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User:EGM6321.f11.team1.pan/HW5

= R*5.2 =

Solution
Part (2) Hermite differential equation $$\displaystyle y''-2xy'+2ny=0$$ is not exact if $$\displaystyle n\neq -1$$. And it is in the power form of L2-ODE-VC: $$\displaystyle \alpha x^ry''+\beta x^sy'+\gamma x^ty$$ with $$\displaystyle \alpha=1;\beta=-2;\gamma=2n.$$ $$\displaystyle r=0;s=1;t=0.$$ Assuming that we can find the integrating factor which is in power form, $$\displaystyle h(x,y)=x^my^n$$. Use the exactness condition to find $$m,n\in \mathbb{E}$$, such that the following N2-ODE is exact:
 * {| style="width:100%" border="0" align="left"

$$ Rewrite Equation(5,2,1) as $$\displaystyle G(x,y,y',y)=\underbrace{2\alpha x^m y^{n+1}-2x^{m+1}y^ny'}_{g(x,y,p)}+\underbrace{x^my^n}_{f(x,y,p)}y$$ Calculate the following terms in the exactness condition: $$\displaystyle f_x=mx^{m-1}y^n;\,f_{xx}=m(m-1)x^{m-2}y^n$$ $$\displaystyle f_{xy}=mnx^{m-1}y^{n-1};\,f_{xp}=0$$ $$\displaystyle f_y=nx^my^{n-1}$$ $$\displaystyle f_{yy}=n(n-1)x^my^{n-2};\,f_{yp}=0$$ $$\displaystyle g_x=2m\alpha x^{m-1}y^{n+1}-2(m+1)x^my^np;\,g_{xp}=-2(m+1)x^my^n$$ $$\displaystyle g_y=2\alpha (n+1)x^{m}y^{n-1}-2nx^{m+1}y^{n-1}p;\,g_{yp}=-2nx^{m+1}y^{n+1}$$ $$\displaystyle g_p=-2x^{m+1}y^n;\,g_{pp}=0$$ Recall the 2nd exactness condition for L2-ODE-VC:
 * $$\displaystyle x^m y^n(y''-2xy'+2\alpha y)=0$$
 * $$\displaystyle (Equation\;5.2.4)
 * $$\displaystyle (Equation\;5.2.4)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle f_{xx}+2pf_{xy}+p^2f_{yy}=g_{xp}+pg_{yp}-g_y$$
 * $$\displaystyle (Equation\;5.2.5)
 * $$\displaystyle (Equation\;5.2.5)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$ Substituting those calculated terms into Equation(5,2,5) and Equation(5.2.6). The 1st exactness condition Equation(5,2,5) is $$\displaystyle m(m-1)x^{m-2}y^n+2p\cdot mnx^{m-1}y^{n-1}+p^2n(n-1)x^my^{n-2}=-2(m+1)x^my^n+p(-2nx^{m+1}y^{n-1})-2\alpha (n+1)x^my^n+2nx^{m+1}y^{n-1}p$$ The 2nd exactness conditionEquation(5.2.6) is $$\displaystyle 0+p\cdot 0+2nx^my^{n-1}=0$$ It is obvious that only if $$\displaystyle n=0$$, the 2nd exactness condition Equation(5.2.6) is satisfied. Substitute $$\displaystyle n=0$$ into Equation(5,2,5), we can obtain $$\displaystyle m(m-1)x^{m-2}=-2(m+1)x^m-2\alpha x^m$$ It means $$\displaystyle m(m-1)=0$$ and $$\displaystyle m+1+\alpha =0$$ We choose $$\displaystyle m=1,\alpha =-2 \,$$ to make the factor h(x,y) not zero.
 * $$\displaystyle f_{xp}+pf_{yp}+2f_y=g_{pp}$$
 * $$\displaystyle (Equation\;5.2.6)
 * $$\displaystyle (Equation\;5.2.6)
 * }
 * }

Part (3) Verify the first three terms of Hermite polynomials. Letting $$\displaystyle y=H_0(x)=1$$ $$\displaystyle y'=H_0'(x)=0$$ $$\displaystyle y=H_0(x)=0$$ Substitute them into the Hermite differential equation Equation(5.2.3): $$\displaystyle 0-2x\cdot 0+0\cdot 1=0$$ Therefore, the first term of Hermite polynomial $$\displaystyle H_0(x)=1$$ is the homogeneous solution of the Hermite differential equation Equation (5.2.3).

Letting $$\displaystyle y=H_1(x)=2x$$ $$\displaystyle y'=H_1'(x)=2$$ $$\displaystyle y=H_1(x)=0$$ Substitute them into the Hermite differential equation Equation(5.2.3): $$\displaystyle 0-2x\cdot 2+2\cdot 2x=0$$ Therefore, the second term of Hermite polynomial $$\displaystyle H_1(x)=2x$$ is the homogeneous solution of the Hermite differential equation Equation (5.2.3).

Letting $$\displaystyle y=H_2(x)=4x^2-2$$ $$\displaystyle y'=H_2'(x)=8x$$ $$\displaystyle y=H_2(x)=8$$ Substitute them into the Hermite differential equation Equation(5.2.3): $$\displaystyle 8-2x\cdot 8x+4\cdot (4x^2-2)=0$$ Therefore, the third term of Hermite polynomial $$\displaystyle H_2(x)=4x^2-2$$ is the homogeneous solution of the Hermite differential equation Equation (5.2.3).

= R*5.4 Find $$y_{xxxxx}$$ in terms of derivatives of y with respect to time= == Given == Transformation of variables: $$\displaystyle x=e^t$$. The derivatives of y with respect to x are: $$\displaystyle y_x=e^{-t} \, y_t$$ $$\displaystyle y_{xx}=e^{-2t} \, (y_{tt}-y_t)$$ $$\displaystyle y_{xxx}=e^{-3t} \, (y_{ttt}-3t_{tt}+2y_t)$$ $$\displaystyle y_{xxxx}=e^{-4t} \,(y_{tttt}-6y_{ttt}+11y_{tt}-6y_t)$$

Find
Find $$\displaystyle y_{xxxxx}$$ in terms of the derivatives of y with respect to t.

Solution
Recall that $$\displaystyle \frac{dt}{dx}=\left(\frac{dx}{dt}\right)^{-1}=e^{(-t)}$$ The derivative is
 * $$\displaystyle y_{xxxxx}=\frac{d}{dx}\,(y_{xxxx})=\frac{dy_{xxxx}}{dt}\cdot \frac{dt}{dx}$$
 * $$\displaystyle =\frac{d}{dt}\left[e^{(-4t)}\,(y_{tttt}-6y_{ttt}+11y_{tt}-6y_t)\right]\cdot e^{-t}$$
 * $$\displaystyle =e^{-t}\cdot\left[\frac{d}{dt}(e^{-4t}\cdot(y_{tttt}-6y_{ttt}+11y_{tt}-6y_t)+e^{-4t}\cdot \frac{d}{dt} (y_{tttt}-6y_{ttt}+11y_{tt}-6y_t)\right]$$
 * $$\displaystyle =e^{-t}\cdot[-4e^{-4t}\cdot (y_{tttt}-6y_{ttt}+11y_{tt}-6y_t)+e^{-4t}\cdot (y_{ttttt}-6y_{tttt}+11y_{ttt}-6y_{tt})$$

 * $$\displaystyle =e^{-5t}(y_{ttttt}-10y_{tttt}+35y_{ttt}-50y_{tt}+24y_t)$$
 * $$\displaystyle$$

title: User:EGM6321.f11.team1.pan/HW5


text:
User:Egm6321.f11.team2.kim/PR5

=R*5.2 - Verify the second exactness for Legendre and Hermite equations =

Given
Legendre equation: $$\displaystyle G = (1 - x^2)y'' - 2xy' + n(n+1)y = 0 $$ (5.2.1)  Hermite equation: $$\displaystyle y'' - 2xy' + 2ny = 0 $$ (5.2.2)

Find
1. Verify the exactness of the designated L2-ODE-VC. 2. If eq (5.2.2) is not exact, check whether it is in power form, and see whether if it can be made exact using IFM $$\displaystyle \begin{align} H_0(x) &=1 \\ H_1(x) &= 2x \\ H_2(x) &= 4x^2-2 \end{align} $$ (5.2.2.2) 3. Verify the above equations are homogeneous solutions of the eq (5.2.2).

First Part
For Legendre equation by using first method: The first exactness condition is that

$$\displaystyle G= g(x,y,p) + f(x,y,p)y'' = 0 $$ $$\displaystyle G = \underbrace {(1 - x^2)}_{f(x,y,p)}y'' - \underbrace {2xp + n(n+1)y}_{g(x,y,p)} = 0 $$ this equation satisties the first exactness condition. In order to satisfy 2nd exactness condition, $$\displaystyle f_{xx} + 2pf_{xy}+p^2f_{yy} = g_{xp} + pg_{yp} - g_{y} $$ (5.2.3) $$\displaystyle f_{xp} + pf_{yp} + 2f_y = g_{pp} $$ (5.2.4) $$\displaystyle f(x,y,p) = (1-x^2) $$  $$\displaystyle g(x,y,p) = -2xp + n(n+1)y $$ $$\displaystyle \begin{align} f_{xx} &=-2 \\ f_{xy} &= 0 \\ f_{yy} &= 0 \\ g_{xp} &= -2 \\ g_{yp} &= 0 \\ g_y   &= n(n+1) \\ f_{xp} &= 0 \\ f_{yp} &= 0 \\ f_y &= 0 \\ g_{pp} &= 0 \\ \end{align} $$ From the above values, eq (5.2.3) and eq (5.2.4) are following that $$\displaystyle -2 = -2 + 0 - n(n+1), \,\,\,\,0=0 $$

