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Plasmas/Magnetohydrodynamics

The word magnetohydrodynamics (MHD) is derived from magneto- meaning magnetic field, and hydro- meaning liquid, and -dynamics meaning movement.

Magnetodynamics
In magnetostatics and magnetodynamics, Gauss's law for magnetism and Maxwell–Faraday equation are respectively:

 * $$\nabla \cdot \mathbf{B} = 0,\quad \frac{ \partial \mathbf{B}}{ \partial t } + \nabla \times \mathbf{E} = 0 $$

which reduce to the Bianchi identity:

 * $$ \partial_\gamma F_{ \alpha \beta } + \partial_\alpha F_{ \beta \gamma } + \partial_\beta F_{ \gamma \alpha } = 0 $$

or using index notation with square brackets for the antisymmetric part of the tensor:

 * $$ \partial_{ [ \alpha } F_{ \beta \gamma ] } = 0 $$

By definition,

 * $$ T_{[abc]} = \frac{1}{3!}(T_{abc} + T_{bca} + T_{cab} - T_{acb} - T_{bac} - T_{cba})$$

So if
 * $$ \partial_\gamma F_{ \alpha \beta } + \partial_\alpha F_{ \beta \gamma } + \partial_\beta F_{ \gamma \alpha } = 0$$

then
 * $$\begin{align}

0 & = \begin{matrix} \frac{2}{6} \end{matrix} ( \partial_\gamma F_{ \alpha \beta } + \partial_\alpha F_{ \beta \gamma } + \partial_\beta F_{ \gamma \alpha }) \\ & = \begin{matrix} \frac{1}{6} \end{matrix} \{ \partial_\gamma (2F_{ \alpha \beta }) + \partial_\alpha (2F_{ \beta \gamma }) + \partial_\beta (2F_{ \gamma \alpha }) \} \\ & = \begin{matrix} \frac{1}{6} \end{matrix} \{ \partial_\gamma (F_{ \alpha \beta } - F_{ \beta \alpha}) + \partial_\alpha (F_{ \beta \gamma } - F_{ \gamma \beta}) + \partial_\beta (F_{ \gamma \alpha } - F_{ \alpha \gamma}) \} \\ & = \begin{matrix} \frac{1}{6} \end{matrix} ( \partial_\gamma F_{ \alpha \beta } + \partial_\alpha F_{ \beta \gamma } + \partial_\beta F_{ \gamma \alpha } - \partial_\gamma F_{ \beta \alpha} - \partial_\alpha F_{ \gamma \beta} - \partial_\beta F_{ \alpha \gamma} ) \\ & = \partial_{[ \gamma} F_{ \alpha \beta ]} \end{align}$$

Hydrodynamics
Def. a branch of science concerned with forces acting on or exerted by fluids (especially liquids) is called hydrodynamics.

Theoretical magnetohydrodynamics
Def. the study of the interaction of electrically conducting fluids with magnetic fields is called magnetohydrodynamics (MHD).

Plasma object entities
Def. "the fraction of a given part [or the whole] of [a stellar surface] covered by magnetic field" is called the magnetic filling factor.

Strong forces
The incompressible MHD equations are

$$ \begin{array}{lcl} \displaystyle\frac{\partial \mathbf{u}}{\partial t} + \mathbf{u} \cdot \nabla \mathbf{u}  & = & -\nabla p + \mathbf{B} \cdot \nabla \mathbf{B} + \nu \nabla^2 \mathbf{u} \\

\displaystyle \frac{\partial \mathbf{B}}{\partial t} + \mathbf{u} \cdot \nabla \mathbf{B}  & = & \mathbf{B} \cdot \nabla \mathbf{u}  + \eta \nabla^2 \mathbf{B} \\

\nabla \cdot \mathbf{u} & = & 0 \\ \nabla \cdot \mathbf{B} & = & 0.

\end{array} $$

where u, B, p represent the velocity, magnetic, and total pressure (thermal+magnetic) fields, $$\nu $$ and $$\eta $$ represent kinematic viscosity and magnetic diffusivity. The third equation is the incompressibility condition. In the above equation, the magnetic field is in Alfvén units (same as velocity units). The total magnetic field can be split into two parts: $$ \mathbf{B} = \mathbf{B_0} + \mathbf{b} $$ (mean + fluctuations).

The above equations in terms of Elsässer variables ($$ \mathbf{z}^{\pm} = \mathbf{u} \pm \mathbf{b} $$) are

$$ \frac{\partial {\mathbf{z}^{\pm}}}{\partial t}\mp\left(\mathbf {B}_0\cdot{\mathbf \nabla}\right){\mathbf z^{\pm}} + \left({\mathbf z^{\mp}}\cdot{\mathbf \nabla}\right){\mathbf z^{\pm}} = -{\mathbf \nabla}p + \nu_+ \nabla^2 \mathbf{z}^{\pm} + \nu_- \nabla^2 \mathbf{z}^{\mp} $$

where $$ \nu_\pm = \nu \pm \eta $$. Nonlinear interactions occur between the Alfvénic fluctuations $$ z^{\mp} $$.

The important nondimensional parameters for MHD are

$$ \begin{array}{lcl} \text{Reynold number } Re & = & U L /\nu \\ \text{Magnetic Reynold number } Re_M & = & U L /\eta \\ \end{array} $$

The Magnetic Reynolds number is a dimensionless group that gives an estimate of the effects of magnetic advection to magnetic diffusion.

$$ \begin{array}{lcl} \text{Magnetic Prandtl number } P_M & = & \nu / \eta. \end{array} $$

The magnetic Prandtl number is an important property of the fluid. Liquid metals have small magnetic Prandtl numbers, for example, liquid sodium's $$ P_M $$ is around $$ 10^{-5} $$. But plasmas have large $$ P_M $$.

The Reynolds number is the ratio of the nonlinear term $$ \mathbf{u} \cdot \nabla \mathbf{u} $$ of the Navier-Stokes equation to the viscous term. While the magnetic Reynolds number is the ratio of the nonlinear term and the diffusive term of the induction equation.

In many practical situations, the Reynolds number $$ Re $$ of the flow is quite large. For such flows typically the velocity and the magnetic fields are random. Such flows are called to exhibit MHD turbulence. Note that $$ Re_M $$ need not be large for MHD turbulence. $$ Re_M $$ plays an important role in dynamo (magnetic field generation) problem.

The mean magnetic field plays an important role in MHD turbulence, for example it can make the turbulence anisotropic; suppress the turbulence by decreasing energy cascade etc. The earlier MHD turbulence models assumed isotropy of turbulence, while the later models have studied anisotropic aspects. In the following discussions will summarize these models. More discussions on MHD turbulence can be found in Biskamp and Verma.

Isotropic turbulence
Iroshnikov and Kraichnan formulated the first phenomenological theory of MHD turbulence. They argued that in the presence of a strong mean magnetic field, $$ z^+ $$  and $$ z^- $$ wavepackets  travel in opposite directions with the phase velocity of $$B_0$$, and interact weakly. The relevant time scale is Alfven time $$(B_0 k)^{-1}$$. As a results the energy spectra is

where $$ \Pi $$ is the energy cascade rate.

Later Dobrowolny et al. derived the following generalized formulas for the cascade rates of $$ z^{\pm} $$ variables:

where $$ \tau^{\pm} $$ are the interaction time scales of $$ z^{\pm} $$ variables.

Iroshnikov and Kraichnan's phenomenology follows once we choose $$ \tau^{\pm} \approx 1/(k V_A) $$.

Marsch chose the nonlinear time scale $$ T_{NL}^{\pm} \approx (k z_k^{\mp})^{-1} $$ as the interaction time scale for the eddies and derived Kolmogorov-like energy spectrum for the Elsasser variables: where $$ \Pi^+ $$  and $$  \Pi^- $$ are the energy cascade rates of $$ z^+ $$  and $$ z^- $$  respectively, and $$ K^{\pm} $$  are constants.

Matthaeus and Zhou attempted to combine the above two time scales by postulating the interaction time to be the harmonic mean of Alfven time and nonlinear time.

The main difference between the two competing phenomenologies (-3/2 and -5/3) is the chosen time scales for the interaction time. The main underlying assumption in that Iroshnikov and Kraichnan's phenomenology should work for strong mean magnetic field, whereas Marsh's phenomenology should work when the fluctuations dominate the mean magnetic field (strong turbulence).

Anisotropic turbulence
"Mean magnetic field makes turbulence anisotropic. This aspect has been studied in last two decades. In the limit $$ \delta z^{\pm} \ll B_0 $$,  Galtier et al. showed using kinetic equations that

$$

E(k) \sim (\Pi B_0)^{1/2} k_{||}^{1/2} k_{\perp}^{-2}

$$

where $$ k_{||} $$ and $$ k_{\perp} $$ are components of the wavenumber parallel and perpendicular to mean magnetic field. The above limit is called the weak turbulence limit.

Under the strong turbulence limit, $$ \delta z^{\pm} \sim B_0 $$, Goldereich and Sridhar argue that $$ k_{\perp} z_{k_{\perp}} \sim k_{||} B_0 $$ ("critical balanced state") which implies that

$$ \begin{array}{lcl} E(k) & \propto & k_{\perp}^{-5/3}; \\ k_{||} & \propto & k_{\perp}^{2/3} \end{array} $$

The above anisotropic turbulence phenomenology has been extended for large cross helicity MHD.

Electromagnetics
The Stuart number (N, also known as magnetic interaction parameter) is a dimensionless number of fluids, i.e. gases or liquids.

It is defined as the ratio of electromagnetic to inertial forces, which gives an estimate of the relative importance of a magnetic field on a flow. The Stuart number is relevant for flows of conducting fluids, e.g. in fusion reactors, steel casters or plasmas.

 * $$ {\it N} = \frac{Ha^2}{Re} = \frac {B^2 \cdot L_{c} \cdot \sigma}{\rho U} $$

 * $$Ha$$ – Hartmann number
 * $$Re$$ – Reynolds number
 * $$B$$ – magnetic flux density
 * $$L_c$$ – characteristic length
 * $$\sigma$$ – electric conductivity
 * $$\mu$$ – dynamic viscosity, sometimes denoted as $$\eta$$

Electrohydrodynamics
Electrohydrodynamics (EHD), also known as electro-fluid-dynamics (EFD) or electrokinetics, is the study of the dynamics of electrically charged fluids. It is the study of the motions of ionised particles or molecules and their interactions with electric fields and the surrounding fluid. The term may be considered to be synonymous with the rather elaborate electrostrictive hydrodynamics. EHD covers the following types of particle and fluid transport mechanisms: Electrophoresis, electrokinesis, dielectrophoresis, electro-osmosis, and electrorotation. In general, the phenomena relate to the direct conversion of electrical energy into kinetic energy, and vice versa.

In the first instance, shaped electrostatic fields create hydrostatic pressure (or motion) in dielectric media. When such media are fluids, a flow is produced. If the dielectric is a vacuum or a solid, no flow is produced. Such flow can be directed against the electrodes, generally to move the electrodes. In such case, the moving structure acts as an electric motor.

Hartmann numbers
A Hartmann number is the ratio of electromagnetic force to the viscous force first introduced by Hartmann. It is defined by:

 * $$H_a = BL\sqrt{\frac{\sigma} {\mu}}$$

where
 * $$B$$ is the magnetic field
 * $$L$$ is the characteristic length scale
 * $$\sigma$$ is the electrical conductivity
 * $$\mu$$ is the viscosity

Computational magnetohydrodynamics
Computational magnetohydrodynamics (CMHD) is a rapidly developing branch of magnetohydrodynamics that uses numerical methods and algorithms to solve and analyze problems that involve electrically conducting fluids. Most of the methods used in CMHD are borrowed from the well established techniques employed in Computational fluid dynamics. The complexity mainly arises due to the presence of a magnetic field and its coupling with the fluid. One of the important issues is to numerically maintain the $$ \nabla \cdot {\mathbf B} = 0$$ (conservation of magnetic flux) condition, from Maxwell's equations, to avoid any unphysical effects.

Numerical resistivity is a problem in computer simulations of ideal magnetohydrodynamics (MHD). It is a form of numerical diffusion. In near-ideal MHD systems, the magnetic field can diffuse only very slowly through the plasma or fluid of the system; it is rate-limited by the resistivity of the fluid. In Eulerian simulations where the field is arbitrarily aligned compared to the simulation grid, the numerical diffusion rate takes the form similar to an additional resistivity, causing non-physical and sometimes bursty magnetic reconnection in the simulation. Numerical resistivity is a function of resolution, alignment of the magnetic field with the grid, and numerical method. In general, numerical resistivity will not behave isotropically, and there can be different effective numerical resistivities in different parts of the computational domain. For current (2005) simulations of the solar corona and inner heliosphere, this numerical effect can be several orders of magnitude larger than the physical resistivity of the plasma.

Weak forces
"[M]otions resulting from [a linear magnetohydrodynamic] instability act as a dynamo to sustain the magnetic field." "Supersonic flows are initially generated by the Balbus-Hawley magnetic shear instability."

A plasma with local magnetohydrodynamic instabilities creates mechanical turbulence, motion, or shear (a dynamo) which in turn generates or sustains the local magnetic field.

"Magnetic field generation by dynamo action is often studied within the theoretical framework of magnetohydrodynamics (MHD)."

