useful Links

 * Egm6321.f10.team03
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title: User:Egm6321.f10.team3.Sudheesh


text:
User:Egm6321.f10.team3.Sudheesh/Homeworks

 * Homework1
 * Homework2
 * Homework3
 * Homework4
 * Homework5
 * Homework6
 * Homework7


text:
User:Egm6321.f10.team3.Sudheesh/Homeworks/Homework1

= Problem 1 =

Solution
= Problem 2 =

Solution
= Problem 3 =

Solution
= Problem 4 =

Solution
= Problem 5 - Verification of Homogenous Solutions of Legendre Differential Equation for n = 1=

Given
Legendre differential equation:
 * {| style="width:100%" border="0" align="left"

{{L}_{2}}\left( y \right) = (1-x^2)y'' - 2xy' +n(n+1)y = 0 $$ $$ Homogenous solutions of the above equation for $$\displaystyle  n = 1 $$ [[media:2010_09_02_14_58_46.djvu|(Mtg 6-page1)]], are given by:
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$ $$
 * $$y_{H}^{1}\left( x \right)=x\equiv P_1(x)
 * $$y_{H}^{1}\left( x \right)=x\equiv P_1(x)
 * $$\displaystyle (Eq. 2)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$ $$ Where $$y_{H}^{1}\left( x \right)$$ is the first homogenous solution and $$y_{H}^{2}\left( x \right)$$ is the second homogenous solution
 * $$y_{H}^{2}\left( x \right)=\frac{x}{2}.\log \left( \frac{1+x}{1-x} \right)-1\equiv Q_1(x)
 * $$y_{H}^{2}\left( x \right)=\frac{x}{2}.\log \left( \frac{1+x}{1-x} \right)-1\equiv Q_1(x)
 * $$\displaystyle (Eq. 3)
 * }
 * }

Find
We need to prove that:
 * {| style="width:100%" border="0" align="left"

 * $${{L}_{2}}\left( y_{H}^{1} \right)={{L}_{2}}\left( y_{H}^{2} \right)=0$$
 * }
 * }
 * }

Solution
Let us consider first $${{L}_{2}}\left( y_{H}^{1} \right)$$:

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

$$ Substitute Eq.2 for $${{L}_{2}}\left( y_{H}^{1} \right)$$ in Eq.4 =>
 * $${{L}_{2}}\left( y_{H}^{1} \right)=\left( 1-{{x}^{2}} \right){{\left( y_{H}^{1} \right)}^{''}}-2x{{\left( y_{H}^{1} \right)}^{'}}+2\left( y_{H}^{1} \right)$$
 * $$\displaystyle (Eq. 4)
 * $$\displaystyle (Eq. 4)
 * }
 * }

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

 * $${{L}_{2}}\left( y_{H}^{1} \right)=\left( 1-{{x}^{2}} \right){{\left( x \right)}^{''}}-2x{{\left( x \right)}^{'}}+2\left( x \right)$$
 * }
 * }
 * }

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

$$ {{L}_{2}}\left( y_{H}^{1} \right)=\left( 1-{{x}^{2}} \right)\left( 0 \right)-2x\left( 1 \right)+2\left( x \right)=0$$ $$
 * style="width:20%; padding:1px; border:2px solid #0000FF" |
 * style="width:20%; padding:1px; border:2px solid #0000FF" |
 * $$\displaystyle (Eq. 5)
 * }
 * }

Now consider $${{L}_{2}}\left( y_{H}^{2} \right)$$:

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

$$ Determine first derivative of $$y_{H}^{2}$$ (Eq.2) using Product rule, Logarithmic derivative rule ($$ (\ln f)'= \frac{f'}{f} \quad$$) and chain rule of differentiation:
 * $${{L}_{2}}\left( y_{H}^{2} \right)=\left( 1-{{x}^{2}} \right){{\left( y_{H}^{2} \right)}^{''}}-2x{{\left( y_{H}^{2} \right)}^{'}}+2\left( y_{H}^{2} \right)$$
 * $$\displaystyle (Eq. 6)
 * $$\displaystyle (Eq. 6)
 * }
 * }

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

 * $${{\left( y_{H}^{2} \right)}^{'}}=\frac{x}{2}.\frac{d}{dx}\left( \log \left( \frac{1+x}{1-x} \right) \right)+\frac{d}{dx}\left( \frac{x}{2} \right).\log \left( \frac{1+x}{1-x} \right)-\frac{d}{dx}\left( 1 \right)$$
 * }
 * }
 * }

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

 * $${{\left( y_{H}^{2} \right)}^{'}}=\frac{x}{2}\left( \frac{1}{\left( \frac{1+x}{1-x} \right)}.\frac{d}{dx}\left( \frac{1+x}{1-x} \right) \right)+\frac{1}{2}\log \left( \frac{1+x}{1-x} \right)+0$$
 * }
 * }
 * }

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

 * $${{\left( y_{H}^{2} \right)}^{'}}=\frac{x}{2}\left( \frac{1-x}{1+x} \right)\left( \frac{\left( 1-x \right)+\left( 1+x \right)} \right)+\frac{1}{2}\log \left( \frac{1+x}{1-x} \right)$$
 * }
 * }
 * }

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

$${{\left( y_{H}^{2} \right)}^{'}}=\frac{x}{1-{{x}^{2}}}+\frac{1}{2}\log \left( \frac{1+x}{1-x} \right)$$ $$
 * style="width:20%; padding:1px; border:2px solid #8888aa" |
 * style="width:20%; padding:1px; border:2px solid #8888aa" |
 * $$\displaystyle (Eq. 7)
 * }
 * }

