text:
User:Egm6321.f11.team2/HW3

= R3.1 - A More General N1-ODE That Can Be Rendered Exact by the IFM=

Given
where

Find
Show that $$-\frac{1}{N} (N_x - M_y)$$ is a function of x only, only if $$\displaystyle k_1 (y) = d_1$$ (a constant).

Show that $$\displaystyle [ a(x) y + k_2(x) ] + \bar{b} (x) y' = 0$$ is a special case of 3.1.1.

Solution
Using 3.1.2 and 3.1.3 to expand the expressions for N and M, then taking their derivatives yields

$$\displaystyle \begin{align} N (x, y) &= \bar{b} (x, y) c(y) \\&= c(y) \int^x b(s) ds + c(y) k_1(y) \\N_x &= c(y) b(x) \end{align} $$

$$ \begin{align} M (x, y) &= a(x) \bar{c} (x, y) \\&= a(x) \int^y c(s) ds + a(x) k_2(x) \\M_y &= a(x) c(y) \end{align} $$

Substituting these expressions into the 2nd Exactness Condition yields

$$ \begin{align} -\frac{1}{N} ( N_x - M_y ) &= - \frac{1}{ c(y) \int^x b(s) ds + c(y) \, k_1(y) } \cdot [ c(y) b(x) - a(x) c(y) ] \\ &= - \frac{1}{c(y) [ \int^x b(s) ds + k_1(y) ] } \cdot c(y) [ \left( b(x) - a(x) \right) ] \\ &= - \frac{1}{ \int^x b(s) ds + k_1(y) } \cdot [ b(x) - a(x) ] \end{align} $$

which is a function of both x and y unless k1 is a constant.

Consider the case $$\displaystyle c(y) = 1$$. Then

$$\bar{c} (x, y) = \int^y ds + k(x) = y + k(x)$$

so 3.1.1 reduces to

$$[ a(x) y + k_2(x) ] + \bar{b} (x, y) y' = 0$$

as required.

In fact, this is still the case if $$\displaystyle c(y)$$ is equal to any constant.

=R*3.2 - Integrating factor method =

Given
$$\displaystyle \left( \frac {1}{3}x^3+d_1\right)\left(y^4\right)y' + (5x^3+2)\left(\frac {1}{5}y^5+ \sin x + d_2\right) = 0$$

Find
Show that the above equation is exact or can be made exact by IFM.

Find the integrating factor h.

Solution
In order to exact condition

First condition of exactness is that

$$ \displaystyle M(x,y) + N(x,y)y' = 0$$3.2.1

$$ \displaystyle (5x^3 +2)\left( \frac {1}{5}y^5 + \sin x + d_2\right) = M(x,y) $$

$$ \displaystyle \left( \frac {1}{3}x^3+d_1\right)(y^4)y' = N(x,y)y' $$

$$\underbrace{ \left( \frac {1}{3}x^3+d_1\right)\left(y^4\right)}_{\displaystyle\color{blue}{N(x,y)}}y' + \underbrace {(5x^3+2)\left( \frac{1}{5}y^5+ \sin x + d_2\right)}_{\displaystyle\color{blue}{M(x,y)}} = 0$$

Thus, this equation satisfies the first condition.

Second condition of exactness is that

$$\displaystyle M_y(x,y) = N_x(x,y)$$3.2.2

$$\displaystyle \frac {\partial M}{\partial y} = \frac {\partial N}{\partial x}$$

$$\displaystyle \frac {\partial M}{\partial y} = 5x^3y^4 + 2y^4 $$

$$\frac {\partial N}{\partial x} = x^2y^4 $$

$$ M_y \neq N_x $$

It shows that this equation does not satisfy the second condition.

