text:
User:Egm6321.f11.team4/HW3

=''' Problem R3.1 - Show Eq. (1) p.13-2 is exact only if $$k_1=d_1$$ (constant) and Eq. (1) p.12-4 is a particular equation '''=

From [[media:Pea1.f11.mtg13.djvu|Mtg 13-3]]

From [[media:Pea1.f11.mtg11.djvu|Mtg 11-2]]

From [[media:Pea1.f11.mtg12.djvu|Mtg 12-4]]

Given
Eq. (1) p.13-2 is $$\displaystyle \bar b(x,y)c(y)y'+a(x)\bar c(x,y)=0$$. Eq. (2) p.11-2 is $$\displaystyle n(x):=(-1/N)(N_x-M_y)$$. Eq. (1) p.12-4 is $$\displaystyle M+Ny'=[a(x)y+k_x(x)]+\bar b(x)y'=0$$.

Find
$$\displaystyle a)$$Show that the N1-ODE (1) p.13-2 satisfies the condition (2) p.11-2 that an integrating factor $$\displaystyle h(x)$$ can be found to render it exact, only if $$\displaystyle k_1(y)=d_1$$ (constant). $$\displaystyle b)$$Show that (1) p.13-2 includes (1) p.12-4 as a particular case.

Solution
Solved independently. $$\displaystyle a)$$Eq. (1) p. 13-2 can be written as $$\displaystyle N(x,y)y'+M(x,y)=0$$ which yields the following: $$\displaystyle N(x,y)=\bar b(x,y)c(y)$$ $$\displaystyle M(x,y)=a(x)\bar c(x,y)$$. These two equations yield the following when differentiated: $$\displaystyle N_x=\bar b(x,y)c_x(y)+\bar b_x(x,y)c(y)$$ $$\displaystyle M_y=a(x)\bar c_y(x,y)+a_y(x)\bar c(x,y)$$. Now we plug these equations into Eq. (2) p. 11-2 to obtain $$\displaystyle n(x)=\frac{-1}{\bar b(x,y)c(y)}*[\bar b_x(x,y)c(y)-a(x)\bar c_y(x,y)]$$ This simplifies to $$\displaystyle n(x)=\frac{a(x)\bar c_y(x,y)}{\bar b(x,y)c(y)}-\frac{\bar b_x(x,y)c(y)}{\bar b(x,y)c(y)}$$. Since $$\displaystyle \bar b_x(x,y)=b(x,y)$$ & $$\displaystyle \bar c_y(x,y)=c(x,y)$$ cancel out, the formula reduces to $$\displaystyle n(x)=\frac{a(x)}{\bar b(x,y)}-1$$. Rearranging the equation produces $$\displaystyle \bar b(x,y)=\frac{a(x)}{n(x)+1}$$. If you differentiate the above equation by $$\displaystyle y $$, you obtain $$\displaystyle \bar b_y(x,y)=0$$ because everything is a function of $$\displaystyle x $$ or a constant. This shows that $$\displaystyle k_1(y)=d_1$$ (constant) and not a function of $$\displaystyle y$$. $$\displaystyle b)$$In order to show Eq. (1) p. 12-4 is a particular case of Eq. (1) p. 13-2, we must make two assumptions. Those assumptions include $$\displaystyle \bar b(x,y)=\bar b(x)$$ and $$\displaystyle c(y)=C$$(constant). Working first with $$\displaystyle M(x,y) $$, we have $$\displaystyle a(x)\bar c(x,y)=a(x)y+k_2(x)$$ (left side is the general case and the right side is the particluar case). With $$\displaystyle c(y)=C $$, Eq. (1) p. 13-3 reduces to $$\displaystyle \bar c(x,y)=Cy+k_2(x)$$. Plug the above equation for $$\displaystyle \bar c(x,y) $$ into the $$\displaystyle a(x)\bar c(x,y)=a(x)y+k_2(x)$$ which yields $$\displaystyle a(x)[Cy+k_2(x)]=a(x)y+k_2(x)$$. The left side simplies becase $$\displaystyle C $$ is absorbed into $$\displaystyle a(x) $$ and $$\displaystyle a(x) $$ is absorbed into $$\displaystyle k_2(x) $$. Working last with $$\displaystyle N(x,y) $$, we have $$\displaystyle \bar b(x,y)c(y)=\bar b(x)$$ (left side is the general case and the right side is the particluar case). Since $$\displaystyle c(y)=C $$, it can be absorbed into $$\displaystyle \bar b(x,y)$$. Also, $$\displaystyle \bar b(x,y)=\bar b(x)$$ because it is not a function of $$\displaystyle y $$. This ultimately yields the particluar case, $$\displaystyle \bar b(x)$$.

Author
Contributed by Allen

=''' Problem R*3.2 - Show Eq. (1) p.13-4 is exact or can be made exact by the IFM. Find the integrating factor h '''=

From [[media:Pea1.f11.mtg13.djvu|Mtg 13-4]]

Given
Eq. (1) p.13-2 is $$\displaystyle \bar b(x,y)c(y)y'+a(x)\bar c(x,y)=0$$ or $$\displaystyle N(x,y)y'+M(x,y)=0$$. Eq. (2) p.13-2 is $$\displaystyle \bar b(x,y) = \int^x b(s)ds+k_1(y) $$. Eq. (1) p.13-3 is $$\displaystyle \bar c(x,y)= \int^y c(s)ds+k_2(x) $$. $$\displaystyle a(x) = 5x^3+2 $$ $$\displaystyle b(x) = x^2 $$ $$\displaystyle c(x) = y^4 $$ $$\displaystyle \bar b(x) = \frac{1}{3}x^3+d_1 $$ $$\displaystyle \bar c(x) = \frac{1}{5}y^5+\sin(x)+d_2 $$ So, after combining the above equations into Eq. (1) p.13-4 we obtain Eq. (1) p.13-4 which is $$\displaystyle (\frac{1}{3}x^3+d_1)y^4y'+(5x^3+2)*(\frac{1}{5}y^5+\sin x + d_2)=0 $$.

Find
Show that Eq. (1) p. 13-4 either is exact or can be made exact by the IFM. Find the integrating factor $$\displaystyle h $$.

Solution
Solved independently. In order for Eq. (1) p. 13-4 to be exact, it must satisfy both conditions of exactness. The first exactness condition requires the formula to be in the following form: $$\displaystyle N(x,y)y'+M(x,y)=0$$. Eq. (1) p. 13-4 already satisfies the above requirement so it meets this condition. The second exactness condition requires the formula to satisfy the following condition: $$\displaystyle M_y(x,y)=N_x(x,y)$$. Differentiate the following equations, $$\displaystyle M(x,y)=(5x^3+2)*(\frac{1}{5}y^5+\sin x + d_2)$$ and $$\displaystyle N(x,y)=(\frac{1}{3}x^3+d_1)y^4 $$, to obtain $$\displaystyle M_y(x,y)=5*(x^3+\frac{2}{5})y^4$$ and $$\displaystyle N_x(x,y)=x^2y^4$$. Clearly, $$\displaystyle M_y(x,y)\ne N_x(x,y)$$. Since the equation is not exact, we must make it exact with the IFM. Using $$\displaystyle h(x)=\exp[\int^x n(s)ds+k] $$ and $$\displaystyle n(x)=\frac{-1}{N}(N_x-M_y) $$ with $$\displaystyle N(x,y)=(\frac{1}{3}x^3+d_1)y^4 $$, $$\displaystyle N_x(x,y)=x^2y^4 $$, and $$\displaystyle M_y(x,y)=(5x^3+2)y^4 $$ we have $$\displaystyle h(x)=\exp \int^x -[\frac{5s^3+s^2+2}{\frac{1}{3}s^3+d_1}]ds $$.

Author
Contributed by Allen

= Problem R*3.3 - Find an N1-ODE of the form (1) p. 13-2 and the First Integral, Phi =

From [[media:Pea1.f11.mtg13.djvu|Mtg 13-4]]

Given
Eq. (1) p.13-2 is $$\displaystyle \bar b(x,y)c(y)y'+a(x)\bar c(x,y)=0$$ or $$\displaystyle N(x,y)y'+M(x,y)=0$$. Eq. (2) p.13-2 is $$\displaystyle \bar b(x,y) = \int^x b(s)ds+k_1(y) $$. Eq. (1) p.13-3 is $$\displaystyle \bar c(x,y)= \int^y c(s)ds+k_2(x) $$. $$\displaystyle a(x) = \sin(x^3) $$. $$\displaystyle b(x) = \cos(x) $$. $$\displaystyle c(x) = \exp(2y) $$.

Find
$$\displaystyle 1) $$ Find an N1-ODE of the form (1) p. 13-2 that is either exact or can be made exact by the IFM. $$\displaystyle 2) $$ Find the first integral $$ \phi (x,y)=k $$.

Solution
Solved independently. $$\displaystyle 1) $$ First solve for $$\displaystyle \bar b(x,y) $$ and $$\displaystyle \bar c(x,y) $$. $$\displaystyle \bar b(x,y) = \int^x \cos(s) ds $$ $$\displaystyle \bar b(x,y) = \sin(x) $$ $$\displaystyle \bar c(x,y) = \int^x \exp(2y) ds $$ $$\displaystyle \bar c(x,y) = \frac{1}{2}*\exp(2y) $$ Fill Eq. (1) p. 13-2 with the known values and you will obtain $$\displaystyle \sin(x)*\exp(2y)*y'+\frac{1}{2}*\sin(x^3)*\exp(2y)=0 $$ This equation simplifies to $$\displaystyle \sin(x)*y'+\frac{1}{2}*\sin(x^3)=0 $$. So, the above equation satisfies the first condition of exactness, $$\displaystyle N(x,y)y'+M(x,y)=0 $$. However, it fails to satisfy the second condition of exactness, as shown below. The second exactness condition requires the formula to satisfy the following form: $$\displaystyle M_y(x,y)=N_x(x,y)$$. Differentiate the following equations, $$\displaystyle M(x,y)= \frac{1}{2}*\sin(x^3)*\exp(2y) $$ and $$\displaystyle N(x,y)= \sin(x)*\exp(2y) $$, to obtain $$\displaystyle M_y(x,y)= \sin(x^3)*\exp(2y) $$ and $$\displaystyle N_x(x,y)= \cos(x)*\exp(2y) $$. Clearly, $$\displaystyle M_y(x,y)\ne N_x(x,y)$$. Since the equation is not exact, we must make it exact with the IFM. Using $$\displaystyle h(x)=\exp[\int^x n(s)ds+k] $$ and $$\displaystyle n(x)=\frac{-1}{N}(N_x-M_y) $$ with $$\displaystyle N(x,y)=\sin(x) $$, $$\displaystyle N_x(x,y)=\cos(x) $$, and $$\displaystyle M_y(x,y)=0 $$ we have $$\displaystyle h(x)=\exp[\int^x [\frac{-1}{\sin(s)}(\cos(s)-0)]ds] $$. This simplies to $$\displaystyle h(x)= \frac{1}{\sin(s)} $$. Multiply $$\displaystyle h(x)= \frac{1}{\sin(s)} $$ and $$\displaystyle \sin(x)*y'+\frac{1}{2}*\sin(x^3)=0 $$ which yields $$\displaystyle y'+\frac{1}{2}\frac{\sin(x^3)}{\sin(x)}=0 $$. $$\displaystyle 2) $$ Find $$ \phi (x,y)=k $$. $$ \phi_x(x,y)=M(x,y) $$ $$ \phi_y(x,y)=N(x,y) $$ Using the above equations, you can integrate with respect to $$\displaystyle x $$ or $$\displaystyle y $$ which yields $$\displaystyle \phi(x,y)=\frac{-1}{2x^2}\cos(x^3)+k_3 $$ $$\displaystyle \phi(x,y)=\sin(x)y+k_4 $$. Equate the equations above and solve for $$\displaystyle k=k_3+k_4 $$ which is $$\displaystyle \phi(x,y)=\sin(x)y+\frac{1}{2x^2}\cos(x^3)=k $$

Author
Contributed by Allen

= Problem 3.4: Construction of N1-ODEs = From [[media:Pea1.f11.mtg13.djvu|Mtg 13-5]]

Given
Given a class of N1-ODEs of the form,

$$\displaystyle \underbrace{\bar b(x,y)c(y)}_{\color{blue}{N(x,y)}}\,y'+\underbrace{a(x)\bar c(x,y)}_{\color{blue}{M(x,y)}}=0$$ (4.1)

where, $$ \bar b(x,y) $$ and  $$ \bar c(x,y) $$ are given by,

$$ \bar b(x,y) := \int^x b(s)ds + k_1(y) $$ (4.2)

$$ \bar c(x,y) := \int^y c(s)ds + k_2(x) $$ (4.3)

Above, $$ \displaystyle a(x), b(x) $$ and $$ \displaystyle c(y) $$ are arbitrary functions.