"For small values of the Hall parameter, the small-scale dynamo is more efficient, displaying faster growth and saturating at larger amplitudes of the magnetic field. For larger values of the Hall parameter, saturation of the magnetic field is reached at smaller amplitudes than in the MHD case."

When this magnetohydrodynamic dynamo occurs between or within radiative layers, a radiative dynamo is operating.

The kinematic approximation becomes invalid when the magnetic field becomes strong enough to affect the fluid motions. In that case the velocity field becomes affected by the Lorentz force, and so the induction equation is no longer linear in the magnetic field. In most cases this leads to a quenching of the amplitude of the dynamo. Such dynamos are sometimes also referred to as hydromagnetic dynamos. Virtually all dynamos in astrophysics and geophysics are hydromagnetic dynamos.

Numerical models are used to simulate fully nonlinear dynamos. A minimum of 5 equations are needed. They are as follows. The induction equation, see above. Maxwell's equation:

 * $$ \nabla \cdot \mathbf{B}=0$$

The (sometimes) Boussinesq conservation of mass:

 * $$ \nabla \cdot \mathbf{u} = 0 $$

The (sometimes) Boussinesq conservation of momentum, also known as the Navier-Stokes equation:

 * $$ \frac{D\mathbf{u}}{Dt} = -\nabla p + \nu \nabla^2 \mathbf{u} + \rho^'\mathbf{g} + 2\mathbf{\Omega} \times \mathbf{u} + \mathbf{\Omega} \times \mathbf{\Omega} \times \mathbf{R} + \mathbf{J} \times \mathbf{B} $$

where $$\nu$$ is the kinematic viscosity, $$\rho^'$$ is the density perturbation that provides buoyancy (for thermal convection $$\rho^'=\alpha\Delta T$$, $$\Omega$$ is the rotation rate of the Earth, and $$\mathbf{J}$$ is the electrical current density.

Finally, a transport equation, usually of heat (sometimes of light element concentration):

 * $$ \frac{\partial T}{\partial t} = \kappa \nabla^2 T +\epsilon $$

where T is temperature, $$\kappa=k/\rho c_p$$ is the thermal diffusivity with k thermal conductivity, $$c_p$$ heat capacity, and $$\rho$$ density, and $$\epsilon$$ is an optional heat source.

Often the pressure is the dynamic pressure, with the hydrostatic pressure and centripetal potential removed. These equations are then non-dimensionalized, introducing the non-dimensional parameters,

 * $$Ra=\frac{g\alpha T D^3}{\nu \kappa}, E=\frac{\nu}{\Omega D^2} , Pr=\frac{\nu}{\kappa} , Pm=\frac{\nu}{\eta} $$

where Ra is the Rayleigh number, E the Ekman number, Pr and Pm the Prandtl and magnetic Prandtl number. Magnetic field scaling is often in Elsasser number units $$B=\rho \Omega/\sigma$$.

Antidynamo theorem
Def. one of several results that restrict the type of magnetic fields that may be produced by dynamo action is called an antidynamo theorem.

Def. no axisymmetric magnetic field can be maintained through a self-sustaining dynamo action by an axially symmetric current is called Cowling's antidynamo theorem.

The impact of the known anti-dynamo theorems is that successful dynamos do not possess a high degree of symmetry.

Plasma objects
Plasma instabilities can be divided into two general groups:
 * 1) hydrodynamic instabilities
 * 2) kinetic instabilities.

Plasma instabilities are also categorised into different modes:

Ideal MHD instabilities driven by current or pressure gradients.

A kink instability, also oscillation or mode, is a class of magnetohydrodynamic instabilities which sometimes develop in a thin plasma column carrying a strong axial current. If a "kink" begins to develop in a column the magnetic forces on the inside of the kink become larger than those on the outside, which leads to growth of the perturbation. As it develops at fixed areas in the plasma, kinks belong to the class of "absolute plasma instabilities", as opposed to convective processes.

Space plasma characteristics
Space plasma pioneers Hannes Alfvén and Carl-Gunne Fälthammar divided the plasmas in the solar system into three different categories:

Classification of Magnetic Cosmic Plasmas λ=Mean free path. ρ= Larmor radius (gyroradius) of electron. λD=Debye length. lc=Characteristic length Adapted From Cosmical Electrodynamics (2nd Ed. 1952) Alfvén and Fälthammar

Gaseous objects
"Supersonic flows are initially generated by the Balbus-Hawley magnetic shear instability."

The magnetorotational instability or MRI is a fluid instability that arises when the angular velocity of a magnetized fluid decreases as the distance from the rotation center increases. It can also be known as the Velikhov-Chandrasekhar instability or Balbus-Hawley instability in the literature; not to be confused with the Velikhov instability which is the electrothermal instability. The MRI is of particular relevance in astrophysics where it is an important part of the dynamics in accretion discs. Balbus and Hawley were the first to realize the astrophysical importance of this instability and explain its physical mechanism.

A rotating hydrodynamic fluid disc (e.g. a nonmagnetic accretion disc) will remain in a laminar flow state as long as the angular momentum (per unit mass) increases outwards. This is also known as the Rayleigh stability criterion:
 * $$ \frac{\partial(r^2\Omega)}{\partial r}>0$$

where $$\Omega$$ is the angular velocity of a fluid element and $$r$$ is its distance to the rotation center. Instabilities leading ultimately to turbulence occur if a magnetic field is present and the angular velocity decreases with radius:
 * $$ \frac{\partial\Omega}{\partial r}<0$$.

The MRI was first noticed in a non-astrophysical context by Evgeny Velikhov in 1959 when considering the stability of Couette flow of an ideal hydromagnetic fluid. His result was later generalized by S. Chandrasekhar in 1960. This mechanism was proposed by Acheson & Hide (1973) to perhaps play a role in the context of the Earth's geodynamo problem. The applicability of this mechanism to the problem of accretion disks was appreciated when S. A. Balbus and J. F. Hawley established that weak magnetic fields can substantially alter the stability character of accretion disks.

The observed accretion rates in astrophysical objects cannot be explained by a molecular viscosity, as the outward angular momentum transport in that case would not be enough to account for the inward flow of mass. The MRI provides a mechanism to account for the additional outward angular momentum transport. It is intrinsically a magnetohydrodynamic (MHD) phenomenon, with no hydrodynamic analog. Dynamo action is usually invoked to justify the existence of the magnetic field required for the instability to set in.

Liquid objects
The resistive skin time is a characteristic time of typical magnetohydrodynamic (MHD) phenomena.

The resistive skin time is defined as: $$\tau_R=\frac{\mu_0 a^2}{\eta}$$ where $$\eta$$ is the resistivity, $$a$$ is a typical radius of the device and $$\mu_0$$ is the magnetic permeability.

Waves
There are several distinct kinds of MHD modes which have quite different dispersive, polarisation, and propagation properties:

 * Kink (or transverse) modes, which are oblique fast magnetoacoustic (also known as magnetosonic waves) guided by the plasma structure; the mode causes the displacement of the axis of the plasma structure. These modes are weakly compressible, but could nevertheless be observed with imaging instruments as periodic standing or propagating displacements of coronal structures, e.g. coronal loops. The frequency of transverse or "kink" modes is given by the following expression:

 * $$\omega_{K}=\sqrt{\frac{2k_{z}B^{2}}{\mu(\rho_{i}+\rho_{e})}}$$

For kink modes the parameter $$m$$ is equal to 1.

 * Sausage modes, which are also oblique fast magnetoacoustic waves guided by the plasma structure; the mode causes expansions and contractions of the plasma structure, but does not displace its axis. These modes are compressible and cause significant variation of the absolute value of the magnetic field in the oscillating structure. The frequency of sausage modes is given by the following expression:

 * $$\omega_{S}=\sqrt{\frac{k_{z}^{2}B^{2}}{\mu\rho_{e}}}$$

For sausage modes the parameter $$m$$ is equal to 0.

 * Longitudinal (or slow, or acoustic) modes, which are slow magnetoacoustic waves propagating mainly along the magnetic field in the plasma structure; these mode are essentially compressible. The magnetic field perturbation in these modes is negligible. The frequency of slow modes is given by the following expression:

 * $$\omega_{L}=\sqrt{k^{2}_{z}\left ( \frac{C_{s}^{2}C_{A}^{2}}{C_{s}^{2}+C_{A}^{2}} \right )}$$

Where we define $$C_{s}$$ as the sound speed and $$C_{A}$$ as the Alfvèn speed.

 * Torsional (Alfvén or twist) modes are incompressible transverse perturbations of the magnetic field along certain individual magnetic surfaces. In contrast with kink modes, torsional modes cannot be observed with imaging instruments, as they do not cause the displacement of either the structure axis or its boundary.

 * $$\omega_{A}=\sqrt{\frac{k_{z}^{2}B^{2}}{\mu\rho_{i}}}$$

Waves in plasmas are an interconnected set of particles and fields which propagates in a periodically repeating fashion. A plasma is a quasineutral, electrically conductive fluid. In the simplest case, it is composed of electrons and a single species of positive ions, but it may also contain multiple ion species including negative ions as well as neutral particles. Due to its electrical conductivity, a plasma couples to electric and magnetic fields. This complex of particles and fields supports a wide variety of waves.

Waves in plasmas can be classified as electromagnetic or electrostatic according to whether or not there is an oscillating magnetic field. Applying Faraday's law of induction to plane waves, we find $$\mathbf{k}\times\tilde{\mathbf{E}}=\omega\tilde{\mathbf{B}}$$, implying that an electrostatic wave must be purely longitudinal. An electromagnetic wave, in contrast, must have a transverse component, but may also be partially longitudinal.

Waves can be further classified by the oscillating species. In most plasmas of interest, the electron temperature is comparable to or larger than the ion temperature. This fact, coupled with the much smaller mass of the electron, implies that the electrons are much faster than the ions. An electron mode depends on the mass of the electrons, but the ions may be assumed to be infinitely massive, i.e. stationary. An ion mode depends on the ion mass, but the electrons are assumed to be massless and to redistribute themselves instantaneously according to the Boltzmann relation. Only rarely, e.g. in the lower hybrid oscillation, will a mode depend on both the electron and the ion mass.

The various modes can also be classified according to whether they propagate in an unmagnetized plasma or parallel, perpendicular, or oblique to the stationary magnetic field. Finally, for perpendicular electromagnetic electron waves, the perturbed electric field can be parallel or perpendicular to the stationary magnetic field.

$$\omega$$ - wave frequency, $$k$$ - wave number, $$c$$ - speed of light, $$\omega_p$$ - plasma frequency, $$\omega_i$$ - ion plasma frequency, $$\omega_c$$ - electron gyrofrequency, $$\Omega_c$$ - proton gyrofrequency, $$\omega_h$$ - upper hybrid frequency, $$v_s$$ - plasma "sound" speed, $$v_A$$ - plasma Alfven speed

Rocky objects
The jump conditions across an MHD shock or discontinuity are referred as the Rankine-Hugoniot equations for MHD. In the frame moving with the shock/discontinuity, those jump conditions can be written:

 * $$ \rho_1 v_{n1} = \rho_2 v_{n2}, $$
 * $$ B_{n1} = B_{n2}, $$
 * $$ \rho_1 v_{n1}^2+ p_1 + \frac{B_{t1}^2}{2 \mu_0}=\rho_2 v_{n2}^2+ p_2 + \frac{B_{t2}^2}{2 \mu_0},$$
 * $$ \rho_1 v_{n1} \mathbf{v_{t1}} - \frac{\mathbf{B_{t1}}B_{n1}}{\mu_0}= \rho_2 v_{n2} \mathbf{v_{t2}} - \frac{\mathbf{B_{t2}}B_{n2}}{\mu_0},$$
 * $$ \left(\frac{\gamma}{\gamma-1}\frac{p_1}{\rho_1}+\frac{v_1^2}{2}\right)\rho_1 v_{n1}+\frac{v_{n1} B_{t1}^2}{\mu_0}-\frac{B_{n1}(\mathbf{B_{t1}}\cdot \mathbf{v_{t1}})}{\mu_0}=\left(\frac{\gamma}{\gamma-1}\frac{p_2}{\rho_2}+\frac{v_2^2}{2}\right)\rho_2 v_{n2}+\frac{v_{n2} B_{t2}^2}{\mu_0}-\frac{B_{n2}(\mathbf{B_{t2}}\cdot \mathbf{v_{t2}})}{\mu_0},$$

 * $$ (\mathbf{v} \times \mathbf{B})_{t1} = (\mathbf{v} \times \mathbf{B})_{t2}, $$

where $$ \rho$$, v, p, B are the plasma density, velocity, (thermal) pressure and magnetic field respectively. The subscripts t and n refer to the tangential and normal components of a vector (with respect to the shock/discontinuity front). The subscripts 1 and 2 refer to the two states of the plasma on each side of the shock/discontinuity.

Shocks
Shocks are transition layers across which there is a transport of particles. There are three types of shocks in MHD: slow-mode, intermediate and fast-mode shocks.

Intermediate shocks are non-compressive (meaning that the plasma density does not change across the shock). A special case of the intermediate shock is referred to as a rotational discontinuity. They are isentropic. All thermodynamic quantities are continuous across the shock, but the tangential component of the magnetic field can "rotate". Intermediate shocks in general however, unlike rotational discontinuities, can have a discontinuity in the pressure.