Therefore, in order to make this equation to exact, integrating factor must be used.

$$ \displaystyle n(x) = - \frac {1}{N}(N_x-M_y) = - \frac {1}{\bar b (x) \cancel{c(y)}}\left[\cancel{c(y)} b(x)- a(x)\cancel{c(y)}\right] = -\frac {1}{\bar b (x)}[b(x)-a(x)]$$3.2.3

where

$$\displaystyle \bar b(x) = \frac {1}{3} x^3 + d_1$$

$$\displaystyle b(x) = x^2 $$

$$\displaystyle a(x) = 5x^3 + 2 $$

Let d1 = 0, $$\displaystyle n(x) = - \frac {1}{\frac {1}{3}x^3}[x^2 - 5x^3 - 2] $$

$$ \displaystyle h(x) = \exp [\int^x n(s)ds+k]$$3.2.4

$$h(x) = \exp [ \int^x - \frac {1}{\frac {1}{3}s^3}(s^2 - 5s^3 - 2)ds+k] $$

$$ h(x) = \exp [ \int^x \frac{3}{s^3}(5s^3 -s^2 + 2)ds + k ] = \exp [ \int^x 3(5 -\frac{1}{s} + \frac{2}{s^3})ds + k ] $$

$$ h(x) = \exp [ 15x - \frac{3}{x^2} - 3\ln (x) + k] = \frac {1}{x^3} \exp [15x - \frac{3}{x^2} + k] $$

=R*3.3 -First Integral =

Given
Consider a class of N1-ODEs of the form in (1)p.13-2:

$$\displaystyle \overline{b}(x,y)c(y)y'+a(x)\overline{c}(x,y)=0$$ (3.3.1)

and

$$\displaystyle a(x)=\sin x^{3}$$

$$\displaystyle b(x)=\cos x$$

$$\displaystyle c(y)=\exp(2y)$$

---

Find
1. Find an N1-ODE of the form (1) that is either exact of can be made exact by IFM.

2. Find the first integral $$ \phi(x,y)=k$$

Solution
I wonder if (2)p.13-4 is right? if $$\displaystyle a(x)=\sin x^{3}$$ is right, I cannot integrate it. But Professor's latex code reads: a(x)= 5x^3+2.I hope Professor could find this problems. In this problems, I treat $$\displaystyle \sin x^{3}= (\sin x)^{3}$$ instead.

Since

$$\displaystyle \overline{b}(x,y)=\int^{x}b(s)ds+k_{1}(y)$$

$$\displaystyle \overline{c}(x,y)=\int^{y}c(s)ds+k_{2}(x)$$

Substitute the given equations in the above two equations, we have

$$\displaystyle \overline{b}(x,y)=\int^{x}\cos sds+k_{1}(y)=\sin x+k_{1}(y)$$

$$\displaystyle \overline{c}(x,y)=\int^{y}\exp(2s)ds+k_{2}(x)=\frac{1}{2}\exp(2y)+k_{2}(x)$$ (3.3.2)

Let $$\displaystyle k_1=k_2=0$$ and substitute the above equations to (3.3.1)

$$\displaystyle \sin x\exp(2y)y'+\frac{1}{2}\exp(2y)\sin x^{3}=0$$

hence

$$\displaystyle M(x,y)= \frac{1}{2}\exp(2y)\sin x^{3}$$ (3.3.3)

$$\displaystyle N(x,y)= \sin x\exp(2y)$$ (3.3.4)

It is easy to know that

$$\displaystyle M_y (x,y) \neq N_x (x,y)$$

Recall IFM

$$\displaystyle \frac{dh}{h}=\frac{M_{y}(x,y)-N_{x}(x,y)}{N(x,y)}=\frac{\exp(2y)(\sin x^{3}-\cos x)}{\sin x\exp(2y)}=\frac{\sin x^{3}-\cos x}{\sin x}$$ (3.3.5)

So, it can be made exact by IFM since the right hand side is a function of x only.

Since $$\displaystyle \exp(\int h(x)dx)=\exp(\frac{x}{2}-\frac{\sin2x}{4}-\ln(\sin x))$$

we have

$$\displaystyle \phi_{x}=M(x,y)\exp(\int h(x)dx)=\frac{1}{2}\exp(\frac{x}{2}-\frac{\sin2x}{4}+2y)\sin x$$

$$\displaystyle \phi_{y}=N(x,y)\exp(\int h(x)dx)=\exp(\frac{x}{2}-\frac{\sin2x}{4}+2y)$$

then

$$\displaystyle \phi=\frac{1}{2}\exp(\frac{x}{2}-\frac{\sin2x}{4}+2y)=k$$ (3.3.6)

=R*3.4- Construct a Class of N1-ODEs=

Given
A class of N1-ODES of the form is as follows:

Euler Integrating factor method (IFM) is written as

For (3.4.2) to be exact, 2nd exactness condition is applied to find h and (3.4.2) becomes

To solve (3.4.3), 2 cases is considered.