In R3.1 (problems given above, on this page), it is shown that this N1-ODE (4.1) satisfies (2) p.11-2 (see lecture notes, XXXXX), i.e.

$$ \frac{1}{N}(N_x-M_y) \, =: \color{blue}{n(x)} $$ (4.4)

which means that the integrating factor, i.e. h, is a function of x alone.

Find
Find a class of N1-ODEs, which is counterpart of above equation, (4.1), that satisfies condition (1) p.11-3. This condition is given by,

$$ \frac{1}{M}(N_x-M_y) \, =: \color{blue}{m(y)} $$ (4.5)

This condition means that an integrating factor, h, which is a function of y alone, can be found for this new class of N1-ODEs that renders them exact.

Solution
We note that in (4.1), $$ \displaystyle N_x = b(x)c(y) $$ and  $$ \displaystyle M_y = a(x)c(y) $$, so we get a common factor, $$  \displaystyle c(y) $$ in the numerator of (4.4) from both $$ \displaystyle N_x $$ and $$ \displaystyle M_y $$, which cancels out with c(y) from $$ \displaystyle N(x,y) $$ in the denominator. What remains after this canceling operation is a function of x alone if and only if $$ \displaystyle k_1(y) = d_1 $$ is a constant.

So, for construction of new class of N1-ODEs, we want $$ \displaystyle N_x $$ and  $$ \displaystyle M_y $$ to give a common factor that's a function of x alone, which will be cancelled by a factor in $$ \displaystyle M(x,y) $$, that will be a function of x alone. Therefore, let,

$$ M(x,y) := a(x) \bar c(x,y) $$ (4.6)

and,

$$ N(x,y) := \bar a(x,y) b(y) $$ (4.7)

Here,

$$ \bar a(x,y) := \int^x a(s)ds + k_1(y) $$ (4.8)

$$ \bar c(x,y) := \int^y c(s)ds + k_2(x) $$ (4.9)

and, now it can be verified that N1-ODE formed from $$ \displaystyle M(x,y) $$ and $$ \displaystyle N(x,y) $$ given by (4.6) and (4.7) respectively, with $$ \displaystyle \bar a(x,y) $$ and $$ \displaystyle \bar c(x,y)  $$ given by (4.8) and (4.9) respectively satisfies,

$$ \frac{h_y}{h} = \frac{1}{M}(N_x-M_y) \, = \frac{1}{\cancel{a(x)} \bar c(x,y)} (\cancel{a(x)}b(y) - \cancel{a(x)}c(y)) =: \color{blue}{m(y)} $$ (4.10)

only when $$ \displaystyle k_2(x) $$ in $$ \displaystyle \bar c(x,y) $$ is a constant.

Author
Contributed by Ankush

= Problem R*3.5 Derivation of Equations of Motion = From [[media:Pea1.f11.mtg14.djvu|Mtg 14-2]]

Given
A particle (e.g., a projectile) moving in the air, with initial velocities, $$ \displaystyle v_x(t=0) $$ and $$ \displaystyle v_y(t=0) $$ given by $$ \displaystyle v_{x0} $$ and $$ \displaystyle v_{y0} $$. The particle travels in air, with two forces acting on the particle, it's weight $$ \displaystyle mg $$, acting vertically downwards (in negative y direction), and $$ \displaystyle kv^n $$ the drag force provided by the fluid, i.e. air, which resists the motion of the particle. $$ \displaystyle m $$ and $$ \displaystyle g $$ are mass of the particle and gravitational acceleration, and $$ \displaystyle k $$ and $$ \displaystyle n $$ are constants.

Fig.5-1: A particle in motion in air, with weight and drag force provided by air to its motion. The initial velocities are also denoted as described above. Angle, $$ \displaystyle \alpha $$ is the angle that velocity of the particle, v makes with the x-axis at a particular time. This image was taken from

Find
1. Derive the equations of motion of the particle (shown above), i.e.

$$ m\frac{dv_x}{dt} = -kv^n\cos \alpha $$  (5.1)

$$ m\frac{dv_y}{dt} = -kv^n\sin \alpha - mg $$  (5.2)

$$ v^2 = (v_x)^2 + (v_y)^2 $$   (5.3)

$$ \tan \alpha\ = \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{v_y}{v_x}  $$   (5.4)

2. For case k = 0, i.e. (no drag force), verify that y(x) is a parabola

3. For case k ≠ 0 and vx0 = 0 (i.e. particle is thrown vertically upwards, i.e. α = 90o), equation of motion (5.2) becomes,

$$ \displaystyle m \frac{dv_y}{dt} = - k(v_y)^n - mg $$  (5.5)

3.1 Find if Eq. (5.5) is exact? If not, find if it can be made exact using IFM [(Integration Factor Method)]?

Find $$ \displaystyle v_y(t) $$ and $$ \displaystyle y(t) $$ for constant $$ \displaystyle m $$.

3.2 For $$ \displaystyle m = m(t) $$, find $$ \displaystyle v_y(t) $$ and $$ \displaystyle y(t) $$. Mass as a function of time, t, is given in the following graph,

Fig. 5-2: Temporal variation of mass, m (in y-aixs) w.r.t. time, t (in x-axis). This graph is taken from

Solution
Solved independently.

Considering forces on the particle shown in Fig. 5-1, we can resolve the force, $$ \displaystyle kv^n $$ in horizontal and vertical directions (x and y respectively), therefore we will have,

$$ \displaystyle (kv^n)_x = -kv^n \cos \alpha $$  (5.6) $$ \displaystyle (kv^n)_y = -kv^n \sin \alpha $$
 (5.7)

Applying Newton's Second Law, which states that the net rate of change of momentum is governed by forces acting on a body, and resolving these forces along x and y directions, we get, (also assuming mass, m to be constant w.r.t. time or rate of change of mass w.r.t. time to be very low when this analysis is performed)

$$ m\frac{dv_x}{dt} = (kv^n)_x $$
 (5.8)

$$ m\frac{dv_y}{dt} = (kv^n)_y - mg $$
 (5.9)

Substituting (5.6) and (5.7) into (5.8) and (5.9) respectively, we will obtain the required equations of motion given by (5.1) and (5.2)

Also, since $$ \displaystyle v_x $$ and $$ \displaystyle  v_y $$ are components of velocity vector, v, (see Fig. 5-1) it is very easy to see that (5.3) is satisfied based on Pythagorus theorem. Also based on definition of tangent definition, and the triangle shown in Fig. 5-1, we can see that,

$$ \tan \alpha\ = \frac{v_y}{v_x} = \frac{\frac{dy}{\cancel dt}}{\frac{dx}{\cancel dt}} = \frac{dy}{dx} $$

2. For particular case, k = 0, equations of motion simplify to,

$$ m\frac{dv_x}{dt} = 0 $$
 (5.10)

$$ \cancel m\frac{dv_y}{dt} = - \cancel mg $$
 (5.11)

Integrating above equations (5.10) and (5.11) we get,

$$ \displaystyle v_x = constant = k_2 $$

$$ \displaystyle v_y(t) = - gt + k_1 $$

Applying initial conditions (at t = 0) for velocities, $$ \displaystyle v_x $$ and $$ \displaystyle v_y $$, we get

$$ \displaystyle k_2 = v_{x0} $$

$$ \displaystyle k_1 = v_{y0} $$

Relating $$ \displaystyle v_x $$ and $$ \displaystyle v_y $$ to x and y respectively,

$$ \displaystyle v_x = \frac{dx}{dt} = v_{x0} $$

$$ \displaystyle v_y = \frac{dy}{dt} = -gt + v_{y0}$$

Integrating above two equations w.r.t. time, t, to obtain x and y, (Note initial location is origin, hence integration constants in both equations given below will be zero)

$$ \displaystyle x(t) = v_{x0} t $$
 (5.12)

$$ \displaystyle y(t) = -\frac{gt^2}{2} + v_{y0} t $$
 (5.13)

Substituting $$ \displaystyle t = \frac{x}{v_{x0}} $$ (from 5.12) into (5.13), we get the parabola form of y in terms of x,

$$ \displaystyle y(x) = -\frac{gx^2}{2(v_{x0})^2} + \frac{v_{y0}}{v_{x0}} x $$
 (5.14)

3. For k ≠ 0, and vx0 = 0, i.e. the particle is thrown vertically upwards, we have (following (5.9)),

$$ \displaystyle \underbrace {m}_{N(t,v_y)}\frac{dv_y}{dt} + \underbrace {k(v_y)^n + mg}_{M(t,v_y)} = 0 $$
 (5.15)

Above equation satisfies 1st exactness condition, as is seen above. Now we check if it satisfies 2nd exactness condition, to see if the equation is exact or not?

$$ \displaystyle \frac{\partial M}{\partial v_y} = nk (v_y)^{n-1} \ne \frac{\partial N}{\partial t} = 0 $$
 (5.16)

Therefore, we see from above that the 2nd exactness condition is not satisfied, therefore Eq. (5.15) is not in exact form.

To check if it can be put into exact form or not, we find that, it satisfies (1) p. 11-3 from the notes, i.e.

$$ \displaystyle \frac{1}{M}(N_t - M_{v_y}) = \frac{1}{(k{v_y}^n + mg)} (0 - nk{v_y}^{n-1}) = F(v_y) $$
 (5.17)

Therefore, equation (1), p.14-3 (in notes) or Eq. 5.15 can be made exact using IFM based on above equation (5.17), which states that the expression on LHS, is a function of $$ \displaystyle v_y $$ alone.

To find $$ \displaystyle v_y(t) $$ and y(t) for constant m case, we rearrange terms in (5.15) and integrate both sides,

$$ \displaystyle \int \frac{1}{(\frac{k}{m} {v_y}^n + g)} dv_y = - \int dt $$
 (5.18)

Above equation is solved using Wolfram Online Integrator, therefore, we have,

$$ \displaystyle v_y(t) = -\frac{gt}{_2F_1(1,\frac{1}{n}; 1 + \frac{1}{n}; - \frac{k}{mg} {v_y}^n)} + C_1 $$
 (5.19)

Note, $$ \displaystyle C_1 $$ above is integration constant, and is equal to vy0 due to the initial condition. Integration of above Eq. (5.19) w.r.t. t, can be performed to get y(t), but in its current form, with n, which can be any value, it is difficult to perform the integration manually. A numerical approach will suit for this purpose for some given n.

3.2 For m = m(t), linear function given in Fig. 5-2, we write the differential equation in following form,

$$ \frac{dv_y}{dt} = - \frac{k}{m} {v_y}^n - g $$
 (5.20)

Above equation is non homogenous ODE, so we can write the solution in terms of homogenous and particular solutions (and take n = 1), i.e. vy = (vy)H + (vy)P. For homogenous ODE, we solve

$$ \frac{dv_{yH}}{dt} = - \frac{k}{m} {v_{yH}}^n $$

Above equation can be solved for vyH for n =1 and we get

$$ v_{yH} = (t)^{-\frac{k t_1}{(m_1 - m_0)}} $$

For particular solution, we use,

$$ \frac{dv_{yP}}{dt} = - g $$

and we get

$$ v_{yP} = - g t $$

Now combining both homogenous and particular portion of vy, and applying initial condition at t = 0, to get the constant of integration to be vy0, we finally get vy to be (for n = 1)

$$ v_{y}(t) = (t)^{-\frac{k t_1}{(m_1 - m_0)}} - gt + v_{y0} $$

Integrating above equation w.r.t. t, we can get y(t), and applying initial condition that y(t = 0) = 0, we get

$$ y(t) = \frac{(t)^{-\frac{k t_1}{(m_1 - m_0)} + 1}}{-\frac{k t_1}{(m_1 - m_0)} + 1} - \frac{gt^2}{2} + v_{y0} t $$

Author
Contributed by Ankush

= Problem R*3.6 Coupled Pendulums = From [[media:Pea1.f11.mtg14.djvu|Mtg 14-6]]

Given
A system of Coupled pendulums shown below,

Fig. 6-1: 2 pendulums connected with a spring of spring constant, k, and point weights of both the pendulums being $$ \displaystyle m_1 $$ and $$ \displaystyle m_2 $$. The rods 1 and 2 (1 is the rod on left and 2 is the rod on right), make angles θ1 and θ2 with the vertical axis respectively. u1 and u2 are two control forces acting on the rods 1 and 2 respectively. Above figure was taken from.