Slow-mode and fast-mode shocks are compressive and are associated with an increase in entropy. Across slow-mode shock, the tangential component of the magnetic field decreases. Across fast-mode shock it increases.

The type of shocks depend on the relative magnitude of the upstream velocity in the frame moving with the shock with respect to some characteristic speed. Those characteristic speeds, the slow and fast magnetosonic speeds, are related to the Alfvén speed, $$ v_A $$ and the sonic speed, $$ c_s $$ as follows:
 * $$a_{\mathrm{slow}}^2 = \frac{1}{2} \left[\left(c_s^2 + V_A^2\right)-\sqrt{\left(c_s^2+V_A^2\right)^2-4c_s^2V_{A}^2 \cos^{2} \theta_{Bn}}\,\right],$$
 * $$ a_{\mathrm{fast}}^2 = \frac{1}{2} \left[\left(c_s^2 + V_A^2\right)+\sqrt{\left(c_s^2+V_A^2\right)^2-4c_s^2V_{A}^2 \cos^{2} \theta_{Bn}}\,\right],$$

where $$V_{A}$$ is the Alfvén speed and $$\theta_{Bn}$$ is the angle between the incoming magnetic field and the shock normal vector.

The normal component of the slow shock propagates with velocity $$a_{\mathrm{slow} }$$ in the frame moving with the upstream plasma, that of the intermediate shock with velocity $$V_{An}$$ and that of the fast shock with velocity $$a_{\mathrm{fast}}$$. The fast mode waves have higher phase velocities than the slow mode waves because the density and magnetic field are in phase, whereas the slow mode wave components are out of phase.

Magnetic tension
The magnetic tension force is a restoring force (SI unit: Pa·m−1) that acts to straighten bent magnetic field lines. It equals:

 * $$\frac{\left(\mathbf{B}\cdot\nabla\right)\mathbf{B}}{\mu_0}$$

It is analogous to rubber bands and their restoring force. The force is directed antiradially. Although magnetic tension is referred to as a force, it is actually a pressure gradient (Pa m−1) which is also a force density (N m−3).

The magnetic pressure is the energy density of the magnetic field and it increases as magnetic field lines convene with each other. In contrast, magnetic tension force is determined by how much the magnetic pressure changes with distance. Magnetic tension forces also rely on vector current densities $$\mathbf{J}$$ and their interaction with the magnetic field $$\mathbf{B}$$. Plotting magnetic tension along adjacent field lines can give a picture as to their divergence and convergence with respect to each other as well as current densities $$\mathbf{J}$$.

Magnetic tension is particularly important in plasma physics and magnetohydrodynamics, where it controls dynamics of some systems and the shape of magnetized structures. In magnetohydrodynamics, the magnetic tension force can be derived from the momentum equation

 * $$\rho\left(\frac{\partial}{\partial t}+ \mathbf{V}\cdot\nabla \right)\mathbf{V} = \mathbf{J}\times\mathbf{B} - \nabla p$$

using the relation $$\mu_0\mathbf{J}=\nabla\times\mathbf{B}$$. The first term on the right hand side of the above equation represents electromagnetic forces and the second term represents pressure gradient forces.

A Taylor state is the minimum energy state of a plasma satisfying the constraint of conserving magnetic helicity.

Consider a closed, simply-connected, flux-conserving, perfectly conducting surface $$S$$ surrounding a plasma with negligible thermal energy ($$\beta \rightarrow 0$$).

Since $$\vec{B}.\vec{ds}=0$$ on $$S$$. This implies that $$\vec{A}_{||}=0$$.

As discussed above, the plasma would relax towards a minimum energy state while conserving its magnetic helicity. Since the boundary is perfectly conducting, there cannot be any change in the associated flux. This implies $$\delta \vec{B}.\vec{ds}=0$$ and $$\delta\vec{A}_{||}=0$$ on $$S$$.

We formulate a variational problem of minimizing the plasma energy $$W=\int d^3rB^2/2\mu_\circ$$ while conserving magnetic helicity $$K=\int d^3r\vec{A}.\vec{B}$$.

The variational problem is $$ \delta W -\lambda \delta K = 0$$.

After some algebra this leads to the following constraint for the minimum energy state

$$ \nabla \times \vec{B} = \lambda \vec{B}$$.

Sun (star)
The solar dynamo is the physical process that generates the Sun's magnetic field. The Sun is permeated by an overall dipole magnetic field, as are many other celestial bodies such as the Earth. The dipole field is produced by a circular electric current flowing deep within the star, following Ampère's law. The current is produced by shear (stretching of material) between different parts of the Sun that rotate at different rates, and the fact that the Sun itself is a very good electrical conductor (and therefore governed by the laws of magnetohydrodynamics).

Any electrically conducting fluid can form a dynamo simply by shear within the fluid itself, because of a consequence of Lenz's law of induction: moving the fluid through a pre-existing magnetic field will induce electrical currents in the fluid that distort the pre-existing magnetic field. The direction of the distortion is such that the existing field lines tend to be dragged along with the fluid, like threads of dye embedded in taffy or syrup. If the flow has a strong shear component then the individual field lines are stretched by the flow, amplifying the existing magnetic field. Such systems are called MHD dynamos.

Depending on the structure of the flow, the dynamo may be self-exciting and stable, self-exciting and chaotic, or decaying. The Sun's dynamo is self-exciting: the direction of the field reverses itself about every 11 years, causing the sunspot cycle as ropes of magnetic field lines rise to the surface of the Sun and manifest as sunspots on the surface. The solar dynamo is thought to be located at the tachocline, a region inside the Sun that exhibits a very large shear profile.

Coronal clouds
In a coronal cloud are magnetohydrodynamic plasma flux tubes along magnetic field lines.

Interplanetary magnetic field
Notation: let the symbol IMF stand for interplanetary magnetic field.

"[T]he IMF switched from northward to southward while Mariner [10] was in the Mercurian magnetosphere."

Magnetospheres
A magnetosphere is formed when a stream of charged particles, such as the solar wind, interacts with and is deflected by the magnetic field of a planet or similar body.

“Planets which generate magnetic fields in their interiors ... are surrounded by invisible magnetospheres. ... [I]n many respects, the magnetosphere of Venus is a scaled-down version of Earth’s. ... Earth’s magnetosphere is 10 times larger [than that of Venus]”

Mercury
Despite its small size and slow 59-day-long rotation, Mercury has a significant, and apparently global, magnetic field. According to measurements taken by Mariner 10, it is about 1.1% as strong as the Earth's. The magnetic field strength at the Mercurian equator is about 300 nT. Like that of Earth, Mercury's magnetic field is dipolar. Unlike Earth, Mercury's poles are nearly aligned with the planet's spin axis. Measurements from both the Mariner 10 and MESSENGER space probes have indicated that the strength and shape of the magnetic field are stable.

During its second flyby of the planet on October 6, 2008, MESSENGER discovered that Mercury's magnetic field can be extremely "leaky." The spacecraft encountered magnetic "tornadoes" – twisted bundles of magnetic fields connecting the planetary magnetic field to interplanetary space – that were up to 800 km wide or a third of the radius of the planet. These 'tornadoes' form when magnetic fields carried by the solar wind connect to Mercury's magnetic field. As the solar wind blows past Mercury's field, these joined magnetic fields are carried with it and twist up into vortex-like structures. These twisted magnetic flux tubes, technically known as flux transfer events, form open windows in the planet's magnetic shield through which the solar wind may enter and directly impact Mercury's surface.

The process of linking interplanetary and planetary magnetic fields, called magnetic reconnection, is common throughout the cosmos. It occurs in Earth's magnetic field, where it generates magnetic tornadoes as well. The MESSENGER observations show the reconnection rate is ten times higher at Mercury. Mercury's proximity to the Sun only accounts for about a third of the reconnection rate observed by MESSENGER.

There are "observed MHD fluctuations in the magnetosphere" of Mercury.

Venus
In 1967, Venera-4 found the Venusian magnetic field is much weaker than that of Earth. This magnetic field is induced by an interaction between the ionosphere and the solar wind, Venus's small induced magnetosphere provides negligible protection to the atmosphere against cosmic radiation. This radiation may result in cloud-to-cloud lightning discharges.

The weak magnetosphere around Venus means the solar wind is interacting directly with the outer atmosphere of the planet. Here, ions of hydrogen and oxygen are being created by the dissociation of neutral molecules from ultraviolet radiation. The solar wind then supplies energy that gives some of these ions sufficient velocity to escape the planet's gravity field. This erosion process results in a steady loss of low-mass hydrogen, helium, and oxygen ions, while higher-mass molecules, such as carbon dioxide, are more likely to be retained.

Earth
Beneath the Earth's mantle, lies the core which is made up of two parts - the solid inner core and liquid outer core - both have significant quantities of iron. The liquid outer core moves in the presence of the magnetic field and eddies are set up into the same due to the Coriolis effect. These eddies develop a magnetic field which boosts Earth's original magnetic field - a process which is self-sustaining and is called as the geomagnetic dynamo.

Based on the MHD equations, Glatzmaier and Paul Roberts have made a supercomputer model of the Earth's interior. After running the simulations for thousands of years in virtual time, the changes in Earth's magnetic field can be studied. The simulation results are in good agreement with the observations as the simulations have correctly predicted that the Earth's magnetic field flips every few thousands of years. During the flips, the magnetic field doesn't vanish altogether - it just gets more complicated.

The Atmospheric dynamo is a pattern of electrical currents that are set up in the Earth's ionosphere by multiple effects, mostly the Sun's solar wind, but also the tides of the Moon and Sun. The currents flow in circuits between the poles and the equator, but they are not well understood.

Magnetite typically carries the dominant magnetic signature in rocks, and so it has been a critical tool in paleomagnetism, a science important in discovering and understanding plate tectonics and as historic data for magnetohydrodynamics and other scientific fields. The relationships between magnetite and other iron-rich oxide minerals such as ilmenite, hematite, and ulvospinel have been much studied, as the complicated [metamorphic] reactions between these minerals and oxygen influence how and when magnetite preserves records of the Earth's magnetic field.

The schematic at right describes the Birkeland or field-aligned currents and the ionospheric current systems they connect to.

A Birkeland current is a set of currents which flow along geomagnetic field lines connecting the Earth’s magnetosphere to the Earth's high latitude ionosphere. Lately, the term Birkeland currents has been expanded by some authors to include magnetic field aligned currents in space plasmas in general. In the Earth’s magnetosphere, the currents are driven by the solar wind and interplanetary magnetic field and by bulk motions of plasma through the magnetosphere (convection which is indirectly driven by the interplanetary environment). The strength of the Birkeland currents changes with activity in the magnetosphere (e.g. during substorms). Small scale variations in the upward current sheets (downward flowing electrons) accelerate magnetospheric electrons which, when they reach the upper atmosphere, create the Aurora Borealis and Australis. In the high latitude ionosphere (or auroral zones), the Birkeland currents close through the region of the auroral electrojet, which flows perpendicular to the local magnetic field in the ionosphere. The Birkeland currents occur in two pairs of field-aligned current sheets. One pair extends from noon through the dusk sector to the midnight sector. The other pair extends from noon through the dawn sector to the midnight sector. The sheet on the high latitude side of the auroral zone is referred to as the Region 1 current sheet and the sheet on the low latitude side is referred to as the Region 2 current sheet.

"A reason why Birkeland currents are particularly interesting is that, in the plasma forced to carry them, they cause a number of plasma physical processes to occur (waves, instabilities, fine structure formation). These in turn lead to consequences such as acceleration of charged particles, both positive and negative, and element separation (such as preferential ejection of oxygen ions). Both of these classes of phenomena should have a general astrophysical interest far beyond that of understanding the space environment of our own Earth."

Auroral Birkeland currents carry about 100,000 amperes during quiet times and more than 1 million amperes during geomagnetically disturbed times. Birkeland had estimated currents "at heights of several hundred kilometres, and strengths of up to a million amperes" in 1908. The ionospheric currents which connect the field-aligned currents heat up the upper atmosphere due to the finite conductivity of the ionosphere. The heat (also known as Joule heat) is transferred from the ionospheric plasma to the gas of the upper atmosphere which rises and increases drag on low-altitude satellites.

Birkeland currents can also be created in the laboratory with multi-terawatt pulsed power generators. The resulting cross-section pattern indicates a hollow beam of electrons in the form of a circle of vortices, a formation called the diocotron instability (similar to, but different from, the Kelvin-Helmholtz instability), that subsequently leads to filamentation. Such vortices can be seen in aurora as "auroral curls".

Birkeland currents are also one of a class of plasma phenomena called a z-pinch, so named because the azimuthal magnetic fields produced by the current pinches the current into a filamentary cable. This can also twist, producing a helical pinch that spirals like a twisted or braided rope, and this most closely corresponds to a Birkeland current. Pairs of parallel Birkeland currents will also interact due to Ampère's force law: parallel Birkeland currents moving in the same direction will attract each other with an electromagnetic force inversely proportional to their distance apart whilst parallel Birkeland currents moving in opposite directions will repel each other. There is also a short-range circular component to the force between two Birkeland currents that is opposite to the longer-range parallel forces.

Electrons moving along a Birkeland current may be accelerated by a plasma double layer. If the resulting electrons approach relativistic velocities (i.e. if they approach the speed of light) they may subsequently produce a Bennett pinch, which in a magnetic field causes the electrons to spiral and emit synchrotron radiation that may include radio, optical (i.e. visible light), x-rays, and gamma rays.