Second derivative of $$y_{H}^{2}$$ (Eq.2) is:

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

 * $${{\left( y_{H}^{2} \right)}^{''}}={{\left( {{\left( y_{H}^{2} \right)}^{'}} \right)}^{'}}=\frac{d}{dx}\left( \frac{x}{1-{{x}^{2}}}+\frac{1}{2}\log \left( \frac{1+x}{1-x} \right) \right)$$
 * }
 * }
 * }

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

 * $${{\left( y_{H}^{2} \right)}^{''}}=\frac{d}{dx}\left( \frac{x}{1-{{x}^{2}}} \right)+\frac{d}{dx}\left( \frac{1}{2}\log \left( \frac{1+x}{1-x} \right) \right)$$
 * }
 * }
 * }

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

 * $${{\left( y_{H}^{2} \right)}^{''}}=\left( \frac{\left( 1-{{x}^{2}} \right)-x.\left( -2x \right)} \right)+\frac{1}{2}\left( \frac{1-x}{1+x} \right)\left( \frac{\left( 1-x \right)+\left( 1+x \right)} \right)$$
 * }
 * }
 * }

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

 * $${{\left( y_{H}^{2} \right)}^{''}}=\frac{1+{{x}^{2}}}+\frac{1}{2}\left( \frac{1-x}{1+x} \right)\left( \frac{2} \right)$$
 * }
 * }
 * }

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

 * $${{\left( y_{H}^{2} \right)}^{''}}=\frac{1+{{x}^{2}}}+\frac{1}$$
 * }
 * }
 * }

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

$${{\left( y_{H}^{2} \right)}^{''}}=\frac{2}$$ $$ Substituting Eqs.3,7,8 into Eq.6 yields,
 * style="width:20%; padding:1px; border:2px solid #8888aa" |
 * style="width:20%; padding:1px; border:2px solid #8888aa" |
 * $$\displaystyle (Eq. 8)
 * }
 * }
 * {| style="width:100%" border="0"

 * $${{L}_{2}}\left( y_{H}^{2} \right)=\left( 1-{{x}^{2}} \right)\left( \frac{2} \right)-2x\left( \frac{x}{1-{{x}^{2}}}+\frac{1}{2}\log \left( \frac{1+x}{1-x} \right) \right)+2\left( \frac{x}{2}\log \left( \frac{1+x}{1-x} \right)-1 \right)$$
 * }
 * }
 * }

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

 * $${{L}_{2}}\left( y_{H}^{2} \right)=\frac{2}{\left( 1-{{x}^{2}} \right)}-\frac{2{{x}^{2}}}{1-{{x}^{2}}}-x\log \left( \frac{1+x}{1-x} \right)+x\log \left( \frac{1+x}{1-x} \right)-2$$
 * }
 * }
 * }

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

$${{L}_{2}}\left( y_{H}^{2} \right)=\frac{2-2{{x}^{2}}-2\left( 1-{{x}^{2}} \right)}{\left( 1-{{x}^{2}} \right)}=0$$ $$
 * style="width:20%; padding:1px; border:2px solid #0000FF" |
 * style="width:20%; padding:1px; border:2px solid #0000FF" |
 * $$\displaystyle (Eq. 9)
 * }
 * }

Eqs.5 and 9 verify that :{| style="width:10%" border="0" $${{L}_{2}}\left( y_{H}^{1} \right)={{L}_{2}}\left( y_{H}^{2} \right)=0$$ Egm6321.f10.team3.Sudheesh 18:11, 14 September 2010 (UTC)
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * }
 * }

=Notes and references=


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User:Egm6321.f10.team2.oztekin/HW1

=Problem 1: Deriving Function Relate to Maglev=

Problem Infrastructure
Any fuction like
 * {| style="width:100%" border="0"

$$ F\left ( x,y(x) \right )=0 \displaystyle  $$ (Eq 1.1) is called a closed fuction.
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

A total differential of F is as follows:
 * {| style="width:100%" border="0"

$$df=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy \displaystyle $$
 * style="width:95%" |
 * style="width:95%" |

(Eq 1.2)
 *
 * }

From that we can write:
 * {| style="width:100%" border="0"

$$\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}\frac{\mathrm{d} y}{\mathrm{d} x}=0 \displaystyle $$
 * style="width:95%" |
 * style="width:95%" |

(Eq 1.3)
 *
 * }

Given
From the maglev train which mentioned in meeting one

$$Y'\left ( t \right )$$:nominal position of the wheel

$$U'\left ( s,t \right )$$:axial deformation of the guide way

$$U^{2}\left ( s,t \right )$$:transverse deformation of the guide way

And we conclude that there is a impilicit and explicit dependence between U and Y(t)

Find

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

$$\frac{\mathrm{d}^2 f(Y'(t),t) }{\mathrm{d} t}=\frac{\partial f(Y'(t),t)}{\partial s}\left ( \ddot{Y'} \right )+\frac{\partial^2 f(Y'(t),t)}{\partial s^2}\left ( \dot{Y'} \right )^2+2\frac{\partial^2 f(Y'(t),t)}{\partial s\partial t}\left ( \dot{Y'} \right )+\frac{\partial^2 f(Y'(t),t)}{\partial t^2} \displaystyle $$ (Eq 1.4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Solution