Find
Derive the equations of motion for the two coupled pendulums,

$$ \displaystyle m_1 l^2 \ddot \theta_1 = -ka^2(\theta_1 - \theta_2) - m_1 g l \theta_1 + u_1 l $$
 (6.1)

$$ \displaystyle m_2 l^2 \ddot \theta_2 = -ka^2(\theta_2 - \theta_1) - m_2 g l \theta_2 + u_2 l $$
 (6.2)

Write the above equations in the matrix form (1) p. 14-4, i.e.

$$ \displaystyle \mathbf{\dot{x}}_{n \times 1}(t) = \mathbf{A}_{n \times n}(t) \, \mathbf{x}_{n \times 1}(t) + \mathbf{B}_{n \times m}(t) \, \mathbf{u}_{m \times 1}(t) $$
 (6.3)

where x and u are written as,

$$ \displaystyle \mathbf{x}:= \left \lfloor \theta_1 \ \dot \theta_1 \ \theta_2 \ \dot \theta_2 \right \rfloor^T \in \mathbb R^{4 \times 1} $$
 (6.4)

$$ \displaystyle \mathbf{u}:= \begin{Bmatrix} u_1| \\u_2| \end{Bmatrix} \in \mathbb R^{2 \times 1} $$
 (6.5)

Find $$ \displaystyle \mathbf{A} \in \mathbb R^{4 \times 4}, \ \mathbf{B} \in \mathbb R^{4 \times 2} $$

Solution
Solved independently.

Since we are considering linear system, we take θ1 and θ2 to be small angles. This assumption will be used in deriving the above equations of motion, (6.1) and (6.2)

Note that the term, $$ \displaystyle \ddot \theta $$ with subscripts 1 and 2 denotes angular acceleration of the rods, i.e.,

$$ \displaystyle \alpha_i = \ddot \theta_i $$
 (6.5)

Angular acceleration, αi, is related to the torque, Ti in following way,

$$ \displaystyle T_i = I_i \alpha_i = I_i \ddot \theta_i $$
 (6.6)

Considering net torque on the rod 1, by forces acting on it, we get three components of the Torque,

(a) $$ \displaystyle T_1a = - m_1 g l \sin \theta_1 = -m_1 g l \theta_1 $$ (Due to weight m1g)

(b) $$ \displaystyle T_1b =  k a (\theta_2 - \theta_1) a \cos \theta_1 = k a (\theta_2 - \theta_1) a $$  (Due to spring force, ka(θ2 - θ1))

(c) $$ \displaystyle T_1c = u_1 l \cos \theta_1 = u_1 l $$ (Due to control force u1)

Note, in above, due to θ1 and θ2 being small angles, we could assume sin(θ1) ~ θ1 and cos(θ1) ~ 1.

Combining all (a), (b) and (c) above, we will get,

$$ \displaystyle T_1 = I_1 \alpha_1 = m_1 l^2 \ddot \theta_1 = - k a^2 (\theta_1 - \theta_2) - m_1 g l \theta_1 + u_1 l $$

Above equation is (6.1) which is derived using Newton's second Law of rotational motion.

Similarly we can also derive (6.2) by applying Newton's second law for rotational motion on the rod 2. Signs of -ka2(θ2 - θ1) is flipped since spring force will pull the rod 2 in -x-direction. Following the steps we will get,

$$ \displaystyle T_2 = I_2 \alpha_2 = m_2 l^2 \ddot \theta_2 = - k a^2 (\theta_2 - \theta_1) - m_2 g l \theta_2 + u_2 l $$

For second part of the problem, we write $$ \displaystyle \dot {x} $$ as follows, (differentiating $$ \displaystyle x $$ in Eq. (6.4) w.r.t. t)

$$ \displaystyle \mathbf{\dot x}:= \left \lfloor \dot \theta_1 \ \ddot \theta_1 \ \dot \theta_2 \ \ddot \theta_2 \right \rfloor^T \in \mathbb R^{4 \times 1} $$

Equations (6.1) and (6.2) now denote rows 2nd and 4th of Eq. (6.3). Rows 1st and 3rd of Eq. (6.3) will have 1 in the 2nd and 4th columns respectively in matrix A. Other columns in these rows will be zero. For 2nd and 4th rows of matrix A, we denote them by unknown variables which we have to find in terms of known variables.

$$ \displaystyle \begin{bmatrix} \dot {\theta}_{1} \\ \ddot{ {\theta}_{1}} \\ \dot {\theta}_{2} \\ \ddot{{\theta}_{2}} \\ \end{bmatrix}

=

\underbrace{ \begin{bmatrix} 0 & 1 & 0 & 0 \\ {a}_{1} & {b}_{1} & {c}_{1} & {d}_{1} \\ 0 & 0 & 0 & 1 \\ {a}_{2} & {b}_{2} & {c}_{2} & {d}_{2} \\ \end{bmatrix} }_{\mathbf{A}(t) =4 \times 4}

\cdot

\begin{bmatrix} {\theta}_{1} \\ \dot{{\theta}_{1}} \\ {\theta}_{2} \\ \dot{{\theta}_{2}} \\ \end{bmatrix} +

\underbrace{ \begin{bmatrix} 0 & 0 \\ {e}_{1} & {f}_{1}  \\ 0 & 0 \\ {e}_{2} & {f}_{2} \\ \end{bmatrix} }_{\mathbf{B}(t) = 4 \times 2}

\cdot

\begin{bmatrix} {u}_{1}l \\ {u}_{2}l \\ \end{bmatrix}

$$
 (6.7)

Above, terms ai - fi, i = 1, 2 are unknowns that can be found by comparing above equation to Eq. (6.1) and (6.2). E.g., a1 is the coefficient of θ1 in Eq. (6.1), which is $$ \displaystyle (-\frac{ka^2}{m_1 l^2} - \frac{g}{l}) $$. Similarly, b1 and d1 will be zero since there is no $$ \displaystyle \dot \theta_1 $$ and $$ \displaystyle \dot \theta_2 $$ in Eq. (6.1). c1 is the coefficient of θ2 in Eq. (6.2) and is thus equal to $$ \displaystyle \frac{ka^2}{m_1 l^2} $$. Note, in calculating all these coefficients above, we have considered dividing Eq. (6.1) and (6.2) by the coefficients of $$ \displaystyle \ddot \theta_1 $$ and $$ \displaystyle \ddot \theta_2 $$ respectively. Also e1 will be $$ \displaystyle \frac{1}{m_1 l^2} $$ and f1 should be zero.

Following the same analysis as highlighted in above paragraph, we can find variables a2 - f2 and thus final system in matrix form is written as,

$$ \displaystyle \begin{bmatrix} \dot {\theta}_{1} \\ \ddot{ {\theta}_{1}} \\ \dot {\theta}_{2} \\ \ddot{{\theta}_{2}} \\ \end{bmatrix}

=

\underbrace{ \begin{bmatrix} 0 & 1 & 0 & 0 \\ -\frac{k{a}^{2}-{m}_{1}lg}{{m}_{1}{l}^{2}} & 0 & \frac{k{a}^{2}}{{m}_{1}{l}^{2}} & 0  \\ 0 & 0 & 0 & 1 \\ \frac {k{a}^{2}}{{m}_{2}{l}^{2}}& 0 & -\frac{k{a}^{2}-m_2gl}{{m}_{2}{l}^{2}} & 0 \\ \end{bmatrix} }_{\mathbf{A}(t) =4 \times 4}

\cdot

\begin{bmatrix} {\theta}_{1} \\ \dot{{\theta}_{1}} \\ {\theta}_{2} \\ \dot{{\theta}_{2}} \\ \end{bmatrix} +

\underbrace{ \begin{bmatrix} 0 & 0 \\ \frac{1}{{m}_{1}{l}^{2}} & 0  \\ 0 & 0 \\ 0 & \frac{1}{{m}_{2}{l}^{2}} \\ \end{bmatrix} }_{\mathbf{B}(t) = 4 \times 2}

\cdot

\begin{bmatrix} {u}_{1}l \\ {u}_{2}l \\ \end{bmatrix}

$$
 (6.8)

Author
Contributed by Ankush

=''' Problem 3.7 - Use IFM to show that the solution of (2) is ... '''= From [[media:Pea1.f11.mtg15.djvu|Mtg 15-1]]

Find
Show that the solution of (7.1) is

Solution
Multiplying Integrating Factor h(t);

h is the function of t, so

Then, put (7.4) into (7.3)

We can rewrite <LHS> as

Then, integrate (7.5) with (7.6);

Author
Contributed by Chung

= Problem 3.8 - Generalize (2)p.15-2 to the case of linear time-variant system = From [[media:Pea1.f11.mtg15.djvu|Mtg 15-3]]

Given
Linear Time Invariant Equation $$\displaystyle \dot x(t) = a(t)x(t) + b(t)u(t) $$

Solution
Integrating factor ;

Multiply integrating factor both side;

As a result,

Then,

Author
Contributed by Chung

= Problem 3.9 - Verify that (1) p.15-4 = From [[media:Pea1.f11.mtg15.djvu|Mtg 15-4]]

Find
Verify 9.1 satisfies 9.2, 9.3

Author
Contributed by Chung

= Problem 3.10 - Free vibration of coupled pendulums = From [[media:Pea1.f11.mtg15.djvu|Mtg 15-5]]

Given
The equations of motion of coupled pendulums: Parameters: $$\displaystyle \begin{align} \mbox{Pendulums: } &a=0.3, \mbox{ } l=1, \mbox{ } k = 0.2 \\ &m_1 g = 3, \mbox{ } m_2 g = 6 \end{align} $$ $$\displaystyle \mbox{No applied forces: } u_1=u_2=0 $$ $$\displaystyle \begin{align} \mbox{Ini}&\mbox{tial conditions: } \\ &\theta_1 (0)=0, \mbox{ } \dot{\theta_1}(0)=-2 \\ &\theta_2 (0)=0, \mbox{ } \dot{\theta_2}(0)=+1 \\ \end{align} $$

Find
1. Integrate the system in matrix form for $$\displaystyle t \in [0,7] $$ 2. Use (2) p.15-2 to find the solution 3. Plot the results

Integrate the system in matrix form
In order to represent the Eq. (10.1) in matrix form (1) From [[media:Pea1.f11.mtg14.djvu|p.14-4]], the following state variables are defined. Then, the Eq. (10.1) can be rewritten as following. Rearrange the terms in the Eq. (10.3), Finally, the follow matrix equation is obtained.

The following is a matlab code to integrate the system.
 * {| style="width:100%" border="0" align="left"

The following figure is the plot of the integration results.
 * style="width:30%; padding:10px; border:1px solid #888888" align="center" |
 * style="width:30%; padding:10px; border:1px solid #888888" align="center" |
 * }
 * }
 * }

Use (2) p.15-2 to find the solution
Since no input is applied to the system, the second term in the Eq. (10.6) is zero. The matlab code shows the problem-solving using the Eq. (10.7).
 * {| style="width:100%" border="0" align="left"

The following figure is the plot of the results solved using the Eq. (10.7).
 * style="width:30%; padding:10px; border:1px solid #888888" align="center" |
 * style="width:30%; padding:10px; border:1px solid #888888" align="center" |
 * }
 * }
 * }

Plot the results
Refer to Figure 10.1 and Figure 10.2 $$\displaystyle x_1 $$ in Figure 10.1 and Figure 10.2 are $$\displaystyle \theta_1(t) $$ $$\displaystyle x_3 $$ in Figure 10.1 and Figure 10.2 are $$\displaystyle \theta_2(t) $$

Author
Contributed by Shin

= Problem 3.11 SC-L1-ODE-CC with State Transition Matrix = From [[media:Pea1.f11.mtg16.djvu|Mtg 16-1]]

Solution
First, we rearrange the equation(11.3), we have Then, we substitute $$\displaystyle \mathbf{x}(t)$$ by $$\displaystyle \mathbf{\Phi}(t,t_0)$$, we can get Therefore, we can derive Euler integrating factor Thus we have So we get We integrate both sides of equation(11.9) in the interval $$\displaystyle t\geqslant{\tau}\geqslant t_{0}$$, then we have

Thus replace $$\mathbf{\Phi} (t,{{t}_{0}})$$ with $$\mathbf{x}(t)$$, and rearrange the equation, we can get

Author
Contributed by Kexin Ren

= Problem 3.12 Roll Control of Rocket = From [[media:Pea1.f11.mtg16.djvu|Mtg 16-3]]

Given
$$ \begin{align} &{\delta} = \text{Aileron angle/deflection}\\ &{\phi} = \text{Roll angle}\\ &{\omega} = \text{Roll angular velocity}\\ &Q = \text{Aileron effectiveness}\\ &{\tau} = \text{Roll time constant}\\ &u = \text{Command signal to Ailerons}\\ \end{align} $$

Find
Put equations(12.1~12.3) in the form of SC-L1-ODE-CC as in equation(12.4) with $$\displaystyle \mathbf{A, B}$$ being constant matrices.