"[T]he currents there are imagined as having come into existence mainly as a secondary effect of the electric corpuscles from the sun drawn in out of space, and thus far come under the second of the possibilities mentioned above". And p. 105, "Fig. 50a represents those in which the current-directions at the storm-centre are directed westwards, and 50b those in which the currents move eastwards".

"[C]urrents there [in the aurora] are imagined as having come into existence mainly as a secondary effect of the electric corpuscles from the sun drawn in out of space,"

Mars
Mars has a magnetosphere.

1963-38C
A U.S. Navy satellite 1963-38C, launched in 1963 and carrying a magnetometer above the ionosphere. ... In 1966 Alfred Zmuda, J.H. Martin, and F.T.Heuring analysed the satellite magnetometer results and reported their findings of magnetic disturbance in the aurora.

Hypotheses

 * 1) Magnetohydrodynamics works even when the electromagnetism occurs first.

title: Plasmas/Magnetohydrodynamics


text:
Instructional design/DS Examples

Audacity 2.0
Audacity allows users to edit multiple layers of audio easily and quickly. Audacity 2.0 and the lame_mp3 plugin can be downloaded from http://audacity.sourceforge.net/. The lame_mp3 plugin allows audacity to export MP3 files. To view an example of an audio story created with Audacity, click here.

Soundcloud
Soundcloud allows users to share their audio stories that also allows listeners to embed comments for others to view within the track. To view an example of an audio story posted on soundcloud, click here.

Images
In addition to using images to enhance video stories, an individual image or a collection of images can be used to express meaning and create a story.

Flickr
Flickr allows users to create collections of images to tell a story. To view a Flickr story, click here.

Video
Videos can tell stories by themselves, as well as through the incorporation of additional text, audio, and images.

YouTube
YouTube allows users to upload, share, edit, comment, and remix videos. To view an example of a YouTube, click here.

Vimeo
Vimeo also allows users to upload and comment on videos, but is more educationally based than YouTube. To view an example of a Vineo story, click here.

Social Media
Many of the tools mentioned could be classified as social media. There are a few platforms that incorporate media from all categories, which are explained below.

Twitter

 * To view a record of a hashtag story, click here.
 * To view a profile story, click here.

Facebook
The new Facebook timeline has enhanced the visual representation of these stories in a chronological fashion, thus making each individual’s Facebook page the digital story of their life. To view an example of a historical use of a Facebook story, click here.

Storify
Storify pulls social media information into one cohesive page. To view an example of a Storify story, click here.

Tumblr
Tumblr allows users to create individual blogs or collaborate on one blog. To view an example of a Tumblr story, click here.

Voicethread
Voicethread allows users to comment on slides using text, video, audio, and can draw on video slides through a comment. To view an example of a Voicethread story, click here.

Prezi
While Prezi provides a “blank canvas” for users to create on and viewers to explore, a path can be created for a viewer to follow. To view an example of a Prezi story, click here.

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

title: Instructional design/DS Examples


text:
TESOL/Guided listening and speaking

Students learning English as a second language should have plenty of opportunities to practice using English with message-focused activities. The focus is not on the form of the language but on communicating a message.

What is it?
A popular technique is called "What is it?". It has basically the following steps:

"'It is flat. It is used in the kitchen. It is made of wood or plastic.  You use it for cutting.'"
 * 1) The teacher describes something to the students.  For example, the teacher says:
 * 1) The students respond with "cutting board" or just by pointing to a picture of a cutting board.
 * 2) Roles are switched and the students describe something to the teacher.
 * 3) Students work in pairs and one student describes something to another who must identify it.

Listening grids
Another popular activity, appropriate for many levels, is "listening grids."

One student gives some information, for example about which sports they like. The other students have a grid in which they check off boxes with the possible answers. For example, one student, let's say Alex, says: "'Noisy people bother me. People who are late bother me. Bad drivers bother me.  People who often complain don't bother me.'"

The other students check off a grid like this:

Surveys
Surveys are very similar but are different in that students ask each other questions rather than students presenting to the whole class without being questioned.

Interviews
Interviews are similar to surveys but involve more freedom concerning which questions are asked, although the interviewer should have a specific topic with specific types of questions. For example, an interviewer may ask about the other student's life with questions like these:
 * Where were you born?
 * Where did you go to school?
 * Did you go to university?
 * What did you study?
 * Who did you work for?

Quizzes
If the teacher gives points for answers, a simple quiz can become a game and become more engaging to the students. For example, if the students have a picture like the one below in their books, the teacher asks everyone, clearly indicating the picture, "What's his name?", and the first person to answer "Steve" gets a point.

Tic-tac-toe
Students work with in pairs with one board. Each square has a picture, and to take a square, the student must name the picture.

Bingo
Each student has a bingo board with vocabulary or pictures. The teacher calls items.

title: TESOL/Guided listening and speaking


text:
User:Georgia gg

My name is Thanachaporn Assawaterakiat, and my nick name is Georgia. I'm 20 years old and my birthday is 5th August, 1991. I have 1 older sister and 2 younger brothers. I'm a 2nd year student at Mahidol University International College, Majoring in Tourism and Hospitality of Management.

title: User:Georgia gg


text:
Common Admission Test

Common Admission Test (CAT) is conducted by the IIMs in India for admitting prospective students for MBA degrees.

General Information

 * 1) The examination is objective type.
 * 2) It is conducted online.
 * 3) The top ten percentile may expect a call from an IIM. Furthermore, there will be Written Ability Test (WAT) and Personal Interview (PI) or Group Discussion or both.
 * 4) In addition to IIMs, many management schools admit students through CAT.

Test Duration and Pattern
Although the pattern of CAT exam is modified frequently. The following pattern was followed in CAT 2015. There will be three separately timed sections in the test. Once a section ends, candidates can no longer go back to it. The sections are (a) Verbal Ability (VA) (b) Data Interpretation & Logical Reasoning (DI/LR) (c) Quantitative Ability (QA). The sections will have 34, 34, 32 questions respectively with a duration of 1 hour each. Including a 15-minute tutorial before the start of the test which candidates are advised to go through, the total duration of the test will be 3 hours.

Mock tests

 * MBA entrance, TCY online]
 * Common Admission Test Questions, JumboTests

title: Common Admission Test


text:
Civil Service Examination

The Civil Services Examination (CSE) is a nationwide competitive examination in India conducted by the Union Public Service Commission for recruitment to the various Civil Services of the Government of India.

Compulsory papers

 * General Studies
 * HIstory
 * Economics
 * Geography
 * Politics
 * International relations
 * Statistics
 * Essay
 * Basic English
 * Basic Indian language

Optionals

 * Agriculture
 * Animal Husbandry & Vetinary Science
 * Anthropology
 * Botany
 * Chemistry
 * Civil Engineering
 * Commerce & Accountancy
 * Economics
 * Electrical Engineering
 * Geography
 * Geology
 * Indian History
 * Law
 * Management
 * Mathematics
 * Mechanical Engineering
 * Medical Science
 * Philosophy
 * Physics
 * Political Science & International Relations
 * Psychology
 * Public Administration
 * Sociology
 * Statistics
 * Zoology

Literature optionals

 * Literature of Arabic
 * Literature of Assamese
 * Literature of Bengali
 * Literature of Bodo
 * Literature of Chinese
 * Literature of Dogri
 * Literature of English
 * Literature of French
 * Literature of German
 * Literature of Gujarati
 * Literature of Hindi
 * Literature of Kannada
 * Literature of Kashmiri
 * Literature of Konkani
 * Literature of Maithili
 * Literature of Malayalam
 * Literature of Manipuri
 * Literature of Marathi
 * Literature of Nepali
 * Literature of Oriya
 * Literature of Pali
 * Literature of Persian
 * Literature of Punjabi
 * Literature of Russian
 * Literature of Sanskrit
 * Literature of Santali
 * Literature of Sindhi
 * Literature of Tamil
 * Literature of Telugu
 * Literature of Urdu

Resources

 * India
 * International relations, Ministry of External Affairs
 * News @ All India Radio

Previous Questions

 * IAS Exam race
 * Jeywin

Mock Tests

 * GK Today

title: Civil Service Examination


text:
Instructional design/ARCS Model Example

◀ ID ARCS Model

Case
Scenario: Now you are going to design a in-person workshop based on the ARCS Model for high school students in a public school district in Alaska. Goal is to introduce students to the basics skills of scientific inquiry. You have formed some general ideas for this two-hour workshop in which you will target main elements of an experiment:
 * Ask a question.
 * Do background research.
 * Construct a hypothesis.
 * Test a hypothesis by doing an experiment.
 * Analyze data and draw a conclusion.
 * Communicate your results.

Your audience is Grade 9 (freshmen) Alaska Native students who are a mix of non-native and native English speakers. With your survey, observation, interviews with the students' former teachers, as well as the research studies you managed to find, it shows the students have low literacy skills yet very strong visual skills.

title: Instructional design/ARCS Model Example


title: File:CORDIC.m.20120420.pdf


text: File:Cordic acc.20120420.pdf

title: File:Cordic acc.20120420.pdf


text: File:Angle resolution.20120420.pdf

title: File:Angle resolution.20120420.pdf


text: File:Cordic ang.20120420.pdf

title: File:Cordic ang.20120420.pdf


text:
University of Florida/Egm4313/s12.team11.R7

Intermediate Engineering Analysis Section 7566 Team 11 Due date: April 25, 2012.

R7.1
Solved by Andrea Vargas

Problem Statement
Verify (4)-(5) pg 19-9: (4) $$\langle \phi_i, \phi_j \rangle = 0\!$$ for $$i \ne j\!$$ (5) $$\langle \phi_j, \phi_j \rangle =\frac{L}{2}\!$$ for $$i = j\!$$

Solution
From the lecture notes on p.19-9 we know that: (2) $$\phi_i=\sin(\omega_i x)\!$$ (3) $$\langle \phi_i, \phi_j \rangle =\int_0^L \phi_i(x) \phi_j(x) dx\!$$ Then, we have $$\langle \phi_i, \phi_j \rangle =\int_0^L \phi_i(x) \phi_j(x) dx=\int_0^L sin(\omega_i x) \sin(\omega_j x)dx\!$$

Since the period of $$\sin(x)\!$$ is $$ 2\pi\!$$ and we know that $$p=2L\!$$, we can assume that $$L=\pi\!$$ for simplicity.

We can compute the result of the previous integral using Wolfram ALpha (Mathematica software). The following is the input given to the software: int from 0 to pi sin(ax)sin(bx)

where a is $$\omega_i\!$$ and b is $$\omega_j\!$$. The software generates the following answer:

$$\int_0^L \sin(ax)sin(bx)= \frac{b \sin(\pi a)\cos(\pi b)-a \cos(\pi a)\sin(\pi b)}{a^2-b^2}\!$$ Substituting for $$ \omega_i \!$$ and $$ \omega_j\!$$: $$\int_0^L \sin(ax)sin(bx)= \frac{\omega_j \sin(\pi \omega_i)\cos(\pi \omega_j)-\omega_i \cos(\pi \omega_i)\sin(\pi \omega_j)}{\omega_i^2-\omega_j^2}\!$$ We also know that $$\sin(c\pi)=0\!$$ where c is any integer. We can cancel any terms with $$\sin(\pi \omega_i)\!$$ or $$\sin(\pi \omega_j)\!$$ as they are equal to zero. Then, we can verify: (4) $$\langle \phi_i, \phi_j \rangle = 0\!$$ for $$i \ne j\!$$

Similarly we can verify (5). We have $$\langle \phi_j, \phi_j \rangle =\int_0^L \phi_j(x) \phi_j(x) dx=\int_0^L sin(\omega_j x) \sin(\omega_j x)dx\!$$

Here, we will keep the integration boundaries as $$0\rightarrow L\!$$ to be consistent with the problem statement.