It shows that eq (5.2.1) satisfies the 2nd exactness condition when n=0 or n= -1 For legendre equation by using second method:  $$\displaystyle g_0 - \frac {dg_1}{dx} + \frac {d^2g_2}{dx^2}=0$$  (5.2.5) $$\displaystyle \begin{align} g_0 &= \frac {\partial G}{\partial y^{(0)}} = n(n+1) \\ g_1 &= \frac {\partial G}{\partial y^{(1)}} = -2x \\ g_2 &= \frac {\partial G}{\partial y^{(2)}} = (1-x^2) \\ \frac {dg_1}{dx} &= -2 \\ \frac {d^2g_2}{dx^2} &= -2 \\ \end{align} $$ From above values, eq 5.2.5 is following that n(n+1) + 2 - 2 = 0 It also shows that this equation satisfies the second exactness when n=0. For Hermite equation by using first method: Hermite equation satisfies the first exactness condition due to following form. $$\displaystyle \underbrace {1}_{f(x,y,p)}\cdot y'' \underbrace {-2xp + 2ny}_{g(x,y,p)} = 0 $$ In order to satisfy 2nd exactness condition, the equation satisfies eq (5.2.3) and eq (5.2.4). $$\displaystyle f(x,y,p) = 1 \,\,\,\, g(x,y,p) = -2xp + 2np $$ $$\displaystyle \begin{align} f_{xx} &= 0 \\ f_{xy} &= 0 \\ f_{yy} &= 0 \\ g_{xp} &= -2 \\ g_{yp} &= 0 \\ g_y   &= 2n \\ f_{xp} &= 0 \\ f_{yp} &= 0 \\ f_y &= 0 \\ g_{pp} &= 0 \\ \end{align} $$ Eq (5.2.3) and Eq(5.2.4) is following that $$\displaystyle 0 + 0 + 0 = -2 + 0 + -2n \,\,\,\,2n = -2 \, \,\,\,\, n= -1 $$ $$\displaystyle 0=0 $$ It shows that eq (5.2.2) satisfies the 2nd exactness condition when n=-1. For Hermite equation by using second method: $$\displaystyle f_0 - \frac {df_1}{dx} + \frac {d^2f_2}{dx^2}=0$$ (5.2.6) $$\displaystyle \begin{align} f_0 &= \frac {\partial F}{\partial y^{(0)}} = 2n \\ f_1 &= \frac {\partial F}{\partial y^{(1)}} = -2x \\ f_2 &= \frac {\partial F}{\partial y^{(2)}} = 1 \\ \frac {df_1}{dx} &= -2 \\ \frac {d^2f_2}{dx^2} &= 0 \\ \end{align} $$ From the above values, eq 5.2.6 is following that 2n + 2 = 0 n= -1 It shows that this equation satisfies 2nd exactness condition when n= -1

Second Part
$$\displaystyle h(x,y) = x^m y^n$$ n in the eq 5.2.2 is replaced with b to avoid some mistakes. $$\displaystyle x^m y^n[y'' - 2xy' + 2ny]= 0 $$ $$\displaystyle \underbrace {x^m y^n}_{f(x,y,p)}y'' \underbrace {-2x^{m+1}y^ny'+2bx^my^{n+1}}_{g(x,y,p)}=0 $$ This equation satisfies the first exactness condition due to form of above equation. $$\displaystyle f(x,y,p) = x^my^n$$ $$\displaystyle g(x,y,p) = -2x^{m+1}y^np + 2bx^my^{n_1}$$ $$\displaystyle \begin{align} f_x &= mx^{m-1}y^n \\ f_{xx} &=m(m-1)x^{m-2}y^n \\ f_{xy} &= nmx^{m-1}y^{n-1} \\ f_{yy} &= n(n-1)x^my^{n-2} \\ g_x &= -2(m+1)x^my^np + 2bmx^{m-1}y^{n+1} \\ g_{xp} &= -2(m+1)x^my^n \\ g_{yp} &= -2nx^{m+1}y^{n-1} \\ g_y   &= -2nx^{m+1}y^{n-1}p+2b(n+1)x^my^n \\ f_{xp} &= 0 \\ f_{yp} &= 0 \\ f_y &= nx^my^{n-1} \\ g_{pp} &= 0 \\ \end{align} $$ From the above values, Eq 5.2.4 is following that $$\displaystyle 2nx^my^{n-1}=0 $$ To satisfy the above equation, n=0. From the above values with n=0, Eq 2.5.3 is following that $$\displaystyle m(m-1)x^{m-2} = -2(m+1)x^m -2bx^m $$ $$\displaystyle \frac {m(m-1)}{x^2}x^m = -2(m+1)x^m - 2bx^m $$ $$\displaystyle m(m-1) = -2(m+1)x^2 - 2bx^2 $$ $$\displaystyle m^2-m = -2mx^2 - 2x^2 - 2bx^2 $$ $$\displaystyle m^2-m+2mx^2+2x^2+2bx^2 $$ $$\displaystyle m^2 + (2x^2-1)m + 2x^2(1+b) = 0 $$ $$\displaystyle m = \frac {-(2x^2-1)\pm \sqrt{(2x^2-1)^2-4\cdot 2x^2(1+b)}}{2} $$ When m satisfies the above relationship, Eq 5.2.2 would be exact. Generic form of power series : $$\displaystyle y(x) = \sum^{oo}_{n=0} a_n\cdot x^n $$ (5.2.7) $$ y'(x) = \sum^{oo}_{n=1} a_n\cdot nx^{n-1}$$  (5.2.8) $$ y''(x) = \sum^{oo}_{n=2} a_n\cdot n\cdot(n-1)x^{n-2}$$ (5.2.9) From eq (5.2.7)~(5.2.9), eq (5.2.2) is following that $$\displaystyle \sum^{oo}_{n=2} a_n\cdot n\cdot(n-1)x^{n-2} -2x\sum^{oo}_{n=1} a_n\cdot nx^{n-1}+2b\sum^{oo}_{n=0} a_n\cdot x^n =0 $$ $$\displaystyle \sum^{oo}_{n=2} a_n\cdot n\cdot(n-1)x^{n-2} -\sum^{oo}_{n=1} 2n\cdot a_n\cdot nx^{n}+\sum^{oo}_{n=0} 2b\cdot a_n\cdot x^n =0 $$ $$ \displaystyle \sum^{oo}_{n=0} (n+2)(n+1)a_{n+2}\cdot x^{n} -\sum^{oo}_{n=0} 2n\cdot a_n\cdot nx^{n}+\sum^{oo}_{n=0} 2b\cdot a_n\cdot x^n =0 $$ (5.2.10) From Eq (5.2.10), $$\displaystyle a_{n+2}= \frac {2(n-b)}{(n+2)(n+1)}a_n = 0$$ (5.2.11) For n=0,1,2,3,...... This is valid only positive integer of b (1,2,3,4,.......)

Third Part
From eq 5.2.2.2 Boundary condition of eq 5.2.2.2 is that where b=0, y(0)= 1, y'(0)=0 where b=1, y(0)= 0, y'(0)=2 where b=2, y(0)= -2, y'(0)= 0 where b=1, and the initial conditions are given as $$\displaystyle y(0)= a_0 = 0, y'(0) = a_1 = 2$$ It shows that all even coefficients will be equal to zero because each coefficient is a multiple of its second predecessor. $$\displaystyle a_0 = a_2 = a_4 = a_6 = a_8 = a_10 ..... = 0$$ From eq (5.2.11), $$\displaystyle a_3 = \frac {2(1-1)}{(1+2)(1+1)}a_1 = 0 $$ It shows that $$\displaystyle a_5 = a_7 = a_9 = .... = 0$$ Only a1 = 2. Thus, from the eq 5.2.7, $$\displaystyle y(x) = H_1(x) = a_0\cdot x^0 + a_1\cdot x^1 + a_2\cdot x^2 + ...... a_n\cdot x^n = x $$ Where b=2, and the initial conditions are given as $$\displaystyle y(0)= a_0 = -2, y'(0) = a_1 = 0$$ eq 5.2.11 is following that $$\displaystyle a_2 = \frac {2(0-2)}{(0+2)(0+1)}a_0 = 4$$ $$\displaystyle a_4 = \frac {2(2-2)}{(2+2)(2+1)}a_2 = 0 $$ It shows that $$\displaystyle a_4 = a_6 = a_8 = .... = 0$$ Only a0 = -2 and a2 = 4

Thus, from the eq 5.2.7 $$\displaystyle y(x) = H_2(x) = a_0 \cdot x^0 + a_1\cdot x^1 + a_2\cdot x^2 + ........ a_n\cdot x^n = 4x^2 -2 $$

Where b=0, and the initial conditions are given as $$\displaystyle y(0)= a_0 = 1, y'(0) = a_1 = 0$$ We can not use eq 5.2.11 because the eq 5.2.11 is only valid for positive integer number. So We can calculate other ways. $$\displaystyle y'' - 2xy' = 0$$ $$\displaystyle y(x)= c1 + \frac {1}{2}\sqrt{\pi}\cdot c2\cdot erfi(x) $$ (5.2.12) From the initial values, c1= 0, c2=0. eq 5.2.12 is that
 $$\displaystyle y(x)=H_0(x) = 1$$ $$\displaystyle \begin{align} H_0(x) &=1 \\ H_1(x) &= 2x \\ H_2(x) &= 4x^2-2\\ \end{align} $$

These results show that the equations in (5.2.2.2) are homogeneos solutions of this Hermite differentail equation.

=R*5.8 - Particular Solution by using Variation Parameters Method =

Given
$$\displaystyle y(x) = A(x)y_H(x)$$
 (5.8.1) $$\displaystyle y' + a_0(x)y = 0 $$ $$\displaystyle y_H(x) = exp[-\int^x a_0(s)ds] $$

Differential eq is that $$\displaystyle y' + P(x)y = Q(x) $$
 (5.8.2) $$\displaystyle y_H(x) = exp[-\int P(x)dx] $$
 (5.8.3)

Find
Find particular solution Yp

Solution
$$\displaystyle y(x) = A(x)y_H(x) $$ $$\displaystyle y(x)' = A(x)'y_H(x) + A(x)y_H(x)' $$
 (5.8.4) From the eq 5.8.3, $$\displaystyle y_H(x)'= -P(x)\cdot exp[-\int P(x)dx]=-P(x)y_H(x)$$ Thus, eq 5.8.4 is following that
 $$\displaystyle y(x)' = A(x)'\cdot exp[-\int P(x)dx] - P(x)\cdot exp[-\int P(x)dx]A(x) = exp[-\int P(x)dx]\cdot[A(x)'-P(x)A(x)]$$
 Plugging the above equation into Eq 5.8.2, $$\displaystyle exp[-\int P(x)dx]\cdot[A(x)'-P(x)A(x)] + P(x)A(x)\cdot exp[-\int P(x)dx]= Q(x) $$ $$\displaystyle exp[-\int P(x)dx](A(x)'-P(x)A(x)+A(x)P(x))=Q(x)$$ $$\displaystyle exp[-\int P(x)dx]\cdot A(x)' = Q(x)$$ $$\displaystyle A(x)' = \frac {Q(x)}{exp[-\int P(x)dx]} = Q(x)\cdot exp[\int P(x)dx] $$ $$\displaystyle A(x) = \int Q(x) \cdot exp[\int P(x)dx]dx $$ Thus, paricular solution is that $$\displaystyle y(x) = \int Q(x) \cdot exp[\int P(x)dx]dx \cdot exp[-\int P(x)dx] $$

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

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Talk:Physics/Essays/Fedosin/Fine structure constant

It seems the article may be deleted if this is necessary - it repeats Wikipedia version, and original or useful information is very small and may be moved to article Stoney scale. Fedosin 18:42, 28 October 2011 (UTC)

The article is prepared now to move to a mainspace page, Fine structure constant. It has some explanatory content, such as a definition, with links to pages that discuss related topics, and reference to Wikipedia. Part of content was distributed to articles Stoney scale and Selfconsistent gravitational constants. Fedosin 17:47, 1 November 2011 (UTC)


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Motivation and emotion/Book/2011/Gambling/Glossary

Affiliation
Affiliation is 'the anxious need to establish, maintain and restore interpersonal relations.' (Reeve, 2009, p. 193.)

