text:
Motivation and emotion/Textbook/Emotion/Psychopathy

Introduction
Psychopathy, otherwise known as Psychopathic Personality Disorder (PPD), has been the most studied personality disorder with much literature and research conducted in an attempt to understand it (Hesse, 2009).

There is a perception that psychopaths are noticeable within our society, that they fit a certain stereotype. This however, is incorrect. Psychopaths are characterised by their ability to be charming, agreeable, have above average intelligence and above all, lie with extreme conviction (Freedman, Verdun-Jones, 2010). Each of these hallmark traits of a psychopath can make them look just like you or I on the surface which is shown throughout history. Infamous psychopaths such as Charles Manson, Jeffery Palmer, (girl one), etc, may have looked “normal” enough on the surface, yet each committed crimes and held thoughts within themselves that are unconscionable to nonpsychopathic individuals.

Robert Hare designed the most commonly used measure to test for psychopathic traits in individuals and is still used today, with various additions to test for such traits in children (PCL- Youth Version) and as a screening measure (PCL: Screening Version) (Hare & Neumann, 2009).

What is Psychopathy?
In his 1982 works Cleckley, a significant player in the field, defined psychopathy as a “psychological construct that describes chronic immoral and antisocial behaviour, a lack of consciousness, and the ability to lie and deceive without feeling guilt or discomfort” (Hesse, 2009).

According to Bishopp and Hare (2008), psychopaths are impulsive sensation seekers with a range of defining features. Such key features include superficial charm, absence of nervousness which would suggest low anxiety, unreliability, intelligence, and most significant of all, psychopaths lack empathy and remorse. Another key feature of psychopaths is their inability to learn from experience which, most troubling of all, means they do not respond to punishment (Bishopp & Hare). In part, this is due to psychopaths failing to accept responsibility for their actions and their lack of remorse (Freedman & Verdun-Jones, 2010). Other key features of psychopaths which often makes them appear like any other “normal” person, is their agreeableness and their ability to lie and be extremely convincing (Freedman, Verdun-Jones, 2010).

Psychopaths are considered as sane individuals in the mental health community according to Freedman and Verdun-Jones (2010). The basis for this being that psychopaths have an accurate sense of reality and “appear to be rational and aware of their actions”. Because of this, psychopathy is considered independent from any mental disorder. Also because psychopathy reflects innate dispositions as opposed to resulting from reactions to trauma or extreme dispositions as is the case in other forms of disorder (Bishopp & Hare, 2008).

The Psychopathy Checklist- Revised (PCL-R) was developed by Robert Hare and is “the most commonly used measure of psychopathic behaviour” (Hesse, 2009) and is used to measure psychopathic traits in individuals. The PCL-R is a rating scale comprised of 20 items, each pertaining to a different trait or symptom of psychopathy (Cooke, Hart, Hare & Michie, 1999). The rating scale is a 3-point scale based on 0= doesn’t apply, 1= item applies somewhat, and 2= item definitely applies (Cooke, Hart, Hare & Michie, 1999). The scores are totalled ranging from 0-40, reflecting how strongly the person matches the traits of a prototypical psychopathic person (Hare and Neumann, 2009). A score of 30 or higher is most often used as the cut off score to diagnose psychopathy (Cooke, et al, 1999).

The 20 items on this scale come under two factors. Factor 1 pertains to the “affective and interpersonal features” of psychopathy and is labelled Selfish, Callous, and Remorseless use of others. Factor 2 pertains to the “social deviance features of psychopathy” and is labelled Chronically Unstable and Antisocial Lifestyle (Cooke, et al, 2009).

Items in the Hare Psychopathy Checklist—Revised
Item /							Factor loading

Glibness/superficial charm/					1

Grandiose sense of self-worth/					1

Need for stimulation/proneness to boredom/			2

Pathological lying/						1

Conning/manipulative/						1

Lack of remorse or guilt/						1

Shallow affect/							1

Callous/lack of empathy/						1

Parasitic lifestyle/						2

Poor behavioural controls/					2

Promiscuous sexual behaviour/					-

Early behaviour problems/					2

Lack of realistic, long-term goals/				2

Impulsivity/							2

Irresponsibility/							2

Failure to accept responsibility for own actions/			1

Many short-term marital relationships/				-

Juvenile delinquency/						2

Revocation of conditional release/				2

Criminal versatility/						-

''Note. Factor 1 = Selfish, Callous, and Remorseless Use of Others; Factor 2 = Chronically Unstable and Antisocial Lifestyle. A dash indicates the item does not load on either factor.''

Glenn, Raine and Schug (2009) suggest there are four factors of psychopathy, each of which are covered in Hare’s PCL-R. These four factors and their associated psychopathic traits are:

Interpersonal Factor

 * Glib, superficially charming
 * Callous, manipulative
 * Pathological lying
 * Grandiose sense of self-worth

Affective Factor

 * Lack of remorse of guilt
 * Shallow affect
 * Callous, lacking empathy
 * Failure to accept responsibility

Lifestyle Factor

 * Stimulation seeking
 * Impulsive
 * Irresponsible
 * Parasitic lifestyle
 * Lack of realistic goals

Antisocial Factor

 * Poor behavioural controls
 * Early behaviour problems
 * Juvenile delinquency
 * Revocation of conditional release
 * Criminal versatility

Psychopathy and Emotion
According to diagnostic criteria, a main hallmark of psychopaths is their lack of emotion. They are unable to feel remorse, shame or empathy, all important personality traits to maintain a civilised society. There are many reasons as to why they cannot experience these emotions, ranging from neurobiological to environmental.

Fear is an emotional response innately experienced by, what should be, all human beings. There is much research however that suggests that psychopaths do not experience the responses of fear that non-psychopaths do. This further supports the notion that psychopaths lack emotion and is a distinctive characteristic of psychopathy. Freedman and Verdun-Jones (2010) discuss various studies that have tested different aspects of this notion that psychopaths do not have the same responses to fear, each resulting in the same findings. One study Freedman and Verdun-Jones (2010) noted as significant found when shown pleasant, neutral and unpleasant images, psychopaths blinking patterns and heart rates remained constant. As this is significantly different to the responses “normal” individuals would have, these findings support the notion that psychopaths are not emotionally affected by unpleasant images as well as not displaying signs of fear (Freedman & Verdun-Jones, 2010). Another study discussed by Freedman and Verdun-Jones (2010) involved psychopaths listening to sentencing designed to elicit fear. The results of this study found that not only did psychopaths not respond to these fear inducing sentences, but they also did not produce the same physiological responses as would be expected by “normal” people. In such a setting, one's muscles would usually tense on an unconscious level however psychopaths failed to show any signs of this. Freedman and Verdun-Jones (2010) suggest these findings indicate that psychopaths have difficulty interpreting emotion, particularly on this level of fear. According to Freedman and Verdun-Jones (2010), this difficulty interpreting emotion also pertains to identifying it being expressed by others. Research has found that psychopaths have difficulty identifying fear and sadness in other people although, interestingly, they can recognise happy facial expressions. This also applies to the psychopathic individual, that is, they can both experience happiness themselves and recognise that emotion in others, however they cannot experience, nor can they recognise fear (Freedman & Verdun-Jones, 2010).