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

$$\frac{\mathrm{d} f(Y'(t),t)}{\mathrm{d} t} =\frac{\partial f(Y'(t),t)}{\partial s}\dot{Y'}+\frac{\partial f(Y'(t),t)}{\partial t} \displaystyle $$ (Eq 1.5)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$\frac{\mathrm{d} }{\mathrm{d} t}\left ( \frac{\partial f(Y'(t),t)}{\partial s}\dot{Y'}+\frac{\partial f(Y'(t),t)}{\partial t}\right )=\left (\frac{\partial^2 f(Y'(t),t)}{\partial s^2}\dot{Y'}+\underbrace{ \frac{\partial^2 f(Y'(t),t)}{\partial s\partial t}}  \right )\dot{Y'}+\frac{\partial f(Y'(t),t)}{\partial s}\ddot{Y'}+\underbrace{\frac{\partial^2 f(Y'(t),t)}{\partial t\partial s}}\dot{Y'}+\frac{\partial^2 f(Y'(t),t) }{\partial t^2}) \displaystyle $$
 * style="width:95%" |
 * style="width:95%" |

(Eq 1.6)
 *
 * }

According to the Schwarz Theorem

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

$$\frac{\partial^2 f}{\partial s\partial t}=\frac{\partial^2 f}{\partial t\partial s} \displaystyle $$ (Eq 1.7)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$\frac{\mathrm{d} }{\mathrm{d} t}\left ( \frac{\partial f(Y'(t),t)}{\partial s}\dot{Y'}+\frac{\partial f(Y'(t),t)}{\partial t}\right )=\frac{\partial f(Y'(t),t)}{\partial s}\left ( \ddot{Y'} \right )+\frac{\partial^2 f(Y'(t),t)}{\partial s^2}\left ( \dot{Y'} \right )^2+2\frac{\partial^2 f(Y'(t),t)}{\partial s\partial t}\left ( \dot{Y'} \right )+\frac{\partial^2 f(Y'(t),t)}{\partial t^2} \displaystyle $$ (Eq 1.8)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

Relation To Engineering Equations
If an object turning around an axis with an angular velocity of '$$\omega$$',its velocity according to a motionless observer can be written as;
 * {| style="width:100%" border="0"

$$\vec{v}=\dot{\vec{r}}+\vec{\omega}*\vec{r} \displaystyle $$
 * style="width:95%" |
 * style="width:95%" |

(Eq 1.9)
 *

And from this we can write acceleration as:
 * }
 * {| style="width:100%" border="0"

$$\vec{a}=\ddot{\vec{r}}+2*\vec{\omega }*\dot{\vec{r}}+\vec{\omega }\left ( \vec{\omega *\vec{r}} \right ) \displaystyle$$
 * style="width:95%" |
 * style="width:95%" |

(Eq 1.10)
 *

For an object moving under gravity force and another force(F), the equation of motion is:
 * }
 * {| style="width:100%" border="0"

$$m\vec{a}=m\vec{g}+\vec{F}\displaystyle$$ (Eq 1.11)
 * style="width:95%" |
 * style="width:95%" |
 *

Substituting $$\vec{a}$$ into the (Eq 1.11) gives:
 * }

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

$$m\ddot{\vec{r}}=m\vec{g}+\vec{F}-2*m*\vec{\omega }*\dot{\vec{r}}-m*\vec{\omega }\left ( \vec{\omega *\vec{r}} \right ) \displaystyle $$ (Eq 1.12)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

We can then show that $$m*\vec{\omega }\left ( \vec{\omega *\vec{r}} \right )$$ is 'centrifugal force'. And the term $$2*m*\vec{\omega }*\dot{\vec{r}}$$ is a 'Coriolis Force'.
 * }

= Problem 2: Boundary Value Problem =

Given
From the note 5-3 and 5-4 we know,

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

$$y(x)=cy_{H}^{1}(x)+dy_{H}^{2}(x)+{{y}_{P}}(x) \displaystyle$$ (Eq 2.1)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$y(a)=\alpha \displaystyle $$ (Eq 2.2)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$\displaystyle y(b)=\beta $$
 * style="width:95%" |
 * style="width:95%" |
 * (Eq 2.3)
 * }
 * }

Find
$$\displaystyle c, d$$ in terms of $$\displaystyle \alpha, \beta. $$

Solution
By substituting $$\displaystyle a $$ and $$\displaystyle b $$ into $$\displaystyle Eq 2.1 $$ and combining with $$\displaystyle Eq 2.2 $$ and $$\displaystyle Eq 2.3 $$, we derive,

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

$$\displaystyle cy_{H}^{1}(a)+dy_{H}^{2}(a)+{{y}_{P}}(a)=\alpha, $$ (Eq 2.4)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * }

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

$$\displaystyle cy_{H}^{1}(b)+dy_{H}^{2}(b)+{{y}_{P}}(b)=\beta. $$ (Eq 2.5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * }

By multipling $$\displaystyle y_{H}^{2}(b) $$ in both sides of $$\displaystyle Eq. 4 $$ and $$\displaystyle y_{H}^{2}(a) $$ of $$\displaystyle Eq 2.5 $$, we obtain,

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

$$\displaystyle y_{H}^{1}(a)y_{H}^{2}(b)*c+y_{H}^{2}(a)y_{H}^{2}(b)*d+{{y}_{P}}(a)y_{H}^{2}(b)=\alpha y_{H}^{2}(b),$$ (Eq 2.6)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * }

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

$$\displaystyle y_{H}^{1}(b)y_{H}^{2}(a)*c+y_{H}^{2}(a)y_{H}^{2}(b)*d+{{y}_{P}}(b)y_{H}^{2}(a)=\beta y_{H}^{2}(a).$$ (Eq 2.7)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * }

Let $$\displaystyle Eq 2.6 $$ minus $$\displaystyle Eq 2.7 $$, then $$\displaystyle d $$ is cancelled, so we get,

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

$$\displaystyle c=\frac{\left[ \alpha -{{y}_{P}}(a) \right]y_{H}^{2}(b)-\left[ \beta -{{y}_{P}}(b) \right]y_{H}^{2}(a)}{y_{H}^{1}(a)y_{H}^{2}(b)-y_{H}^{1}(b)y_{H}^{2}(a)}. $$ (Eq 2.8)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }
 * }

Note the structrue of $$\displaystyle Eq 2.4 $$ and $$\displaystyle Eq 2.5 $$, we find that $$\displaystyle c $$ is bound to $$\displaystyle y_{H}^{1} $$ while $$\displaystyle d $$ is bound to $$\displaystyle y_{H}^{2} $$, that is, when we switch the superscript between $$\displaystyle y_{H}^{1} $$ and $$\displaystyle y_{H}^{2} $$, $$\displaystyle d $$ will become the former $$\displaystyle c $$ thus the same result of $$\displaystyle Eq 2.8 $$. After that we switch back the superscript between $$\displaystyle y_{H}^{1} $$ and $$\displaystyle y_{H}^{2} $$ then we get the right answer of $$\displaystyle d $$. By doing this, simply exchange the superscript of right side of $$\displaystyle Eq 2.8 $$, we have,

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

$$\displaystyle d=\frac{\left[ \alpha -{{y}_{P}}(a) \right]y_{H}^{1}(b)-\left[ \beta -{{y}_{P}}(b) \right]y_{H}^{1}(a)}{y_{H}^{2}(a)y_{H}^{1}(b)-y_{H}^{2}(b)y_{H}^{1}(a)}. $$ (Eq 2.9)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }
 * }

= Problem 3: Verifying 2 homogeneous solutions =

Given

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

$$\displaystyle {{L}_{2}}(y)=(1-{{x}^{2}})\ddot{y}-2x\dot{y}+2y$$ (Eq 3.1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * }

where $$\displaystyle {{L}_{2}} $$ is an operator of functions.

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

$$\displaystyle y_{H}^{1}(x)=x,$$ (Eq 3.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * }

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

$$\displaystyle y_{H}^{2}(x)=\frac{x}{2}\log (\frac{1+x}{1-x})-1,$$ (Eq 3.3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * }

where $$\displaystyle y_{H}^{1}(x) $$ and $$\displaystyle y_{H}^{2}(x) $$ are homogeneous solutions of the equation $$\displaystyle {{L}_{2}}(y)=0$$.

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

$$\displaystyle {{L}_{2}}(y_{H}^{1})={{L}_{2}}(y_{H}^{2})=0$$
 * style="width:95%" |
 * style="width:95%" |
 *

 * }.
 * }.

Solution
From $$\displaystyle Eq 3.2 $$, we have,

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

$$\displaystyle \dot{y}_{H}^{1}(x)=1,$$ (Eq 3.4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * }

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

$$\displaystyle \ddot{y}_{H}^{1}(x)=0.$$ (Eq 3.5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * }

Then by sustituting $$\displaystyle \dot{y}_{H}^{1}(x)$$ and $$\displaystyle \ddot{y}_{H}^{1}(x)$$ into $$\displaystyle {{L}_{2}}(y)$$ we have,

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

$$\displaystyle \begin{align} {{L}_{2}}(y_{H}^{1})& =0-2x+2x \\ & =0. \end{align}$$ (Eq 3.6)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }
 * }

From $$\displaystyle Eq 3.3 $$, we derive,

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

$$\displaystyle \begin{align} \dot{y}_{H}^{2}(x)& ={{\left[ \frac{x}{2}\log \left( \frac{2}{1-x}-1 \right)-1 \right]}^{\prime }} \\ & ={{\left( \frac{x}{2} \right)}^{\prime }}*\log \left( \frac{2}{1-x}-1 \right)+\frac{x}{2}*{{\left[ \log \left( \frac{2}{1-x}-1 \right) \right]}^{\prime }} \\ & =\frac{1}{2}\log \left( \frac{2}{1-x}-1 \right)+\frac{x}{1-{{x}^{2}}} \end{align}$$ (Eq 3.7)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * }

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

$$\displaystyle \begin{align} \ddot{y}_{H}^{2}(x) & ={{\left[ \frac{1}{2}\log \left( \frac{2}{1-x}-1 \right)+\frac{x}{1-{{x}^{2}}} \right]}^{\prime }} \\ & ={{\left[ \frac{1}{2}\log \left( \frac{2}{1-x}-1 \right) \right]}^{\prime }}+{{\left( \frac{x}{1-{{x}^{2}}} \right)}^{\prime }} \\ & =\frac{1}{1-{{x}^{2}}}+\frac{1+{{x}^{2}}} \end{align}$$ (Eq 3.8)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * }

Then by substituting $$\displaystyle \dot{y}_{H}^{2}(x)$$ and $$\displaystyle \ddot{y}_{H}^{2}(x)$$ into $$\displaystyle {{L}_{2}}(y)$$ we have,