Solution
Integrate both sides of equation(12.3), we have

So we can find equation(12.2) become

Then we integrate both sides of equation(12.6)

Similar to equation(12.1), we get

So equation(12.7) becomes

Therefore, we can get

where,

Author
Contributed by Kexin Ren

= Problem 3.13 - The 2nd exactness condition = From [[media:Pea1.f11.mtg16.djvu|Mtg 16-6]]

Given
2nd exactness condition for N2-ODEs in the class note [[media:Pea1.f11.mtg16.djvu|16-5]] And, the N2-ODE equation

Find
1. Derive the Eq. (13.2), i.e., the 2nd relation in the 2nd exactness condition. 2. Derive the Eq. (13.1), i.e., the 1st relation in the 2nd exactness condition. 3. Verify that the Eq. (13.3) satisfies the 2nd exactness condition.

Derive the 2nd relation in the 2nd exactness condition.
From the equation (3), (4) in the class note [[media:Pea1.f11.mtg16.djvu|16-4]], g(x,y,p) and f(x,y,p) are as below. Differentiating the Eq. (13.4) with respect to p, Differentiating the Eq. (13.6) with respect to p, Rewriting the Eq. (13.7) using the symmetric property of mixed 2nd partial derivatives of the function $$\displaystyle \phi(x,y,p) $$ as below, Substituting the Eq. (13.5) into the Eq. (13.8),

- The Eq. (13.9) shows the derivation of the 2nd relation in the 2nd exactness condition.

Derive the 1st relation in the 2nd exactness condition.
To solve this problem, 3 relations (Eq. 13.10) for the symmetry of mixed 2nd partial derivatives are exploited. From the first relation in the Eq. (13.10), Express $$\displaystyle \phi_y $$ the Eq. (13.12) in terms of $$\displaystyle f, g, p $$, From the second relation in the Eq. (13.10), Rewrite the Eq. (13.14) using the Eq. (13.4), (13.5), Express $$\displaystyle \phi_x $$ the Eq. (13.15) in terms of $$\displaystyle f, g, p $$, Take the partial derivative of the Eq. (13.13) with respect to x, Take the partial derivative of the Eq. (13.16) with respect to y, Use the third relation in the Eq. (13.10), Move the terms with f to the LHS and move the terms with g to the RHS, - The Eq. (13.20) shows the derivation of the 1st relation in the 2nd exactness condition.

Verify that the Eq. (13.3) satisfies the 2nd exactness condition.
From the 1st exactness condition for N2-ODEs in the class note [[media:Pea1.f11.mtg18.djvu|Mtg 18-3]], The 1st relation in the 2nd exactness condition is satisfied as below. Also, the 2nd relation in the 2nd exactness condition is satisfied as below.

Author
Contributed by Shin

= Problem 3.14 - Find h(x,y) = From [[media:Pea1.f11.mtg18.djvu|Mtg 18-1]]

Given
The equation (1) in the class note [[media:Pea1.f11.mtg18.djvu|17-1]]. where $$\displaystyle p(x) := y^{\prime}(x) $$. The equation (1) in the class note [[media:Pea1.f11.mtg18.djvu|18-1]]

Find
Find $$\displaystyle h(x,y) $$ without assuming a priority that h=constant.

Solution
The Eq. (14.2) can be rewritten in the following form. Since y is the function of x, the Eq (14.3) is the partial derivative of the function h(x,y) with respect to x. So, the LHS term in the Eq. (14.3) is rewritten as follows.

Then, the following is true from the Eq. (14.3), (14.4).

The Eq. (14.5) implies that h(x,y) is the function of only y.

Substituting the Eq. (14.5) into the Eq. (14.2) would result the following. It is concluded that $$\displaystyle h_y = 0 $$ because if $$\displaystyle y^{\prime} = 0 $$ the solution would be trivial.

Author
Contributed by Shin

= Problem 3.15 - Verification of the 2nd exactness condition = From [[media:Pea1.f11.mtg18.djvu|Mtg 18-3]]

Given
The equation (1) in the class note [[media:Pea1.f11.mtg18.djvu|18-3]].

2nd exactness condition for N2-ODEs

Find
Verify whether the Eq. (15.1) satisfies the 2nd exactness condition (Eq. 15.2).

Solution
As proven in the class note [[media:Pea1.f11.mtg18.djvu|Mtg 18-3]], the Eq. (15.1) satisfies the 1st exactness condition for N2-ODEs.

The first condition (Eq. 15.2)
Computing each term in the Eq. (15.2) $$\displaystyle f_{xx} :$$ $$\displaystyle 2pf_{xy} :$$ $$\displaystyle f_{xy} = 0$$, since $$\displaystyle f_x$$ is a function of $$\displaystyle x$$ and $$\displaystyle y^{\prime}$$. $$\displaystyle p^2 f_{yy} :$$ $$\displaystyle f_{y} = 0, f_{yy} = 0 $$, since $$\displaystyle f$$ is a function of $$\displaystyle x$$ and $$\displaystyle y^{\prime}$$. $$\displaystyle g_{xp} :$$ $$\displaystyle g_{yp} :$$ $$\displaystyle g_{y} :$$ Substituting the Eq. (15.5)-(15.8) into the LHS of the Eq. (15.2), Substituting the Eq. (15.9)-(15.12) into the RHS of the Eq. (15.2), - Since the Eq. (15.13) and the Eq. (15.14) are the same equations, the first condition is satisfied.

The second condition (Eq. 15.3)
Computing each term in the Eq. (15.3) $$\displaystyle f_{xp} :$$ $$\displaystyle f_{yp} :$$ $$\displaystyle f_{y} :$$ $$\displaystyle g_{pp} :$$ Substituting the Eq. (15.13)-(15.15) into the LHS of the Eq. (15.3), Substituting the Eq. (15.16) into the RHS of the Eq. (15.3), - Since the Eq. (15.17) and the Eq. (15.18) are the same equations, the second condition is satisfied.

Author
Contributed by Shin

=Problem 3.16 - Solve the Second Choice of Solution=

Given
Remove the cancelled terms:

The first choice for solution assumes:

thus:

so:

So we get:

Then:

and:

Find
Finish the other solution assuming:

Solution
Equation (16.9) is actually:

Integrating both sides yields:

Take derivative of equation (16.12) with respect to $$\,x$$ gives:

It can be seen from (16.2) and (16.9) that $$\,{{h}_{x}}$$ is also equal to:

So we get:

Substitute $$\displaystyle y' $$ for $$\displaystyle \frac{y}{3x}$$ in equation (16.9) yields:

That means:

So we have:

and:

which is the same as the first choice of solution.

Author
Contributed by YuChen

=Problem 3.17 - Prove the Simple Form of Reynolds Transport Theorem=

Given
The Reynolds transport theorem:

Let:

And the conservation of mass equation:

Find
Prove the simple form of the Reynolds transport theorem:

Solution
Replace $$\displaystyle \mathbf f(x,t) $$ with $$\displaystyle \rho (x,t) \mathbf u (x,t) $$ in equation (17.1) yields:

As equation (17.3) shows $$\displaystyle \frac{\partial \rho}{\partial t}+{\rm div}(\rho \mathbf u) = 0$$:

Author
Contributed by YuChen

=Problem 3.18 - Obtain 1-dimensional Reynolds Transport Theorem from n-dimensional Reynolds Transport Theorem=

Given
N-dimensional RTT:

Find
Obtain 1-dimensional RTT:

Solution
In 1-D case, $$\displaystyle \mathcal B_t $$ is actually $$\displaystyle s $$. So:

and

From equations (17.1) and (18.1) we know that $$\displaystyle \int_{\partial \mathcal B_t} \mathbf n (\mathbf f \mathbf u) \, d \partial \mathcal B_t = \int_{\mathcal B_t} {\rm div}(\mathbf f \mathbf u) \, d \mathcal B_t$$, so:

Since in 1-D case $$\displaystyle \mathbf u=u= \frac{ds}{dt}$$, then:

Add equations (18.3), (18.4) and (18.6) will get:

Author
Contributed by YuChen

=Contributing Members=

{| cellspacing=0 align=center cellpadding=5px width=60% style="background: white; border: 2px solid black;"
 * colspan="7" style="background: yellow;border-bottom:2px solid gray" | Team Contribution Table
 * Problem Number||Lecture(Mtg)||Assigned To||Solved By||Typed By||Signature(Author)
 * R3.1||[[media:Pea1.f11.mtg13.djvu|Mtg 13-3]]|| Allen || Allen || Allen || Allen 03:51, 4 October 2011 (UTC)]]
 * R*3.2||[[media:Pea1.f11.mtg13.djvu|Mtg 13-4]]|| Allen || Allen || Allen || Allen 04:50, 5 October 2011 (UTC)
 * R*3.3||[[media:Pea1.f11.mtg13.djvu|Mtg 13-4,5]]|| Allen || Allen || Allen || Allen 04:00, 5 October 2011 (UTC)]]
 * R3.4||[[media:Pea1.f11.mtg13.djvu|Mtg 13-5]]|| Ankush || Ankush || Ankush || Ankush 10:51, 4 October 2011 (UTC)
 * R*3.5||[[media:Pea1.f11.mtg14.djvu|Mtg 14-2]]|| Ankush || Ankush || Ankush || Ankush 20:51, 4 October 2011 (UTC)
 * R*3.6||[[media:Pea1.f11.mtg14.djvu|Mtg 14-6]]|| Ankush || Ankush || Ankush || Ankush 10:51, 5 October 2011 (UTC)
 * R*3.7||[[media:Pea1.f11.mtg15.djvu|Mtg 15-1]]|| Chung || Chung .|| Chung || Chung  3 Oct. 2011 at 23:36(UTC)
 * R*3.8||[[media:Pea1.f11.mtg15.djvu|Mtg 15-3]]|| Chung || Chung || Chung || Chung  2 Oct. 2011 at 21:55(UTC)
 * R*3.9||[[media:Pea1.f11.mtg15.djvu|Mtg 15-4]]|| Chung || Chung || Chung || Chung  2 Oct. 2011 at 21:26(UTC)
 * R3.10||[[media:Pea1.f11.mtg15.djvu|Mtg 15-5]]|| Shin || Shin || Shin || Shin 5 Oct. 2011 at 17:32(UTC)
 * R*3.11||[[media:Pea1.f11.mtg16.djvu|Mtg 16-1]]|| Ren || Ren || Ren || Ren 4 Oct. 2011 at 01:52(UTC)
 * R*3.12||[[media:Pea1.f11.mtg16.djvu|Mtg 16-3]]|| Ren || Ren || Ren || Ren 4 Oct. 2011 at 02:45(UTC)
 * R*3.13||[[media:Pea1.f11.mtg16.djvu|Mtg 16-6]]|| Shin || Shin || Shin || Shin 30 Sep. 2011 at 03:54(UTC)
 * R*3.14||[[media:Pea1.f11.mtg18.djvu|Mtg 18-1]]|| Shin || Shin || Shin || Shin 29 Sep. 2011 at 18:00(UTC)
 * R*3.15||[[media:Pea1.f11.mtg18.djvu|Mtg 18-3]]|| Shin || Shin || Shin || Shin 29 Sep. 2011 at 20:18(UTC)
 * R*3.16||[[media:Pea1.f11.mtg19-20.djvu|Mtg 19-2]]|| YuChen || YuChen || YuChen || YuChen 04 Oct. 2011 at 20:00(UTC)
 * R*3.17||[[media:Pea1.f11.mtg19-20.djvu|Mtg 19-10]]|| YuChen || YuChen || YuChen || YuChen 04 Oct. 2011 at 22:15(UTC)
 * R*3.18||[[media:Pea1.f11.mtg19-20.djvu|Mtg 19-12]]|| YuChen || YuChen || YuChen || YuChen 05 Oct. 2011 at 01:25(UTC)
 * -|}
 * R3.10||[[media:Pea1.f11.mtg15.djvu|Mtg 15-5]]|| Shin || Shin || Shin || Shin 5 Oct. 2011 at 17:32(UTC)
 * R*3.11||[[media:Pea1.f11.mtg16.djvu|Mtg 16-1]]|| Ren || Ren || Ren || Ren 4 Oct. 2011 at 01:52(UTC)
 * R*3.12||[[media:Pea1.f11.mtg16.djvu|Mtg 16-3]]|| Ren || Ren || Ren || Ren 4 Oct. 2011 at 02:45(UTC)
 * R*3.13||[[media:Pea1.f11.mtg16.djvu|Mtg 16-6]]|| Shin || Shin || Shin || Shin 30 Sep. 2011 at 03:54(UTC)
 * R*3.14||[[media:Pea1.f11.mtg18.djvu|Mtg 18-1]]|| Shin || Shin || Shin || Shin 29 Sep. 2011 at 18:00(UTC)
 * R*3.15||[[media:Pea1.f11.mtg18.djvu|Mtg 18-3]]|| Shin || Shin || Shin || Shin 29 Sep. 2011 at 20:18(UTC)
 * R*3.16||[[media:Pea1.f11.mtg19-20.djvu|Mtg 19-2]]|| YuChen || YuChen || YuChen || YuChen 04 Oct. 2011 at 20:00(UTC)
 * R*3.17||[[media:Pea1.f11.mtg19-20.djvu|Mtg 19-10]]|| YuChen || YuChen || YuChen || YuChen 04 Oct. 2011 at 22:15(UTC)
 * R*3.18||[[media:Pea1.f11.mtg19-20.djvu|Mtg 19-12]]|| YuChen || YuChen || YuChen || YuChen 05 Oct. 2011 at 01:25(UTC)
 * -|}
 * R*3.15||[[media:Pea1.f11.mtg18.djvu|Mtg 18-3]]|| Shin || Shin || Shin || Shin 29 Sep. 2011 at 20:18(UTC)
 * R*3.16||[[media:Pea1.f11.mtg19-20.djvu|Mtg 19-2]]|| YuChen || YuChen || YuChen || YuChen 04 Oct. 2011 at 20:00(UTC)
 * R*3.17||[[media:Pea1.f11.mtg19-20.djvu|Mtg 19-10]]|| YuChen || YuChen || YuChen || YuChen 04 Oct. 2011 at 22:15(UTC)
 * R*3.18||[[media:Pea1.f11.mtg19-20.djvu|Mtg 19-12]]|| YuChen || YuChen || YuChen || YuChen 05 Oct. 2011 at 01:25(UTC)
 * -|}
 * R*3.17||[[media:Pea1.f11.mtg19-20.djvu|Mtg 19-10]]|| YuChen || YuChen || YuChen || YuChen 04 Oct. 2011 at 22:15(UTC)
 * R*3.18||[[media:Pea1.f11.mtg19-20.djvu|Mtg 19-12]]|| YuChen || YuChen || YuChen || YuChen 05 Oct. 2011 at 01:25(UTC)
 * -|}
 * -|}

title: User:Egm6321.f11.team4/HW3


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

title: File:Pea1.f11.mtg18.djvu


title: File:HW03.pdf


text: File:DSP HW04.pdf

title: File:DSP HW04.pdf


text: File:DSP HW05.pdf

title: File:DSP HW05.pdf


text: File:DSP HW06.pdf

title: File:DSP HW06.pdf


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User talk:Jblack69

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You don't need to be an educator to edit. You only need to be bold to contribute and to experiment with the sandbox or your userpage. See you around Wikiversity! --Ottava Rima (talk) 04:03, 28 September 2011 (UTC)

By the way, thanks for this edit. Ottava Rima (talk) 04:03, 28 September 2011 (UTC)

Editing Templates
I'm the guy who's template you so kindly were trying to protect at Critique of article template. I wasn't logged in at the time (I accidentally forgot to login). I did it because I was noticing that more legitimate changes that I had made in the past to my template were not showing up when I used the template. So I made a more obvious change, tested it immediately, found it didn't change, and planned to try it again in an hour to see if that would work.

Can you tell me why changes aren't reflecting when I paste ? Areil314 21:59, 17 January 2012 (UTC)

title: User talk:Codeks


text:
User talk:Hmw8522

 Hello Hmw8522, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. All users are expected to abide by our Privacy policy, Civility policy, and the Terms of Use while at Wikiversity.

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You don't need to be an educator to edit. You only need to be bold to contribute and to experiment with the sandbox or your userpage. See you around Wikiversity! --Ottava Rima (talk) 04:05, 28 September 2011 (UTC)

title: User talk:Hmw8522


text:
User talk:Collapse~enwikiversity

 Hello Collapse, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. All users are expected to abide by our Privacy policy, Civility policy, and the Terms of Use while at Wikiversity.

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You don't need to be an educator to edit. You only need to be bold to contribute and to experiment with the sandbox or your userpage. See you around Wikiversity! --Ottava Rima (talk) 04:05, 28 September 2011 (UTC)

Your account will be renamed
Hello,

The developer team at Wikimedia is making some changes to how accounts work, as part of our on-going efforts to provide new and better tools for our users like cross-wiki notifications. These changes will mean you have the same account name everywhere. This will let us give you new features that will help you edit and discuss better, and allow more flexible user permissions for tools. One of the side-effects of this is that user accounts will now have to be unique across all 900 Wikimedia wikis. See the announcement for more information.

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

Your account will still work as before, and you will be credited for all your edits made so far, but you will have to use the new account name when you log in.

Sorry for the inconvenience.

Yours, Keegan Peterzell Community Liaison, Wikimedia Foundation 23:33, 17 March 2015 (UTC)

Renamed
 This account has been renamed as part of single-user login finalisation. If you own this account you can |log in using your previous username and password for more information. If you do not like this account's new name, you can choose your own using this form after logging in: . -- Keegan (WMF) (talk) 06:05, 19 April 2015 (UTC)

title: User talk:Collapse~enwikiversity


text:
User talk:Thanawat lek

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You don't need to be an educator to edit. You only need to be bold to contribute and to experiment with the sandbox or your userpage. See you around Wikiversity! --Ottava Rima (talk) 04:05, 28 September 2011 (UTC)

title: User talk:Thanawat lek


text:
User talk:Panpicha Prukophas

 Hello Panpicha Prukophas, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. All users are expected to abide by our Privacy policy, Civility policy, and the Terms of Use while at Wikiversity.

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You don't need to be an educator to edit. You only need to be bold to contribute and to experiment with the sandbox or your userpage. See you around Wikiversity! --Ottava Rima (talk) 04:05, 28 September 2011 (UTC)

title: User talk:Panpicha Prukophas


text:
User talk:Thidarath Bhew

 Hello Thidarath Bhew, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. All users are expected to abide by our Privacy policy, Civility policy, and the Terms of Use while at Wikiversity.

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You don't need to be an educator to edit. You only need to be bold to contribute and to experiment with the sandbox or your userpage. See you around Wikiversity! --Ottava Rima (talk) 04:06, 28 September 2011 (UTC)

title: User talk:Thidarath Bhew


text:
User talk:Telecomtom

 Hello Telecomtom, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. All users are expected to abide by our Privacy policy, Civility policy, and the Terms of Use while at Wikiversity.

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You don't need to be an educator to edit. You only need to be bold to contribute and to experiment with the sandbox or your userpage. See you around Wikiversity! --Ottava Rima (talk) 04:06, 28 September 2011 (UTC)

title: User talk:Telecomtom


text:
User talk:ELMO-Y-QUENOCHE

 Hello ELMO-Y-QUENOCHE, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. All users are expected to abide by our Privacy policy, Civility policy, and the Terms of Use while at Wikiversity.

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You don't need to be an educator to edit. You only need to be bold to contribute and to experiment with the sandbox or your userpage. See you around Wikiversity! --Ottava Rima (talk) 04:06, 28 September 2011 (UTC)

title: User talk:ELMO-Y-QUENOCHE


text:
Template:EnglishLesson

Key sentences

Drills Say these sentences without looking.

Example Practice this conversation.

Change partners. Use your imagination and change the sentences. Exercise Use your imagination. Make your own conversation.

Change partners. Use your imagination and make another conversation. Practice Further study
 * English lesson topics

title: Template:EnglishLesson


text:
Template:(

{

title: Template:(


text:
Template:)

}

title: Template:)


text:
Template:(/doc

Usage
This template can be used in place of a literal "{" when the latter would be interpreted as a syntactic marker.

For example, $($ gives "{".

title: Template:(/doc


text:
Template:)/doc

Usage
This template can be used in place of a literal "}" when the latter would be interpreted as a syntactic marker.

For example, $)$ gives "}".

ja:Template:)

title: Template:)/doc


text:
Template talk:=

Hi all, I tried using this template and it didn't work for me, so I compared it to the one at WP and found that this one has a newline after the equals sign, whereas the WP template has none. I made another version without an equal sign at Template:=2 which has no newline, and that one worked better for me. I don't know if the other one is broken per se, but I thought I should mention it. Thanks Haplology 05:53, 28 September 2011 (UTC)

title: Template talk:=


text:
Engineering Projects/Bridges/Howard Community College/fall2011/501 DAS/Week 2 (Sept. 26 - Sept. 30, 2011)

This week our team worked on actually building the bridge. We started with gluing skinny long spaghettis as the arch part of the bridge. We failed, because it was really hard to cut them the exact same size. In addition, it took a long time for epoxy to dry up and the spaghettis didn't stick one another since we didn't hold them tied until the glue completely dry. We realized that it was almost impossible to build the bridge without an exact measurement of each part of the bridge. Therefore, we decided to figure out the exact measurement and sketch it on a big piece of paper. Then we can follow the bridge on the paper to build the real one. It will be more acurate and much easier. Besides, we decided to use Ditalini pasta instead of the type we used before to build the arch. For the glue part, we wanted to try glue gun because it dries faster than epoxy. However, in the presentation, Professor Foerster suggested to use epoxy not glue gun. We still want to test it using glue gun. And maybe code epoxy on top. At the same time, we were also looking for bridge test software. We couldn't find free helpful testing software.


text:
User:Mahendra~enwikiversity/mahendra

test

title: User:Mahendra~enwikiversity/mahendra


text: Literature/Tables/Subject-meaning-trigon-full


text:
Template:EnglishLesson/doc

This is the documentation page for the template at Template:EnglishLesson

Usage
This is a template that takes key phrases and produces an interactive English lesson.

You can copy and paste the fields below.

 * q stands for "question"
 * a stands for "answer"
 * alt stands for "alternate"
 * ex stands for "example"
 * pos stands for "part of speech", such as "adjective"
 * The numbering is for row numbers. The first row should have the lesson's most essential language as it will be used again in the course of the lesson.

The alternate fields and the rows following the first row may be left out. The first row of essential language is required.

By default, the value of role1 (e.g. customer) appears on the left, and role2 appears on the right. To reverse them, set the corresponding reverse value to any non-null value. For example, if the roles should be reversed for the second row of language, set reverse2 to any value. Otherwise, it may be omitted.

The instructions are for the students in the final stage. They may be roles for the students to act out in a role play (e.g. student 1: you are a customer in a restaurant, student 2: you are a waiter) or instructions to set up an information gap activity (e.g. student 1: choose a picture, student 2: ask student 1 questions to guess which picture was chosen). Those are only examples and many forms of review are possible.

This template produces the bulk of a lesson but other material, such as pictures, may be helpful to establish context or to be used as lesson material, especially in the final stage, such as in an information gap activity where one student chooses a picture and another must guess which picture was chosen.