Neurobiological Processing
The lack of emotion psychopaths display has been argued to be a result of neurobiological deficits. Gao, Glenn, Raine, Schug and Yang (2009) discuss research which suggests deficits in the prefrontal cortex or in the amygdala are the main neurobiological areas that are associated with psychopathic-like traits. It is suggested reduced grey matter volume in the prefrontal cortex is associated with “poor decision-making, emotional dysregulation, and impaired moral judgment in psychopathic people” (Gao, Glenn, Raine, Schug & Yang, 2009). In terms of lack of remorse, it is suggested this could be due to deficits in the amygdala (Gao, Glenn, Raine, Schug & Yang, 2009). Cantani, Craig, Daley, Deely, Fahy, Kanaan, Latham, McGuire, Murphy and Picchioni (2009) suggest this effect of amygdala dysfunction is supported by research in which psychopaths performance in tasks sensitive to amygdala damage reflects that dysfunction. This research also suggests that amygdala volume is significantly reduced in psychopaths, as well as decreased brain activity in those areas controlled by the amydala during various learning and emotive tasks (Cantani, et al, 2009).

In support of reduced amygdala activity being the “core deficit in psychopathy Glenn, Raine and Schug (2009) suggest there is a strong association between the four factors of psychopathy and reduced activity in the amygdala, particularly during emotional moral decision making. Further research indicates the reduction in amygdala activity disrupts moral decision making in psychopaths and, as a result, making them unable to identify distress etc. in other individuals (Glenn, Raine and Schug, 2009). As a result of this disruption to amygdala functioning, psychopathic individuals may be undeterred from “conning and manipulating others, making impulsive, irresponsible decisions, and engaging in criminal behavior without feeling guilt or remorse”, all key characteristics of psychopathy (Glenn, Raine and Schug, 2009).

Response Modulation Hypothesis (RMH)
According to the RMH, a failure to process the meaning of information they consider unimportant in terms of its effect on them personally, could explain psychopathic individuals “impulsivity, poor passive avoidance, and emotion-processing deficits” (Lorenz and Newman, 2002). Glass and Newman (2009) state that the based on RMH, psychopathic individuals can display normal emotional responses when it is their primary attentional focus, that is, when it relates to them specifically. In contrast, when faced with information that is secondary to their primary attentional focus, psychopathic individuals have difficulty processing this affective information, suggestibly because in the latter, they are not the focus (Glass and Newman, 2009). Response modulation is defined by Lorenz and Newman (2002) as “a brief and highly automatic shift of attention that enables individuals to monitor and, if relevant, use information that is peripheral to their dominant response set (i.e., deliberate focus of attention)”. According to Lorenz and Newman (2002), RMH is evident in such situations as when the psychopathic individual is avoiding punishment, a primary task as it directly affects them, and thus they display appropriate responses. In contrast, when attention needs to be on processing secondary information, that is, does not directly affect them, psychopathic individuals experience performance deficits (Lorenz and Newman, 2002).

Gray’s Theory of Behavioural Inhibition System (BIS)
Gray’s BIS pertains to the inhibition of goal-directed behaviour, as in the RMH, as a result of potential threats or when inconsistent stimuli is detected within the environment (Baskin-Sommers, Curtin, MacCoon, Newman & Wallace, 2009). As a result of this inhibition, attention is redirected in an attempt to process the threatening stimuli (Baskin-Sommers, et al., 2009). According to Baskin-Sommers, Curtin, MacCoon, Newman and Wallace (2009), this process is based on the individuals past experiences with punishment and nonreward and, according to Glas and Newman (2009), when operating effectively, the BIS “enables a person to use environmental cues to anticipate future significant negative events and inhibit behaviour accordingly”. This theory is of importance specifically to psychopathy as according to Fowles (1980), “psychopathy could be understood as a consequence of weak BIS functioning”. Baskin-Sommers, et al. (2009) discuss supporting research which indicates and association between psychopathy and BIS-related deficiencies including poor passive avoidance learning and “weak skin conductance responses in anticipation of aversive events”.


text:
User:Egm6321.f10.team2.dube/HW1

Test Page for HW1 Team 2. Sandbox.

$$\alpha = \beta = \gamma$$

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

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


text:
User:Egm6321.f10.team4

Welcome to the public Wikiversity Web site for Team 4 of the University of Florida's Fall 2010 Principles of Engineering Analysis course (EGM 6321).

=Homework Assignments= Homework 1

Homework 2

Homework 3

Homework 4

Homework 5

Homework 6

Homework 7

=Links= Course Lecture Plan Course Report Table

title: User:Egm6321.f10.team4


text: Portal:Biophysical Chemistry

title: Portal:Biophysical Chemistry


text:
User:Wbeadle3

William Frank Beadle born October 9th, 1985 in Atlanta, Georgia.

Background
Undergraduate degrees from the [| Georgia Institute of Technology] in:
 * Biology
 * Biochemistry

Research

 * Studied the Nitrogen cycle in Eddies off Western Australia.
 * Studied the effects that pyocyanin has on RAW-1 macrophage cells BAI-1 binding and uptake of apoptotic Jurkat Cells

Interests
All fields of Chemistry and Biochemisty.

Future
To recieve a doctorate in biochemistry field at [| UGA] graduate school.

title: User:Wbeadle3


text:
User:Egm6321.f10.team1.allison

Problem 1: Derive
This solution is based on the one previously found by Vasquez et al. The primary improvement is the inclusion of a background section, where discussion of why these equations are important is included.

Background
A very common use for the time derivative is in consideration of applied forces. Newton's 2nd law is based in an inertial frame, whereas, whereas most physical systems studied are referenced to a non-inertial coordinate system (the rotating earth). This gives rise to several inertial forces, which go to zero when an inertial reference frame is considered. Amongst these are the Coriolis force and the centrifugal force.

The Coriolis force is calculated as

$${{F}_{c}}=-2m\Omega \times v$$

where m represents the mass of the body, $${\Omega}$$ is the angular velocity of the rotating coordinate system, and v is the linear velocity of the body referenced to the rotating frame. The Coriolis effect of the Earth's rotation can often be ignored, but there are certain engineering applications. Long range ballistic missiles tend to veer off from their intended path if the Coriolis effect is not considered. The Earth's Coriolis effect also causes certain types of waves, the rotation of cyclones, and creates the jet streams and western boundary currents in the oceans. Various other systems may have important Coriolis contributions including rolling jets and rotating missile/gun turrets.

Another important force caused by a rotating frame of reference is the centrifugal force. This force can be defined mathematically as:

$$F=-m\Omega \times (\Omega \times r)$$

where $$ \Omega $$ is defined as above and r is the radius from the center of rotation of the frame. The centrifugal force acts outward from the center of rotation. Some examples of use for this force are: slings, artificial gravity through rotation, spin casting where liquid metal is placed inside of a tube and then the tube spun until the metal is dry thus creating a hollow metal cylinder, and centrifuges to separate materials based on their mass.

Given
The equation

$$f({{y}^{1}}(t),t)$$.

can be used to model the motion of a high speed train.