Eudaimonic well-being
The experience of seeking out challenges,exerting effort, being fully engaged and experiencing flow in what one is doing, acting on one's true values, and feeling fully alive and authentic (Ryan & Deci, 2001; Reeve, 2009, p. 443).

External Locus of Control (ELOC)
Simply put, individuals with an ELOC believe that the outcome of their gambling activity is under the control of some outside force. That force could be chance, fate, God or luck. This delusory thinking is reinforced because gamblers believe that, as their losses outweigh their wins, there is some external reason confounding the outcome of their efforts (de Stadelhofen, Aufrère,  Besson, & Rossier, 2009).

Extrinsic motivation
Extrinsic motivation (EM) is 'an environmentally created reason (i.e. to get money or to end criticism) to initiate or persist in an action.' (Reeve,2009, p. 113.)

EM is closely identified with an external locus of control i.e. the activity is being undertaken following an instruction or direction to perform the activity, or, for example, to impress someone other than the self. EM has four sub-types: External regulation (where the importance of the reward or punishment, is paramount), Introjection (involves the ego and is about approval from others - and to an extent from the self), Identification (being aware of the value of the activity and includes some level of self endorsement of goals), and  Integration (where the regulations governing extrinsic activity are understood/integrated in relation to the other needs and values of the individual) (Ryan & Deci, 2000).

Gambling
1. a. to play games of chance for money, esp. for unduly high stakes; to stake money (esp. to an extravagant amount) on some fortuitous event. As the word is (at least in serious use) essentially a term of reproach, it would not ordinarily be applied to the action of playing for stakes of trifling amount, except by those who condemn playing for money altogether (Oxford English Dictionary (OED) 2011). This definition treads a delicate line between the two types of gambling considered in this chapter, Pathological gambling and Recreational (or Social) gambling. The American Psychiatric Association (APA), (2008) includes Pathological gambling as an 'Impulse-Control Disorder Not Elsewhere Classified' (pp. 671 - 674). Hence, (i) Recreational or Social gambling is not considered a disorder, and (ii) this perhaps raises questions about the veracity of assertions that Pathological gambling is an addiction.

Hedonic well-being
The experience of pleasure, the absence of problems and the living of a relaxed and good life (Reeve, 2009, p. 443.

Intrinsic motivation
Intrinsic motivation is 'a natural motivation that emerges spontaneously out of peoples psychological needs for autonomy, competence and relatedness.' (Reeve, 2009, p. 112.) A person is moved to act for the fun or challenge of doing so, not because of any external pressure (Ryan & Deci, 2000).

Pathological gambling
Pathological gambling (PG) refers to persistent and recurrent maladaptive gambling behaviour that disrupts, personal, family or vocational pursuits. (American Psychiatric Association (APA), 2008, p. 671.)

Pitfalls and treatment of pathological gambling
The problem lies in establishing whether the individual has control over the situation in which she or he places themself. Gambling in a semi-organised manner with friends is one way to avoid excessive gambling. For example, Friedman, Deci,  Elliot, Moller, and Aarts (2009) found that when extrinsically motivated individuals observed intrinsically motivated individuals engaging in a task, the extrinsically motivated individuals adopted  an intrinsic approach to the same activity when presented with that activity. That type of priming is a favourable sign, however this may not prove successful on every occasion, and it is unlikely that a problem gambler will successfully cease gambling for any prolonged period of time without continued support.

Some researchers e.g. Chóliz (2009) have suggested Operant Conditioning to change gambling behaviour, that however requires changing e.g. the reward frequency of slot machines.

By far the most difficult decision for people in this category is to acknowledge that they have a problem with gambling. The psychological community recognises that significantly more research is needed to identify the different reasons why some people become problem gamblers and other do not (Ladd & Pertry, 2002). It is also recognised that a better understanding of gender specific factors is necessary to enable more suitable and adaptive methods for treatment and prevention of pathological gambling (Ladd & Petry, 2002). Research suggests that, in the Australian Capital Territory (ACT), only a small proportion of gamblers (21%) seek help for their gambling problem, rather they seek assistance with relationship problems, for financial problems, and access welfare and food subsidies, for example (Carroll, Davidson, Marsh & Rogers, 2011).

Recreational gambling
Recreational gambling (RG) typically occurs with friends or colleagues and lasts for a limited period of time, with predetermined acceptable losses. (APA, 2008, p. 673.)

Reactance
The psychological and behavioural attempt at re-establishing ("reacting" against) an eliminated or threatened freedom. (Reeve, 2009, p. 256.)

Risk
Risk has a number of interpretations but primarily refers to the choices an individual makes. For example (i) Risk as in, jeopardising something of value e.g. money, relationship; or (ii) Risk as in, the potential for positive or negative outcomes. The quality of Risk when used in this chapter should be evident to the reader.

South Oaks Gambling Screen
For Self administered test, click http://www.in.gov/judiciary/ijlap/docs/south-oaks-gambling-screen.pdf.

Urge
'The urge to gamble is a physiological, psychological and emotional motivational state often associated with continued gambling.' (Raylu & Oei, 2004, p. 100.)

References (Glossary)
American Psychiatric Association (2008). Pathological Gambling, Diagnostic and Statistical Manual of Mental Disorders, (4th ed., Text Revision). Arlington, Va: APA Publishing.

Chóliz, M. (2010). Experimental analysis of the game in pathological gamblers: Effect of the immediacy of the reward in slot machines. Journal of Gambling Studies, 26, 2, 249-256.

Carroll, A., Davidson, T., Marsh, R., & Rodgers, B. (2011). Help-seeking and uptake of services amongst people with gambling problems in the ACT, 54 & 68. Retrieved from http://lyceum.anu.edu.au/wp-content/blogs/3/uploads//ServiceUseReport_26%20oct%202011%20%20prepared%20for%20printing.pdf

de Stadelhofen, F. M., Aufrère, L. Besson, J., and Rossier, J. (2009). Somewhere between illusion of control and powerlessness: Trying to situate the pathological gambler's locus of control. International Journal of Clinical and Health Psychology, 9, 1, 117-126.

Friedman, R., Deci, E. L., Elliot, A. J., Moller, A. C., & Aarts, H. (2010). Motivational synchronicity: Priming motivational orientations with observations of others' behaviors. Motivation and Emotion, 34, 34-38.

Ladd, G. T., & Petry, N. M. (2002). Gender differences among pathological gamblers seeking treatment. Experimental and Clinical Psychopharmacology, 10, 3, 302-309.

Oxford English Dictionary, (2011). Gambling: Definition. Oxford University Press, online. Retrieved from: http://www.oed.com.ezproxy1.canberra.edu.au/view/Entry/76447

Reeve, J. (2009). Understanding motivation and emotion, (5th ed.). Danvers, MA: John Wiley & Sons.

Ryan, R. M., & Deci, E. L. (2001). On happiness and human potentials: A review of research on hedonic and eudaimonic well-being. Annual Review of Psychology, 52, 141-166.

Ryan, R. M., & Deci, E. L. (2000). Intrinsic and extrinsic motivations: Classic definitions and new directions. Contemporary Educational Psychology, 25, 54-67.

Raylu, N., & Oei, T. P. S. (2004). The gambling urge scale: Development, confirmatory factor validation, and psychometric properties. Psychology of Addictive Behaviors, 18, 2, 100-105.


text:
University of Canberra/RCC2012/Participants/Scholarship

Scholarship information
Wikimedia Australia has allocated $10,000 to grants for attendance to the RecentChangesCamp in Canberra in January 2012. While the dates are not confirmed yet, scholarship applications are open.

The decision can be found in the committee meeting minutes: http://wikimedia.org.au/wiki/Meeting:Committee_%282011-10-16%29#RecentChangesCamp_2012

Anyone can apply by emailing comm[@]wikimedia[.]org[.]au, providing a bit of background about yourself (wiki bio and/or real world bio, as applicable), why you would like to attend, and how much financial assistance you are requesting.

We are especially interested in assisting people from states and nearby countries who, without a grant, would be unlikely to attend due to travel costs. (WA, Tas, NT, NZ, etc)

Preference will be given to members, but please pass this on to non-members who you think would benefit from attendance.

Early applications will be more likely to be successful, as that allows flight costs to be locked in early & more cheaply.

Back to RCC2012 main page...


text:
Talk:Motivation and emotion/Book/Self-efficacy

Comment
Great plan alicia :) I like what youve included and looks to be a very interesting chapter that everyone can relate too. cant wait to read more! Courtney.reis 07:41, 29 October 2011 (UTC)

This chapter has great potential. I think you've done a great job in covering the important questions that relate to the topic. I dont know much about self-efficacy, so i am looking forward to what you write! Atice 01 03:54, 30 October 2011 (UTC)

Hey Alicia made some minor spelling edits on ur page, feel free to undo if I am wrong! I like the new bits youve put on, should be great once you expand! Courtney.reis 04:51, 2 November 2011 (UTC)

Hi Alicia. Your outline looks really comprehensive. Looking like it's going to be a good chapter. I like how you have main points in the boxes on the side, these are a great (and helpful!) addition.

Hi Alicia, Looking good. This article may be of use for the section 'Self efficacy and motivation' http://web.ebscohost.com.ezproxy1.canberra.edu.au/ehost/pdfviewer/pdfviewer?sid=89530e19-7a30-411e-9ec5-5c9756487980%40sessionmgr114&vid=4&hid=108

I have a copy if that link does not work AlexMC 05:05, 4 November 2011 (UTC)

Hey Alicia, great planning and execution :) Do you need any feedback or help in particular for your chapter? --Jaybay 08:26, 5 November 2011 (UTC)

Thanks - yours look really fab. Think I stole the shape of one of your little boxes - they were good. Just wish mine was finished just now though. :) Jeanette 12:33, 6 November 2011 (UTC)

Hi Alicia. I know it's late - I just thought I would drop through and congratulate you! The page looks great, well done :) (Psych 125 21:47, 6 November 2011 (UTC))

Hi - this would make an excellent topic to adapt for school children - very well researched - well done - Magnolia

-- Jtneill - Talk - c 12:04, 28 November 2011 (UTC)

-- Jtneill - Talk - c 10:26, 10 December 2011 (UTC)


text:
User:Jessica Perkins/The Olympic Games as a Political Driver and its Impacts on the Australian Sports System Presentation

=Outline= This presentation uses audio and slides to discuss how politics within the Olympic Games have the potiential to impact sport at every level. This presentation looks into the development of the Australian sport system and how political events that have occurred in the history of the Olympic Games have determined where sport in Australia is today. When discussing the nature of sport as big business and considering the politics that exist in modern sport, I have chosen to focus on the Olympic Games as we approach the 2012 Games in London.

=Links=

http://www.archive.org/details/TheOlympicGamesAsAPoliticalDriverAndTheImpactsOnTheAustralianSport


text:
Building model

A building model is either a physical (real) or virtual (computer) representation of a building. Very often, the physical model is smaller than the original (scale model).

There are three basic types of building models: architectural, structural and mathematical (virtual).