In the cases, the second case is defined as follows:

Case 2:suppose h(x,y), thus h is a function of y only, and then (3.4.3) becomes

Find
Construct a class of N1-ODEs,which is the counterpart of (3.4.1),and satisfies the condition (3.4.4) that an integrating factor h(y) can be found to render it exact.

Solution
This problem was solved without referring to past solutions.

Based on (3.4.4), consider:

Thus:

If  $$\displaystyle k_1(y)=d_1 $$ (constant), then (3.4.4) is satisfied as follows:

=R*3.5 Motion of a particle in the air. =

Given

We know that:

Find
Part 1 : Derive the [[media:pea1.f11.mtg14.djvu|equation of motion]]

$$\displaystyle m\frac{dv_x}{dt}=-kv^n\cos\alpha$$

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

Part 2 : Particular case when $$\begin{align}k=0\end{align}$$, verify $$\begin{align}y(x)\end{align}$$ is a parabola

Part 3 : Consider $$k\ne0$$, $$\begin{align}v_{xo}=0\end{align}$$ for constant mass.

Part 3.1 : Consider $$k\ne0$$, $$\begin{align}v_{xo}=0\end{align}$$ for mass as function of time.

Solution
Part 1:

Newton's Second Law of Motion: Rate of change of momentum is directly proportion to force applied and this change takes place in the direction of force. Change in momentum in the X-direction: Using 3.5.4 we can write Putting the value of $$\displaystyle v $$ and $$\displaystyle \alpha $$ from Equation 3.5.1 and 3.5.2 into Equation 3.5.5, we get: Change in momentum in Y-direction: Using 3.5.7 we can write Putting the value of $$\displaystyle v $$ and $$\displaystyle \alpha $$ from Equation 3.5.1 and 3.5.2 into Equation 3.5.8, we get: Equation 3.5.6 and 3.5.9 forms a System of coupled N1-ODE.

Part 2:

When $$\displaystyle k=0 $$ Equation 3.5.6 reduces to: At $$\displaystyle t=0\,\,,v_x=v_{x0} $$

Therefore $$\displaystyle v_x=v_{x0} $$ Or Where x_0 is the initial X-coordinate of projectile.

When $$\displaystyle k=0 $$ Equation 3.5.9 reduces to: At $$\displaystyle t=0\,,v_y=v_{y0} \,; t=t\,,v_y=v_y $$

Integrating Equation 3.5.13 using the above boundary conditions we get: At $$\displaystyle t=0\,,y=y_0 \,; t=t\,,y=y $$

Integrating Equation 3.5.14 using the above boundary conditions we get: From Equation 3.5.12 Substituting the value of time from Equation 3.5.17 into equation 3.5.16 we get: Equation 3.5.18 is clearly an equation of parabola.

Part 3:

When $$\displaystyle k \ne 0 $$ and $$\displaystyle v_x=0 $$

The SC-N1-ODE reduces to: Equation 3.5.20 satisfies the first exactness condition but doesn't satisfy the second exactness condition. So we multiply it by the Integrating factor $$\displaystyle h(t,v_y) $$ to make it exact. It should satisfy the second exactness condition i.e Equation 3.5.21 becomes Mass is constant so $$\displaystyle\frac{\partial m}{\partial t}=0$$. Let's also consider that $$\displaystyle h_t(t,v_y)=0 $$. So equation 3.5.23 transforms to: Solving for Integrating factor from Equation 3.5.24 we get: Now Equation 3.5.21 becomes: Integrating both sides and solving for $$\displaystyle v_y $$ For n=1 Integrating both sides we get: At $$\displaystyle t=0, \, v_y=v_{y0} $$

Solving for $$\displaystyle v_y $$ using above condition we get: Solution for $$\displaystyle v_y $$ is possible when $$\displaystyle n=1 $$ but for higher value of $$\displaystyle n=2,3,4... $$ integration is not possible and solution cannot be found out.