Find

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

 * style="width:95%" |
 * style="width:95%" |

$$\frac{\partial }{\partial t}f({{y}^{1}}(t),t)=f{{,}_{s}}{{y}^{1}}{{,}_{t}}+f{{,}_{t}}$$

(p1-3.3)
 *
 * }

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

 * style="width:95%" |
 * style="width:95%" |

$$\frac{d{{t}^{2}}}({{y}^{1}},t)=f{{,}_{s}}{{y}^{1}}{{,}_{tt}}+f{{,}_{ss}}{{\left( {{y}^{1}}{{,}_{t}} \right)}^{2}}+2f{{,}_{st}}{{y}^{1}}{{,}_{t}}+f{{,}_{tt}}$$

(p1-3.4)
 *
 * }

Equation 3
First, for compactness define:

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

$$s:={{y}^{1}}(t)$$ (1.1) so that the given term can be re-written as:
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

 * style="width:95%" |
 * style="width:95%" |

$$f(s,t)$$

(1.2)
 *
 * }

Next, take a derivative with respect to time of this term. This gives the equation:

$$\frac{\partial }{\partial t}f(s,t)=\frac{\partial f}{\partial s}\frac{\partial s}{\partial t}+\frac{\partial f}{\partial t}\frac{\partial t}{\partial t}$$

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

 * style="width:95%" |
 * style="width:95%" |

$$\frac{\partial }{\partial t}f(s,t)=\frac{\partial f}{\partial s}\frac{\partial s}{\partial t}+\frac{\partial f}{\partial t}$$

(1.3)
 *
 * }

Then substituting back in the definition of s (1):

$$ \frac{d}{dt}f(y^1(t),t)= \frac{\partial f(y^1(t),t)}{\partial s} \frac{\partial y^1}{\partial t}+ \frac{\partial f(y^1(t),t)}{\partial t} $$

Finally, using the notation:

$${{y}^{1}}{{,}_{t}}=\frac{d}{dt}{{y}^{1}}(t)$$

The final equation can be expressed as:

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

$$\frac{\partial }{\partial t}f({{y}^{1}}(t),t)=f{{,}_{s}}{{y}^{1}}{{,}_{t}}+f{{,}_{t}}$$
 * style="width:95%" |
 * style="width:95%" |

(p1-3.3)
 *
 * }

Equation 4
Begin with equation 1.3:

$$\frac{\partial }{\partial t}f(s,t)=\frac{\partial f}{\partial s}\frac{\partial s}{\partial t}+\frac{\partial f}{\partial t}$$

Then, taking the time derivative term-wise will produce:

$$\frac{\partial }{\partial t}\left( \frac{\partial f}{\partial t} \right)=\frac{{{\partial }^{2}}f}{\partial {{s}^{2}}}\frac{\partial s}{\partial t}\frac{\partial s}{\partial t}+\frac{{{\partial }^{2}}f}{\partial t\partial s}\frac{\partial s}{\partial t}\frac{\partial t}{\partial t}+\frac{\partial f}{\partial s}\frac{\partial {{s}^{2}}}{\partial t\partial s}+\frac{\partial f}{\partial s}\frac{{{\partial }^{2}}s}{\partial {{t}^{2}}}\frac{\partial t}{\partial t}$$

$$\frac{{{\partial }^{2}}f}{\partial {{t}^{2}}}=\frac{{{\partial }^{2}}f}{\partial {{s}^{2}}}\frac{\partial s}{\partial t}\frac{\partial s}{\partial t}+\frac{{{\partial }^{2}}f}{\partial t\partial s}\frac{\partial s}{\partial t}+\frac{\partial f}{\partial s}+\frac{\partial f}{\partial s}\frac{{{\partial }^{2}}s}{\partial {{t}^{2}}}$$

and again substituting in (1) and switching to comma notation will give the final equation:

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

$$\frac{d^2f}{dt^2}=f,_s(y^1,t) \frac{d^2y^1}{dt^2}+f,_{ss}\left(\frac{dy^1}{dt} \right)^2+2f,_{st} \frac{dy^1}{dt}+f,_{tt} $$ (p1-3.4)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Equation 1

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

$${{c}_{o}}({{y}^{1}},t)=-{{F}^{1}}[1-\overline{R}({{u}^{2}}{{,}_{ss}})({{y}^{1}},t)-{{F}^{2}}{{u}^{2}}{{,}_{s}}-\frac{T}{R}+M[[1-\overline{R}({{u}^{2}}{{,}_{ss}})][{{u}^{1}}{{,}_{tt}}-\overline{R}({{u}^{2}}{{,}_{stt}})]+{{u}^{2}}{{,}_{s}}{{u}^{2}}{{,}_{tt}}]$$     (p2-2.1)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Problem 3: Prove
Prove that the equation:

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

 * style="width:95%" |
 * style="width:95%" |

$${{c}_{3}}({{y}^{1}},t)\frac{d{{t}^{2}}}$$

(p*-*.*)
 *

 * }

is nonlinear with respect to $$ y^1 $$.

Problem 4: Boundary Value Problem
Solve the equation:

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

 * style="width:95%" |
 * style="width:95%" |

$$y(x)=cy_{H}^{1}(x)+dy_{H}^{2}+{{y}_{p}}(x)$$

(p*-*.*) using the boundary values:
 *

 * }

$$\begin{align} & y(a)=\alpha \\ & y(b)=\beta \\ \end{align}$$.

Problem 5: Prove
Show that:

$${{L}_{2}}(y_{H}^{1})={{L}_{2}}(y_{H}^{2})=0$$.

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

plugging these into the equation:

$$(1-{{x}^{2}})y''-2xy'+2y=0$$

results in:

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

similarly, the second solution

$$y=\frac{x}{2}\ln \left( \frac{1+x}{1-x}-1 \right)$$

can be shown to be a homogeneous solution to the solution by:

title: User:Egm6321.f10.team1.allison


text:
User:Kyb3R

Current Position: PhD Student in the University of Tampere, The Department of Information Studies and Interactive Media (INFIM) Research pages: http://www.uta.fi/~jm60697/ Blog: http://extreme.ajatukseni.net/ Interests: FOSS Hacktivism Hacker Culture Hackerspaces

title: User:Kyb3R


text:
User:Egm6321.f10.team03.schulze

 * Now Eq.2.4 and Eq.2.5 can be put in the form of Eq.2.3 as:

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

 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |

\left[ \begin{matrix} {{\overset{\centerdot }{\mathop{\theta }}\,}_{1}} \\ {{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{1}} \\ {{\overset{\centerdot }{\mathop{\theta }}\,}_{2}} \\ {{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{2}} \\ \end{matrix} \right]=\underbrace{\left[ \begin{matrix} 0 & 1 & 0 & 0 \\   \frac{-\left( k{{a}^{2}}+{{m}_{1}}gl \right)}{m{}_{1}{{l}^{2}}} & 0 & \frac{k{{a}^{2}}}{m{}_{1}{{l}^{2}}} & 0  \\ 0 & 0 & 0 & 1 \\   \frac{k{{a}^{2}}}{m{}_{2}{{l}^{2}}} & 0 & \frac{-\left( k{{a}^{2}}+{{m}_{2}}gl \right)}{m{}_{2}{{l}^{2}}} & 0  \\ \end{matrix} \right]}_{\underset{-}{\mathop{A}}\,}\left[ \begin{matrix} {{\theta }_{1}} \\ {{\overset{\centerdot }{\mathop{\theta }}\,}_{1}} \\ {{\theta }_{2}} \\ {{\overset{\centerdot }{\mathop{\theta }}\,}_{2}} \\ \end{matrix} \right]+\underbrace{\left[ \begin{matrix} 0 & 0 \\   \frac{1}{m{}_{1}{{l}^{2}}} & 0  \\ 0 & 0 \\   0 & \frac{1}{m{}_{2}{{l}^{2}}}  \\ \end{matrix} \right]}_{\underset{-}{\mathop{B}}\,}\left[ \begin{matrix} {{u}_{1}}l \\ {{u}_{2}}l \\ \end{matrix} \right]</math== Solution ==