Architectural building model
An architectural model is a type of a physical model of a structure to study an architectural design or to share design ideas to clients, committees, and the general public. Architectural models are a tool which may be used for show, presentation, fundraising, getting permits, and sale purposes. Such models are a good tool for three-dimensional understanding of a design, used by architects, interior designers and exhibit designers.

Structural building model
Scale structural building model is the best choice for the engineers who need to test the likely performance of a designed structure at an early stage of development without paying for a full-sized prototype.

Virtual building model
Virtual model is a digital model of the object (typically greatly simplified) that can be used in a computer simulation or virtual reality.

The most common examples of virtual models are those created in 3D for the purpose of visualisation - seeing how the structure looks before it is built. The field of architecture has greatly popularized the use of virtual models to animate fly-throughs of yet-to-be-built buildings.

title: Building model


text:
User talk:KaryAnca

Speak Math Now
Hello KaryAnca! I do apologize for taking so long to respond to your question. My Speak Math Now! course is open for enrollment, however, my laptop is currently being worked on, and thus it may take a little while for us to actually begin. I am very happy that you are interested in the course, but I would like to first ask you these questions that I hope you can respond to:

 * What is your prior experience with math?
 * Is there a specific topic(s) that you would like to improve on?
 * Will you devote yourself to the course? (I do know that this is a very long course, which is why I have the option for all of my students to start in any particular semester that you would like, and end with any semester. However, please know: You cannot quit in the middle of a specific semester.)

If you can answer these questions confidently, then I would love to set up an orientation session with you once my computer is repaired. When this occurs, I will notify you immediately and we can get started. Until then, please bear with me and thank you again for taking interest in my course. I hope to be working with you very soon! --MrABlair23 13:01, 29 October 2011 (UTC)

No hurry. I'm currently taking another (certification) course which completes in December. To answer your questions:

What is your prior experience with math? I've taken some math courses in university, but it's been a few years.

Is there a specific topic(s) that you would like to improve on? No. Mainly, I'd like to be more precise when using mathematical language.

Will you devote yourself to the course? That's a challenging question. It's not the length of the course that concerns me, but the ability to schedule around other obligations. It's hard to judge the workload from the course outline, but it's certainly my intention to complete the course. KaryAnca 13:45, 29 October 2011 (UTC)

Thank you very much for responding so quickly, Kary. I completely understand as there are always other obligations to attend to, and I do need to update my course page a bit. If you would like, I could send you a course syllabus that completely outlines everything that will be going on throughout the course, and you can decide from there. Please let me know if you are interested in one, and if so, the correct method of sending it to you. Thank you very much, and I look forward to hearing from you! --MrABlair23 21:00, 29 October 2011 (UTC)

Hello KaryAnca! First of all, I would like to thank you for being so patient with me. My laptop has been repaired, and I am ready to go if you are. If you would like, we can set up an orientation session whenever you would like. I just need to know your time zone along with dates and times that you may be available. I live in the Eastern Standard Time Zone of the United States, and my hours are from 3:00 PM - 10:00 PM EST from Monday - Friday, and 10:00 AM - 10:00 PM from Saturday - Sunday. This week is a bit hectic for me, and thus, I was wondering if you would be comfortable for any of these times on Monday, November 7. If you have any other questions, please do not hesitate to ask and I am looking forward to hearing from you! --MrABlair23 23:51, 1 November 2011 (UTC)

title: User talk:KaryAnca


text:
User:Andyn6361

These are the projects I have been in for the Introduction to Engineering Design Class at Howard Community College during the Fall Semester of 2011

Poppit
Link to Poppit userpage

Car Hack
Link to Car Hack userpage

Helicopter
Link to Helicopter userpage

title: User:Andyn6361


text:
User:Egm6321.f11.team2/HW5

=R*5.1 - Equivalence of Two Forms of the Second Exactness Conditions for N2-ODEs, Method 2=

Given
The second exactness condition for N2-ODEs is given by

$$g_0 - \frac{dg_1}{dx} + \frac{d^2g_2}{dx^2} = 0$$

where

$$g_j := \frac{\partial G}{\partial y^{(j)}}, \quad y^{(i)} := \frac{\partial^i y}{\partial x^i}, \quad\text{and}\quad G := g + fy''$$

Find
Show that the form above is equivalent to the form of the second exactness condition for N2-ODEs given by

$$\displaystyle \begin{align} f_{xx} + 2pf_{xy} + p^2f_{yy} &= g_{xp} + pg_{yp} - g_y\\ f_{xp} + pf_{yp} + 2f_y &= g_{pp} \end{align} $$

Solution
Let $$p := y'$$ and $$q:= y''$$.

Then,

$$\displaystyle g_1 = \frac{\partial}{\partial p} (g + fy'') = g_p + f_pq$$

$$\displaystyle g_2 = \frac{\partial}{\partial q} (g + fy'') = f$$

$$\displaystyle \begin{align} \frac{d^2g_2}{dx^2} =& \frac{d}{dx} (f_x + f_yp + f_pq) \\ =&\qquad f_{xx} + f_{xy}p + f_{xp}q\\ &+ f_{yx}p + f_{yy}p^2 + f_{yp}pq + f_yq \\ &+ f_{px}q + f_{py}pq + f_{pp}q^2 + f_pq' \end{align} $$

Plugging 5.1.1, 5.1.2, and 5.1.3 into

$$\displaystyle g_0 - \frac{dg_1}{dx} + \frac{d^2g_2}{dx^2} = 0 $$

yields

$$\displaystyle ( f_{xx} + 2pf_{xy} + p^2f_{yy} - g_{xp} - pg_{yp} + g_y ) + ( f_{xp} + pf_{yp} + 2f_y - g_{pp} ) q = 0$$

As noted in the lecture notes, since 1 and q are linearly independent, the only way for the above equation to be true is for their coefficients to be equal to zero.

Thus,

$$\displaystyle \begin{align} f_{xx} + 2pf_{xy} + p^2f_{yy} &= g_{xp} + pg_{yp} - g_y \\ f_{xp} + pf_{yp} + 2f_y &= g_{pp} \end{align} $$

as required.

=R*5.2 - Verify the second exactness for Legendre and Hermite equations =

Given
Legendre equation: $$\displaystyle G = (1 - x^2)y'' - 2xy' + n(n+1)y = 0 $$ (5.2.1)  Hermite equation: $$\displaystyle y'' - 2xy' + 2ny = 0 $$ (5.2.2)

Find
1. Verify the exactness of the designated L2-ODE-VC. 2. If eq (5.2.2) is not exact, check whether it is in power form, and see whether if it can be made exact using IFM $$\displaystyle \begin{align} H_0(x) &=1 \\ H_1(x) &= 2x \\ H_2(x) &= 4x^2-2 \end{align} $$ (5.2.2.2) 3. Verify the above equations are homogeneous solutions of the eq (5.2.2).

First Part
For Legendre equation by using first method: The first exactness condition is that

$$\displaystyle G= g(x,y,p) + f(x,y,p)y'' = 0 $$ $$\displaystyle G = \underbrace {(1 - x^2)}_{f(x,y,p)}y'' - \underbrace {2xp + n(n+1)y}_{g(x,y,p)} = 0 $$ this equation satisties the first exactness condition. In order to satisfy 2nd exactness condition, $$\displaystyle f_{xx} + 2pf_{xy}+p^2f_{yy} = g_{xp} + pg_{yp} - g_{y} $$ (5.2.3) $$\displaystyle f_{xp} + pf_{yp} + 2f_y = g_{pp} $$ (5.2.4) $$\displaystyle f(x,y,p) = (1-x^2) $$  $$\displaystyle g(x,y,p) = -2xp + n(n+1)y $$ $$\displaystyle \begin{align} f_{xx} &=-2 \\ f_{xy} &= 0 \\ f_{yy} &= 0 \\ g_{xp} &= -2 \\ g_{yp} &= 0 \\ g_y   &= n(n+1) \\ f_{xp} &= 0 \\ f_{yp} &= 0 \\ f_y &= 0 \\ g_{pp} &= 0 \\ \end{align} $$ From the above values, eq (5.2.3) and eq (5.2.4) are following that $$\displaystyle -2 = -2 + 0 - n(n+1), \,\,\,\,0=0 $$

It shows that eq (5.2.1) satisfies the 2nd exactness condition when n=0 or n= -1 For legendre equation by using second method:  $$\displaystyle g_0 - \frac {dg_1}{dx} + \frac {d^2g_2}{dx^2}=0$$  (5.2.5) $$\displaystyle \begin{align} g_0 &= \frac {\partial G}{\partial y^{(0)}} = n(n+1) \\ g_1 &= \frac {\partial G}{\partial y^{(1)}} = -2x \\ g_2 &= \frac {\partial G}{\partial y^{(2)}} = (1-x^2) \\ \frac {dg_1}{dx} &= -2 \\ \frac {d^2g_2}{dx^2} &= -2 \\ \end{align} $$ From above values, eq 5.2.5 is following that n(n+1) + 2 - 2 = 0 It also shows that this equation satisfies the second exactness when n=0. For Hermite equation by using first method: Hermite equation satisfies the first exactness condition due to following form. $$\displaystyle \underbrace {1}_{f(x,y,p)}\cdot y'' \underbrace {-2xp + 2ny}_{g(x,y,p)} = 0 $$ In order to satisfy 2nd exactness condition, the equation satisfies eq (5.2.3) and eq (5.2.4). $$\displaystyle f(x,y,p) = 1 \,\,\,\, g(x,y,p) = -2xp + 2np $$ $$\displaystyle \begin{align} f_{xx} &= 0 \\ f_{xy} &= 0 \\ f_{yy} &= 0 \\ g_{xp} &= -2 \\ g_{yp} &= 0 \\ g_y   &= 2n \\ f_{xp} &= 0 \\ f_{yp} &= 0 \\ f_y &= 0 \\ g_{pp} &= 0 \\ \end{align} $$ Eq (5.2.3) and Eq(5.2.4) is following that $$\displaystyle 0 + 0 + 0 = -2 + 0 + -2n \,\,\,\,2n = -2 \, \,\,\,\, n= -1 $$ $$\displaystyle 0=0 $$ It shows that eq (5.2.2) satisfies the 2nd exactness condition when n=-1. For Hermite equation by using second method: $$\displaystyle f_0 - \frac {df_1}{dx} + \frac {d^2f_2}{dx^2}=0$$ (5.2.6) $$\displaystyle \begin{align} f_0 &= \frac {\partial F}{\partial y^{(0)}} = 2n \\ f_1 &= \frac {\partial F}{\partial y^{(1)}} = -2x \\ f_2 &= \frac {\partial F}{\partial y^{(2)}} = 1 \\ \frac {df_1}{dx} &= -2 \\ \frac {d^2f_2}{dx^2} &= 0 \\ \end{align} $$ From the above values, eq 5.2.6 is following that 2n + 2 = 0 n= -1 It shows that this equation satisfies 2nd exactness condition when n= -1