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

$$\displaystyle \begin{align} {{L}_{2}}(y_{H}^{2}) & =(1-{{x}^{2}})\left( \frac{1}{1-{{x}^{2}}}+\frac{1+{{x}^{2}}} \right)-2x\left( \frac{1}{2}\log M+\frac{x}{1-{{x}^{2}}} \right)+2\left( \frac{x}{2}\log M-1 \right) \\ & =1+\frac{1+{{x}^{2}}}{1-{{x}^{2}}}-x\log M-\frac{2{{x}^{2}}}{1-{{x}^{2}}}+x\log M-2 \\ & =0, \end{align}$$ (Eq 3.9)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }
 * }

where $$\displaystyle M=\frac{2}{1-x}-1$$

By $$\displaystyle Eq 3.6$$ and $$\displaystyle Eq 3.9$$, we have proved $$\displaystyle {{L}_{2}}(y_{H}^{1})={{L}_{2}}(y_{H}^{2})=0$$.

=Problem 4: Perform Dimensional analysis on Governing Equation of Force acting along Maglev Deformed Guideway=

Given
Governing Equation of force acting along a Maglev deformed guideway
 * {| style="width:100%" border="0"

$$C_0\left ( Y^1,t \right)=-F^1\left [ 1-\overline{R}{u^2_{,ss}}\left ( Y^1,t \right) \right] -F^2 {u^2_{,s}}\left ( Y^1,t \right) - \frac{T}{R} +M\left [ \left [ 1-\overline{R}{u^2_{,ss}}\left ( Y^1,t \right) \right ] \left [ {u^1_{,tt}}\left ( Y^1,t \right)-\overline{R}{u^2_{,stt}}\left ( Y^1,t \right) \right ]+{u^2_{,s}}\left ( Y^1,t \right){u^2_{,tt}}\left ( Y^1,t \right) \right ]
 * style="width:95%" |
 * style="width:95%" |

$$     (Eq 4.1)
 *

 * }

Note: Argument $$\left ( Y^1, t \right)$$ will be omitted for clarity in subsequent steps.

The following are the known variables of (Eq 4.1):

$$F^{1}$$:vertical force

$$F^{2}$$:horizontal force

$$F^{1}$$:vertical force

$$\overline{R}$$:distance from deformed guide way centerline to wheel centerline

$$ u^2_{,s}$$:slope of deform guide way

$$T$$:Torque exerted on wheel

$$R$$:radius of wheel

$$M$$:mass of wheel

The following are the derived variables of (Eq 4.1):

$$ u^2_{,ss}$$

$$ u^1_{,tt}$$

$$ u^2_{,stt}$$

$$ u^2_{,tt}$$

Find
Verify validity of (Eq 4.1) via “dimensional analysis”

Solution
Dimensional analysis of each variable must be completed for each variable in the equation prior to determining the overall dimension of the equation.

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

$$[F^{1}]\longrightarrow$$ $$F$$ (Eq 4.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$[F^{2}]\longrightarrow$$ $$F$$ (Eq 4.3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$[\overline{R}]\longrightarrow$$ $$L$$ (Eq 4.4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$ [u^2_{,s}]=\frac{\partial{u}^2}{\partial s}\longrightarrow $$ $$\frac{L}{L}$$ $$ \longrightarrow [1] $$ (Eq 4.5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$[T] \longrightarrow$$ $$F\cdot L$$ (Eq 4.6)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

||
 (Eq 4.7)
 * $$[R]\longrightarrow$$ $$L$$
 * $$[R]\longrightarrow$$ $$L$$
 * }

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

(Eq 4.8)
 * $$[M]\longrightarrow $$ $$L$$
 *

 *

 * }

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

(Eq 4.9)
 * $$ [u^2_{,ss}]=\frac{\partial^2{u}^2}{\partial s^2}\longrightarrow $$$$\frac{L}{{L}^2}$$ $$ \longrightarrow \frac{1}{L}$$
 *

 *

 * }

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

(Eq 4.10)
 * $$ [u^1_{,tt}]=\frac{\partial^2{u}^1}{\partial s^2}\longrightarrow $$$$\frac{L}{T}^2$$
 *

 *

 * }

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

(Eq 4.11)
 * $$ [u^2_{,stt}]=\frac{\partial {u}^2}{\partial s\partial^2 {t}}\longrightarrow $$$$\frac{L}{L\cdot{T}^2}$$ $$ \longrightarrow \frac{1}{T}^2$$
 *

 *

 * }

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

(Eq 4.12)
 * $$ [u^2_{,tt}]=\frac{\partial^2 {u}^2}{\partial {t}^2} \longrightarrow $$$$\frac{L}{T^2}$$
 *

 *

 * }

The dimensional analysis of the entire equation can now be applied to (Eq 4.1) by substitution of (Eq 4.2) to (Eq 4.12).

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

$$ [C_0\left ( Y^1,t \right)]=F\left [ 1-L \frac{1}{L}\right] -F \cdot\frac{L}{L} -\frac{F \cdot L}{L} +M\left [ \left [ 1-L \cdot \frac{1}{L} \right] \left [ \frac{L}{T^2} - \frac{L}{T^2} \right] +1 \cdot \frac{L}{T^2} \right]
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

$$     (Eq 4.13)
 *

 * }

In conclusion, it can be deduced by simplification of fractions that the components of equation (Eq 4.13) will reduce down to force ($$F$$) dimensions and mass x acceleration ($$M \cdot \tfrac{L}{T^2}$$), which further reduces to force ($$F$$) dimensions.