$$\begin{align} & \Rightarrow \overset{\centerdot }{\mathop{\omega }}\,=\frac{-1}{\tau }\omega +\frac{Q}{\tau }ut \\ & \Rightarrow \omega =\frac{-1}{\tau }\omega t+\frac{Q}{2\tau }u{{t}^{2}}\to eq.B \\ & \overset{\centerdot }{\mathop{\phi }}\,=\omega \to eq.2\Rightarrow \phi =\omega t \\ & \Rightarrow \omega =\frac{-1}{\tau }\phi +\frac{Q}{2\tau }u{{t}^{2}}\to eq.B \\ & \Rightarrow \overset{\centerdot }{\mathop{\phi }}\,=\underbrace{\frac{-1}{\tau }}_{\underset{-}{\mathop{A}}\,}\phi +\underbrace{\frac{Q{{t}^{2}}}{2\tau }}_{\underset{-}{\mathop{B}}\,}u \\ \end{align}$$

= Problem 7 - Finding Integrating Factor and solve for y =

Given

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

y' + {x}y = 2x + 3 $$ $$
 * $$\displaystyle
 * $$\displaystyle (Eq. 1)
 * }
 * }

Find
1. Find the integrating factor 2. Solve for $$y$$

Solution
1. With an integrating factor, $$h(x)$$, we should get the equation
 * {| style="width:100%" border="0" align="left"

h = e^{ \int a0(x) dx} $$
 * $$ \displaystyle
 * $$ \displaystyle

where a0(x) = x
(hy)' = h x^2 $$ $$ Using $$h(x) = x$$ yields
 * $$\displaystyle
 * $$\displaystyle (Eq. 2)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

xy' + y = x^3 $$ $$ Dividing the above equation with $$x$$ yields the original equation
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 3)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

y' + \frac{1}{x} y = x^2 $$ $$ Hence, $$h(x) = x$$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

2. Substituting $$h(x) = x$$ into equation 2 yields
 * {| style="width:100%" border="0" align="left"

(xy)' = x^3 $$ $$ Next, we integrate the above equation
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 4)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

xy = \frac{x^4}{4} + C $$ $$ where $$C$$ is a constant. The final form of this equation is
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 5)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

 * style="width:50%; padding:10px; border:2px solid #8888aa" |
 * $$ \displaystyle
 * $$ \displaystyle

y = \frac{x^3 }{4} + \frac{C}{x} $$ $$
 * $$\displaystyle (Eq. 6)
 * }
 * }

Solution
Integrate equation 6
 * {| style="width:100%" border="0" align="left"

\int \frac{h_y}{h} dy = - \int g(y) dy $$ $$ This yields
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 7)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\ln h = - \int^y g(y) dy $$ $$ The final equation is
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 8)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

 * style="width:50%; padding:10px; border:2px solid #8888aa" |
 * $$ \displaystyle
 * $$ \displaystyle

h = e^{- \int^y g(y) dy} $$ $$
 * $$\displaystyle (Eq. 9)
 * }
 * }

title: User:Egm6321.f10.team03.schulze


text:
User:Egm6321.f10.team6.csquared

=Problem 3 - Proof of Non-Linearity in EOM of Train Wheel on Flexible Guide Rail=

Problem Statement
The equation of motion of a wheel on flexible guide rail type problem that is consistent with what occurs when the wheel of a high speed train interacts with its guiderail is given as :

$${{c}_{3\left( {{Y}^{1}},t \right)}}{{{\ddot{Y}}}^{1}}+{{c}_{2\left( {{Y}^{1}},t \right)}}{{\left( {{{\dot{Y}}}^{1}} \right)}^{2}}+{{c}_{1\left( {{Y}^{1}},t \right)}}{{{\dot{Y}}}^{1}}+{{c}_{o\left( {{Y}^{1}},t \right)}}=0$$$$\displaystyle (Eq. 1) $$

where the coefficients may be expressed as:

$$\begin{align} & {{c}_{o}}_{\left( {{Y}^{1}},t \right)}\approx -{{F}^{1}}\left[ 1-\bar{R}u{{_{,ss}^{2}}_{\left( {{Y}^{1}},t \right)}} \right]-{{F}^{2}}u{{_{,s}^{2}}_{\left( {{Y}^{1}},t \right)}}-\frac{T}{R}+ \\ & M\left[ \left[ 1-1-\bar{R}u{{_{,ss}^{2}}_{\left( {{Y}^{1}},t \right)}} \right]\left[ u{{_{,tt}^{1}}_{\left( {{Y}^{1}},t \right)}}-\bar{R}u{{_{,stt}^{2}}_{\left( {{Y}^{1}},t \right)}} \right]+u{{_{,s}^{2}}_{\left( {{Y}^{1}},t \right)}}u{{_{,tt}^{2}}_{\left( {{Y}^{1}},t \right)}} \right] \\ \end{align}$$ $$\displaystyle (Eq. 2) $$

$${{c}_{1\left( {{Y}^{1}},t \right)}}\approx 2M\left[ \left[ 1-\bar{R}u{{_{,ss}^{2}}_{\left( {{Y}^{1}},t \right)}} \right]\left[ u{{_{,st}^{1}}_{\left( {{Y}^{1}},t \right)}}-\bar{R}u{{_{,sst}^{2}}_{\left( {{Y}^{1}},t \right)}} \right]+u{{_{,s}^{2}}_{\left( {{Y}^{1}},t \right)}}u{{_{,st}^{2}}_{\left( {{Y}^{1}},t \right)}} \right]$$ $$\displaystyle (Eq. 3) $$

$${{c}_{2}}_{\left( {{Y}^{1}},t \right)}\approx M\left[ \left[ 1-\bar{R}u{{_{,ss}^{2}}_{\left( {{Y}^{1}},t \right)}} \right]\left[ u{{_{,ss}^{1}}_{\left( {{Y}^{1}},t \right)}}-\bar{R}u{{_{,sss}^{2}}_{\left( {{Y}^{1}},t \right)}} \right]+u{{_{,s}^{2}}_{\left( {{Y}^{1}},t \right)}}u{{_{,ss}^{2}}_{\left( {{Y}^{1}},t \right)}} \right]$$ $$\displaystyle (Eq. 4) $$

$${{c}_{3\left( {{Y}^{1}},t \right)}}\approx M{{\left[ 1-\bar{R}u{{_{,ss}^{2}}_{\left( {{Y}^{1}},t \right)}} \right]}^{2}}+\frac$$ $$\displaystyle (Eq. 5) $$

Required
Prove the following is non-linear:

$${{c}_{3\left( {{Y}^{1}},t \right)}}{{{\ddot{Y}}}^{1}}$$ $$\displaystyle (Eq. 6) $$

Solution
In general a function may be considered linear if the following conditions are met :

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$$f\left( \alpha x \right)=\alpha f\left( x \right)$$ $$\displaystyle (Eq. 7) $$
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$$f\left( {{x}_{1}}\pm {{x}_{2}} \right)=f\left( {{x}_{1}} \right)\pm f\left( {{x}_{2}} \right)$$ $$\displaystyle (Eq. 8) $$
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Consider for a moment the differential operator. The differential operator is a linear operator, and as such the above conditions will be met if the the differential operator is applied to an arbitrary function. For example:

$$\text{let }{{f}_{\left( y \right)}}=5y$$ $$\displaystyle (Eq. 9) $$

First let us examine the first condition of linearity given in eq. 7. According to eq. 7 the result if we multiply the differential operator by some constant C should be exactly the same as if we multiply function by some constant C and then take the derivative.

$$\begin{align} & C\frac{d{{f}_{\left( y \right)}}}{dy}=\frac{d{{f}_{\left( C\cdot y \right)}}}{dy} \\ & C\frac{d\left( 5y \right)}{dy}=C\left( 5 \right)=5C \\ & \frac{d\left( 5Cy \right)}{dy}=5C \\ \end{align}$$

Clearly the placement of the scalar multiple C has no effect on the result, and therefore the differential operator is linear. If we were to apply the second condition of linearity it would also hold.