Second Part
$$\displaystyle h(x,y) = x^m y^n$$ n in the eq 5.2.2 is replaced with b to avoid some mistakes. $$\displaystyle x^m y^n[y'' - 2xy' + 2ny]= 0 $$ $$\displaystyle \underbrace {x^m y^n}_{f(x,y,p)}y'' \underbrace {-2x^{m+1}y^ny'+2bx^my^{n+1}}_{g(x,y,p)}=0 $$ This equation satisfies the first exactness condition due to form of above equation. $$\displaystyle f(x,y,p) = x^my^n$$ $$\displaystyle g(x,y,p) = -2x^{m+1}y^np + 2bx^my^{n_1}$$ $$\displaystyle \begin{align} f_x &= mx^{m-1}y^n \\ f_{xx} &=m(m-1)x^{m-2}y^n \\ f_{xy} &= nmx^{m-1}y^{n-1} \\ f_{yy} &= n(n-1)x^my^{n-2} \\ g_x &= -2(m+1)x^my^np + 2bmx^{m-1}y^{n+1} \\ g_{xp} &= -2(m+1)x^my^n \\ g_{yp} &= -2nx^{m+1}y^{n-1} \\ g_y   &= -2nx^{m+1}y^{n-1}p+2b(n+1)x^my^n \\ f_{xp} &= 0 \\ f_{yp} &= 0 \\ f_y &= nx^my^{n-1} \\ g_{pp} &= 0 \\ \end{align} $$ From the above values, Eq 5.2.4 is following that $$\displaystyle 2nx^my^{n-1}=0 $$ To satisfy the above equation, n=0. From the above values with n=0, Eq 2.5.3 is following that $$\displaystyle m(m-1)x^{m-2} = -2(m+1)x^m -2bx^m $$ $$\displaystyle \frac {m(m-1)}{x^2}x^m = -2(m+1)x^m - 2bx^m $$ $$\displaystyle m(m-1) = -2(m+1)x^2 - 2bx^2 $$ $$\displaystyle m^2-m = -2mx^2 - 2x^2 - 2bx^2 $$ $$\displaystyle m^2-m+2mx^2+2x^2+2bx^2 $$ $$\displaystyle m^2 + (2x^2-1)m + 2x^2(1+b) = 0 $$ $$\displaystyle m = \frac {-(2x^2-1)\pm \sqrt{(2x^2-1)^2-4\cdot 2x^2(1+b)}}{2} $$ When m satisfies the above relationship, Eq 5.2.2 would be exact. Generic form of power series : $$\displaystyle y(x) = \sum^{oo}_{n=0} a_n\cdot x^n $$ (5.2.7) $$ y'(x) = \sum^{oo}_{n=1} a_n\cdot nx^{n-1}$$  (5.2.8) $$ y''(x) = \sum^{oo}_{n=2} a_n\cdot n\cdot(n-1)x^{n-2}$$ (5.2.9) From eq (5.2.7)~(5.2.9), eq (5.2.2) is following that $$\displaystyle \sum^{oo}_{n=2} a_n\cdot n\cdot(n-1)x^{n-2} -2x\sum^{oo}_{n=1} a_n\cdot nx^{n-1}+2b\sum^{oo}_{n=0} a_n\cdot x^n =0 $$ $$\displaystyle \sum^{oo}_{n=2} a_n\cdot n\cdot(n-1)x^{n-2} -\sum^{oo}_{n=1} 2n\cdot a_n\cdot nx^{n}+\sum^{oo}_{n=0} 2b\cdot a_n\cdot x^n =0 $$ $$ \displaystyle \sum^{oo}_{n=0} (n+2)(n+1)a_{n+2}\cdot x^{n} -\sum^{oo}_{n=0} 2n\cdot a_n\cdot nx^{n}+\sum^{oo}_{n=0} 2b\cdot a_n\cdot x^n =0 $$ (5.2.10) From Eq (5.2.10), $$\displaystyle a_{n+2}= \frac {2(n-b)}{(n+2)(n+1)}a_n = 0$$ (5.2.11) For n=0,1,2,3,...... This is valid for only positive integer of b (1,2,3,4,.......)

Third Part
From eq 5.2.2.2 Boundary condition of eq 5.2.2.2 is that where b=0, y(0)= 1, y'(0)=0 where b=1, y(0)= 0, y'(0)=2 where b=2, y(0)= -2, y'(0)= 0 where b=1, and the initial conditions are given as $$\displaystyle y(0)= a_0 = 0, y'(0) = a_1 = 2$$ It shows that all even coefficients will be equal to zero because each coefficient is a multiple of its second predecessor. $$\displaystyle a_0 = a_2 = a_4 = a_6 = a_8 = a_10 ..... = 0$$ From eq (5.2.11), $$\displaystyle a_3 = \frac {2(1-1)}{(1+2)(1+1)}a_1 = 0 $$ It shows that $$\displaystyle a_5 = a_7 = a_9 = .... = 0$$ Only a1 = 2. Thus, from the eq 5.2.7, $$\displaystyle y(x) = H_1(x) = a_0\cdot x^0 + a_1\cdot x^1 + a_2\cdot x^2 + ...... a_n\cdot x^n = x $$ Where b=2, and the initial conditions are given as $$\displaystyle y(0)= a_0 = -2, y'(0) = a_1 = 0$$ eq 5.2.11 is following that $$\displaystyle a_2 = \frac {2(0-2)}{(0+2)(0+1)}a_0 = 4$$ $$\displaystyle a_4 = \frac {2(2-2)}{(2+2)(2+1)}a_2 = 0 $$ It shows that $$\displaystyle a_4 = a_6 = a_8 = .... = 0$$ Only a0 = -2 and a2 = 4

Thus, from the eq 5.2.7 $$\displaystyle y(x) = H_2(x) = a_0 \cdot x^0 + a_1\cdot x^1 + a_2\cdot x^2 + ........ a_n\cdot x^n = 4x^2 -2 $$

Where b=0, and the initial conditions are given as $$\displaystyle y(0)= a_0 = 1, y'(0) = a_1 = 0$$ We can not use eq 5.2.11 because the eq 5.2.11 is only valid for positive integer number. So We can calculate other ways. $$\displaystyle y'' - 2xy' = 0$$ $$\displaystyle y(x)= c1 + \frac {1}{2}\sqrt{\pi}\cdot c2\cdot erfi(x) $$ (5.2.12) From the initial values, c1= 1, c2=0. eq 5.2.12 is that
 $$\displaystyle y(x)=H_0(x) = 1$$ $$\displaystyle \begin{align} H_0(x) &=1 \\ H_1(x) &= 2x \\ H_2(x) &= 4x^2-2\\ \end{align} $$

These results show that the equations in (5.2.2.2) are homogeneos solutions of this Hermite differentail equation.

Given
Given L4-ODE-CC,

$$\displaystyle X^{(4)} - K^4 X = 0 $$
 (5.3.1)

Assuming

$$\displaystyle X(x) = e^{(rx)} $$
 (5.3.2)

Where

$$\displaystyle r_{1,2} = \pm K $$

$$\displaystyle r_{3,4} = \pm i \, K $$
 (5.3.3)

Find
In terms of $$ \displaystyle \cos Kx, \sin Kx, \cosh Kx, \sinh kh$$

Find expression for $$ \displaystyle X(x)$$

Solution
According to the (5.3.3),

$$\displaystyle X(x) = C_1 e^{Kx} + C_2 e^{-Kx} + C_3 e^{iKx} + C_4 e^{-iKx} $$
 (5.3.4)

Since

$$ \displaystyle e^{iKx} = \cos Kx +i\sin Kx$$

$$ \displaystyle e^{Kx} = \cosh Kx + \sinh Kx$$

Substituting the above relations to (5.3.4)

$$\displaystyle X(x) = C_1 (\cosh Kx \ + \sinh Kx) + C_2 (\cosh Kx \ - \sinh Kx) + C_3 (\cos Kx + i \sin Kx) + C_4 (\cos Kx - i\sin Kx) $$
 (5.3.5) $$\displaystyle X(x) = c_1 \cosh Kx \ + c_2 \sinh Kx + c_3 \cos Kx  + c_4 \sin Kx $$
 (5.3.6) Where

$$\displaystyle c_1 = C_1 + C_2 \in \mathbb R $$

$$\displaystyle c_2 = C_1 - C_2 \in \mathbb R $$

$$\displaystyle c_3 = C_3 + C_4 \in \mathbb R $$

$$\displaystyle c_4 = i(C_3 - C_4) \in \mathbb R$$

=R*5.4- Find $$\displaystyle y_{xxxxx} $$=

Given
A new variable (i.e. t) is introduced from current variable (i.e. x) in Euler Ln-ODE-CC. The transformation of variables is

Find
Find $$\displaystyle y_{xxxxx} $$ in terms of the derivatives of y with respect to t.

Solution
With the transformation of variables in Euler Ln-ODE-CC

We can obtain the derivatives of y with respect to x in terms of the derivatives of y with respect to t.

Finally,

=R*5.5 - Solve L2-ODE-VC with Method 2 Trial Solution =

Given
With Boundary Conditions as:

Find
Find the solution assuming trial solution as: where r = constant and plot the solution.

Solution
Substituting the value of $$ \displaystyle y' $$ and $$ \displaystyle y'' $$ in equation 5.5.1. Now $$ \displaystyle x^r \neq 0 $$ therefore : Equation 5.5.6 is the characteristic equation whose roots are: Hence the solution can be expressed as: Where $$ \displaystyle C_1 $$ and $$ \displaystyle C_2 $$ are constants need to be determined by applying boundary conditions. On applying boundary conditions we get: Solving these simultaneous equations we get: Hence the solution becomes as: Plotting the solution with 'x' varying from [-10:10]

=R5.6 - Equivalence of the Two Trial Solutions for an Euler Ln-ODE-VC=

Given
An Euler Ln-ODE-VC takes the form
 * $$\displaystyle a_n x^n y^{(n)} + a_{(n-1)} x^{(n-1)} y^{(n-1)} + \cdots + a_1 x y' + a_0 y = 0$$

(5.6.1) Or equivalently,
 * $$\displaystyle \sum_{i=0}^n a_i x^i y^{(i)} = 0$$

(5.6.2) There are two methods for solving an Euler Ln-ODE-VC. Method 1: Use a transformation of variables to transform the Euler Ln-ODE-VC into an Euler Ln-ODE-CC
 * $$\displaystyle x = e^t$$

(5.6.3) Then use the trial solution
 * $$\displaystyle y = e^{rt}$$

(5.6.4) Where r is a constant. Method 2: Use the trial solution
 * $$\displaystyle y = x^r$$

(5.6.5)

Find
Show that the trial solution for Method 2 is equivalent to the combined trial solution for Method 1.

Solution
It can be seen that the two methods are equivalent by combining equations (5.6.3) and (5.6.4) to form equation (5.6.5).
 * $$\displaystyle y = \left ( e^t \right )^r$$
 * $$\displaystyle y = \left ( x \right )^r$$

Which is equivalent to equation (5.6.5).