There is no need to add formatting such as italics or parentheses as the template does this automatically.

ja:テンプレート:EnglishLesson

title: Template:EnglishLesson/doc


title: Literature/Tables/Magritte-1929-300px


text:
User:Damien.oconnor

Human Computer Interaction or HCI

The Impact on Culture

"Bhutan, the last country in the world to have TV" http://news.bbc.co.uk/1/hi/world/south_asia/358230.stm

Other links to reports and articles concerning the introduction of the internet, mobile phones, home computers, multinational IT companies setting up in Bhutan.

Articles and notes needed for the following:

The culture of Bhutan The population The employment opportunites The number of IT millionaries in Bhutan Number of Mobile phones The user friendly nature of HCI in different devices that are being brought into the country Reports of any adverse effects of HCI on the traditionalway of life Discussions/reports of

title: User:Damien.oconnor


text:
User:Moortooza

I am just trying to learn how to use the editing of wikiuniversity. Moortooza

title: User:Moortooza


text: File:Probability Discussion Distribution Function and Its Inverse.svg


text: File:Probability Discussion Uniform Distribution.svg


text: File:Probability-Mapping.svg

title: File:Probability-Mapping.svg


text: File:Probability-Vector.svg

title: File:Probability-Vector.svg


text:
User talk:Tfalak3666/enes100/illuminate clothing

10/10/30 ... good job 1sfoerster 22:18, 28 September 2011 (UTC)

Interesting designs on the arrangement of the lights by Tfalak. Hope we can get more lights on the dress. Epcole 18:26, 02 October 2011 (UTC)

very nice blue print i much say, Tfalak have done really well in understanding how the lilypad workedTruongdat120891 03:45, 3 October 2011 (UTC)

10/10/10 .... need better description of what was done .. more pictures .. why is the picture repeated? .. is there something needed? What? 1sfoerster 19:13, 3 October 2011 (UTC)

0/0/0 ... nothing to grade 1sfoerster 18:12, 11 October 2011 (UTC)

10/0/10 .. don't see anything other than some pictures .. needs to be a narrative that describes the pictures and connects them together. no comments on team mates pages, no commitments for next weeks work .. 1sfoerster 20:28, 17 October 2011 (UTC)


text: Literature/Tables/I-mean-it

title: Literature/Tables/I-mean-it


text: File:Probability-Venn Diagram.svg

title: File:Probability-Venn Diagram.svg


text:
User:Egm6321.f11.team4/HW4

= Problem R4.1: Direct Derivation of RTT = From [[media:Pea1.f11.mtg19-20.djvu|Mtg 19-12]]

Given
A control volume, denoted by $$ \mathcal B_t $$. Its boundary is denoted by $$ \partial \mathcal B_t $$, as seen in figure below. This figure (shown below) was taken from Lecture Notes of Dr Vu-Quoc, for his course, EGM 6321 - Principles of Engineering Analysis - I. The outward normal to boundary is denoted by $$ \mathbf n $$

Find
Show the direct derivation of RTT, given by following equation,

$$\displaystyle \frac{D}{Dt} \int_{\mathcal B_t} \mathbf f (x,t) \, d \mathcal B_t = \int_{\mathcal B_t} \frac{\partial \mathbf f}{\partial t} \, d \mathcal B_t + \int_{\mathcal \partial \mathcal B_t} \mathbf n \cdot (\mathbf f \mathbf u) \, d (\partial \mathcal B_t) $$  (4.1)

Solution
- Solved on our own Consider the material property f(x,t) (given per unit volume), where x is the spatial variable. Principle of conservation of this material property governs that net rate of change of total amount of f(x,t) associated with material present within the control volume, $$ \mathcal B_t $$ (Term 1) is equal to instantaneous change of this property f(x,t) within the control volume, integrated over the complete domain (Term 2), and the net influx and outflux of the property f(x,t) happening at the surface (or boundary) of the control volume (Term 3). The above statement can be written mathematically as,

Term 1 = Term 2 + Term 3

These terms are given by,

Term 1: Net rate of change of total amount of f(x,t) associated with material currently present within control volume, (integrated over $$ \mathcal B_t $$)

$$\displaystyle \frac{D}{Dt} \int_{\mathcal B_t} \mathbf f (x,t) \, d \mathcal B_t $$

Term 2: Integration over the complete volume of the instantaneous change of the property f(x,t) associated with the material present instantaneously within the control volume,

$$\displaystyle \int_{\mathcal B_t} \frac{\partial \mathbf f}{\partial t} \, d \mathcal B_t $$

Term 3: Influx and outflux of the property f(x,t) happening at the surface (or boundary) of the control volume ($$ \partial \mathcal B_t $$), at a particular point (given by $$\displaystyle \mathbf n \cdot (\mathbf f \mathbf u) $$), and integrated over the complete boundary, $$ \partial \mathcal B_t $$,

$$\displaystyle \int_{\mathcal \partial \mathcal B_t} \mathbf n \cdot (\mathbf f \mathbf u) \, d (\partial \mathcal B_t) $$

Note in above term (Term 3), if $$\displaystyle (\mathbf f \mathbf u) $$ has same direction as n, then it is the outflux of the material, and it will add on the term 2 for conservation of total change of f(x,t) associated with material currently present in the control volume at given time.

However, when a fluid is coming into the control volume, it is not the material that we identified being associated with being present in the control volume at current time, and hence that has to be subtracted from term 2. This happens above since, in that case u has opposite direction to n, and hence we get negative sign with the dot product.

Add all the terms to get Eq. (4.1). Hence the above derivation is proven.

Author
Contributed by Ankush

= Problem R*4.2 Verifying Exactness of L2-ODE-VC = From [[media:Pea1.f11.mtg21.djvu|Mtg 21-1]]

Given
Given a L2-ODE-VC,

$$\displaystyle \sqrt{x} \,y'' + 2xy' + 3y = 0 $$  (2.1)

Find
Verify the exactness of Eq. (2.1)

Solution
- Solved on our own We can identify terms, f(x,y,p) and g(x,y,p) corresponding to the particular form for a L2-ODE to be exact (Eq. (1), p. 16-4 in the Lecture Notes).

$$\displaystyle \underbrace {\sqrt{x}}_{f(x,y,p)} \,y'' + \underbrace {2xy' + 3y}_{g(x,y,p)} = 0 $$  (2.2)

Therefore above equation satisfies first exactness condition. Now we will verify second exactness condition, which is Eqns. (1) - (2), p. 16-5 of the Lecture Notes, i.e.

$$ \displaystyle f_{xx} + 2pf_{xy} + p^2 f_{fyy} = g_{xp} + p g_{yp} - g_y $$                     (2.3)

$$ \displaystyle f_{xp} + pf_{yp} + 2 f_y = g_{pp} $$                     (2.4)

We list all required partial derivatives of f(x,y,p) and g(x,y,p) to verify Eq. (2.3) and (2.4) for checking 2nd exactness condition.

$$ \displaystyle g_y = 3, \, \, g_x = 2p, \, \, g_{xp} = 2, \, \, g_{yp} = 0, \, \, g_{p} = 2x, \, \, g_{pp} = 0, \, \, $$

$$ f_x = \frac{1}{2 \sqrt{x}}, \, \, f_{xx} = - \frac{1}{4 (x)^{\frac {3}{2}}}, \, \, f_{xy} = 0, \, \, f_{yy} = 0, \, \, f_{xp} = f_{yp} = f_y = 0 $$  (2.5)

Using above values we get for LHS of Eq. (2.3),

$$ \displaystyle f_{xx} + 2p \cancel{f_{xy}} + p^2 \cancel{f_{yy}} = - \frac{1}{4 (x)^{\frac {3}{2}}} $$                     LHS (2.3)

The RHS of Eq. (2.3) is given as,

$$ \displaystyle g_{xp} + p \cancel{g_{yp}} - g_y = 2 + 0 - 3 = -1 $$                     RHS (2.3)

Hence we clearly see that RHS and LHS of Eq. (2.3) do not match up thus violating the second exactness condition for L2-ODE.

For Eq. (2.4), LHS and RHS both come out to be zero, since all the terms are zero as can be seen from Eq. (2.5).

$$ \displaystyle f_{xp} + pf_{yp} + 2 f_y = 0 $$     LHS (2.4)

$$ \displaystyle g_{pp} = 0 $$     RHS (2.4)

Therefore, even though Eq. (2.4) is satisfied, but for we need both (2.3) and (2.4) to be satisfied for second exactness condition to be true for given L2-ODE. Hence we prove that Eq. (2.1) is not exact.

Author
Contributed by Ankush

= Problem R*4.3 Solve L2-ODE-VC via the Power Form = From [[media:Pea1.f11.mtg21.djvu|Mtg 21-3]]

Given
$$\displaystyle (x^my^n)[\sqrt{x}y''+2xy'+3y]=0$$

Find
$$\displaystyle \mathbf a)$$ Find $$\displaystyle m, n $$ to render the above equation, equation (1) p.21-3, exact. $$\displaystyle \mathbf b)$$ Show that the first integral is a L1-ODE-VC as shown below, equation (2) p.21-3. $$\displaystyle \phi(x,y,p)=xp+(2x^\frac{3}{2}-1)y=k $$ with $$\displaystyle p(x):=y'(x) $$. $$\displaystyle \mathbf c)$$ Solve equation (2) p.21-3 for $$\displaystyle y(x) $$.

Solution Part A
Solved independently. $$\displaystyle \mathbf a)$$ In order to determine $$\displaystyle m $$ and $$\displaystyle n $$, we must analyze the two conditions of exactness for N2-ODEs. The first condition requires the N2-ODE to fit the following particular form of equation (2) p.16-4. $$\displaystyle G(x,y,y',y)=g(x,y,p)+f(x,y,p)y=0 $$ The second condition requires the N2-ODE to satisfy the following equations, (1)-(2) p.16-5. $$\displaystyle f_{xx}+2pf_{xy}+p^2f_{yy}=g_{xp}+pg_{yp}-g_y $$ $$\displaystyle f_{xp}+pf_{yp}+2f_y=g_{pp} $$ First, we must determine the new $$\displaystyle f(x,y,p) $$ and $$\displaystyle g(x,y,p) $$ that include the Integrating Factor, $$\displaystyle h(x,y)=x^m y^n $$. Factor through by the integrating factor to yield the following: $$\displaystyle g(x,y,p)=2x^{m+1}y^np+3x^m y^{n+1} $$ $$\displaystyle f(x,y,p)=x^{0.5+m}y^n $$ This satisfies the first exactness condition. Since $$\displaystyle f(x,y,p) $$ is not a function of $$\displaystyle p $$, the second equation of the second exactness condition significantly simplifies to $$\displaystyle \cancel {f_{xp}}+\cancel {pf_{yp}}+2f_y=\cancel {g_{pp}} $$ $$\displaystyle 2f_y=0 $$. Differentiate $$\displaystyle f(x,y,p)=x^{0.5+m}y^n $$ with respect to $$\displaystyle y $$ to obtain $$\displaystyle f_y(x,y,p)=nx^{0.5+m}y^{n-1} $$. Plugging in $$\displaystyle f_y(x,y,p) $$ into $$\displaystyle 2f_y=0 $$ yields $$\displaystyle 2f_y=2nx^{0.5+m}y^{n-1}=0 $$. The only way to solve the problem is to set $$\displaystyle n=0 $$. Since we solved for $$\displaystyle n $$, we can simplify $$\displaystyle f(x,y,p) $$ and $$\displaystyle g(x,y,p) $$ to $$\displaystyle f(x,y,p)=x^{0.5+m} $$. $$\displaystyle g(x,y,p)=2x^{m+1} p+3x^m y $$. Now we turn our attention to the first equation of the second exactness condition after we simply. $$\displaystyle f_{xx}+\cancel {2pf_{xy}}+\cancel {p^2f_{yy}}=g_{xp}+\cancel {pg_{yp}}-g_y $$ Next, solve for the partial differentials of $$\displaystyle f(x,y,p) $$ and $$\displaystyle g(x,y,p) $$ with the results shown below. $$\displaystyle f_{xx}(x,y,p)=(m+0.5)(m-0.5)x^{m-0.5} $$. $$\displaystyle g_{xp}(x,y,p)=2(m+1)x^m $$. $$\displaystyle g_y(x,y,p)=3x^m $$. Plug these three equations back into the first equation of the second exactness condition to obtain $$\displaystyle (m+0.5)(m-0.5)x^{m-0.5}=2(m+1)x^m-3x^m $$. After a visual inspection, one can deduce the LHS to be zero due to no other like terms. $$\displaystyle 2(m+1)\cancel {x^m}-3 \cancel {x^m}=0 $$ The $$\displaystyle x^m $$ terms cancel out leaving a simple algegra problem. $$\displaystyle 2(m+1)-3=0 $$ So $$\displaystyle m=0.5 $$. The integrating factor for this L2-ODE-VC is $$\displaystyle h(x,y)=x^{0.5} $$.