For this particular problem we will apply the first condition of linearity to prove that the function given in eq. 6 is non-linear. By applying the first condition of linearity using an arbitrary constant α we obtain:

$$\alpha \left( M{{\left[ 1-\bar{R}u{{_{,ss}^{2}}_{\left( {{Y}^{1}},t \right)}} \right]}^{2}}+\frac \right)\ddot{Y}_{\left( {{Y}^{1}},t \right)}^{1}\overset{?}{\mathop{=}}\,\left( M{{\left[ 1-\bar{R}u{{_{,ss}^{2}}_{\left( \alpha {{Y}^{1}},t \right)}} \right]}^{2}}+\frac \right)\ddot{Y}_{\left( \alpha {{Y}^{1}},t \right)}^{1}$$

$$\displaystyle (Eq. 10) $$

Since the differential operator is linear we can take the scalar multiply from within the derivative terms to obtain:

$$\alpha \left( M{{\left[ 1-\bar{R}u{{_{,ss}^{2}}_{\left( {{Y}^{1}},t \right)}} \right]}^{2}}+\frac \right)\ddot{Y}_{\left( {{Y}^{1}},t \right)}^{1}\overset{?}{\mathop{=}}\,\left( M{{\left[ 1-\alpha \bar{R}u{{_{,ss}^{2}}_{\left( {{Y}^{1}},t \right)}} \right]}^{2}}+\frac \right)\alpha \ddot{Y}_{\left( {{Y}^{1}},t \right)}^{1}$$ $$\displaystyle (Eq. 11) $$

Doing some simple algebraic manipulation we obtain the following:

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$$\left( M{{\left[ \alpha -\alpha \bar{R}u{{_{,ss}^{2}}_{\left( {{Y}^{1}},t \right)}} \right]}^{2}}+\alpha \frac \right)\ddot{Y}_{\left( {{Y}^{1}},t \right)}^{1}\ne \left( M{{\left[ \alpha -{{\alpha }^{2}}\bar{R}u{{_{,ss}^{2}}_{\left( {{Y}^{1}},t \right)}} \right]}^{2}}+\alpha \frac \right)\ddot{Y}_{\left( {{Y}^{1}},t \right)}^{1}$$ $$\displaystyle (Eq. 12) $$
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Clearly the left and right sides of eq. 12 don't match (there is second power scalar on the right, but only a first power on the left), therefore eq. 6 is non-linear.

title: User:Egm6321.f10.team6.csquared


text:
User:Selket

 Most of my editing work is at the English Wikipedia You should also see my userpage there.

title: User:Selket


text: File:2010 09 07 14 59 22.djvu

title: File:2010 09 07 14 59 22.djvu


text:
User talk:Egm6321.f10.team4.Yoon

Team4, Try to use Joho writer.

http://writer.zoho.com/index.do

Even if you may need to sign up for using Equation function, Joho writer provides strong latex math code generator.

Also, refer this web page that is in the email from Dr. VC.

https://docs.google.com/Doc?docid=0AY5m1lfNbLroZGM4Mm5mZ2JfNThqdjlrZnBocw&hl=en

= HW Problem 1 - First and Second Order Derivative Derivations =

Given
 The nominal motion of the Maglev train can be modeled by the following equation

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$$\displaystyle f(Y^1 (t),t) $$
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Find
Derive the given modeling equations through first and second order derivatives.

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Axial deformation under moving wheel/guideway $$ \frac{d}{dt} f \left(Y^1(t),t\right) = \frac{\partial f}{\partial s} \left(Y^1(t),t\right) \dot Y^1(t) + \frac{\partial f}{\partial t} \left(Y^1(t),t\right) $$

 * (Eq.1)

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Transverse deformation of wheel/guideway $$\frac{d^2}{dt^2} f\left(Y^1(t),t\right) = f_{,s} \left(Y^1(t),t\right) \ddot Y^1 + f_{,ss} \left(Y^1(t),t\right) (\dot Y^1)^2 + 2 f_{,st} \left(Y^1(t),t\right) \dot Y^1 + f_{,tt} \left(Y^1(t),t\right) $$

 * (Eq.2)

 * }
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Solution for (Eq.1)
First, equating the function to a dummy variable $$s$$ and taking the time derivative

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$$\displaystyle Y^1(t)=s $$
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\frac{d}{dt} f \left(Y^1(t),t\right) = \frac{d}{dt} f \left(s,t\right) $$

By the chain rule, the equation becomes
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$$\displaystyle \frac{d}{dt} f \left(s,t\right) = \frac{\partial f}{\partial s} \underbrace {\frac{\partial s}{\partial t}}_{=\dot Y^1(t)}+ \frac{\partial f}{\partial t} \underbrace{\frac{\partial t}{\partial t}}_{=1} $$
 * (Eq.3)

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(Equation 3) can be rewritten by reverting function $$f$$ back to its original form:

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$$ \frac{d}{dt} f \left(Y^1(t),t\right) = \frac{\partial f}{\partial s} \left(Y^1(t),t\right) \dot Y^1(t) + \frac{\partial f}{\partial t} \left(Y^1(t),t\right) $$ (Eq.4)
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Solution for (Eq.2)
Substituting the left side of (Eq.2) to the solution of (Eq.1) yields

\frac{d^2}{dt^2} f\left(Y^1(t),t\right) = \frac{d}{dt}\left( \frac{d}{dt} f\left(Y^1(t),t\right)\right) $$

Applying the chain rule again ahead of (Eq.2) and the dummy s

= \frac{\partial}{\partial s}\left( \frac{d}{dt} f\left(Y^1(t),t\right)\right) + \frac{\partial}{\partial t}\left( \frac{d}{dt} f\left(Y^1(t),t\right)\right) = \frac{\partial}{\partial s}\left( \frac{\partial f}{\partial s} \frac{\partial s}{\partial t}+ \frac{\partial f}{\partial t} \right) + \frac{\partial}{\partial t}\left( \frac{\partial f}{\partial s} \frac{\partial s}{\partial t} +  \frac{\partial f}{\partial t} \right) $$

=\frac{\partial^2 f}{\partial s^2} \left(\frac{\partial s}{\partial t}\right)^2 + \frac{\partial^2 f}{\partial s \partial t} \frac{\partial s}{\partial t} \underbrace{\frac{\partial t}{\partial t}}_{=1} + \frac{\partial f}{\partial s} \underbrace{\frac{\partial^2 s}{\partial s \partial t}}_{=0}\frac{\partial s}{\partial t} + \frac{\partial f}{\partial s} \frac{\partial^2 s}{\partial t^2}\underbrace{\frac{\partial t}{\partial t}}_{=1} + \frac{\partial^2 f}{\partial s\partial t}\frac{\partial s}{\partial t} + \frac{\partial^2 f}{\partial t^2} \underbrace{\frac{\partial t}{\partial t}}_{=1} $$

Collecting terms in the above equation by reduced order of y fucntion.