=R5.7 - Derivation of Homogeneous Solutions to the Euler-Cauchy Differential Equation with Repeated Real Roots of the Characteristic Equation — Variable & Constant Coefficient Cases=

Find

Find a2, a1, and a0 such that 5.7.1 is the characteristic equation of 5.7.2.

Show that $$\displaystyle y_1 = x^\lambda$$ is a solution to 5.7.2 with the coefficients determined in part 1.

Consider y = c(x) y1(x). Determine c(x) so that y is a solution of 5.7.2

Use the results above to determine the second homogeneous solution of 5.7.2

Repeat the above for 5.7.3 with $$\displaystyle y_1 = e^{\lambda x}$$

Solution
We begin by determining the characteristic equation of 5.7.2. Suppose, for some r yet to be determined,

$$\displaystyle \begin{align} \mathrm{Let}\quad y &= x^r \\ \mathrm{then}\quad y' &= r\, x^{r-1} \\ y'' &= r(r-1)\,x^{r-2} \end{align} $$

Substituting these into 5.7.2 yields

$$\displaystyle a_2\, x^2 \left[ r(r-1)\,x^{r-2} \right] + a_1\, x \left[r x^{r-1} \right] + a_0\, x^r = \left[ a_2\, r^2 + (a_1 - a_2)\, r + a_0 \right] x^r$$

The expression in brackets on the right hand side is the characteristic equation of 5.7.2. Expanding 5.7.1 as

$$\displaystyle r^2 - 2\lambda\, r + \lambda^2$$

and equating coefficients of r yields

$$\displaystyle \begin{align} a_2 &= 1 \\ a_1 &= 1 - 2\lambda \\ a_0 &= \lambda^2 \end{align} $$

Substituting these values back into 5.7.2 gives us our equation

The standard procedure for solving the Euler-Cauchy differential equations with distinct roots $$\lambda_1$$ and $$\lambda_2$$ is to use two solutions of the form $$y_1 = x^{\lambda_1}$$ and $$y_2 = x^{\lambda_2}$$. The general solution to the homogeneous equation would then be a linear combination of these two solutions. When there are real, repeated roots to the characteristic equation, as in this example, we must determine another homogeneous solution since $$\lambda_1 = \lambda_2$$ and the two solutions above are not linearly independent (indeed, they are identical!).

We begin by showing that $$\displaystyle y_1 = x^\lambda$$ is a solution of 5.7.2b.

$$\displaystyle \begin{align} y_1 &= x^\lambda \\ y_1' &= \lambda\, x^{\lambda - 1} \\ y_1'' &= \lambda\, (\lambda - 1)\, x^{\lambda - 2} \end{align} $$

Upon substituting these values into 5.7.2b, we get

$$\displaystyle \left[ (\lambda^2 - \lambda) + (1 - 2\lambda)\, \lambda + \lambda^2 \right] x^\lambda = \cancelto{0}{(\lambda^2 - 2\lambda^2 + \lambda^2 - \lambda + \lambda)}\, x^\lambda = 0$$

so y1 is indeed a solution of 5.7.2b.

We next seek a solution of the form y = c(x) y1(x).

$$\displaystyle \begin{align} y = c(x)\, y_1(x) &= c\, x^\lambda \\ y' = c'\, y_1 + c\, y_1' &= c'\, x^\lambda +c\, \lambda\,x^{\lambda - 1} \\ y = c\, y_1 + 2c'\,y_1' + c\,y_1 &= c\,x^\lambda + 2c'\, \lambda\, x^{\lambda-1} + c\,(\lambda^2 - \lambda)\, x^{\lambda - 2} \end{align} $$

Substituting these into 5.7.2b and simplifying gives

$$\displaystyle \begin{align} &x^2 \left[ c''\,x^\lambda + 2c'\, \lambda\, x^{\lambda-1} + c\,(\lambda^2 - \lambda)\, x^{\lambda - 2} \right] + (1 - 2\lambda)\,x\left[ c'\, x^\lambda +c\, \lambda\,x^{\lambda - 1} \right] + \lambda^2\,\left[ c\,x^\lambda\right] \\ &= c''\,x^{\lambda+2} + \left[ 2\lambda\, c' + (1 - 2\lambda)\,c'\right]x^{\lambda+1} +\cancelto{0}{\left[(\lambda^2-\lambda)\, c + (1-2\lambda)\,\lambda\, c + \lambda^2\,c\right]} x^\lambda \\ &= x^{\lambda+2}\, c'' + x^{\lambda+1}\, c' = 0 \end{align} $$

which is a L2-ODE-VC for c.

Since c does not appear in this equation, we can set p = c' to reduce the order of the equation by one. We are left with

5.7.4 is not exact, since

$$\displaystyle M_p = x^{\lambda+1} \neq (\lambda+2)\,x^{\lambda+1} = N_x$$

It can, however, be made exact with the IFM:

$$\displaystyle \begin{align} \frac{h_x}{h} &= \frac{1}{N}(M_p - N_x) \\ &= x^{-(\lambda+2)}\left[ x^{\lambda+1} - (\lambda+2)\,x^{\lambda+1}\right] \\ &= x^{-(\lambda+2)} (-\lambda - 1)\, x^{\lambda+1} \\ \frac{h_x}{h} &= -(\lambda + 1)\, x^{-1} \\ \int \frac{dh}{h} &= -\int (\lambda + 1)\, x^{-1}\,dx \\ \ln h &= -(\lambda + 1)\,\ln x \\ h &= x^{-(\lambda+1)} \end{align} $$

Multiplying 5.7.4 by h yields

$$\displaystyle \begin{align} p + x\, p' &+ 0 \\ \frac{d}{dx}(xp) &= 0 \\ xp &= k_1 \\ p &= k_1\,x^{-1} = c' \\ \therefore\, c &= k_1 \ln x + k_2 \end{align} $$

Plugging this value of c into the definition of y gives

$$\displaystyle y(x) = (k_1\ln x + k_2)\,x^\lambda = k_1\, x^\lambda \ln x + k_2\, x^\lambda$$

Inspection of this expression reveals that it is composed of a linear combination of the functions xλ ln x and xλ. But the second function is the first homogeneous solution y1, which means that the first function must be the second homogeneous solution we were seeking.

$$\displaystyle y_2 = x^\lambda\ln x$$

Since the method of variation of parameters has proven to be such a success with the second-order Euler-Cauchy differential equation with varying coefficients, we shall now turn our attention to the second order Euler differential equation with constant coefficients, where we hope to have similar success.

As before, we will consider the case where the characteristic equation has repeated real roots and attempt to determine a second homogeneous solution via the method of variation of parameters. As we are following the same steps as above, the exposition will be brief.

We begin by determining the values of b2, b1, and b0 such that the characteristic equation of 5.7.3 is equal to 5.7.1, the definition of repeated real roots.

$$\displaystyle \begin{align} \mathrm{Let}\quad y &= e^{rx} \\ \mathrm{then}\quad y' &= r\,e^{rx} \\ y'' &= r^2\,e^{rx} \end{align} $$

Plugging these into 5.7.3, expanding 5.7.1 as before, and matching coefficients

$$\displaystyle b_2\,r^2\,e^{rx} + b_1\, r\, e^{rx} + b_0\,e^{rx} = \left( b_2\,r^2 + b_1\, r + b_0 \right) e^{rx}$$

Thus

$$\displaystyle \begin{align} b_2 &= 1 \\ b_1 &= -2\lambda \\ b_0 &= \lambda^2 \end{align} $$

which, upon substitution into 5.7.3 gives

Let $$\displaystyle y_1 = e^{\lambda x}$$

y1 satisfies 5.7.3b since

$$\lambda^2\,e^{\lambda x} - 2\lambda^2\, e^{\lambda x} + \lambda^2\,e^{\lambda x} = 0$$

We now begin the method of variation of parameters to determine a second homogeneous solution of 5.7.3b.

$$\displaystyle \begin{align} y &= c\,e^{\lambda x} \\ y' &= c'\,e^{\lambda x} + \lambda\, c\, e^{\lambda x} \\ y &= c\, e^{\lambda x} + 2\lambda\,c'\, e^{\lambda x} + \lambda^2\,c\, e^{\lambda x} \end{align} $$

Plugging these into 5.7.3b gives

$$\displaystyle \begin{align} &\left[ c''\, e^{\lambda x} + 2\lambda\,c'\, e^{\lambda x} + \lambda^2\,c\, e^{\lambda x} \right] - 2\lambda\left[c'\,e^{\lambda x} + \lambda\, c\, e^{\lambda x}\right] + \lambda^2\,c\,e^{\lambda x} \\ &= c''\,e^{\lambda x} + \cancel{2\lambda\,c'\,e^{\lambda x}} + \cancel{\lambda^2\,c\, e^{\lambda x}} - \cancel{2\lambda\,c'\,e^{\lambda x}} - \cancel{2\lambda^2\,c\, e^{\lambda x}} + \cancel{\lambda^2\,c\,e^{\lambda x}} \\ &= c''\,e^{\lambda x} = 0 \end{align} $$

The only way this expression can be true for all x is if $$\displaystyle c'' = 0$$. Thus,

$$\displaystyle c = k_1\,x + k_2$$

Thus,

$$\displaystyle y = (k_1\,x + k_2)\,e^{\lambda x} = k_1\,x\,e^{\lambda x} + k_2\,e^{\lambda x}$$

Using the same logic as in part 1, we conclude that

$$\displaystyle y_2 = x\,e^{\lambda x}$$

is the second homogeneous solution we are seeking.