Part 3.1

When mass is variable with respect to time:

Equation 3.5.23 is difficult to solve for Integrating Factor because it is a Partial Differential Equation containing terms $$\displaystyle m_t, h_v , h_t $$ Hence no definite solution is possible to find using Euler Integration factor Method.

=R*3.6 - Deriving the Equations of Motion for Coupled Pendulums=

Given

The equations of motion for the coupled pendulums shown above are

$$m_1 l^2 \ddot \theta_1 = -k a^2 (\theta_1 - \theta_2) - m_{1} g l \theta_1 + u_1 l $$ (1)

$$m_2 l^2 \ddot \theta_2 = -k a^2 (\theta_2 - \theta_1) - m_{2} g l \theta_2 + u_2 l $$ (2)

Find
1. Derive the equation of motion (1)-(2). 2. Write equations (1)-(2) in matrix form

$$\mathbf{\dot{x}}(t) = \mathbf{A}(t)\,\mathbf{x}(t) + \mathbf{B}(t) \,\mathbf{u}(t)$$ (3)

with

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

$$\mathbf{u} := \begin{Bmatrix} u_1l \\ u_2l \end{Bmatrix} \in \mathbb R^{2 \times 1}$$

Solution
This problem was solved without referring to past solutions.

Deriving the Equations of Motion
Using Newton's Second Law as it relates to rotation

 * $$\displaystyle \sum \tau := I \alpha$$ (4)

Where,

 * $$\displaystyle \tau $$ = torque about a singular principal axis.
 * $$\displaystyle I = r l^2$$ = Moment of Inertia.
 * $$\displaystyle \alpha = \ddot \theta$$ = angular acceleration

Using the definition of torque.

 * $$\displaystyle \tau := \mathbf r \ x \ \mathbf F = r F sin(\theta)$$(4)

Where r and F are the magnitudes of $$ \mathbf r$$ and $$ \mathbf F$$ respectively. We can use equation (4) to obtain the effect of each force, and then sum them up to equal the total torque.

Hooke's Law,

 * $$\displaystyle \mathbf F = -k x$$

where x is the displacement of the spring's end from its equilibrium position and k is the spring constant.

 * $$\displaystyle \tau_{s1} = -k a^2 (sin(\theta_1) - sin(\theta_2)) = -k a^2 (\theta_1 - \theta_2)$$

Since the spring force acting on the second pendulum is of equal magnitude and opposite direction of the spring force acting on the first pendulum the torque created by the spring for the second pendulum is,

 * $$\displaystyle \tau_{s2} = -k a^2 (\theta_2 - \theta_1)$$

The torque due to gravity,

 * $$\displaystyle \tau_{g1} = - m_1 g l sin(\theta_1) = - m_1 g l \theta_1$$
 * $$\displaystyle \tau_{g2} = - m_2 g l sin(\theta_2) = - m_2 g l \theta_2$$

The torque due to the applied forces,

 * $$\displaystyle \tau_{u1} = u_1 l cos(\theta_1) = u_1 l$$
 * $$\displaystyle \tau_{u2} = u_2 l cos(\theta_2) = u_2 l$$

Summing all the forces,

 * $$\displaystyle \sum \tau_1 = -k a^2 (\theta_1 - \theta_2) - m_1 g l \theta_1 + u_1 l$$
 * $$\displaystyle \sum \tau_2 = -k a^2 (\theta_2 - \theta_1) - m_2 g l \theta_2 + u_2 l$$

Using equation (1),

 * $$\displaystyle m_1 l^2 \ddot \theta_1 = -k a^2 (\theta_1 - \theta_2) - m_{1} g l \theta_1 + u_1 l $$ (5)
 * $$\displaystyle m_2 l^2 \ddot \theta_2 = -k a^2 (\theta_2 - \theta_1) - m_{2} g l \theta_2 + u_2 l $$ (6)

Which are equivalent to equations (1) and (2).