=Problem 5:Proof of Nonlinearity=

Given
Show why $${{c}_{3}}\left( {{Y}^{1}},t \right){{\ddot{Y}}^{1}}$$ is nonlinear with respect to Y1

Solution
A nonlinear ordinary differential equation (ODE) is as an ODE that is not linear. There are few properties of ODEs that distinguish them as linear. First, the dependent variable and all of the derivatives of the dependent variable are of a degree one. Second, nonlinear functions of the dependent variable or its derivatives aren’t allowed.

In general, linear ODEs must satisfy the following statement:
 * {| style="width:100%" border="0"

$$F\left( \alpha u+\beta v \right)=\alpha F\left( u \right)+\beta F\left( v \right)\displaystyle$$ (5.1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Evaluating the left hand side (LHS) of the equation for $$F\left( {{Y}^{1}},t \right)={{c}_{3}}\left( {{Y}^{1}},t \right){{\ddot{Y}}^{1}}\left( t \right)$$ yields:

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

$$\begin{align} & F\left( \alpha u+\beta v \right)={{c}_{3}}\left( \alpha u+\beta v,t \right){{{\ddot{Y}}}^{1}}\left( t \right) \\ & \text{                  }=M\left[ 1-\overline{R}u_{ss}^{2}\left( \alpha u+\beta v,t \right) \right]{{{\ddot{Y}}}^{1}}\left( t \right) \\ \end{align}\displaystyle$$ (5.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

An evaluation of the right hand side (RHS) of the equation yields:

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

$$\begin{align} & \alpha F\left( u \right)+\beta F\left( v \right)=\alpha {{c}_{3}}\left( u,t \right){{{\ddot{Y}}}^{1}}\left( t \right)+\beta {{c}_{3}}\left( v,t \right){{{\ddot{Y}}}^{1}}\left( t \right) \\ & \text{                        }=\alpha M\left[ 1-\overline{R}u_{ss}^{2}\left( u,t \right) \right]{{{\ddot{Y}}}^{1}}\left( t \right)+\beta M\left[ 1-\overline{R}u_{ss}^{2}\left( v,t \right) \right]{{{\ddot{Y}}}^{1}}\left( t \right) \\ & \text{                        }=M\left[ \alpha \left( 1-\overline{R}u_{ss}^{2}\left( u,t \right) \right)+\beta \left( 1-\overline{R}u_{ss}^{2}\left( v,t \right) \right) \right]{{{\ddot{Y}}}^{1}}\left( t \right) \\ \end{align}\displaystyle$$ (5.3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

It is evident that the RHS does not equal the LHS and the expression is nonlinear because

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

$$F\left( \alpha u+\beta v \right)\ne \alpha F\left( u \right)+\beta F\left( v \right)\displaystyle$$ (5.4)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

= References =

=Contrubituons Of Team Members=
 * Problem 1              $$\Rightarrow$$ Oztekin,Dube
 * Problem 2/Problem 3    $$\Rightarrow$$ Zou,Abudaram,Sahin (See ZOU'S PAGE)
 * Problem 4              $$\Rightarrow$$ Rasul (See RASUL'S PAGE)
 * Problem 5              $$\Rightarrow$$ Patterson (See PATTERSON'S PAGE)
 * Final Page Designer    $$\Rightarrow$$ Oztekin

title: User:Egm6321.f10.team2.oztekin/HW1


text:
Mr. Danoff's FWE 8A Lesson 12

Lesson

 * identify: opinion / fact and argument / conversation
 * Briefly review context.

Focuses

 * Listening Set Skill = 6. Listen for gist in passages and conversations.
 * Listening Curriculum Statement = c) Follow discussion to identify fact/opinion.
 * Speaking Set Skill = A. Awareness of context.

Definition

 * Opinion = What a person thinks.
 * Fact = What a person knows.
 * Context = What a person or people is/are talking about.

Why?

 * Because knowing the difference between fact and opinion is important for making decisions.

Example

 * Hold up a book. Say this book is terrible. fact or opinion? this book is great! fact or opinion? this book is (color of the book). fact or opinion?

Lesson Question

 * Is this fact or opinion?

Word(s) of the Day

 * king/queen
 * The boss of a country.
 * servant
 * Someone who works for the family of a king.
 * Demonstrate, by having 2 students be the king and queen, and order the servant around (Mr. Brown).
 * roar
 * The sound a lion makes when it yells.

Contact Information

 * QQ ID 867996874
 * QQ Name Danoff
 * e-mail danoff.charles@gmail.com

Teaching Materials

 * FUN WITH ENGLISH 8A Textbook
 * Lessons Notebook
 * Stopwatch

Chinese English

 * If the following things are said by a student, address the whole class.
 * No why.
 * Give me!
 * I know! I know!

English Tips

 * If a student uses the adjective interesting, try to push them to use other words, as well.

0) Before the Bell

 * In classroom at least 30 minutes before my first class of the day.
 * Before class, make sure chairs are set up correctly
 * Write Lesson plan, Definitions, my contact info, and that it's OK to say I don't know or No reason on the board.
 * Ask if everyone has their textbooks and pens. If they don't tell them to go get them.
 * Stopwatch ready to go before class.