=R*5.8 - Particular solution by using variation parameters method =

Given
$$\displaystyle y(x) = A(x)y_H(x)$$
 (5.8.1) $$\displaystyle y' + a_0(x)y = 0 $$ $$\displaystyle y_H(x) = exp[-\int^x a_0(s)ds] $$

Differential eq is that $$\displaystyle y' + P(x)y = Q(x) $$
 (5.8.2) $$\displaystyle y_H(x) = exp[-\int P(x)dx] $$
 (5.8.3)

Find
Find particular solution Yp

Solution
$$\displaystyle y(x) = A(x)y_H(x) $$ $$\displaystyle y(x)' = A(x)'y_H(x) + A(x)y_H(x)' $$
 (5.8.4) From the eq 5.8.3, $$\displaystyle y_H(x)'= -P(x)\cdot exp[-\int P(x)dx]=-P(x)y_H(x)$$ Thus, eq 5.8.4 is following that
 $$\displaystyle y(x)' = A(x)'\cdot exp[-\int P(x)dx] - P(x)\cdot exp[-\int P(x)dx]A(x) = exp[-\int P(x)dx]\cdot[A(x)'-P(x)A(x)]$$
 Plugging the above equation into Eq 5.8.2, $$\displaystyle exp[-\int P(x)dx]\cdot[A(x)'-P(x)A(x)] + P(x)A(x)\cdot exp[-\int P(x)dx]= Q(x) $$ $$\displaystyle exp[-\int P(x)dx](A(x)'-P(x)A(x)+A(x)P(x))=Q(x)$$ $$\displaystyle exp[-\int P(x)dx]\cdot A(x)' = Q(x)$$ $$\displaystyle A(x)' = \frac {Q(x)}{exp[-\int P(x)dx]} = Q(x)\cdot exp[\int P(x)dx] $$ $$\displaystyle A(x) = \int Q(x) \cdot exp[\int P(x)dx]dx $$ Thus, paricular solution is that $$\displaystyle y(x) = \int Q(x) \cdot exp[\int P(x)dx]dx \cdot exp[-\int P(x)dx] $$

=R*5.9 -Special IFM=

Given
Nonhomogeneous L2-ODE-CC

$$\displaystyle a_2 y'' + a_1 y' +a_0 y = f(t)$$
(5.9.1)

Find
Using special IFM to solve (5.9.1) for general $$\displaystyle f(t)$$

Solution
Suppose the integrating factor has the form $$\begin{align} h(t,y) \end{align}$$

Multiply the integrating factor to the 5.9.1.

$$\displaystyle h(t,y) \cdot \left[ {{a}_{2}}y''+{{a}_{1}}y'+{{a}_{0}}y \right] = h(t,y) \cdot \left[{{f}_{\left( t \right)}}\right] $$
 (5.9.2)

rewrite the 5.9.2 to check the exactness condition.

$$\displaystyle F(x,y,y',y) = \underbrace{h(t,y){a}_{2}}_{f(x,y,p)}y+ \underbrace{h(t,y){{a}_{1}}p+ h(t,y) {{a}_{0}}y - h(t,y)\cdot f(t)}_{g(x,y,p)} $$
 (5.9.3)

where $$\begin{align} p:= y' \end{align} $$

therefore,

$$\displaystyle f(x,y,p) = h(t,y){a}_{2} $$

$$\displaystyle g(x,y,p) = h(t,y){{a}_{1}}p+ h(t,y) {{a}_{0}}y - h(t,y)\cdot f(t)$$

There are two relations for the 2nd condition of exactness.

$$\displaystyle {f}_{tt}+2p{f}_{ty}+{p}^{2}{f}_{yy}={g}_{tp}+p{g}_{yp}-{g}_{y} $$
 (5.9.4)

$$\displaystyle {f}_{tp}+p{f}_{yp}+2{f}_{y}={g}_{pp} $$
 (5.9.5)

From (5.9.5), we have

$$\displaystyle {h}_{y}=0$$

 (5.9.6) and from 5.9.4

$$\displaystyle {h}_{tt}{a}_{2}-{h}_{t}{a}_{1}+h{a}_{0}=0 $$
 (5.9.7) So

$$\displaystyle {a}_{2}h''-{a}_{1}h'+{a}_{0}h=0 $$
 (5.9.8)

thus, assuming $$ \displaystyle h(t)={e}^{\alpha t} $$

From 5.9.8, we have

$$\displaystyle {a}_{2}{\alpha}^{2}-{a}_{1}\alpha +{a}_{0}=0 $$
 (5.9.10)

Then, substituting (5.9.8) to (5.9.2) and integrating it yields

$$\displaystyle y(t)=\frac{{e}^{-\alpha t}}{2\alpha a_{2}-a_{1}}\left({e}^{(\frac{a_{1}}{a_{2}}-2\alpha)t}\,\int{e}^{-\alpha t}\, f\left(t\right)dt-\int{e}^{\alpha t}\, f\left(t\right)dt\right)$$
 ( 5.9.11) If $$\displaystyle a_{1}^{2}-4a_{0}a_{2}\neq0$$, substitute the solution of (5.9.10) to (5.9.11), we have

$$\displaystyle y(t)={C}_{1}\,{e}^{-\frac{\left(\sqrt{{a}_{1}^{2}-4\,{a}_{0}\,{a}_{2}}+{a}_{1}\right)\, t}{2\,{a}_{2}}}+{C}_{2}\,{e}^{\frac{\left(\sqrt{{a}_{1}^{2}-4\,{a}_{0}\,{a}_{2}}-{a}_{1}\right)\, t}{2\,{a}_{2}}+}$$

$$\displaystyle \frac{{e}^{-\frac{\left(\sqrt{{a}_{1}^{2}-4\,{a}_{0}\,{a}_{2}}+{a}_{1}\right)\, t}{2\,{a}_{2}}}}{\sqrt{{a}_{1}^{2}-4\,{a}_{0}\,{a}_{2}}}\left({e}^{\frac{\sqrt{{a}_{1}^{2}-4\,{a}_{0}\,{a}_{2}}\, t}{{a}_{2}}}\,\int{e}^{-\frac{\left(\sqrt{{a}_{1}^{2}-4\,{a}_{0}\,{a}_{2}}-{a}_{1}\right)\, t}{2\,{a}_{2}}}\, f\left(t\right)dt-\int{e}^{\frac{\left(\sqrt{{a}_{1}^{2}-4\,{a}_{0}\,{a}_{2}}+{a}_{1}\right)\, t}{2\,{a}_{2}}}\, f\left(t\right)dt\right)$$

=Contributing Members= Egm6321.f11.team2.epps I solved problems R5.1 and R*5.7 02:11, 2 November 2011 (UTC) Egm6321.f11.team2.kim 15:11, 2 November 2011 (UTC) I solved problems R*5.2 and R*5.8 and proofread problems R*5.3, R*5.4, R*5.7, and R*5.9. Egm6321.f11.team2.rho 16:50, 2 November 2011 (UTC) I solved Problem R*5.4 and proofread problems R*5.2, R*5.3, R*5.8, and R*5.9. Egm6321.f11.team2.johri.y 17:36, 2 November 2011 (UTC) I solved R*5.5 and proofread R*5.4 Egm6321.f11.team2.Xia I solved problems R*5.3 and R*5.9, proofread R*5.1, R*5.5 04:08, 2 November 2011 (UTC)

title: User:Egm6321.f11.team2/HW5


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Computer model of the DC Motor Shaft that we intend to use with the project.

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Category:Binary matrix templates

These templates allow to create small permutation matrices with a short source code.

gives

This arrangement copied and pasted gives: [1,0,0,0, 0,0,0,0; 0,0,0,0, 1,0,0,0; 0,0,0,0, 0,0,1,0; 0,0,1,0, 0,0,0,0;

0,0,0,0, 0,0,0,1; 0,0,0,1, 0,0,0,0; 0,1,0,0, 0,0,0,0; 0,0,0,0, 0,1,0,0]

See: Table of 3-bit Walsh permutations

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User:Amcilrick/Rodeo in Australia - Welfare of Animals Used in Rodeo Events

''“Professional rodeo is a unique and spectacular sports entertainment. It is colourful, exciting and rich in the traditions of the Australian outback”'' (APRA, n.d.).

Presentation
http://www.youtube.com/watch?v=Z2d2zxE4Hrg

Rodeo in Australia – Animal Welfare
Australian rodeo has had to deal with Animal Welfare groups and negativity from the RSPCA since rodeo commenced. These groups are naive to the fact that there are Codes of Practices and rules in place for the welfare of rodeo stock. No one enforces these rules more than the rodeo associations.

History
Rodeo in Australia can be traced back to the arrival of the first horses in New South Wales on January 26 1788 (Hayes, 1982). The First Fleet en route to Sydney Cove took on board a small number of horses (Hayes, 1982). These wild horses were broken in, trained for the New South Wales climate and terrain and used mainly for transport and security.

By the 1800’s the need for horse-breakers was growing, creating buck-jumping. Soon informal matches between stations best horsemen and horses were created and the Bushman’s Carnival contests began (Hayes, 1982).

Bushman’s Carnivals were the bushman’s annual holiday. Stockmen would demonstrate their skill, city slickers would venture out to the bush and family celebrations would occur. Bushman’s Carnivals were not only a time of fun and joy but also the challenge of determining who the best buckjump rider was, which carried the highest honour (Poole, 1977).

In December 1944 the first official rodeo association was formed, the Australian Rough Riders Association (ARRA) (Poole, 1977).

In March 1946 the Bushman’s Carnival became an official association, the Northern NSW Bushmen’s Carnival Association. Today it is known as the Australian Bushman’s Campdraft and Rodeo Association (ABCRA), one of the largest Rodeo and Campdraft associations in Australia (ABCRA, n.d.).

Since then there have been numerous Rodeo organisations and associations formed.

One of the first rules the ARRA put in place was fairness to the animals used in the rodeo arena. Anyone who mistreated an animal would face some sort of penalty (Poole, 1977).

Rodeo Events
Rough Stock

 * Bull Ride
 * Saddlebronc
 * Bareback

Timed Events:

 * Rope & Tie
 * Steerwrestling
 * Ladies Breakaway
 * Ladies Barrel Racing
 * Ladies Steer Undecorating
 * 14 - U18 Juvenile Barrel
 * 11-U14 Junior Barrel
 * U11 Junior Barrel
 * Junior Breakaway

Team Draw

 * Team Roping

Chute Draw

 * 14-U18 Juvenile Steer
 * 11-U14 Junior Steer
 * 8-U11 Poddy Ride
 * Novice Bull Ride
 * Novice Bareback
 * Novice Saddlebronc

Animal Welfare
“The Australian Bushman’s Campdraft and Rodeo Association recognises that animals used in competition must be treated in a humane manner and that the ABCRA does not condone the ill treatment or abuse of any animals whilst competing at campdrafts or rodeos” (ABCRA)

The National Consultative Committee on Animal Welfare (NCCAW), part of The Department of Agriculture, Fisheries and Forestry, are responsible on advising the Australian Government on animal welfare issues. With the help of rodeo associations they have produced a national standard document for managing and controlling rodeos. “The NCCAW Rodeo Standards are designed to address issues specific to rodeos, and complement legislation and national Codes of Practice” (NCCAW, 2008)

Each state has used the national standard and produced a more detailed Code of Practice. The Animal Welfare Advisory Council is responsible for the NSW Code of Practice for the Welfare of Animals used in Rodeo events. The Australian Professional Rodeo Association, Australian Bushman’s Campdraft and Rodeo Association and the National Rodeo Association strongly enforce this code.

Misconceptions
Barrel Racing

A report done in 1991 by the Senate Select Committee on Animal Welfare titled Equine Welfare in Competitive Events other than Racing found horses used in barrel racing were excessively being whipped and spurred during their run. The ABCRA implanted the rule two whips/contacts between barrels and four whips/contact coming home. An over under action and overuse of the whip would result in disqualification.

The International Professional Rodeo Association (IPRA) will fine a barrel racing cowgirl if she is dressed inappropriately. Appropriate western attire is tucked in long sleeved shirts, with collars, no jumpers or t-shirts, non worn out jeans, western hats and boots (IPRA) IPRA barrel racers will be disqualified if they recross the finish line before the completion of a cloverleaf pattern. They will be fined $100 for going around a barrel more than once.

Rope and Tie

One of the most controversial events in rodeo is the Rope and Tie. This is where a mounted rider ropes a calf, jumps off and ties three of its legs together. Animal welfare groups hate this event as there is a chance the calf may be victim to a jerk down (pulling the calf backward off its feet) or dragging.