We can compute the result of the previous integral using Wolfram ALpha (Mathematica software). The following is the input given to the software: int 0 to L (sin(ax))^2

where a is $$\omega_j\!$$. The software generates the following answer:

$$\int_0^L \sin(ax)^2=\frac{L}{2} - \frac{\sin(2aL)}{4a}\!$$ Substituting for $$ \omega_j \!$$: $$\int_0^L \sin(ax)^2=\frac{L}{2} - \frac{\sin(2L\omega_j)}{4\omega_j}\!$$ We know from the previous explanation that $$L=\pi\!$$.So,we can apply the same assumption as before that $$\sin(c\pi)=0\!$$ where c is any integer. This allows us to cancel any terms with $$sin(2L\omega_j)\!$$ as they are equal to zero. Then, we can verify: (5) $$\langle \phi_j, \phi_j \rangle = \frac{L}{2}\!$$ for $$i = j\!$$

R7.2
Solved by Francisco Arrieta

Problem Statement
Plot the truncated series $$ u(x,t)=\sum_{j=1}^{n}a_j\cos C\omega_j t \sin \omega_j x \!$$ with $$ n=5 \!$$ and for:

$$ t=\alpha P_1=\alpha \frac{2\pi}{C\omega_1}=\alpha \frac{2L}{C} \!$$

$$ \alpha=0.5, 1, 1.5, 2 \!$$

Solution
Using:

$$ \left\{\begin{matrix} f(x)=x(x-2) \\ g(x)=0 \\ C=3 \\ L=4

\end{matrix}\right. \!$$

Then:

$$ a_j=\frac{2}{L}\int_{0}^{L}f(x)\sin \omega_j x dx \!$$

 * $$ =2\left [ \frac{(-1)^j-1}{\pi^3j^3} \right ] \!$$

$$ \therefore a_j=0 \!$$ for all even values of j

Plugging back to the truncated series:

$$ u(x,t)=\sum_{j=1}^{n}2\left [ \frac{(-1)^j-1}{\pi^3j^3} \right ]\cos [C \frac{j\pi}{L} \alpha \frac{2L}{C}] \sin[\frac{j\pi}{L}x] \!$$

 * $$ =\sum_{j=1}^{n}2\left [ \frac{(-1)^j-1}{\pi^3j^3} \right ]\cos (\alpha j2\pi) \sin(\frac{j\pi}{2}x) \!$$

For $$ n=5 \!$$ :

$$ u(x,t)=[\frac{-4}{\pi^3}\cos (2\pi \alpha) \sin(\frac{\pi x}{2})]+[\frac{-4}{27\pi^3}\cos (6\pi \alpha) \sin(\frac{3\pi x}{2})]+[\frac{-4}{125\pi^3}\cos (10\pi \alpha) \sin(\frac{5\pi x}{2})] \!$$

When $$ \alpha=0.5 \!$$ :

When $$ \alpha=1 \!$$ :

When $$ \alpha=1.5 \!$$ :

When $$ \alpha=2 \!$$ :

--Egm4313.s12.team11.arrieta (talk) 06:20, 22 April 2012 (UTC)

Problem Statement
Find (a) the scalar product, (b) the magnitude of $$f\!$$ and $$g\!$$ ,(c) the angle between $$f\!$$ and $$g\!$$ for:

1) $$f(x)=cos(x), \ g(x)=x \ for -2 \le x \le 10\!$$

2) $$f(x)=\frac{1}{2}(3x^2-1), \ g(x)=\frac{1}{2}(5x^3-3x) \ for -1 \le x \le 1\!$$

Part 1
solved by Kyle Gooding

Scalar Product
$$=\int_a^bf(x)g(x) \ dx\!$$

$$=\int_{-2}^{10}x\cos(x) \ dx\!$$

Using integration by parts;

$$=[x\sin(x)+\cos(x)]_{-2}^{10} $$

$$=-7.68\!$$

Magnitude
$$\| f \|=^{1/2}=\int_{a}^{b} f^2(x) \ dx\!$$

$$=\int_{-2}^{10} [\cos(x)]^2 \ dx\!$$

$$=[.5(x+\sin(x)\cos(x)|_{-2}^{10}]^{1/2}$$

$$\| f \| =2.457\!$$

$$\| g \|=\int_{a}^{b} g^2(x) \ dx\!$$

$$=\int_{-2}^{10}x^2 \ dx $$ $$=[x^{3}/3]_{-2}^{10}$$

$$\| g \| =\frac{1008}{3}\!$$

Angle Between Functions
$$cos(\theta)=\frac{}{\| f \| \| g \|}\!$$

$$cos(\theta)=\frac{-7.68}{\frac{1008}{3}(2.457)}$$

$$\theta=89.47$$ The two functions are nearly orthogonal.

Part 2
solved by Luca Imponenti

Scalar Product
$$=\int_a^bf(x)g(x) \ dx\!$$

$$=\int_{-1}^{1}[\frac{1}{2}(3x^2-1)][\frac{1}{2}(5x^3-3x)] \ dx\!$$
 * $$=\int_{-1}^{1}\frac{1}{4}(15x^5-4x^3+3x) \ dx\!$$
 * $$=\left. \frac{1}{4}(\frac{15}{6}x^6-x^4+\frac{3}{2}x^2)\right|_{-1}^{1}\!$$
 * $$=\frac{1}{4}[(\frac{15}{6}1^6-1^4+\frac{3}{2}1^2)-(\frac{15}{6}(-1)^6-(-1)^4+\frac{3}{2}(-1)^2)]\!$$

Since all exponents are even, everything in brackets cancels out

$$=0\!$$

Magnitude
$$\| f \|=^{1/2}=\int_{a}^{b} f^2(x) \ dx\!$$
 * $$=\int_{-1}^{1} [\frac{1}{2}(3x^2-1)]^2 \ dx\!$$
 * $$=\int_{-1}^{1} \frac{1}{4}(9x^4-6x^2+1) \ dx\!$$
 * $$=\left. \frac{1}{4}(\frac{9}{5}x^5-2x^3+x)\right|_{-1}^{1}\!$$
 * $$=\frac{1}{4}[(\frac{9}{5}1^5-2(1)^3+1)-(\frac{9}{5}(-1)^5-2(-1)^3+(-1))]\!$$
 * $$=\frac{1}{4}[\frac{4}{5}-(-\frac{4}{5})]\!$$

$$\| f \| =\frac{2}{5}\!$$

$$\| g \|=\int_{a}^{b} g^2(x) \ dx\!$$
 * $$=\int_{-1}^{1} [\frac{1}{2}(5x^3-3x)]^2 \ dx\!$$
 * $$=\int_{-1}^{1} \frac{1}{4}(25x^6-30x^4+9x^2) \ dx\!$$
 * $$=\left. \frac{1}{4}(\frac{25}{7}x^7-6x^5+3x^3)\right|_{-1}^{1}\!$$
 * $$=\frac{1}{4}[(\frac{25}{7}1^7-6(1)^5+3(1)^3)-(\frac{25}{7}(-1)^7-6(-1)^5+3(-1)^3)]\!$$
 * $$=\frac{1}{4}[\frac{4}{7}-(-\frac{4}{7})]\!$$

$$\| g \| =\frac{2}{7}\!$$

Angle Between Functions
$$cos(\theta)=\frac{}{\| f \| \| g \|}\!$$

Since $$=0\!$$ the two functions are orthogonal

$$\theta=90\!$$

R7.4
Solved by Gonzalo Perez

Problem Statement
Sketch or graph $$ f(x) \! $$ which for $$ -\pi < x < \pi \! $$ is given as follows:

$$ f(x) = \left | x \right | \! $$

Solution
The MATLAB code shown below was used to developed the graph of $$ f(x) = \left | x \right | \! $$:

Problem Statement
Sketch or graph $$ f(x) \! $$ which for $$ -\pi < x < \pi \! $$ is given as follows:

$$ f(x)=\left\{\begin{matrix} x, if: -\pi<x<0\\

\pi - x, if: 0<x<\pi \end{matrix}\right. \! $$

Solution
The MATLAB code shown below was used to developed the graph of the piecewise function $$ f(x)=\left\{\begin{matrix} x, if: -\pi<x<0\\

\pi - x, if: 0<x<\pi \end{matrix}\right. \! $$:

Solved by Jonathan Sheider

Problem Statement
Find the Fourier series of the given function which is assumed to have a period of $$ 2 \pi \!$$. Show the details of your work. Sketch or graph the partial sums up to that including $$ cos(5x) \!$$ and $$ sin(5x) \!$$

Given: $$f(x) = |x| \!$$

Solution
The Fourier series of a function with a period of $$ p = 2\pi \!$$ is defined: $$ f(x) = a_{0} + \sum_{n=1}^{\infty } (a_{n}cos(nx) + b_{n}sin(nx))\!$$ Where: $$ a_{0} = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(x) dx \!$$ $$ a_{n} = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) cos(nx) dx \!$$ $$ b_{n} = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) sin(nx) dx \!$$ This particular function given in the problem can be also defined in a piecewise manner over this interval, namely: $$ f(x) = \left\{\begin{matrix} -x \text{ if } -\pi \leq x \leq 0\\x \text{ if } 0 \leq x \leq \pi \end{matrix}\right. \!$$ Calculating the first term $$ a_{0} \!$$: $$ a_{0} = \frac{1}{2\pi} \left ( \int_{-\pi}^{0} -x dx + \int_{0}^{\pi} x dx \right) \!$$ $$ a_{0} = \frac{1}{2\pi} \left ( \left [\frac{-x^2}{2} \right ]_{-\pi}^{0} + \left [\frac{x^2}{2} \right ]_{0}^{\pi} \right)\!$$ $$ a_{0} = \frac{1}{2\pi} \left ( - \left(\frac{-\pi^2}{2}\right ) + \left (\frac{\pi^2}{2}\right)\right) \!$$ $$ a_{0} = \frac{1}{2\pi} (\pi^2) \!$$ $$ a_{0} = \frac{\pi}{2} \!$$ Calculating the coefficient $$ a_{n} \!$$: $$ a_{n} = \frac{1}{\pi} \left( \int_{-\pi}^{0} -x cos(nx) dx + \int_{0}^{\pi} x cos(nx) dx \right ) \!$$ $$ a_{n} = \frac{1}{\pi} \left( - \int_{-\pi}^{0} x cos(nx) dx + \int_{0}^{\pi} x cos(nx) dx \right ) \!$$ Using integration by parts with the following substitutions: $$ u = x \!$$ and therefore $$ du = dx \!$$ $$ dv = cos(nx)dx \!$$ and therefore $$ v = \frac{1}{n}sin(nx) \!$$ This yields for the integral: $$ a_{n} = \frac{1}{\pi} \left( -\left[\frac{1}{n} x sin(nx) - \int \frac{1}{n}sin(nx)dx \right]_{-\pi}^{0} + \left[\frac{1}{n} x sin(nx) - \int \frac{1}{n}sin(nx)dx \right]_{0}^{\pi} \right ) \!$$ $$ a_{n} = \frac{1}{\pi} \left( -\left[\frac{1}{n} x sin(nx) + \frac{1}{n^2}cos(nx) \right]_{-\pi}^{0} + \left[\frac{1}{n} x sin(nx) + \frac{1}{n^2}cos(nx) \right]_{0}^{\pi} \right ) \!$$ $$ a_{n} = \frac{1}{\pi} \left( -\left[\frac{1}{n^2} - \left (\frac{1}{n} (-\pi) sin(-n\pi) + \frac{1}{n^2}cos(-n \pi) \right) \right] + \left[\frac{1}{n} \pi sin(n\pi) + \frac{1}{n^2}cos(n\pi) - \frac{1}{n^2} \right] \right ) \!$$ Note that for all n = 1,2,3... : $$ sin(n\pi) = 0 \!$$ as $$ sin(\pi) = sin(2\pi) = sin(3\pi) ... = 0 \!$$ therefore these terms are evaluated as zero, which yields: $$ a_{n} = \frac{1}{\pi} \left( -\left[\frac{1}{n^2} - \left (0 + \frac{1}{n^2}cos(-n \pi) \right) \right] + \left[0 + \frac{1}{n^2}cos(n\pi) - \frac{1}{n^2} \right] \right ) \!$$ $$ a_{n} = \frac{1}{\pi} \left( -\frac{1}{n^2} + \frac{1}{n^2}cos(-n \pi) +  \frac{1}{n^2}cos(n\pi) - \frac{1}{n^2} \right ) \!$$ Note that $$ cos(-x) = cos(x) \!$$ therefore: $$ a_{n} = \frac{1}{\pi} \left( -\frac{1}{n^2} + \frac{1}{n^2}cos(n \pi) +  \frac{1}{n^2}cos(n\pi) - \frac{1}{n^2} \right ) \!$$ $$ a_{n} = \frac{1}{\pi} \left( -\frac{2}{n^2} + \frac{2}{n^2}cos(n \pi) \right ) \!$$ $$ a_{n} = \frac{1}{\pi} \left (\left ( -\frac{2}{n^2} \right )(1 - cos(n \pi) \right )\!$$ $$ a_{n} = -\frac{2}{n^2\pi} \left (1 - cos(n \pi) \right )\!$$ To evaluate the term  $$ (1-cos(n\pi)) \!$$: Note that $$ cos(n\pi) = -1 \!$$ for odd n values as $$ cos(\pi) = cos(3\pi) = cos(5\pi) ... = -1 \!$$ And that $$ cos(n\pi) = 1 \!$$ for even n values as $$ cos(2\pi) = cos(4\pi) = cos(6\pi) ... = 1 \!$$. Therefore, it can be concluded that for odd n values: $$ (1-cos(n\pi)) = 1-(-1) = 2 \!$$ And for even n values: $$ (1-cos(n\pi) = 1-(1) = 0 \!$$ Therefore, for the coefficient $$ a_{n} \!$$, all even terms will equal zero while all odd terms will have a multiplier of 2, this yields (for all odd n values): $$ a_{n} = -\frac{4}{n^2\pi} \text{ for  } n = 1,3,5... \!$$ Calculating the coefficient $$ b_{n} \!$$: $$ b_{n} = \frac{1}{\pi} \left( \int_{-\pi}^{0} -x sin(nx) dx + \int_{0}^{\pi} x sin(nx) dx \right ) \!$$ $$ b_{n} = \frac{1}{\pi} \left( - \int_{-\pi}^{0} x sin(nx) dx + \int_{0}^{\pi} x sin(nx) dx \right ) \!$$ Using integration by parts with the following substitutions: $$ u = x \!$$ and therefore $$ du = dx \!$$ $$ dv = sin(nx)dx \!$$ and therefore $$ v = \frac{-1}{n}cos(nx) \!$$ This yields for the integral: $$ b_{n} = \frac{1}{\pi} \left( -\left[\frac{-1}{n} x cos(nx) - \int \frac{-1}{n}cos(nx)dx \right]_{-\pi}^{0} + \left[\frac{-1}{n} x cos(nx) - \int \frac{-1}{n}cos(nx)dx \right]_{0}^{\pi} \right ) \!$$ $$ b_{n} = \frac{1}{\pi} \left( -\left[\frac{-1}{n} x cos(nx) + \frac{1}{n^2}sin(nx) \right]_{-\pi}^{0} + \left[\frac{-1}{n} x cos(nx) + \frac{1}{n^2}sin(nx) \right]_{0}^{\pi} \right ) \!$$ $$ b_{n} = \frac{1}{\pi} \left( -\left[0 - \left (\frac{-1}{n} (-\pi) cos(-n\pi) + \frac{1}{n^2}sin(-n \pi) \right) \right] + \left[\frac{-1}{n} \pi cos(n\pi) + \frac{1}{n^2}sin(n\pi) - 0 \right] \right ) \!$$ Note that for all n = 1,2,3... : $$ sin(n\pi) = 0 \!$$ as $$ sin(\pi) = sin(2\pi) = sin(3\pi) ... = 0 \!$$ therefore these terms are evaluated as zero, which yields: $$ b_{n} = \frac{1}{\pi} \left( -\left[0 - \left (\frac{-1}{n} (-\pi) cos(-n\pi) + 0 \right) \right] + \left[\frac{-1}{n} \pi cos(n\pi) + 0 \right] \right ) \!$$ $$ b_{n} = \frac{1}{\pi} \left( \left[\frac{1}{n} (\pi) cos(-n\pi) \right] + \left[\frac{-1}{n} \pi cos(n\pi) \right] \right ) \!$$ Note that $$ cos(-x) = cos(x) \!$$ therefore: $$ b_{n} = \frac{1}{\pi} \left( \frac{1}{n} \pi cos(n\pi) + \frac{-1}{n} \pi cos(n\pi)  \right ) \!$$ $$ b_{n} = \frac{1}{\pi} \left( 0 \right ) \!$$ $$ b_{n} = 0 \!$$ Therefore there will be no $$ sin(nx) \!$$ terms in the Fourier representation. In conclusion, the Fourier series representation for the given function is as follows: $$ f(x) = \frac{\pi}{2} - \frac{4}{\pi} cos(x) - \frac{4}{3^2 \pi} cos(3x) - \frac{4}{5^2 \pi} cos(5x) + ... \!$$                                     $$ f(x) = \frac{\pi}{2} - \frac{4}{\pi}\left (cos(x) + \frac{1}{9} cos(3x) + \frac{1}{25} cos(5x) + ... \right ) \!$$ A graph of the function, and the Fourier series for $$ n = 1,3,5 \!$$ is shown below: --Egm4313.s12.team11.sheider (talk) 06:00, 22 April 2012 (UTC)