=\frac{\partial^2 f}{\partial s^2} \left(\frac{\partial s}{\partial t}\right)^2 + 2\frac{\partial^2 f}{\partial s \partial t} \frac{\partial s}{\partial t} + \frac{\partial f}{\partial s} \frac{\partial^2 s}{\partial t^2} +  \frac{\partial^2 f}{\partial t^2} $$

By transforming the Leivniz expression to Lagrange expression such as X,y instead of dx/dy,

\frac{d^2}{dt^2} f\left(Y^1(t),t\right) =f_{,ss} \underbrace{\left(\frac{\partial s}{\partial t}\right)^2}_{\dot {Y^1}^{2}} + 2f_{,st} \underbrace{\frac{\partial s}{\partial t}}_{\dot Y^1} + f_{,s}\underbrace{\frac{\partial^2 s}{\partial t^2}}_{\ddot Y^1} + f_{,tt} $$ Finally, we can complete the desired solution by displacing the dummy variable s to the original function.
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$$ \frac{d^2}{dt^2} f\left(Y^1(t),t\right) =f_{,ss} \left(\dot Y^1\right)^2 + 2f_{,st} \dot Y^1 + f_{,s}\ddot Y^1 + f_{,tt} $$
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Similaritry to/Difference from Derivation of Coriolis Force
in Rotating coordinate systems(i.e. the earth) To compare second order differential equation to derivation of Coriolis effect, the idea of fictitious forces is used. If the rotation of frame B can be described as a vector Ω pointed along the axis of rotation

 * $$ |\boldsymbol{\Omega} | = \frac {d \theta }{dt} = \omega (t) \, $$

then the time derivative of the three unit vectors consisting the vector Ω, defined as an observed frame with respect to frame A is

 * $$ \frac {d \mathbf{u}_j (t)}{dt} = \boldsymbol{\Omega} \times \mathbf{u}_j (t) \, $$

and

 * $$\frac {d^2 \mathbf{u}_j (t)}{dt^2}= \frac{d\boldsymbol{\Omega}}{dt} \times \mathbf{u}_j+ \boldsymbol{\Omega} \times \underbrace{\left(  \boldsymbol{\Omega} \times  \mathbf{u}_j (t) \right)}_{\frac {d \mathbf{u}_j (t)}{dt}}, $$

as is verified using the properties of the vector cross product. These derivative formulas now are applied to the relationship between acceleration in an inertial frame, and that in a coordinate frame rotating with time-varying angular velocity ω ( t ). From the previous section, where subscript A refers to the inertial frame and B to the rotating frame, setting aAB = 0 to remove any translational acceleration, and focusing on only rotational properties

 * $$ \frac {d^2 \mathbf{x}_{A}}{dt^2}=\mathbf{a}_B + 2\ \sum_{j=1}^3 v_j \ \frac{d \mathbf{u}_j}{dt} $$ $$+ \sum_{j=1}^3 x_j \ \frac{d^2 \mathbf{u}_j}{dt^2}\ ,$$
 * $$\mathbf{a}_A=\mathbf{a}_B +\ 2\ \sum_{j=1}^3 v_j \boldsymbol{\Omega} \times \mathbf{u}_j (t)\ $$ $$+\   \sum_{j=1}^3 x_j \frac{d\boldsymbol{\Omega}}{dt} \times \mathbf{u}_j  \  + \sum_{j=1}^3 x_j \boldsymbol{\Omega} \times \left(  \boldsymbol{\Omega} \times  \mathbf{u}_j (t) \right)\ $$
 * $$=\mathbf{a}_B $$ $$+ 2\ \boldsymbol{\Omega} \times\sum_{j=1}^3 v_j \mathbf{u}_j (t) \  $$ $$+ \frac{d\boldsymbol{\Omega}}{dt} \times \sum_{j=1}^3 x_j \mathbf{u}_j  \   $$ $$+\ \boldsymbol{\Omega} \times \left(  \boldsymbol{\Omega} \times \sum_{j=1}^3 x_j  \mathbf{u}_j (t) \right)\ .$$

Collecting terms,

 * $$\mathbf{a}_A=\mathbf{a}_B + 2\ \boldsymbol{\Omega} \times\mathbf{v}_B\ $$ $$+ \frac{d\boldsymbol{\Omega}}{dt} \times \mathbf{x}_B  \  + \boldsymbol{\Omega} \times \left(  \boldsymbol{\Omega} \times \mathbf{x}_B \right)\ .$$

Technically, the aceleration aA in the inertial frame A is not exact with the acceleration aB seen by observers in the rotational frame B but has several additional geometric acceleration terms associated with the rotation of B. Rearranging the above equation by the acceleration B

\mathbf{a}_{B} = \mathbf{a}_A - 2 \boldsymbol\Omega \times \mathbf{v}_{B} - \boldsymbol\Omega \times (\boldsymbol\Omega \times  \mathbf{x}_B )  - \frac{d \boldsymbol\Omega}{dt} \times \mathbf{x}_B \. $$

The force upon the object in the B coordinate system with respect to observers in the rotating frame A is FB = m aB. If their observations are to result in the correct force on the object when using Newton's laws, they must consider that the additional force Ffict is present, so the end result is FB = FA + Ffict. Thus, the fictitious force used by observers in B to get the correct behavior of the object from Newton's laws equals:

\mathbf{F}_{\mathrm{fict}} = \underbrace{- 2 m \boldsymbol\Omega \times \mathbf{v}_{B}}_{Coriolis force} \underbrace{- m \boldsymbol\Omega \times (\boldsymbol\Omega \times \mathbf{x}_B )}_{centrifugal force} $$ $$ \underbrace{\ - m \frac{d \boldsymbol\Omega }{dt} \times \mathbf{x}_B}_{Euler force=0}  \. $$

When the rate of rotation doesn't change, as is typically the case for a planet, the Euler force is zero.