At the Equine Welfare in Competitive Events other than Racing inquiry this was the most criticised event due to the abrupt stopping of the calves. This has been reduced due to the introduction of neck ropes and jerk lines. Evidence shown to the inquiry proved the roping devices “improved the situation” (Senate Select Committee on Animal Welfare, 1991).

The amount of time cowboys spend practicing and the amount of money spent on livestock, equipment, and travel the chances of a jerk down occurring decrease every time they step into a rodeo arena. If a jerk down does occur the rider is disqualified and fined $50. If the rider drags the calf they are disqualified (NSW Code of Practice, 1988).

Use of Stock

All stock used in a rodeo, whether it be rough stock or timed event stock, must be examined by a veterinarian or a person with livestock experience at least two hours before being used and at the end of the event, before being loaded for transport. If the examiner deems the animal unhealthy, injured, having poor eye-sight, over-fat or pregnant they will not be used.

Contract cattle used for steer wrestling and roping may be used no more than three times a day. Non-contract cattle may only be used twice per day (NSW Code of Practice, 1988). Bucking animals can be used no more than four times a day. Bucking animals must be supplied by an approved contractor (ABCRA, n.d.).

 * Cattle used in calf roping events must be 100kg-140kg
 * Steer wrestling and team roping steers must be 180kg-300kg
 * Steer Undecorating must be a minimum of 300kg
 * Cattle used in riding events minimum weight is 200kg with a maximum rider weight of 40kg

Stock contractors need their stock to be healthy. Their livelihood comes from competing stock and if their animals aren’t healthy and meeting the minimum standard they will not be used and the contractor won’t be paid. No contractor wants stock to be treating poorly. As professional stock contractor Matt Adams puts it, “these animals are my livelihood – they’re valuable to me and I love them. Most of them are like pets and there’s no way known I’d mistreat them or deliberately subject them to any cruelty.” (Cowley, 2010)

Electric Prod

The electric prod is used mainly for the transportation on livestock. It is not allowed in the arena at any rodeo in Australia. The electric prod must be powered by dry cell batteries and be of low amperage. It can only be used on parts of an animal covered by hair (DAFF, 2008). The electric prod can only be used by the stock contractor or by a person nominated by the stock contractor (ABCRA & APRA).

“The prod produces 5,000-6,000 volts of electricity, but virtually no current or amperage. Because amperage, not voltage, causes burns, the cattle prod causes a mild shock, but no injury” (APRA).

The electric shock an animal feels from a prod is the same as what they would feel if they touched an electric fence sitting in their paddock.

Violation of this rule results in a $200 fine to the stock contractor (ABCRA).

Flank Strap

Rough stock are bred to buck. Putting a flank strap on them does not make them buck. The flank strap enhances their bucking style but causes no pain (APRA). Dr Ian Gollan, a large animal veterinarian stated “It is untrue to suggest that kick straps are pulled so tightly that they cause pain or restrict the free movement of the animal – such a procedure would inhibit the animal’s bucking prowess not enhance it.”

Flank straps must be lined with sheepskin or another suitable soft flexible material and must be fitted with a quick release (DAFF, 2008). No sharp or cutting objects may be used. If competitors are caught cheating they are fined and disqualified. Flank straps must be placed over the flank and belly of the animal.

Animal Welfare groups claim an animal will only buck because the flank strap has been placed over their genital area and deliberately tightened to cause them pain. Dr Gollan said “It is impossible to position these straps to interfere with the horse’s genitalia” (APRA) A flank strap is just like a human belt: Sitting in about the same position and easy to release.

Spurs

Spurs are used to assist the rider’s timing and riding position on the animals in rough stock events (DAFF, 2008). They are one of the most closely monitored rules in rodeo. Spur rowels must not lock. They must be blunt and free running. They must be three millimetres thick and two centimetres in diameter. Spurs in bull riding are allowed to have a restricted movement but must not fully lock (DAFF, 2008). This is allowed as more grip is needed on a bull than on a horse.

Spurs that are free running allow the rowel to roll over the horse’s hide rather than cutting or bruising them, as animal welfare groups believe (APRA). Dr Gollan in an article for the Australian Equine Veterinarian named Rodeos said “…having regard to the blunt surfaces of the rowels the thickness of the animals skin, and the absence of any visible damage to the animal, it does not seem likely that significant pain is inflicted."

Riders who violate the spur rule get fined $100, automatically disqualified and may even face expulsion (ABCRA)

Penalties for Breaking Codes of Practice (ABCRA)
The NCCAW has not ruled that a violation of the code will lead to a punishable action. It is the rodeo associations that have chosen to punish anyone caught mistreating an animal. Penalties include (if not already mentioned);

 * Any contestant, stock contractor, pick up man guilty of mistreatment of livestock, may be fined by the Board of the Association with the fine not to exceed $500.

 * Contestants will be disqualified for any mistreatment of livestock.

 * All contest stock must be numbered by either ear tag or freeze, fire brand after being tried and proven with a minimum of three runs with paint brand numbering. Failure of contractors in possession of livestock to meet this criteria, will result in $10 fine per animal. Fine to progressively double. After 3 infringements in one rodeo year (per rodeo), contractor will be suspended.

 * If animals remain in yards for more than 24 hrs before loading, or if animals are to travel for 24 hrs, feed as well as water, must be provided. Failure to abide by this rule shall subject the stock contractor or person involved to a $250 fine for first offence and a progressive doubling for offences thereafter

 * If a member abuses an animal by any unnecessary non-competitive or competitive action they may be disqualified for the remainder of the rodeo and fined $100 for the first offence with that fine progressively doubling with each offence thereafter.

 * In the event of stock shortage the stock contractor or rodeo committee shall be fined as follows. First offence of the rodeo year - $25 for each animal short. Second offence - $50 for each animal short. Third offence - $100 for each animal short.

 * An event representative may declare particular animals unsatisfactory. Upon notification, either written or verbal, the stock contractor or rodeo committee shall eliminate such animal(s) from competition draw. Continued use of said animal(s) after notification shall result in a $50 fine per competition levied against offending party.

 * An animal used in the contest events of an ABCRA rodeo may not be used in any other way until after the last time that animal has been used in the contest events of that rodeo. Failure to abide by this rule shall result in $25 fine per head per competition, levied against the party concerned.

The following provisions regarding conduct shall apply to any event whether in competition or practice:

 * Inhumane treatment or abuse of a horse in any manner whilst in the arena or on the grounds is prohibited. Inhumane treatment includes, but is not limited to, the following:

 * Riding of a crippled, injured or lame animal

 * Riding of a horse with a health abnormality, which could result in the horse’s undue discomfort or distress

 * Abuse of an animal which includes excessive jerking, spurring, whipping or any other act intended to cause trauma or injury to the animal.

 * Any act of abuse or intent to abuse an animal whilst in the campdraft or rodeo arena or on the grounds or the vicinity which could also endanger the safety of other persons or animals will not be tolerated and the persons may be subject to disciplinary action.

International
“The International Professional Rodeo Association acknowledges the valuable role of animals in rodeo and our responsibility to provide a proper environment for the animals in competition, in transport and at rest” (IPRA).

Rodeos will always have to deal with animal welfare groups whether it is on a national or international scale. The IPRA promotes the same rulings as Australian rodeo associations. The difference is the degree of fine; eg – Australian Rope & Tie $50 fine, IPRA Calf Roping minimum $200 fine.

Cowboys and Cowgirls spend a lot of time, money and effort on their chosen profession. They are not going to be cruel to an animal for risk of getting fined, disqualified or even banned from a rodeo arena. Rodeo is a huge part of their livelihood and life in general.

The Future of Rodeo in Australia
Rodeo is a traditional part of Australian outback life. The cowboy is one of the most universally known figures in history (Poole, 1977). Australian rodeo is a multimillion dollar industry (Hicks, 2000) that attracts corporate sponsorship and raises hundreds of thousands of dollars for local communities and charities.

Rodeo associations will always have to deal with animal welfare groups but it is the distinct lack of knowledge and research of these groups and the support of the government, veterinaries, communities and the growing population of rodeo participants that gives Australian rodeo a strong future.


text:
Talk:Motivation and emotion/Book/2011/Competitiveness

Comments
getting my content together just figuring out this coding can be difficult though. Any ideas would be good! eg pics etc and boxes how to get everything together Rednoodles 00:36, 30 October 2011 (UTC)

 * Hey Rednoodles, Yea im having problems uploading pics lol... but on the other hand, I got some interesting stuff on sport competitiveness and testosterone (physiologically important for competition). Great amount of research in relation to men ;)Lucas K1 10:34, 30 October 2011 (UTC)

Hey! Really like the video and layout of your page! the coloured boxes are a great idea! Tashc 08:38, 31 October 2011 (UTC)

How can I remove the word template from my page? They are there and I'm not sure how to remove it I put it there by mistake.Rednoodles 05:02, 1 November 2011 (UTC)
 * Hey there, I have removed those for you - I hope you don't mind. You had just mis-spelt some of the commands and also put one too many in. The chapter is looking good! ShaunaB 06:17, 1 November 2011 (UTC)

Hey Your page is looking really good so far and sounds like an interesting topic, maybe you could add in some tips on how to feel more competitive if your the competition aviodance type like you mentioned :) Kayla 01:39, 3 November 2011 (UTC)

Hey thanks that's a great idea I'll definitely have a poke around the literature and find something thanks for the suggestion Rednoodles 01:42, 3 November 2011 (UTC)

Hey, I just read through your page, it's looking great! All of your information is really well written and well referenced. I have a couple of questions. Firstly, in the table where you show competitiveness, you have headings of 'Col.1, Col.2 & Col.3', i don't really understand the headings, is there another way you could set up the table or have different headings? Also in the same section, the pictures with one word under them don't really seem to fit there either. Is there a specific reason for the pictures or can they be spaced out throughout the page with further explanation as to why they are there? Are you able to find a quiz or questionnaire that can determine how competitive an individual is? I personally hate losing when i play sport so it would be interesting to see if i am considered overly competitive or not. =] I am jealous you have found so many photo's and pictures for your page! I am struggling to even know where to look to find the appropriate photos. Good luck with the finishing touches! Ashleetronerud 1:20pm, 3 November 2011 (UTC)

Comment by Magnolia - update your completion status - you did a great job on this topic - loved the video of the running man

Rednoodles' topic
Hey james just wondering why you changes my topic?Rednoodles 21:00, 30 October 2011 (UTC)
 * I've replied to your email, with details of the book chapter contents edit history, showing what you signed up for and unsigned from. Let me know what you'd like to do. Sincerely, James. -- Jtneill - Talk - c 22:14, 30 October 2011 (UTC)

-- Jtneill - Talk - c 11:22, 25 November 2011 (UTC)

A very interesting topic that has great potential to be expanded for another paper to be published - really well done - Magnolia28

-- Jtneill - Talk - c 00:51, 9 December 2011 (UTC)