Problem Statement
Find the Fourier series of the given function which is assumed to have a period of $$ 2 \pi \!$$. Show the details of your work. Sketch or graph the partial sums up to that including $$ cos(5x) \!$$ and $$ sin(5x) \!$$

Given: $$ f(x) = \left\{\begin{matrix} -x \text{ if } -\pi < x < 0\\\pi - x \text{ if } 0 < x < \pi \end{matrix}\right. \!$$

Solution
The Fourier series of a function with a period of $$ p = 2\pi \!$$ is defined: $$ f(x) = a_{0} + \sum_{n=1}^{\infty } (a_{n}cos(nx) + b_{n}sin(nx))\!$$ Where: $$ a_{0} = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(x) dx \!$$ $$ a_{n} = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) cos(nx) dx \!$$ $$ b_{n} = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) sin(nx) dx \!$$ Calculating the first term $$ a_{0} \!$$: $$ a_{0} = \frac{1}{2\pi} \left ( \int_{-\pi}^{0} x dx + \int_{0}^{\pi} (\pi - x) dx \right) \!$$ $$ a_{0} = \frac{1}{2\pi} \left ( \left [\frac{x^2}{2} \right ]_{-\pi}^{0} + \left [\pi x - \frac{x^2}{2} \right ]_{0}^{\pi} \right)\!$$ $$ a_{0} = \frac{1}{2\pi} \left ( - \left(\frac{\pi^2}{2}\right ) + \left (\pi^2 - \frac{\pi^2}{2}\right)\right) \!$$ $$ a_{0} = \frac{1}{2\pi} \left ( -\frac{\pi^2}{2} + \pi^2 - \frac{\pi^2}{2}\right) \!$$ $$ a_{0} = \frac{1}{2\pi} \left (\pi^2 - \pi^2 \right) \!$$ $$ a_{0} = 0 \!$$ Calculating the coefficient $$ a_{n} \!$$: $$ a_{n} = \frac{1}{\pi} \left( \int_{-\pi}^{0} x cos(nx) dx + \int_{0}^{\pi} (\pi - x) cos(nx) dx \right ) \!$$ Using integration by parts with the following substitutions for the integral $$ \int_{-\pi}^{0} x cos(nx) dx \!$$: $$ u = x \!$$ and therefore $$ du = dx \!$$ $$ dv = cos(nx)dx \!$$ and therefore $$ v = \frac{1}{n}sin(nx) \!$$ Using integration by parts with the following substitutions for the integral $$ \int_{0}^{\pi} (\pi - x) cos(nx) dx \!$$: $$ u = \pi - x \!$$ and therefore $$ du = -dx \!$$ $$ dv = cos(nx)dx \!$$ and therefore $$ v = \frac{1}{n}sin(nx) \!$$ This yields for the overall expression: $$ a_{n} = \frac{1}{\pi} \left( \left[\frac{1}{n} x sin(nx) - \int \frac{1}{n}sin(nx)dx \right]_{-\pi}^{0} + \left[\frac{1}{n} (\pi - x) sin(nx) - \int -\frac{1}{n}sin(nx)dx \right]_{0}^{\pi} \right ) \!$$ $$ a_{n} = \frac{1}{\pi} \left( \left[\frac{1}{n} x sin(nx) + \frac{1}{n^2}cos(nx) \right]_{-\pi}^{0} + \left[\frac{1}{n} (\pi - x) sin(nx) - \frac{1}{n^2}cos(nx) \right]_{0}^{\pi} \right ) \!$$ $$ a_{n} = \frac{1}{\pi} \left( \left[\frac{1}{n^2} - \left (\frac{1}{n} (-\pi) sin(-n\pi) + \frac{1}{n^2}cos(-n \pi) \right) \right] + \left[-\frac{1}{n^2}cos(n\pi) + \frac{1}{n^2} \right] \right ) \!$$ Note that for all n = 1,2,3... : $$ sin(n\pi) = 0 \!$$ as $$ sin(\pi) = sin(2\pi) = sin(3\pi) ... = 0 \!$$ therefore these terms are evaluated as zero, which yields: $$ a_{n} = \frac{1}{\pi} \left( \left[\frac{1}{n^2} - 0 - \frac{1}{n^2}cos(-n \pi) \right] + \left[-\frac{1}{n^2}cos(n\pi) + \frac{1}{n^2} \right] \right ) \!$$ $$ a_{n} = \frac{1}{\pi} \left(\frac{1}{n^2} - \frac{1}{n^2}cos(-n \pi) - \frac{1}{n^2}cos(n\pi) + \frac{1}{n^2} \right ) \!$$ Note that $$ cos(-x) = cos(x) \!$$ therefore: $$ a_{n} = \frac{1}{\pi} \left(\frac{1}{n^2} - \frac{1}{n^2}cos(n \pi) - \frac{1}{n^2}cos(n\pi) + \frac{1}{n^2} \right ) \!$$ $$ a_{n} = \frac{1}{\pi} \left(\frac{2}{n^2} - \frac{2}{n^2}cos(n \pi) \right ) \!$$ $$ a_{n} = \frac{1}{\pi} \left( \left (\frac{2}{n^2} \right)(1 - cos(n \pi)) \right ) \!$$ $$ a_{n} = \frac{2}{n^2\pi} \left (1 - cos(n \pi) \right )\!$$ To evaluate the term $$ (1-cos(n\pi)) \!$$: Note that $$ cos(n\pi) = -1 \!$$ for odd n values as $$ cos(\pi) = cos(3\pi) = cos(5\pi) ... = -1 \!$$ And that $$ cos(n\pi) = 1 \!$$ for even n values as $$ cos(2\pi) = cos(4\pi) = cos(6\pi) ... = 1 \!$$. Therefore, it can be concluded that for odd n values: $$ (1-cos(n\pi)) = 1-(-1) = 2 \!$$ And for even n values: $$ (1-cos(n\pi)) = 1-(1) = 0 \!$$ Therefore, for the coefficient $$ a_{n} \!$$, all even terms will equal zero while all odd terms will have a multiplier of 2, this yields (for all odd n values): $$ a_{n} = \frac{4}{n^2\pi} \text{ for  } n = 1,3,5... \!$$ Calculating the coefficient $$ b_{n} \!$$: $$ b_{n} = \frac{1}{\pi} \left( \int_{-\pi}^{0} x sin(nx) dx + \int_{0}^{\pi} (\pi - x) sin(nx) dx \right ) \!$$ Using integration by parts with the following substitutions for the integral $$ \int_{-\pi}^{0} x sin(nx) dx \!$$: $$ u = x \!$$ and therefore $$ du = dx \!$$ $$ dv = sin(nx)dx \!$$ and therefore $$ v = \frac{-1}{n}cos(nx) \!$$ Using integration by parts with the following substitutions for the integral $$ \int_{0}^{\pi} (\pi - x) sin(nx) dx \!$$: $$ u = \pi - x \!$$ and therefore $$ du = -dx \!$$ $$ dv = sin(nx)dx \!$$ and therefore $$ v = \frac{-1}{n}cos(nx) \!$$ This yields for the overall expression: $$ b_{n} = \frac{1}{\pi} \left( \left[\frac{-1}{n} x cos(nx) - \int \frac{-1}{n}cos(nx)dx \right]_{-\pi}^{0} + \left[\frac{-1}{n}(\pi - x) cos(nx) - \int -\frac{-1}{n}cos(nx)dx \right]_{0}^{\pi} \right ) \!$$ $$ b_{n} = \frac{1}{\pi} \left( \left[\frac{-1}{n} x cos(nx) + \frac{1}{n^2}sin(nx) \right]_{-\pi}^{0} + \left[\frac{-1}{n}(\pi- x) cos(nx) - \frac{1}{n^2}sin(nx) \right]_{0}^{\pi} \right ) \!$$ $$ b_{n} = \frac{1}{\pi} \left( \left[0 - \left (\frac{-1}{n} (-\pi) cos(-n\pi) + \frac{1}{n^2}sin(-n \pi) \right) \right] + \left[- \frac{1}{n^2}sin(n\pi) - \left( \frac{-1}{n}\pi \right) \right] \right ) \!$$ $$ b_{n} = \frac{1}{\pi} \left( - \frac{1}{n} \pi cos(-n\pi) - \frac{1}{n^2}sin(-n \pi) - \frac{1}{n^2}sin(n\pi) + \frac{1}{n}\pi   \right ) \!$$ Note that for all n = 1,2,3... : $$ sin(n\pi) = 0 \!$$ as $$ sin(\pi) = sin(2\pi) = sin(3\pi) ... = 0 \!$$ therefore these terms are evaluated as zero, which yields: $$ b_{n} = \frac{1}{\pi} \left( - \frac{1}{n} \pi cos(-n\pi) -0 - 0 + \frac{1}{n}\pi   \right ) \!$$ $$ b_{n} = \frac{1}{\pi} \left(\frac{1}{n}\pi - \frac{1}{n} \pi cos(-n\pi) \right ) \!$$ Note that $$ cos(-x) = cos(x) \!$$ therefore: $$ b_{n} = \frac{1}{\pi} \left(\frac{1}{n}\pi - \frac{1}{n} \pi cos(n\pi) \right ) \!$$ $$ b_{n} = \frac{1}{\pi} \left( \left(\frac{\pi}{n} \right)(1-cos(n\pi)) \right ) \!$$ $$ b_{n} = \frac{1}{n} \left(1-cos(n\pi) \right ) \!$$ To evaluate the term $$ (1-cos(n\pi)) \!$$: Note that $$ cos(n\pi) = -1 \!$$ for odd n values as $$ cos(\pi) = cos(3\pi) = cos(5\pi) ... = -1 \!$$ And that $$ cos(n\pi) = 1 \!$$ for even n values as $$ cos(2\pi) = cos(4\pi) = cos(6\pi) ... = 1 \!$$. Therefore, it can be concluded that for odd n values: $$ (1-cos(n\pi)) = 1-(-1) = 2 \!$$ And for even n values: $$ (1-cos(n\pi)) = 1-(1) = 0 \!$$ Therefore, for the coefficient $$ b_{n} \!$$, all even terms will equal zero while all odd terms will have a multiplier of 2, this yields (for all odd n values): $$ b_{n} = \frac{2}{n} \text{ for  } n = 1,3,5... \!$$ In conclusion, the Fourier series representation for the given function is as follows: $$ f(x) = 0 + \left (\frac{4}{\pi} cos(x) + \frac{4}{3^2 \pi} cos(3x) + \frac{4}{5^2 \pi} cos(5x) + ... \right ) + \left (2sin(x) + \frac{2}{3} sin(3x) + \frac{2}{5} sin(5x) + ... \right ) \!$$

$$ f(x) = \frac{4}{\pi} \left (cos(x) + \frac{1}{9} cos(3x) + \frac{1}{25} cos(5x) + ... \right ) + 2 \left (sin(x) + \frac{1}{3} sin(3x) + \frac{1}{5} sin(5x) + ... \right ) \!$$ A graph of the function, and the Fourier series for $$ n = 1,3,5 \!$$ is shown below: --Egm4313.s12.team11.sheider (talk) 06:00, 22 April 2012 (UTC)