Write the Equations of Motion in Matrix Form
We need to find an equation that takes the form:

$$\begin{Bmatrix} \dot \theta_1 \\ \ddot \theta_1 \\ \dot \theta_2 \\ \ddot \theta_2 \end{Bmatrix} = \begin{Bmatrix} ... \ \ \ \\ \ ... \ \ \\ \ \ ... \ \\ \ \ \ ... \end{Bmatrix} \ \begin{Bmatrix} \theta_1 \\ \dot \theta_1 \\ \theta_2 \\ \dot \theta_2 \end{Bmatrix} + \begin{Bmatrix} ... \ \\ \ \ \\ \ ... \\ \ \ \end{Bmatrix} \ \begin{Bmatrix} u_1 l \\ u_2 l \end{Bmatrix}$$ (7)

Reordering equations (5) and (6) we obtain,

 * $$\displaystyle \ddot \theta_1 = \frac{-(k a^2 + m_{1} g l)}{m_1 l^2} \theta_1 + \frac{k a^2}{m_1 l^2} \theta_2 + \frac{1}{m_1 l^2} u_1 l $$ (8)

 * $$\displaystyle \ddot \theta_2 = \frac{-(k a^2 + m_{2} g l)}{m_2 l^2} \theta_2 + \frac{k a^2}{m_2 l^2} \theta_1 + \frac{1}{m_2 l^2}u_2 l $$ (9)

Entering equations (8) and (9) into equation (7) we obtain:

$$\displaystyle \begin{Bmatrix} \dot \theta_1 \\ \ddot \theta_1 \\ \dot \theta_2 \\ \ddot \theta_2 \end{Bmatrix} = \begin{Bmatrix} 0 \ \ 1 \ \ 0 \ \ 0 \\ \frac{-(k a^2 + m_{1} g l)}{m_1 l^2} \ 0 \ \frac{k a^2}{m_1 l^2} \ 0 \\ \ 0 \ \ 0 \ \ 0 \ \ 1 \\ \frac{k a^2}{m_2 l^2} \ 0 \ \frac{-(k a^2 + m_{2} g l)}{m_2 l^2} \ 0 \end{Bmatrix} \ \begin{Bmatrix} \theta_1 \\ \dot \theta_1 \\ \theta_2 \\ \dot \theta_2 \end{Bmatrix} + \begin{Bmatrix} 0 \ \ 0 \\ \frac{1}{m_1 l^2} \ 0 \\ 0 \ 0 \\ 0 \ \frac{1}{m_2 l^2} \end{Bmatrix} \ \begin{Bmatrix} u_1 l \\ u_2 l \end{Bmatrix}$$ (7)

= R*3.7 - Solving Linear Systems with the Integrating Factor Method =

Find
Show that the solution to 3.7.1 is

$$x(t) = \exp\{ a(t-t_0)\} \, x(t_0) + \int^t_{t_0} \exp\{a(t - \tau)\} \, b \, u(\tau) d\tau$$

and identify the integrating factor, the homogenous solution, and the particular solution.

Next, show that the solution to 3.7.2 is

$$x(t) = \left[\exp\int^t_{t_0} a(\tau) d\tau\right] x(t_0) + \int^t_{t_0}\left[\exp\int^t_\tau a(s) ds \right] b(\tau) u(\tau) d\tau$$

Solution
This problem was solved without referring to past solutions.

We first put 3.7.1 into the particular form, suppressing the functional dependence notation for now

$$0 = \underbrace{ a x + b u }_M + \underbrace{(- 1) }_N \dot x$$

In contrast to the normal notation for this course, in this problem t is the independent variable and x is the dependent variable. Thus, following the normal logic of multiplying the above equation through by h(t, x), requiring that $$\displaystyle (hM)_x = (hN)_t$$ and then assuming that h is a function of t only yields

Now,

$$ \begin{align} N_t &= 0 \\M_x &= a \end{align} $$

Plugging these values into 3.7.3 and then solving for h

$$ \begin{align} \frac{h_t}{h} &= -a \\ \int\frac{dh}{h} &= - \int^t_{t_0} a\, d\tau \\ \log h &= - a (t - t_0) \\ h &= \exp\{- a (t - t_0) \} \end{align} $$

This is the integrating factor.