1) Coversation
6 Minutes
 * If the kids don't respond, ask them the questions directly.
 * Are you tired?
 * Are you hungry?
 * Are you cold?
 * What is happening in the news these days?
 * What did you do in your clubs after school yesterday?
 * Have you watched any good [movies, cartoons, tv shows, ....] recently?
 * Have you read any good [books, comics, articles ....] recently?
 * What class did you just have? What did you learn? Do you like it?
 * what club did you have yesterday? what will you do in your club today?
 * Is it anyone's birthday today in class?
 * Any questions for me? Add that the longer we talk here, the less time we have to spend on the textbook.

2) Joke
2 Minutes
 * Blind mice joke.
 * write - blind means can't see - on the board

3) Introduction
2 Minutes
 * Point out the Lesson Plan, mention if we move fast, we can get to the game quicker.
 * Tell them my contact info, and say that I am busy. I want to talk to you, but if I can't it is not because I don't like them, it is because I am busy. I like all of you. Please say hello again or e-mail me. E-mail is much better.
 * Words of the day.
 * Mention that if they ask a question and they don't know the answer, it's fine to say I don't know. Write No why and No reason ask which is correct if I ask Why? cross out No reason explain that it is Chinese, not English.
 * Ask kids to please push in their chairs before they leave class when the bell rings.

4) Page 47 Let's listen 1 A & B
5 - 10 Min
 * Walk around the class, help the students who are struggling.
 * Tell the students if they have any questions, to raise their hands.
 * Do not read the Let's listen activities on page 47, just have the students answer the questions.
 * After the 47 A answers, ask if the T/F statements are facts or opinions?
 * After reviewing all the answers, try to get a dialogue going, ask
 * What do you think about lions?
 * Are lions our friends?
 * Have you watched any movies with lions?
 * Who is stronger/better/faster lions or tigers?
 * Do you think it is OK for lions to live in zoos?
 * Are zoos good or bad?
 * Are animals happy in zoos?
 * Ask what is the context of our conversation?

5) Page 48 Let's listen 2 - Tapescript Page 93
10 Minutes
 * I read it through.
 * For advanced classes, try having three students act out the part of the lion.
 * Try to get a conversation going afterwards.
 * Do you like the King? Servant? Lion? Why?
 * Is the lion scary?

6) Show & Tell Activity
15 - 30 Minutes (Depending on how many kids share.)
 * Ask the students if they remembered the homework?
 * Tell the students they have to write three sentences about what they brought. Give them 5 minutes.
 * If they don't understand, write three sentences about something I brought on the board.
 * Walk around and help, if a kid forgot to bring something, have them choose something they are wearing, or a pen or something like that.
 * After the writing time is finished, ask for volunteers. Who wants to share?
 * If no one does, start selecting students.
 * After the speaker has come forward, chose two students to ask questions.
 * Possibly have the speaker give their speech again, if they only speak quietly looking down into their book the entire time or if they mumble.
 * After the chosen kids ask the questions, see if there are any other questions.
 * At the end, ask the kids who do speak to bring their objects again, and those who forgot to please bring an object next week.

Extra Time

 * Competitive Hangman
 * Is 'Simon says' too easy for you?

g8 c2
Anqing Foreign Language School Teaching Journal 4 Page 13
 * again, asking discussion questions is like pulling teeth
 * acting out the lion's tale worked well
 * what kind of activity can I do that they'll like?
 * they are absurdly fast about getting together to cheat through all the answers
 * overall attitude today was decent, no volunteers

g8 c6
AFLS TJ4 Page 17
 * Normally friendly girl with a Japanese English name was very sarcastic about the email & qq. maybe because I haven't been using them
 * Super hero named rambunctuous boy just has too much energy to control. he simply cannot help himself.
 * textbook went well actually, show and tell writing did not ... next week
 * ki ki was some sort of word that made them all laugh
 * great class overall
 * drawing the lion on the board was good

g8 c2
AFLS TJ4 Page 13
 * have some girls who haven't spoken go
 * do something about all the girls calling a girlm named for a month - 'man' ?

Time

 * class g8c6
 * plan
 * 1 - 14:04
 * 2 - 2:48
 * 3 - 2:46
 * 4 - 15:05
 * 5 - 9:54
 * 6 - 11:53

Materials Used Whose Copyright is Owned by Others

 * I used these because it was allowed within the rights given by the people who made the materials.
 * The copyright of this lesson plan does not reflect the copyright of these materials.
 * Please contact the creators to see how it is OK to use their materials.

 * Copyright Holder
 * East Meets West China
 * Website
 * http://emwchina.com/about/about.htm
 * Materials Used
 * Focuses, taken from the curriculum section of the 2009/2010 East Meets West China Information Booklet.
 * Copyright Holder
 * The Internet TESL Journal
 * Website
 * http://iteslj.org/
 * Copyright Notice
 * http://iteslj.org/copyright.html
 * Contact Information
 * http://iteslj.org/mail/
 * Materials Used
 * Joke was slightly modified from one I got from the ITESLJ.
 * http://iteslj.org/c/jokes.html
 * Copyright Holder
 * Two of my students at AFLS who gave me the idea for Competitive Hangman.
 * Materials Used
 * Competitive Hangman
 * Copyright Holder
 * Dave Sperling of Dave's ESL Cafe
 * Website
 * http://daveseslcafe.com
 * Contact Information
 * http://www.eslcafe.com/contact.html
 * Material Used
 * Is 'Simon says' too easy for you? submitted by David Kendrick from New Zealand.
 * http://www.eslcafe.com/idea/index.cgi?display:1063330109-41948.txt

Materials Used Which I Made

 * I will likely only use materials I created which I entered into the Public Domain, but read carefully in case I licensed one thing different which may be what you want to re-use.