Solution Part B
$$\displaystyle \mathbf b) $$ We can solve for the first integral using equation (3) p.16-5, $$\displaystyle \phi(x,y,p)=h(x,y)+\int f(x,y,p) dp $$. Since $$\displaystyle f(x,y,p)=x $$ we can simply equation (3) p.16-5 to $$\displaystyle \phi(x,y,p)=h(x,y)+xp $$. Solve this for the partials as shown below. $$\displaystyle \phi_x=h_x+p $$. $$\displaystyle \phi_y=h_y $$. Next, we use the first equation of the second exactness condition, $$\displaystyle g(x,y,p):=\phi_x+\phi_yp $$, to obtain the following: $$\displaystyle 3x^{0.5}y+2x^{1.5}p=h_x+p+h_yp $$. One can equate the following: $$\displaystyle h_x=3x^{0.5} $$. $$\displaystyle h_y+1=2x^{1.5} $$. Integrating these yields the following: $$\displaystyle h=2x^{1.5}y+k_1(y) $$. $$\displaystyle h=(2x^{1.5}-1)y+k_2(x) $$. After differentiating the equation you obtain $$\displaystyle h_y=\cancel {2x^{1.5}}+k_1'(y)=\cancel {2x^{1.5}}-1 $$. $$\displaystyle h_x=\cancel {3x^{1.5}y}+k_x'(x)=\cancel {3x^{0.5}y} $$. Integrating yields the next result: $$\displaystyle h_1(y)=-y+k_1 $$. $$\displaystyle h_2(x)=k_2 $$. Plug these back into the integrating function equation to yield $$\displaystyle h=2x^{1.5}y-y+k_1 $$. $$\displaystyle h=(2x^{1.5}-1)y+k_2 $$. These equation are the same. Now that we have solved for the integrating function equation, we can input this into the original equation for the final solution. $$\displaystyle \phi(x,y,p)=xp+(2x^{1.5}-1)y=k $$ with $$\displaystyle k=k_3-k_1 $$.

Solution Part C
$$\displaystyle \mathbf c) $$ Next we need to solve for $$\displaystyle y(x) $$ starting from the first integral equation above. The N1-ODE-VC can be rearranged to the following form: $$\displaystyle P(x)y'+Q(x)y=R(x) $$. $$\displaystyle xy'+(2x^{1.5}-1)y=k $$. Divide through by $$\displaystyle x $$ to obtain $$\displaystyle y'+\frac{Q(x)}{P(x)}y=\frac{R(x)}{P(x)} $$. $$\displaystyle y'+(2x^{0.5}-\frac{1}{x})y=\frac{k}{x} $$. Using equation (3) p.11-4, we can solve for the integrating factor. Knowing $$\displaystyle a_0(x)=\frac{Q(x)}{P(x)} $$ and solving yields $$\displaystyle h(x)=\exp[\int^x a_0(s)ds+k_1] $$ $$\displaystyle h(x)=\exp[\int^x \frac{Q(s)}{P(s)} ds+k_1] $$ $$\displaystyle h(x)=\exp[\int^x (2s^{0.5}-\frac{1}{s})ds+k_1] $$ $$\displaystyle h(x)=\exp[\frac{4}{3}x^{1.5}-\ln{x}+k_1] $$ $$\displaystyle h(x)=\frac{1}{x}\exp[{\frac{4}{3}x^{1.5}}+k_1] $$. Next we use equation (1) p. 11-5 to solve for $$\displaystyle y(x) $$. $$\displaystyle y(x)=\frac{1}{h(x)}\left[\int^x h(s)b(s)ds+k_2\right] $$ $$\displaystyle b(x)=\frac{R(x)}{P(x)}=\frac{k}{x} $$ $$\displaystyle y(x)=\frac{1}{\frac{1}{x}\exp[{\frac{4}{3}x^{1.5}}+k_1]}*\left[\int^x (\frac{1}{s}\exp[{\frac{4}{3}s^{1.5}}+k_1])(\frac{k}{s})ds+k_2 \right]$$

Author
Egm6341.f11.team4.allen 04:50, 19 October 2011 (UTC)

= Problem R*4.4 - Derivation of a form of exact L2-ODE-VC = From [[media:Pea1.f11.mtg13.djvu|Mtg 13-5]]

Given
The equation (2),(3) in the class note [[media:Pea1.f11.mtg13.djvu|21-4]]. The equation (1) in the class note [[media:Pea1.f11.mtg13.djvu|21-5]].

Find
Show that (2),(3) in the class note [[media:Pea1.f11.mtg13.djvu|21-4]] leads to the equation (1) in the class note [[media:Pea1.f11.mtg13.djvu|21-5]].

Solution
- Solved on our own The first exactness condition for L2 ODE VC: where $$\displaystyle p = y^{\prime} $$. By comparing the Eq. (4.4) and the Eq. (4.1), the following is obtained. Integrate the Eq. (4.5), By differentiating the Eq. (4.6) with respect to x and y, the following is obtained. Substituting the Eq. (4.7) into the Eq. (4.4) yields the following. From the last term of the Eq. (4.8), the following is true. Integrate the Eq. (4.9) with respect to $$\displaystyle x $$. Substitute the Eq. (4.10) into the Eq. (4.6). Take the partial derivative of the Eq. (4.11) with respect to $$\displaystyle y $$. Also, the following can be proven to be true using the problem statement. Hence, by comparing the Eq. (4.12) and the Eq. (4.13), the following can be concluded. Therefore, the following is justified.

Author
Contributed by Shin

=Problem 4.5 - Solve a L2-ODE-VC=

Given
A L2-ODE-VC:

Find
1. Show equation (5.1) is exact. 2. Find $$\displaystyle \phi $$ for (5.1). 3. Solve for $$\displaystyle y(x) $$.

Solution
- Solved on our own

1. Show Exactness
First exactness condition requires the equation to be in the form of: $$\displaystyle g(x,y,p)+f(x,y,p)y''=0 $$, where $$\displaystyle p=y' $$ , $$\displaystyle g(x,y,p)= \phi_x+ \phi_y p $$ and $$\displaystyle f(x,y,p) = \phi_p $$. In equation (5.1),

So (5.1) satisfies the first exactness condition. The second exactness condition requires the following two equations to be satisfied:

In (5.1),

(5.4) is satisfied.

(5.5) is satisfied.

So (5.1) satisfies the second exactness condition. Then equation is exact.

2. Find $$\displaystyle \phi $$
Since (5.1) is exact, then:

So:

Compare (5.11) and (5.2), assume $$\displaystyle h_x = 2xy $$.Then:

So $$\displaystyle h_y = x^2 +k_{1}^{'} (y) $$. Then (5.11) becomes:

We get:

So

3. Solve for $$\displaystyle y(x) $$
Equation (5.16) is a L1-ODE-VC. It can be transformed into:

In order to solve (5.17), we need to find a $$\displaystyle \bar{h}(x,y) $$ to make (5.17) exact. Assume $$\displaystyle \bar{h}_y=0 $$.So:

Solve for $$\displaystyle \bar{h} $$:

And:

where $$\displaystyle k_2, k_3 $$ are constants.

Author
Contributed by YuChen

=Problem 4.6 - Show the Equivalence=

Given
When $$\displaystyle n=2$$, we have:

Find
Use (6.1) to show that:

Solution
- Solved on our own Since:

So (6.1) becomes:

We get:

The LHS of (6.9) is:

And the RHS of (6.9) is:

From equations (6.10) (6.11) (6.12) (6.13) we can tell that:

Author
Contributed by YuChen

=Contributing Members=

title: User:Egm6321.f11.team4/HW4


text:
User:Bjburrows1121

History major, poli sci minor. Love to learn!

title: User:Bjburrows1121


text:
User talk:Giftbasketsforall

Gift baskets Include Numerous Sweets But They Should Be Handled Thoroughly
Thanks for your contributions to Wikiversity. Unfortunately, content that is largely promotional or an advertisement is unlikely to become an educational resource and is outside our scope. Accordingly, I have deleted Gift baskets Include Numerous Sweets But They Should Be Handled Thoroughly. Please consider contributing to educational resources. --Draicone (talk) 03:49, 29 September 2011 (UTC)

Blocked
It appears that you are also User talk:Agegiftbaskets. Given the promotional editing and the use of multiple accounts, I'm blocking this account. You may use Agegiftbaskets, it is not blocked, and probably won't be unless you continue to create inappropriate pages. You may also still edit this Talk page, in case we have made some mistake. Thanks. --Abd 17:45, 29 September 2011 (UTC)

title: User talk:Giftbasketsforall


text:
User:Hosborne~enwikiversity

/BPS2011 Review on the effect of pre commitment gambling on support of sport/

/BPS2011 Presentation on Poker Machine Reform/

title: User:Hosborne~enwikiversity


text:
User talk:Hosborne~enwikiversity

 Hello Hosborne, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. All users are expected to abide by our Privacy policy, Civility policy, and the Terms of Use while at Wikiversity.

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Your account will be renamed
Hello,

The developer team at Wikimedia is making some changes to how accounts work, as part of our on-going efforts to provide new and better tools for our users like cross-wiki notifications. These changes will mean you have the same account name everywhere. This will let us give you new features that will help you edit and discuss better, and allow more flexible user permissions for tools. One of the side-effects of this is that user accounts will now have to be unique across all 900 Wikimedia wikis. See the announcement for more information.

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

Your account will still work as before, and you will be credited for all your edits made so far, but you will have to use the new account name when you log in.

Sorry for the inconvenience.

Yours, Keegan Peterzell Community Liaison, Wikimedia Foundation 23:38, 17 March 2015 (UTC)

Renamed
 This account has been renamed as part of single-user login finalisation. If you own this account you can |log in using your previous username and password for more information. If you do not like this account's new name, you can choose your own using this form after logging in: . -- Keegan (WMF) (talk) 06:11, 19 April 2015 (UTC)

title: User talk:Hosborne~enwikiversity


title: File:DigitalDesign.20110929.pdf


text:
User:Stewart.boyd/Business, Politics and Sport – The expansion of the Australian Football League

Australian Football is one of the most skillful and entertaining football codes played anywhere in the world. Yet it does not enjoy the same international exposure or recognition as soccer or even rugby codes. These other codes, began in England and spread internationally with the expansion of the British Empire in the 19th century. Indeed, until just a few decades ago, Australian Football was mostly a game played by expat 'southerners'. From its humble beginnings on the Victorian goldfields in the 1850s this is now changing. Driven by the vision of the Australian Football League (AFL) Commission, created in 1985, the game is expanding rapidly throughout Australia and internationally AFL Commission, 2007. At the elite level, a new AFL club has been established in southern Queensland - the Gold Coast Suns, and in NSW - Greater Western Sydney or GWS Giants. More importantly, the AFL is investing heavily in junior and womens programs throughout Australia. It is also expanding internationally with outreach into China and South Africa and the establishment of the AFL International Cup competition. This growth has been underpinned by two important factors: the political and financial support of the state and federal governments in Australia and the enormous revenue being generated by television broadcast rights. In a tough competitive market against the other football codes, Australian Football's future for further expansion is optimistic.

Background

The origins of Australian Football are a little murky. There are records of the early settlers playing a form of 'foot-ball' in the 1840s. There are also records of Irish, English and Scottish immigrant miners entertaining themselves with a hybrid Rugby and of Gaelic football on the gold fields of Victoria in the 1850s. Over time, rules were created and the first recorded game of Australian Football was played on 7 August 1858 between Scotch College and Melbourne Grammar.