= HW1.problem 6 =

Given

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$$ \left(1-x^2 \right)\cdot y''-2x\cdot y'+2y=0$$
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Verify

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$$L_{2} \left( {y}^{1}_{H}\right)=L_{2} \left( {y}^{2}_{H}\right)=0$$
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Solution

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To find $$L_{2} \left( {y}^{1}_{H}\right)=0$$
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Differntiating the first given solution, and differentiating again to find

$${y}^{1}_{H}\left(x \right)=x$$

$$\frac{dy}{dx}=1,$$

$$ \frac{d^2y}{dx^2}=0$$

These can then be substituted into the given differential equation (1) to find

$$ L_{2} \left( {y}^{1}_{H}\right)=\left(1-x^2 \right)\cdot \left( 0 \right)-2x\left( 1 \right)+2x=0$$

To find $$L_{2} \left( {y}^{2}_{H}\right)=0$$

Repeatly differentiated the second solution to find

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

$$\frac{dy}{dx}= \frac{1}{2}log \left( \frac{1+x}{1-x} \right)+ \left( \frac{1}{1+x}+ \frac{1}{1-x}\right)=\frac{1}{2}log \left( \frac{1+x}{1-x} \right)+ \left(  \frac{x}{1-x^2}\right),  $$

$$\frac{d^2y}{dx^2}= \frac{1}{2} \left( \frac{1}{1+x}+ \frac{1}{1-x} \right)+ \frac{1\cdot \left( 1-x^2\right)-x\cdot \left( -2x\right)}{\left( 1-x^2 \right)^2}=\frac{1}{ \left(1-{x}^{2} \right)}+ \frac{1+x^2}{\left( 1-x^2 \right)^2}$$

Again, substituted into the given differential equation (1) to find

$$L_{2} \left( {y}^{2}_{H}\right)=\left(1-x^2 \right)\cdot \left( \frac{1}{ \left(1-{x}^{2} \right)}+ \frac{1+x^2}{\left( 1-x^2 \right)^2} \right)-2x\left( \frac{1}{2}log \left( \frac{1+x}{1-x} \right)+ \left( \frac{x}{1-x^2}\right) \right)+2\left(\frac{x}{2}log \left( \frac{1+x}{1-x} \right)-1\right)=1+\frac{1-x^2}{1-x^2}-2=0$$

Thus,
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$$L_{2} \left( {y}^{1}_{H}\right)=L_{2} \left( {y}^{2}_{H}\right)=0$$
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==(Referred from Wikipedia)Mathematical derivation of fictitious forces for understanding of Coriolis effect ["Fictitious force" http://en.wikipedia.org/w/index.php?title=Fictitious_force&action=submit] ==

General derivation
Many problems require use of noninertial reference frames, for example, those involving satellites and particle accelerators. Figure 2 shows a particle with mass m and position vector xA(t) in a particular inertial frame A. Consider a non-inertial frame B whose origin relative to the inertial one is given by XAB(t). Let the position of the particle in frame B be xB(t). What is the force on the particle as expressed in the coordinate system of frame B?

To answer this question, let the coordinate axis in B be represented by unit vectors uj with j any of { 1, 2, 3 } for the three coordinate axes. Then

 * $$ \mathbf{x}_{B} = \sum_{j=1}^3 x_j\ \mathbf{u}_j \ . $$

The interpretation of this equation is that xB is the vector displacement of the particle as expressed in terms of the coordinates in frame B at time t. From frame A the particle is located at:

 * $$\mathbf{x}_A =\mathbf{X}_{AB} + \sum_{j=1}^3 x_j\ \mathbf{u}_j \ . $$

As an aside, the unit vectors { uj } cannot change magnitude, so derivatives of these vectors express only rotation of the coordinate system B. On the other hand, vector XAB simply locates the origin of frame B relative to frame A, and so cannot include rotation of frame B.

Taking a time derivative, the velocity of the particle is:

 * $$ \frac {d \mathbf{x}_{A}}{dt} =\frac{d \mathbf{X}_{AB}}{dt}+ \underbrace{\sum_{j=1}^3 \frac{dx_j}{dt} \mathbf{u}_j}_{\mathbf{v}_B} + \sum_{j=1}^3 x_j\ \frac{d \mathbf{u}_j}{dt} \ . $$

The second term summation is the velocity of the particle, say vB as measured in frame B. That is:

 * $$ \frac {d \mathbf{x}_{A}}{dt} =\mathbf{v}_{AB}+ \mathbf{v}_B + \sum_{j=1}^3 x_j \ \frac{d \mathbf{u}_j}{dt} \ . $$

The interpretation of this equation is that the velocity of the particle seen by observers in frame A consists of what observers in frame B call the velocity, namely vB, plus two extra terms related to the rate of change of the frame-B coordinate axes. One of these is simply the velocity of the moving origin vAB. The other is a contribution to velocity due to the fact that different locations in the non-inertial frame have different apparent velocities due to rotation of the frame; a point seen from a rotating frame has a rotational component of velocity that is greater the further the point is from the origin.

To find the acceleration, another time differentiation provides:

 * $$ \frac {d^2 \mathbf{x}_{A}}{dt^2} = \mathbf{a}_{AB}+\frac {d\mathbf{v}_B}{dt} + \sum_{j=1}^3 \frac {dx_j}{dt} \ \frac{d \mathbf{u}_j}{dt} + \sum_{j=1}^3 x_j \ \frac{d^2 \mathbf{u}_j}{dt^2}\ . $$

Using the same formula already used for the time derivative of xB, the velocity derivative on the right is:

 * $$\frac {d\mathbf{v}_B}{dt} =\sum_{j=1}^3 \frac{d v_j}{dt} \ \mathbf{u}_j+ \sum_{j=1}^3 v_j \ \frac{d \mathbf{u}_j}{dt} =\mathbf{a}_B + \sum_{j=1}^3 v_j \ \frac{d \mathbf{u}_j}{dt} \ . $$

Consequently,
 * $$ \frac {d^2 \mathbf{x}_{A}}{dt^2}=\mathbf{a}_{AB}+\mathbf{a}_B + 2\ \sum_{j=1}^3 v_j \ \frac{d \mathbf{u}_j}{dt} + \sum_{j=1}^3 x_j \ \frac{d^2 \mathbf{u}_j}{dt^2}\ . $$   (Eq. 1)

The interpretation of this equation is as follows: the acceleration of the particle in frame A consists of what observers in frame B call the particle acceleration aB, but in addition there are three acceleration terms related to the movement of the frame-B coordinate axes: one term related to the acceleration of the origin of frame B, namely aAB, and two terms related to rotation of frame B. Consequently, observers in B will see the particle motion as possessing "extra" acceleration, which they will attribute to "forces" acting on the particle, but which observers in A say are "fictitious" forces arising simply because observers in B do not recognize the non-inertial nature of frame B.

The factor of two in the Coriolis force arises from two equal contributions: (i) the apparent change of an inertially constant velocity with time because rotation makes the direction of the velocity seem to change (a dvB / dt term) and (ii) an apparent change in the velocity of an object when its position changes, putting it nearer to or further from the axis of rotation (the change in Σ xj $$\stackrel{\frac{d}{dt}}{}$$uj due to change in x j ).

To put matters in terms of forces, the accelerations are multiplied by the particle mass:

 * $$\mathbf{F}_A = \mathbf{F}_B + m\ \mathbf{a}_{AB}+ 2m\ \sum_{j=1}^3 v_j \ \frac{d \mathbf{u}_j}{dt} + m\ \sum_{j=1}^3 x_j \ \frac{d^2 \mathbf{u}_j}{dt^2}\ . $$

The force observed in frame B, FB = m aB is related to the actual force on the particle, FA, by:

 * $$\mathbf{F}_B = \mathbf{F}_A + \mathbf{F}_{\mbox{fictitious}} \ . $$

where:

 * $$ \mathbf{F}_{\mbox{fictitious}} =-m\ \mathbf{a}_{AB} -2m\ \sum_{j=1}^3 v_j \ \frac{d \mathbf{u}_j}{dt} - m\ \sum_{j=1}^3 x_j \ \frac{d^2 \mathbf{u}_j}{dt^2}\ . $$

Thus, we can solve problems in frame B by assuming that Newton's second law holds (with respect to quantities in that frame) and treating Ffictitious as an additional force.