 * Show & Tell Activity, which I entered into the Public Domain
 * http://mr.danoff.org/show-and-tell-v0.2.txt

Acknowledgments

 * I would like to thank all the 8th grade classes at Anqing Foreign Language School who had this lesson for their participation and feedback. I would also like to thank the Anqing Foreign Language School for allowing me to teach.
 * Explanation of the servant by having the kids act it out is from Mr. Brown.

Page 46 Let's get ready A & B
5 Min
 * Walk around the class, help the students who are struggling.
 * Try to get a dialogue going, ask What do you think about lions? Are lions our friends? Have you watched any movies with lions? Who is stronger/better/faster lions or tigers?
 * Ask what is the context of our conversation?

Page page 47 Let's listen 1 A - Tapescript Page 92
3 Min
 * Students read it through.
 * Ask if the statements are facts or opinions.
 * Try to get a dialogue going, ask do you think it is OK for lions to live in zoos? Are zoos good or bad? Are animals happy in zoos?

Page 47 Let's listen 1 B - Tapescript Page 93
4 Min
 * Students read it through.
 * Perhaps have 1 or 2 kids at the board drawing a picture of a lion and labeling it?

Page 49 Let's listen 3 A - Tapescript Page 93
6 Minutes
 * I read it through.

Page 49 Let's listen 3 B - Tapescript Page 93
6 Minutes
 * I read it through.

Version 0.1 Nov 15, 2009

 * Chose the textbook questions on an afternoon train ride from Hefei to Anqing with Mr. Brown.

Version 0.11 November 26, 2009 5:55 PM

 * Crafted the lesson plan around the questions using the lesson skeleton from 8A.Lesson11ver0.4.

Version 0.2 November 29, 2009 7:51 AM to November 29, 2009 9:11 AM

 * Added a copyright notice to the beginning of the lesson plan.
 * Added Version information to the beginning of the lesson plan.
 * Took out a few items that don't need to be on the board from step 0.
 * Took out the Thanksgiving Speech.
 * Added Show & Tell Activity.
 * Changed the joke, because the kids don't know the word blind.
 * Took out Words of the Day from the Into because this lesson has no Words of the Day at the moment.
 * Added Ask kids to please push in their chairs before they leave class when the bell rings. to the introduction.
 * Moved original page 46 & 47 instructions to Additional Questions & Instructions for This Lesson.
 * Combined 46 & 47 into one set of instructions, because the kids can do both pages of exercises on their own, without listening to the Tapescripts.
 * Added time estimates for page 48 & 49.
 * Added the game Is Simon says too easy for you? to Extra Time.
 * Took out the Copyright Owners of Materials and incorporated the information into Materials Used Whose Copyright is Owned by Others.
 * Added Show and Tell Activity Copyright Information.
 * Took out quotation marks in the Is Simon says too easy for you? instructions.
 * Added Lesson Associated Blog Post section.
 * Took out my email address from the lesson header, because it is already at the bottom of the document and doesn't need to be there twice.
 * Created another version of the lesson plan minus all copyright and contact information that I will also post for teachers to print off and use who are only interested in the lesson itself. The naked version.

Version 0.3 December 1st, 2009
Summary - Changed the Show & Tell instructions and moved it behind a few textbook activities, because in class on Monday we didn't do any textbook.
 * slowly making the lesson plan more wiki friendly with = for section heads and *'s instead of *'s
 * added servant to word of the day
 * added clarifications for the joke
 * moved Show & Tell Activity after a few activities from the textbook, because if I don't then, we will not spend any time on the textbook.
 * took out Page 46 Let's get ready A & B because the activities are not good, and our classes time can be better spent in other ways.
 * capped Page 47 Let's listen 1 A & B at 10 minutes to make sure we have time for the other activities.
 * moved Page 49 Let's listen 3 A & Page 49 Let's listen 3 B to = Additional Questions & Instructions for This Lesson = because there is not time for them.
 * Added = Notes = to BEFORE CLASS and the note Always ask if students have questions or be patient and what for volunteers to answer my quesitons before forcing kids to respond.

Version 0.31 December 2nd, 2009

 * added roar, king and queen to words of the day and added some instructions
 * added words of the day to introduction
 * added instructions to Page 48 Let's listen 2
 * added instruction to show & tell
 * added Tell the students if they have any questions, to raise their hands. to the Page 47 Let's listen 1 A & B.
 * added = Time = section to AFTER CLASS, for recording how long different sections take.
 * added = English Tips = with If a student uses the adjective interesting, try to push them to use other words, as well.

Version 0.32 26 August 2010
finished @ 01:20 USA CT
 * Made the formatting Mediawiki friendly.

Version 0.33 4 September 2010

 * Took out the copyright notice, because it changes depending on where things are published online.

0.34 17 Feb 2011

 * Wikignomed for print publication, added detail box.

Contact Information
I am actively seeking feedback on how to improve this lesson, please email me at, danoff.charles@gmail.com

Lesson Associated Blog Post

 * The following links did work when I wrote this plan, but as of September 2010, they do not work.
 * http://www.danoff.org/leftinfront/?p=389
 * Please visit the link above to leave comments, criticisms, or thoughts about this lesson in the comment section.

title: Mr. Danoff's FWE 8A Lesson 12