The game blossomed throughout Australia. By the 1860s the game had expanded nationally to Western Australia, South Australia, New South Wales, Queensland, and internationally to New Zealand. Australian Football's heartland remained Victoria. In 1897 the Victorian Football League (VFL) was formed, initially comprising seven teams from Melbourne and one from Geelong. However, despite growing quickly in southern Australia, by the 1890s the game was in decline in NSW, QLD and NZ due to competition from other codes and colonial rivalry. Throughout the 20th century, Australian Football was by far the most popular code in Victoria, South Autralia, Western Australia and Tasmania. Acknowledging the need to become an Australia wide game to survive against the expansion of rugby and soccer, the VFL renamed itself the Australian Football League in 1985 and began its expansion into the other states AFL.com.au Network, 2007

In the 2011 season, the Gold Coast Suns joined the competition and in 2012 the Greater Western Sydney Giants will become the 18th team in the competition (Official website of the AFL: AFL.com.au Network, 2000). The expansion makes the AFL Commission a unique business, operating in all Australian states and territories. The immediate goal of the AFL Commission now is to consolidate this expansion and to expand its presence internationally.

The AFL's Broadcasters and Government Relations Underpin Financial Performance and Growth
Despite only eleven of the sixteen AFL clubs trading profitably in 2010, the league as a whole recorded a revenue of $335.9 million. This record profit was $32.4 million more than in 2009(Fitzpatrick, M., 2010: pg. 15). Furthermore, the operating surplus before grants and distributions was $230.4 million. After payments from reserves of $35.1 million, the net profit stood at $4.8 million (Anderson, I., 2010: pg. 57). The AFL clubs earn most of their own income from membership fees, gate takings and merchandise. This is an important source of revenue but clubs could not survive without the disbursements of cash from the AFL Commission, which the Commission receives for managing the broadcasting rights on behalf of the clubs. In 2011, the AFL Commission sealed a five year $1.235 billion broadcasting rights deal with Channel 7 and pay-TV giant Foxtel [http://www.foxsports.com.au/afl/afl-poised-to-announce-historic-new-tv-deal-with-foxtel-channel-7-worth-an-estimated-1-billion/story-e6frf33l-1226046219071 Fox Sports Staff,. 2011]. The AFL can command such huge broadcasting sums due to the popularity of the game. During the premiership season, for example, broadcasters showcased the game to an average audience of 4.16 million each week. Digital media partner, Telstra BigPond attracted 3.7 million unique browsers per month [http://www.afl.com.au/portals/0/afl_docs/afl_hq/annual_reports/2010/AR2010_04_ExecSummary.pdf Fitzpatrick, M., 2010: pg. 15]. Furthermore, the 2010 Grand Final between Collingwood and St Kilda was the most-watched sporting program of the year, with an average audience of 3.639 million people on the channel Seven Network, while the 2010 Grand Final replay attracted an average audience of 3.522 million and was the second most-watched sporting event in Australia, contributing a further $16 million in revenue (Anderson, I., 2010: pg. 57). The broadcasting rights revenue has allowed the AFL to ensure the financial stability of clubs. The AFL also invests nearly $24 million into the Gold Coast and GWS regions, resulting in additional revenue from these two new priority growth markets [http://www.afl.com.au/portals/0/afl_docs/afl_hq/annual_reports/2010/AR2010_09_FinanceAdminCorporate.pdf Anderson, I., 2010: pg. 57].

Australian governments at all levels have been quick to capitalize on the game’s popularity, often making election promises related to improving major AFL infrastructure. For example, according to the AFL’s 2010 Annual Report, the NSW government has contributed $45 million to redevelop the Royal Agricultural Society’s show grounds at Sydney Olympic Park, so they could become the home of the GWS Giants AFL club. In addition, the South Australian government has committed $535 million to redevelop the Adelaide Oval, and the Victorian government has agreed to contribute $30 million to the redevelopment of the Great Southern Stand at the Melbourne Cricket Ground [http://www.afl.com.au/portals/0/afl_docs/afl_hq/annual_reports/2010/AR2010_04_ExecSummary.pdf Fitzpatrick, M., 2010: pg. 13]. At the community level, local governments have also supported the game through the provision of new and improved suburban grounds and club rooms. In emerging markets, governments have also recognized the economic benefits – particularly to the hospitality sector – by hosting AFL games. The ACT government, for example, made a 10 year agreement for the GWS giants to play 4 games per year at Manuka Oval in Canberra, The financial support provided by the federal, state and local governments was phenomenal and exceeded $100 million. This form of funding ensures AFL’s future financial stability [http://www.afl.com.au/portals/0/afl_docs/afl_hq/annual_reports/2010/AR2010_04_ExecSummary.pdf Fitzpatrick, M., 2010: pg. 13].

Flourishing of AFL Queensland, Gold Coast and the licensing of GWS
In Recognizing the Gold Coast region as a priority area for expansion, in 2000, AFL Queensland implemented a state development plan on behalf of the AFL Commission to increase grass root participation in the state Demetriou, A., 2010: pg.19. Initiatives included expansion of the South-east Queensland competition, support for regional bodies especially in the Cape York region of far north Queensland, refinement of talent pathway programs, new administration and talent development headquarters in Brisbane, the development of community facilities in Townsville and a multi-sports complex on the sunshine coast Demetriou, A., 2010: pg.19. The development plan has been highly successful. In 1999, 23,000 people were actively playing AFL in Queensland. By 2010, participation had increased to 112,447 people[http://www.afl.com.au/portals/0/afl_docs/afl_hq/annual_reports/2010/AR2010_04_ExecSummary.pdf Demetriou, A., 2010: pg. 19]. On 31 March 2009, Queensland welcomed its second team to the AFL competition when the Gold Coast Suns was granted a provisional license[http://www.afl.com.au/portals/0/afl_docs/afl_hq/annual_reports/2010/AR2010_04_ExecSummary.pdf Demetriou, A., 2010: pg. 20]. In 2011, the Gold Coast Suns became the 17th club in the AFL competitionOfficial Website of Gold Coast Football Club., 2011.

Apart from south Queensland, the AFL Commission has also identified the suburbs of Western Sydney as an area it must penetrate in order to consolidate the game in NSW. This will be a huge challenge given the popularity of rugby league in this area. On 29 July 2010, the GWS Giants were licensed and became the AFL’s 18th club. They will enter the AFL competition in 2012. The license to GWS, is the seventh issued by the AFL, since it was first established in 1985 [http://www.afl.com.au/portals/0/afl_docs/afl_hq/annual_reports/2010/AR2010_04_ExecSummary.pdf Demetriou, A., 2010: pg. 21]. In credit to its broadcasting revenue, the AFL Commission is cashed up and prepared to make the long term investment required until the Gold Coast and GWS are financially viable.

AFL: A global game
Kevin Sheedy: “The smaller the world gets the more I see other games such as soccer, rugby and American Football, the more I am convinced that we have the greatest game on the planet and we should be sharing it with the rest of the world” 2008: AFL.com.au Network.

For more than a decade Australian Football has been played by teams across the world. The first international competition was in 1995 at the Arafura Games in the Northern Territory capital of Darwin. Played annually, these Games bring together Asia Pacific nations and were instrumental in the formation of the International Australian Football Council. The success of the Arafura Games ignited the first International Cup in 2002, making the AFL Commission the governing body of Australian Football internationally. In 2002, the first official International Cup was held in Melbourne, with eleven nations competing. Ireland took out the top spot. By 2011, the International Cup competition expanded to include 18 men’s teams and 5 women’s teams 2011: AFL.com.au Network. The International Cup competition has helped expose AFL internationally. There are currently 14 recognized international bodies across four regional delegates – America, Africa, Europe and Asia Pacific 2011: International Football. Global participation has rapidly increased and now exceeds 60,000 registered players outside Australia [http://www.afl.com.au/portals/0/afl_docs/afl_hq/annual_reports/2010/AR2010_10_CommunityDevelopment.pdf Mathews, D., 2010: pg. 89].

The AFL Commission has deliberately focused its expansion efforts on two countries – South Africa and China. In South Africa, the AFL Commission no doubt hopes that the game will prove as popular with the black population as it is with indigenous Australians. In April 2007, AFL South Africa launched Footy Wild, a development program for youth players. In one year, the program has grown to include 8,000 Footy Wild players and over 500 volunteer coaches, umpires and administrators http://www.afl.com.au/development/international/internationalleagues/southafrica/tabid/10346/default.aspx, 2011: AFL.com.au]. In April 2010, The AIS-AFL Academy toured South Africa competing against the South African Lions at Sahara Park Newlands in Cape Town. Furthermore, the Australian Under-23 amateur side also successfully toured South Africa during the 2010 South Africa National Championships in Cape town (Mathews, D. 2010: pg. 74).

The second key international growth market, identified by the AFL, is China. In part, efforts to expand the game there resulted from the 'China Strategy’ implemented by the Melbourne Demons AFL club. The Club’s objective was to exploit links between Melbourne's large Chinese community and promote interest in the game abroad. The strategy gave two Chinese players, Zhao Wei and Zhao Yong Gen, the opportunity to train in Australia with the Melbourne Football Club on a two-week pre-season development scholarship 2011 AFL China – AFL.com.au. A major highlight was the exhibition match between two AFL Clubs, the Melbourne Demons and the Brisbane Lions at Jiangwan Stadium on October 17, 2010 (AFL China – AFL.com.au, 2011). More than 7000 people attended this match as part of the Kaspersky Cup, AFL Shanghai Showdown. The match was televised live in English and Mandarin on International Channel Shanghai and G-Sports Channel with a total average audience of 300,000 viewers. Furthermore, in 2010 a full-time development officer has been appointed in China, introducing grass root level development program to more than 300 children in Shanghai schools (Mathews, D. 2010: pg. 78).

Conclusion
Since the first game in 1858, Australian Football has expanded throughout Australia and internationally. Today, it is played by over 751,015 people across Australia. It has also become a highly successful business enterprise run by the Australian Football League Commission. Its success is underpinned by two key factors. The first factor is the Commission’s adept handling of broadcast rights agreements which have reaped a financial windfall. These funds are used to keep several of the clubs financially viable and to finance its development and expansion program at both the elite, state and community levels across Australia. The second factor has been the favourable political treatment AFL has received from governments. Once again, the AFL Commission can take some credit for astutely managing these relationships. They have resulted in significant government investment in AFL facilities to support the game, both at the elite level and in the community. Within Australia, the AFL has eyed south east Queensland and western Sydney, as high population areas in which it must penetrate and capture market share, to ensure its status as the pre eminent national football code. As a result of its expansion and development initiatives in both Queensland and New South Wales, membership numbers for newly developed AFL clubs, Gold Coast Suns and GWS Giants has generated further revenue for the AFL Commission. This has fostered the financial stability of the sport in the heartland of rugby league. Beyond these two focal areas it allows the AFL Commission to continue to its wider national expansion objectives. Outside Australia, the dominance of soccer as the ‘World Game’ means the AFL Commission faces a tough environment to expand the game. Its strategy of focusing on South Africa and China, as well as support for the International Cup competition is bearing fruit. The game appears to have a bright future.


text:
Talk:Virtues/Temperance

virtues as moderation
Plato or prehaps Socraties conceived of each virtue as a balance between vices at the extremes. For example courage is a balance between cowardice and recklessness. Try to find a good reference for this and include the list here. --Lbeaumont 13:41, 29 September 2011 (UTC)

title: Talk:Virtues/Temperance


text:
Literature/Tables/Blackburn-1984-trigon

Comments
Blackburn's trigon may be an interpretation of the triangle of reference put forward by Ogden & Richards (1923), who essentially assumes that words mean nothing by themselves but by way of the user. They pay far more attention to the hearer's than speaker's thought as being influenced by language, hence their subtitle. Thus Blackburn's "speakers (psychology)" is a surprise. Such is his "language (meaning)" as they do suspect language of meaning by itself. Problematic also is his "world (metaphysics)" (ontology?) rather than "world (physics)" since "water" should mean H2O, as per Putnam (1975).

The breakdown of the holistic theory or philosophy as shown on the left appears untenable, especially since the (theory of) meaning matters above all and after all, and along the base line between the word and the world. Either truth or science may matter in addition to that. Either may be distilled through the meaning rather than vice versa. No science without meaning. Logicism such as logic first of all may be mistaken. The base line should be the orthogonal projection of both sides, the verbal or virtual life (language & user) and the real life (user & world - being in the world), as suggested by:
 * Hayakawa (1949) Language in Thought and Action,
 * Austin (1955) How to Do Things with Words, and so on, not to mention Ogden & Richards (1923).

title: Literature/Tables/Blackburn-1984-trigon