R7.5
Solved by Daniel Suh

Problem Statement
Consider the following, $$\left \langle \phi_{2j-1}, \phi_{2k-1} \right \rangle=\int_{0}^{p}\phi _{2j-1}(x) \cdot \phi _{2k-1}(x)dx\!$$

$$\left \langle \phi_{2j-1}, \phi_{2k-1} \right \rangle=\int_{0}^{p}\sin {j\omega x} \cdot \sin {k\omega x} \; dx\!$$

with $$j\neq k\; and\; j,k = 1,2,...\!$$, and $$p=2\pi, j=2, k=3\!$$

1. Find the integration with the given data.

2. Confirm the results with Matlab's trapz command for the trapezoidal rule.

Trigonometric Identities
Angle Sum and Difference Identities

$$(1) \cos{(a+b)} = \cos{a}\cos {b} - \sin {a}\sin{b}\!$$ $$(2) \cos{(a-b)} = \cos{a}\cos {b} + \sin {a}\sin{b}\!$$

Rearrange

$$(1) \sin {a}\sin{b} = \cos{a}\cos {b}-\cos{(a+b)}\!$$ $$(2) \cos{a}\cos {b} = \cos{(a-b)} - \sin{a}\sin {b}\!$$

Substitute and Combine

$$\sin {a}\sin{b} = \cos{(a-b)} - \sin{a}\sin {b} - \cos{(a+b)}\!$$ $$2\sin {a}\sin{b} = \cos{(a-b)} - \cos{(a+b)}\!$$ $$\sin {a}\sin{b} = \frac{1}{2}\cos{(a-b)} - \frac{1}{2}\cos{(a+b)}\!$$

Utilize Trig Identities
$$\left \langle \phi_{2j-1}, \phi_{2k-1} \right \rangle=\int_{0}^{p}\sin {jwx} \cdot \sin {kwx} \; dx\!$$

$$\left \langle \phi_{2j-1}, \phi_{2k-1} \right \rangle=\int_{0}^{2\pi}\sin {2\omega x} \cdot \sin {3\omega x} \; dx\!$$

$$\left \langle \phi_{2j-1}, \phi_{2k-1} \right \rangle=\int_{0}^{2\pi}\frac{1}{2}\cos {(2\omega x - 3\omega x)} - \frac{1}{2}\cos{(2\omega x + 3\omega x)} \; dx\!$$

$$\left \langle \phi_{2j-1}, \phi_{2k-1} \right \rangle=\frac{1}{2}\int_{0}^{2\pi}\cos {(-\omega x)} \; dx - \frac{1}{2}\int_{0}^{2\pi}\cos{(5\omega x)} \; dx\!$$

$$\left \langle \phi_{2j-1}, \phi_{2k-1} \right \rangle=\frac{1}{2}\sin {(-\omega x)}|_{0}^{2\pi} - \frac{1}{2}\sin{(5\omega x)}|_{0}^{2\pi}\!$$

$$\left \langle \phi_{2j-1}, \phi_{2k-1} \right \rangle= (0-0) - (0-0) \!$$

$$\left \langle \phi_{2j-1}, \phi_{2k-1} \right \rangle= 0 \!$$

Part 2
>> X = 0:2*pi/100:2*pi;

>> Y = sin(2*X).*sin(3*X);

>> Z = trapz(X,Y)

Z =

2.9490e-017


text:
Helicopter/Howard Community College/HelicopterSummary

/Summary of the Helicopter Teams/


text:
Helicopter/Howard Community College/HelicopterSummary/Summary of the Helicopter Teams

So far most of the teams in the Fall semester havent really gone far trying to build their own quadcopters. But most of them managed to play around with the Parrot Drone and fly it around.


text:
User:Jsunday9/enes100/My Work3

Write problem/project Goal
What is your wording of the overall problem/project goal?

My First Task
What are you going to do for the team this first weekend?

Summary of actual work over first weekend
How is what you did different than what you planned?

Week1 Narrative
Tell a detailed story describing what you did for your team over the weekend.

My Second Task
Record what you are planning on doing for the team during the second weekend.

Summary of actual work over second weekend
How is what you did different than what you planned?

Week2 Narrative
Tell a detailed story describing what you did for your team over the weekend.

My Third task
Record what you are planning on doing for the team during the third weekend.

Summary of actual work over third weekend
How is what you did different than what you planned?

Week3 Narrative
Tell a detailed story describing what you did for your team over the weekend.

My Fourth task
''Record what you are planning on doing for the team during the final weekend of project work. ''

Summary of actual work over fourth weekend
How is what you did different than what you planned?

Week4 Narrative
Tell a detailed story describing what you did for your team over the weekend.

Complete Team Page
Fill out the Team Form (should have already copied the form, created the team page, linked to it and started filling it out).

title: User:Jsunday9/enes100/My Work3


text:
User:Egm4313.s12.team8.koester/R7.4

Problem
K 2011 page 482 # 6, 9, 12, 13

-for 6 and 9 graph f(x) on $$ \displaystyle -\pi $$ to $$ \displaystyle \pi $$

-for 12 and 13 find the Fourier series of the given function f(x) and sketch or graph the partial sums up to that including $$ \displaystyle \cos(5x) $$ and $$ \displaystyle \sin(5x) $$.

6.
$$ \displaystyle f(x)=|x| $$

Matlab Code:

>> x=linspace(-pi,pi)

>> y=abs(x)

>> plot(x,y)

9.
$$ \displaystyle f(x)=x $$ on $$ \displaystyle (-\pi,0) $$

$$ \displaystyle f(x)=-x+\pi $$ on $$ \displaystyle (0,\pi) $$

Matlab Code:

>> x1=linspace(-pi,0)

>> y1=x1

>> x2=linspace(0,pi)

>> y2=pi-x2

>> plot(x1,y1,x2,y2)

12.
$$ \displaystyle f(x)=|x| $$

This function is an even function.

This is proven by the fact that $$ \displaystyle f(-x)=f(x) $$

Since this is an even function and of period $$ \displaystyle 2\pi $$ its Fourier series reduces to the Fourier cosine series shown here:

$$ \displaystyle f(x)=a_0 + \sum_{n = 1}^{\infty} a_n\cos(nx) $$

Where the coefficients are equal to:

$$ \displaystyle a_0=\frac{1}{\pi}\int_{0}^{\pi}f(x)dx $$

$$ \displaystyle a_n=\frac{2}{\pi}\int_{0}^{\pi}f(x)\cos(nx)dx $$

$$ \displaystyle a_0=\frac{1}{\pi}\int_{0}^{\pi}xdx $$

$$ \displaystyle a_0=\frac{\pi}{2} $$

$$ \displaystyle a_n=\frac{2}{\pi}\int_{0}^{\pi}x\cos(nx)dx $$

Using integration by parts:

$$ \displaystyle a_n=\frac{2}{\pi}(\frac{nx\sin(nx)+\cos(nx)}{n^2})$$ evaluated on x=$$ \displaystyle (0,\pi) $$

Evaluating, reduces to:

$$ \displaystyle a_n=\frac{2}{\pi}(\frac{\cos(\pi n)-1}{n^2}) $$

Plugging this back into the function equation:

$$ \displaystyle f(x)=\frac{\pi}{2} + \sum_{n = 1}^{\infty} \frac{2}{\pi}(\frac{\cos(\pi n)-1}{n^2}) \cos(nx) $$

Expanding this, one obtains:

$$ \displaystyle f(x)=\frac{\pi}{2} - \frac{4}{\pi}(\cos x+ \frac{1}{9}\cos3x+ \frac{1}{25}\cos5x+...) $$

Matlab Code:

>> x=linspace(-pi,pi)

>> y=pi/2-(4/pi)*(cos(x)+(1/9)*cos(3*x)+(1/25)*cos(5*x))

>> plot(x,y)

13.
$$ \displaystyle f(x)=x $$ on $$ \displaystyle (-\pi,0) $$

$$ \displaystyle f(x)=-x+\pi $$ on $$ \displaystyle (0,\pi) $$

Since the function has a period of $$ \displaystyle 2\pi $$ its Fourier series is:

$$ \displaystyle f(x)=a_0 + \sum_{n = 1}^{\infty}(a_n\cos(nx)+ b_n\sin(nx)) $$

Where the coefficients equal:

$$ \displaystyle a_0=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)dx $$

$$ \displaystyle a_n=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\cos(nx)dx $$

$$ \displaystyle b_n=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\sin(nx)dx $$

$$ \displaystyle a_0=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)dx $$

$$ \displaystyle a_0=\frac{1}{2\pi}(\int_{-\pi}^{0}xdx+ \int_{0}^{\pi}(\pi-x)dx) $$

$$ \displaystyle a_0=\frac{1}{2\pi}(\frac{-\pi^2}{2}+\pi^2-\frac{\pi^2}{2}) $$

$$ \displaystyle a_0=0 $$

$$ \displaystyle a_n=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\cos(nx)dx $$

$$ \displaystyle a_n=\frac{1}{\pi}(\int_{-\pi}^{0}x\cos(nx)dx + \int_{0}^{\pi}(\pi-x)\cos(nx)dx) $$

$$ \displaystyle a_n=\frac{1}{\pi}(\int_{-\pi}^{0}x\cos(nx)dx + \int_{0}^{\pi}\pi\cos(nx)dx - \int_{0}^{\pi}x\cos(nx)dx) $$

Using integration by parts and then cancelling like terms:

$$ \displaystyle a_n=\frac{1}{\pi}(\frac{2-2\cos(\pi n)}{n^2})=\frac{2-2\cos(\pi n)}{\pi n^2} $$

$$ \displaystyle b_n=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\sin(nx)dx $$

$$ \displaystyle b_n=\frac{1}{\pi}(\int_{-\pi}^{0}x\sin(nx)dx + \int_{0}^{\pi}(\pi-x)\sin(nx)dx) $$

$$ \displaystyle b_n=\frac{1}{\pi}(\int_{-\pi}^{0}x\sin(nx)dx + \int_{0}^{\pi}\pi\sin(nx)dx - \int_{0}^{\pi}x\sin(nx)dx) $$

Using integration by parts and then cancelling like terms:

$$ \displaystyle b_n=\frac{1}{\pi}(\frac{\pi-\pi cos(\pi n)}{n})=\frac{1-cos(\pi n)}{n} $$

Plugging this back into the function equation:

$$ \displaystyle f(x)=\sum_{n = 1}^{\infty}(\frac{2-2\cos(\pi n)}{\pi n^2}\cos(nx)+ \frac{1-cos(\pi n)}{n}\sin(nx)) $$

Expanding this, one obtains:

$$ \displaystyle f(x)=\frac{4}{\pi}(\cos x+\frac{1}{9}\cos3x+\frac{1}{25}\cos5x) + 2(\sin x + \frac{1}{3}\sin3x + \frac{1}{5}\sin5x) $$

Matlab Code:

>> x=linspace(-pi,pi)

>> y=(4/pi)*(cos(x)+(1/9)*cos(3*x)+(1/25)*cos(5*x))+2*(sin(x)+(1/3)*sin(3*x)+(1/5)*sin(5*x))

>> plot(x,y)

title: User:Egm4313.s12.team8.koester/R7.4


text:
User:Zbennett/enes100/My Work (Rube Goldberg 2)

Write problem/project Goal
Create a Rube Goldberg machine out of printer parts, that effectively demonstrates how printer parts move and function together.

My First Task
Figure out how to program solenoid with the arduino to activate, and impact the chimes.

Summary of actual work over first weekend
I figured out that the solenoids would have to be powered by a PSU, because the arduinos cannot put out enough voltage to cause the solenoids to activate.

Week1 Narrative
In order to cause the solenoids to activate in a timed manner, with any level of precision, I'll need to devise a mechanism that turns the PSU on and off rather quickly. Because this may damage the PSU, I realized that I can make a circuit from the PSU to the solenoid that has a switch in it. This switch can be controlled by the arduino, which CAN be precisely timed. This way, the arduino can complete and break the circuit in perfect timing, causing the solenoid to impact the chime in an audibly pleasing manner, rather than a spastic mess. Here is a video, showing what the solenoid looks like when impacting the chime.

My Second Task
I will prepare the stand to bring in.

Summary of actual work over second weekend
I made final modifications to the stand to prepare to bring the stand for the rube goldberg machine in.

Week2 Narrative
I realized that, due to the use of scrap pieces of wood, the stand was approximately 8 feet tall. I used a pencil to mark an appropriate length, and cut them with a hand saw.

My Third task
Figure out why one of the solenoids is maintaining its magnetic field.

Summary of actual work over third weekend
After attaching both solenoids to the board and wiring them into the power supply, I found that one of the solenoids seems to "stick" after the power source has been deactivated.