Below are a number of examples applying this result for fictitious forces. More examples can be found in the article on centrifugal force.

Rotating coordinate systems
A common situation in which noninertial reference frames are useful is when the reference frame is rotating. Because such rotational motion is non-inertial, due to the acceleration present in any rotational motion, a fictitious force can always be invoked by using a rotational frame of reference. Despite this complication, the use of fictitious forces often simplifies the calculations involved.

To derive expressions for the fictitious forces, derivatives are needed for the apparent time rate of change of vectors that take into account time-variation of the coordinate axes. If the rotation of frame B is represented by a vector Ω pointed along the axis of rotation with orientation given by the right-hand rule, and with magnitude given by

 * $$ |\boldsymbol{\Omega} | = \frac {d \theta }{dt} = \omega (t) \, $$

then the time derivative of any of the three unit vectors describing frame B is:

 * $$ \frac {d \mathbf{u}_j (t)}{dt} = \boldsymbol{\Omega} \times \mathbf{u}_j (t) \, $$

and

 * $$\frac {d^2 \mathbf{u}_j (t)}{dt^2}= \frac{d\boldsymbol{\Omega}}{dt} \times \mathbf{u}_j +\boldsymbol{\Omega} \times \frac{d \mathbf{u}_j (t)}{dt} $$ $$= \frac{d\boldsymbol{\Omega}}{dt} \times \mathbf{u}_j+ \boldsymbol{\Omega} \times \left(  \boldsymbol{\Omega} \times  \mathbf{u}_j (t) \right)\, $$

as is verified using the properties of the vector cross product. These derivative formulas now are applied to the relationship between acceleration in an inertial frame, and that in a coordinate frame rotating with time-varying angular velocity ω ( t ). From the previous section, where subscript A refers to the inertial frame and B to the rotating frame, setting aAB = 0 to remove any translational acceleration, and focusing on only rotational properties (see Eq. 1):

 * $$ \frac {d^2 \mathbf{x}_{A}}{dt^2}=\mathbf{a}_B + 2\ \sum_{j=1}^3 v_j \ \frac{d \mathbf{u}_j}{dt} $$ $$+ \sum_{j=1}^3 x_j \ \frac{d^2 \mathbf{u}_j}{dt^2}\ ,$$
 * $$\mathbf{a}_A=\mathbf{a}_B +\ 2\ \sum_{j=1}^3 v_j \boldsymbol{\Omega} \times \mathbf{u}_j (t)\ $$ $$+\   \sum_{j=1}^3 x_j \frac{d\boldsymbol{\Omega}}{dt} \times \mathbf{u}_j  \  + \sum_{j=1}^3 x_j \boldsymbol{\Omega} \times \left(  \boldsymbol{\Omega} \times  \mathbf{u}_j (t) \right)\ $$
 * $$=\mathbf{a}_B $$ $$+ 2\ \boldsymbol{\Omega} \times\sum_{j=1}^3 v_j \mathbf{u}_j (t) \  $$ $$+ \frac{d\boldsymbol{\Omega}}{dt} \times \sum_{j=1}^3 x_j \mathbf{u}_j  \   $$ $$+\ \boldsymbol{\Omega} \times \left(  \boldsymbol{\Omega} \times \sum_{j=1}^3 x_j  \mathbf{u}_j (t) \right)\ .$$

Collecting terms, the result is the so-called acceleration transformation formula:

 * $$\mathbf{a}_A=\mathbf{a}_B + 2\ \boldsymbol{\Omega} \times\mathbf{v}_B\ $$ $$+ \frac{d\boldsymbol{\Omega}}{dt} \times \mathbf{x}_B  \  + \boldsymbol{\Omega} \times \left(  \boldsymbol{\Omega} \times \mathbf{x}_B \right)\ .$$

The physical acceleration aA due to what observers in the inertial frame A call real external forces on the object is, therefore, not simply the acceleration aB seen by observers in the rotational frame B, but has several additional geometric acceleration terms associated with the rotation of B. As seen in the rotational frame, the acceleration aB of the particle is given by rearrangement of the above equation as:

\mathbf{a}_{B} = \mathbf{a}_A - 2 \boldsymbol\Omega \times \mathbf{v}_{B} - \boldsymbol\Omega \times (\boldsymbol\Omega \times  \mathbf{x}_B )  - \frac{d \boldsymbol\Omega}{dt} \times \mathbf{x}_B \. $$

The net force upon the object according to observers in the rotating frame is FB = m aB. If their observations are to result in the correct force on the object when using Newton's laws, they must consider that the additional force Ffict is present, so the end result is FB = FA + Ffict. Thus, the fictitious force used by observers in B to get the correct behavior of the object from Newton's laws equals:

\mathbf{F}_{\mathrm{fict}} = - 2 m \boldsymbol\Omega \times \mathbf{v}_{B} - m \boldsymbol\Omega  \times (\boldsymbol\Omega \times \mathbf{x}_B ) $$ $$\ - m \frac{d \boldsymbol\Omega  }{dt} \times \mathbf{x}_B \. $$

Here, the first term is the Coriolis force, the second term is the centrifugal force, and the third term is the Euler force. When the rate of rotation doesn't change, as is typically the case for a planet, the Euler force is zero.

Reference http://en.wikipedia.org/w/index.php?title=Fictitious_force&action=submit

File:Evenness of legendre.png
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title: User talk:Egm6321.f10.team4.Yoon


text:
User:Felicityrai

I am Felicity Rai, PhD candidate at the University of Canberra. My topic is the South China Morning Post, an English-language newspaper in Hong Kong. =Contact= My email address is felicityraiATgmailDOTcom. My mobile number is 61419261539. =Projects=

Current
/PhD/

title: User:Felicityrai


text:
User talk:Felicityrai

Hi Felicity, thanks for joining the session Wikis for research. If you could go to that page and sign the Participants section. Leighblackall 02:09, 8 September 2010 (UTC)

Hi Felicity - glad you managed to get this info. I find it extremely useful. It has opened a few doors, I now want to further push open. Interesting reserach topic you have. My research area is also media. I'll galdly follow up on what you are doing. Joelle

title: User talk:Felicityrai


text:
User:Felicityrai/PhD

My research in Hong Kong ranges in time from 1941 until 2007. I will be looking for incidences of agenda-setting at this broadsheet English-language newspaper.

title: User:Felicityrai/PhD


text:
User talk:MAS100

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title: User talk:MAS100


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User talk:Rotorhead15

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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 11:51, 9 September 2010 (UTC)

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title: User talk:Rotorhead15


text:
User talk:Jerome0012

 Hello Jerome0012, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

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title: User talk:Jerome0012


text:
User talk:Egm6321.f10.team03.schulze

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title: User talk:Egm6321.f10.team03.schulze


text:
User talk:Kyb3R

 Hello Kyb3R, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

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title: User talk:Kyb3R


text:
User talk:Egm6321.f10.team1.allison

 Hello Egm6321.f10.team1.allison, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

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And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Ottava Rima (talk) 02:03, 8 September 2010 (UTC)

title: User talk:Egm6321.f10.team1.allison


text:
User talk:Pinpinpim

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

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And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Ottava Rima (talk) 02:03, 8 September 2010 (UTC)

title: User talk:Pinpinpim