Week3 Narrative
I wired both the solenoids into the power supply, and found that one if them experiences residual magnetism after the current is no longer running through the solenoid. In the video, this is clearly demonstrated by the solenoid on the right. Both of the solenoids are connected in a parallel circuit to the power supply, and both are unplugged at the exact same time. Through experimentation, I found that the longer the current runs through the solenoid, the longer it maintains the magnetic field after the current is cut. Inversely, when the current is run through the solenoid in very quick, short bursts, it functions identically to the one on the left. While taking a break from this problem to let it process in my brain, I attached the laser printer motor to the printer board that was previously attached to the plywood. It functions off of a 5V current.

My Fourth task
Mount the entire device on wheels.

Summary of actual work over fourth weekend
I mounted the Rube Goldberg machine on wheels, and encountered a few problems while doing so. I also attempted to replace the painters tape belt that was attaching the various printer boards together with other materials.

Week4 Narrative
In order to attach the wheels to the board, I had to drill holes in the bottom of the stand. Since the bottom of the stand had no large flat spots on it where I could drill holes for the wheels, I had to cut four short lengths of 2X4, and attach those to the bottom of the stand, so then I could drill holes in those. After drilling the holes and inserting the wheels (the holes were ever so slightly smaller than the wheel's axle to assure a tight fit), I inserted the wheels and gave it a push. The stand moved just fine, but I noticed the legs of the stand wobbled severely when moving across the room. I then measured and cut another length of 2X4 to use as a cross-brace between the two legs of the stand. I first tried to screw this onto the legs of the stand, but realized the screws were too short. I then attempted to countersink the screws which worked until I applied pressure to the cross-brace, which caused the screws to strip out the wood, and fall off. I ended up just hammering two long nails into each leg which held the cross-brace on securely. As an aside, I attempted to replace the painter's tape belt which was connecting the printer boards together with another material. I decided to use string with knots tied in it, to act like the teeth of a rubber belt that was designed to go with the system, but was too short to serve our purpose. The knots ended up getting caught on the metal parts of the printer board, so I decided to try just using plain string. This actually worked the best, seeing as there was no crunching sound of tape, and no sticking of tape which slowed the system down. Attached is a photo of the entire assembly mounted on the stand. I also created this tutorial on how to countersink a screw.

Complete Team Page
Fill out the Team Form (should have already copied the form, created the team page, linked to it and started filling it out).


text:
Instructional design/CPBL/Practice Reflection

Reflection
Question: Based on the checklist, discuss how you would improve the project and your project process/activities. Direction: Click "edit" for this section and add your response. Sign your name by inserting four ~'s.
 * Example: I want to add a workshop on social skills to my activity design as a way to reduce intragroup conflict. Dablee (talk) 23:57, 20 April 2012 (UTC)


text:
Instructional design/Reducing cognitive load in multimedia instruction/Overload Scenario 2

Essential processing (in both channels) > cognitive capacity
Problem:

Too many pictures and/or words and audio at one time leads to Dual Channel Overload

Multimedia Principal

Modality- best use of visual and auditory channels

Instructional Rxs:

Segmenting:

Break up the topic or concept into to smaller bites.

How: Break up concepts in the most logical places. Provide navigation to move forward and backward within the instruction (learner control).

What if you can’t segment the instruction for some reason? Try this:

Pretraining:

Helping the learner construct a mental model of the components of a system before learning how those components operate within the system. How:

Provide training in names and characteristics of components before training on the system itself.

Why this Works: Helps to reduce overload by introducing components before the entire system instead of asking the learner to absorb both at the same time.

Why this Works:

Better transfer occurs when a lesson is presented in learner-controlled segments rather than as a continuous unit. (It will be important at some point to bring all the bits back together.)


text: File:Real-life problem.png

title: File:Real-life problem.png


text:
Instructional design/Reducing cognitive load in multimedia instruction/Find and Fix

 * Remember the YouTube video you watched at the beginning of this lesson? Let's re-visit the video now that you are aware of the Cognitive Theory of Multimedia Learning.    Follow the steps below:

 * If you want to re-watch the video, Ctrl+click on the link to open the video in a new window.| Cognitive Theory of MML Video
 * Pull out the notes you made when you first watched the video.
 * Identify the types of cognitive overload you see and recommend a solution based on what you've learned. Put this information in the this page's Discuss tab at the top left of the page. If you need a "cheat sheet" open the Overload Rx Chart(Remember to use Ctrl+click to open in a new window.)

 * Once you're done with your "Find and Fix", save it, and click Next for possible answers.


text:
Instructional design/Practice Now/Sample answer

This is a sample PBL problem that is generated to the group of IT students:
New Concept Website Company(NCWC) is now facing a problem, their newly designed website for a marketing companyX cannot attract enough customers. What’s more, some CompanyX’s customers also complain about the difficulties in navigation and finding important information on this site. Now as a group of website interaction design experts, please identify the detailed problems of this newly designed website, and try to fix these problems. During this process, you may need to conduct interviews with Company X’s clients, managers and staff of CompanyX and NCWC to know their needs and reactions to your modifications. You need to take turns to be group leader, and group leader should report weekly progress.

Grading Results
Final score: There are 6 items in the checklist. This sample answer only correctly covers 4 of them. So its final score is 67%, less than 80%. It fails this test.

Please click here to return to "Practice now" session of generating PBL problems.


text:
CA Case Study

Case Description
Title: Reciprocal Teaching of Reading

Objectives: Students are expected to develop the strategic skills of:
 * Formulating questions based on the text
 * Summarizing the text
 * Making predictions about what will come next
 * Clarifying difficulties with the text

Conditions: The method has been used with groups of two to seven students, as well as individual students.

Activities:
 * 1. Both the teacher and students read a paragraph silently.

 * 2. Whoever is playing the role of teacher formulates a question based on the paragraph, constructs a summary, and makes a prediction or clarification, if any come to mind. Initially, the teacher models this process and then turns the role of teacher over to the students.

 * 3. When students first undertake the process, the teacher coaches them extensively on how to construct good questions and summaries, offering prompts and critiquing their efforts. In this way, the teacher provides scaffolding for the students, enabling them to take on whatever portion of the task they are able to.

 * 4. As the students become more proficient, the teacher fades, assuming the role of monitor and providing occasional hints or feedback.

Analysis
1. The method engages students in asset of activities that help them form a new conceptual model of the task of reading.
 * In traditional schooling, students learn to identify reading with the subskills of recognizing and pronouncing words and with the activities of scanning text and saying it aloud. Under the new conception, students recognize that reading requires constructive activities, such as formulating questions and making summaries and predictions, as well as evaluative ones, such as analyzing and clarifying the pints of difficulty.

2. The teacher models expert strategies in a shared problem context of knowing that they will soon undertake the same task.
 * After they have tried to do it themselves, and perhaps had difficulties, they listen with new knowledge about the task. That is, they can compare their own questions or summaries generated by the group. They can reflect on any differences, trying to understand what led to those differences.

3. The technique of providing scaffolding is a significant characteristic of reciprocal teaching.
 * Most importantly, it decomposes the task as necessary for the students to carry it out, thereby helping them to see how, in detail, to go about it. For example, in formulating questions, the teacher might want to see if the student can generate a question on his or her own; if not, she might suggest starting with a "Why" question about the agent in the story. If that fails, she might generate one herself and ask the student to reformulate it in his or her own words. In this way, it gets students started in the new skills, giving them a "feel" for the skills and helping them develop confidence that they can do them. With successful scaffolding techniques, students get as much support as they need to catty out the task, but no more. Hints and modeling are then gradually faded out, with students taking on more and more of the task as they become more skillful. These techniques of scaffolding and fading slowly build students' confidence that they can master the skills required.

4. The final critical aspect of reciprocal teaching is having students assume the dual roles of producer and critic.
 * They not only must produce good questions and summaries, but they also learn to evaluate the summaries or questions of others. By becoming critics as well as producers, students are forced to articulate their knowledge about what makes a good question, prediction, or summary. This knowledge then becomes more readily available for application to their own summaries and questions, thus improving a crucial aspect of their metacognitive skills. Moreover, once articulated, this knowledge can no longer simply reside in tacit form. It becomes more available for performing a variety of tasks; that is, it is freed from its contextual binding and can be used in many different contexts.

title: CA Case Study


text:
Instructional design/Generate PBL Problems/1.1 Problems reflect students' learning interests

Click here to return to the page of "Generate PBL Problems"

Case
Monica, a new Master's student majoring in Instructional Design System at XXX university, was asked by Professor K to work in a team to solve an instructional problem for a real client. Monica has a habit of writing a diary.

Monica’s diary: Jan. 23rd, 2012 I am so excited to take this class; the final project sounds so appealing to me, I love group work and I need to solve a problem for a real client, I even can imagine my future business card. “Monica - Instructional Designer”, aha, sounds so professional. Way to go, Monica~ Feb. 2nd, 2012 I am a little bit worried about this project; due to “gender imbalance”, I was assigned with two other male classmates who have a lot of experience working in a corporate setting, while, for me I am  interested in designing instruction for K-12 setting. Additionally, Professor K has such a strong background in bussiness, what if they insist on designing instruction for a business setting? How should I balance our interests? Feb. 16th, 2012 I could be a fortune teller, as I “expected” Professor K asked us to design instruction for a company. Sigh, corporation, ok, business, fine….. Never mind, Monica, you can do it!!! Mar.2nd, 2012 First interview with our client, it was so-so. I feel my team members have a special bond with our client, what is the term they are laughing about today? “Alden's Laws”? No…What is that, I cannot remember… Mar.14th, 2012 I cannot believe this, how I can miss such an important requirement for our client? Oh, I was so lost in their glorious business titles… I felt so stupid today at our team meeting…… Mar.21st, 2012 Second round interview, unhappy…take notes…invisible… …… Apr. 30th, 2012 Yeah, final report!!! No more “MBA” course!!!

Reflection
In this case, Monica started with a great passion towards the final project but it ended badly. The reason for that is obvious. Even though the format of this problem (design real instruction) generated Monica's interest, the specific problem was not appealing to Monica. Professor K didn’t fully take his/her student’s learning interest into consideration, this is a process that should not be neglected by instructors and instructional designers. By framing learning with student interests in mind, instructors can increase student motivation and better learning outcomes.

Tips for generating problems that would reflect students' interests.

 * Gain information:

-Icebreaker: At the beginning of the class, list several questions from general to specific, such as from “where you have been” to “What course you have taken so far”, to have students answer (raise hands/stand up/write). In this way, instructors could have general ideas of his/her students’ backgrounds, so that activities in aligning with students' interests could be designed. -Student Survey: Ask students to write a response to a series of open-ended questions about the upcoming class. The questions can assess their interests within the topic.
 * Alternatives:

-Provide students with 2-3 case scenarios that they can choose from. The alternatives should also be set up within students’ interests.

Quiz
{Which one is the IRRELEVANT understanding of Monica’s Case? -Start generating problems based on students’ interest. -Pay attention to the format as well as the content of the problems that would interest students. +Instructor should assign students problem based on his/her understanding about the teaching content. -Professor K could conduct an Icebreaker activity before assigning students into different groups.
 * type="[]"}
 * Based on Monica's case, instructor neglected her interests. C is irrelevant.

{Professor Hodgins is incorporating PBL methods into his language teaching class; students should work in a small team to help a real client to solve a curriculum problem. Four of his students decided to work together since they are all curious about building English Vocabulary for non native English speakers. Which problem should Professor Hodgins generate to this group? - 7-day pronuciation teaching towards business vocabulary for non native English speakers. + 10-day glossary teaching towards business vocabulary for non native English speakers.
 * type=""}
 * Students'interests lies in building English vocabulary. “7-day pronuciation teaching towards business vocabulary for non native English speakers” is a curriculum that emphasize on teaching pronuciation, it is not process of building vocabulary. "10-day glossary teaching towards business vocabulary for non native English speakers" is a process of building students' vocabulary in bussiness field.

Click here to return to the page of "Generate PBL Problems"


text:
Instructional design/ARCS Evaluation

Rubrics
Once you complete the final task, please refer to the rubric below for quality assurance. You need to score at least 75% (3 out of 4) or higher in order to pass.

Reference
Clark, R. & Mayer, R. E. (2008). E-learning and the science of instruction (2nd ed.). San Francisco: Jossey-Bass.

Dick, W., & Carey, L. (1996). The systematic design of instruction (4th ed.). New York, NY: Harper Collin.

Fleming, N. (2001). VARK: A guide to learning styles. Retrieved from

http://www.vark-learn.com/english/index.asp

Keller, J. M. (1983). Motivational design of instruction. In C. M. Reigeluth (Ed.), Instructional-design theories and models: an overview of their current status (pp. 386-434). Hillsdale, NJ : Lawrence Erlbaum Associates.

Kemp, J. E., Morrison, G. R., & Ross, S. M. (2001). Designing effective instruction (6th ed.). Upper Saddle River, NJ: Prentice-Hall.

Rothwell, W. J., & Kazanas, H. C. (1998). Mastering the instructional design process: A systematic approach. San Francisco, CA: Jossey-Bass Publishers.

Willingham, D. T. (2005). Do visual, auditory, and kinesthetic learners need visual, auditory, and kinesthetic instruction? American Educator, 29(2), 31-35.

Wlodkowski, R. J. (1985). Enhancing adult motivation to learn. San Francisco, CA: Jossey-Bass.

title: Instructional design/ARCS Evaluation