We procede to solve for x(t) by multiplying 3.7.1 by the integrating factor and integrating

$$ \begin{align} \exp\{ -a (t-t_0) \} \, \dot{x} (t) &= \exp \{ -a (t-t_0) \} \, a\, x(t) + \exp \{ -a (t-t_0) \} \, b\, u(t) \\ \exp\{ -a (t-t_0) \} \, \dot{x} (t) - a\, \exp \{ -a (t-t_0) \} \, x(t) &= \exp \{ -a (t-t_0) \} \, b\, u(t) \\ \frac{d}{dt} \left( \exp\{ -a (t-t_0) \}\, x(t) \right) &= \exp \{ -a (t-t_0) \} \, b\, u(t) \\ \int^t_{t_0} d \left( \exp \{ -a (t-t_0) \}\, x(t) \right) &= \int^t_{t_0} \exp \{ -a (\tau-t_0) \}\, b \, u(\tau) \, d\tau \\ \exp \{ -a(t-t_0) \}\, x(t) - x(t_0) &= \int^t_{t_0} \exp \{ -a (\tau-t_0) \}\, b \, u(\tau) \, d\tau \\ x(t) &= \exp \{ a(t-t_0) \}\, x(t_0) + \int^t_{t_0} \exp \{ a(t-\tau) \} \, b \, u(\tau) \, d\tau \end{align} $$

The first term on the right-hand side is the homogenous solution (known in linear systems theory as the zero-input response) while the second term is the particular solution (know in linear systems theory as the zero-state response).

To solve 3.7.2, we simply note that, since we don't know the functional form of a(t), we cannot arrive at a closed-form representation of its integral. Thus, the integrating factor when a is a function of t is

$$h = \exp \{ - \int^t_{t_0} a(\tau) d\tau \}$$

The rest of the solution proceeds exactly the same as above, where we need to change the dummy integration variable in the particular solution since $$\tau$$ is being used in the outer integral, and of course we must add in the functional dependence of b on t.

=R*3.8 - Linear Time Variant System =

Given
$$\displaystyle \mathbf {x}(t) = exp [{\mathbf {A}(t-t_0)}]\mathbf{x}(t_0) + \int^t_{t0} exp[{ \mathbf{A}(t-\tau)}]\mathbf{B}\mathbf{u}(\tau)d\tau $$3.8.1

Find
Generalize $$\displaystyle \mathbf {x}(t) = exp [{\mathbf {A}(t-t_0)}]\mathbf{x}(t_0) + \int^t_{t0} exp[{ \mathbf{A}(t-\tau)}]\mathbf{B}\mathbf{u}(\tau)d\tau $$ to the case of linear time varient system.

Verify that your expression is indded the solution for

$$\displaystyle \dot\mathbf{X}(t)= \mathbf{A}(t)\mathbf{x}(t)+\mathbf{B}(t)\mathbf{u}(t) $$

Solution
$$\displaystyle \mathbf{X}(t)= \mathbf{A}(t)\mathbf{x}(t)+\mathbf{B}(t)\mathbf{u}(t) $$

Let A(t)= A and B(t) = B Where $$\mathbf{x} $$ is a n x 1 vector, $$\mathbf{u}$$ is 'm x 1' vector, $$\mathbf{A}$$ is a 'n x n' matrix and $$\mathbf{B}$$ is a 'n x m' matrix.

The equation is following that

$$\displaystyle \dot x(t) = Ax(t) + Bu(t)$$

Rearrange the above equation. Thus,

$$\displaystyle \underbrace{1}_{N}\cdot \dot x(t) - \underbrace{Ax(t)}_{M} = Bt(u) $$

$$\displaystyle h(t) = e^{-\int \frac{1}{N}(N_t - M_x)dt}$$3.8.2

Where :

$$\displaystyle N =1, N_t = 0, M_x = - A$$ From the above equation, intergrating factor is that $$\displaystyle h(t) = e^{\int^t-Ads}=e^{-At}$$3.8.3

$$\displaystyle e^{-At}(\dot x(t)-Ax(t)) = \mathbf{B}\mathbf{u}(t)e^{-At}$$

[LHS] : $$\displaystyle e^{-At}(\dot x(t)-Ax(t)) = [e^{-At}x(t)]'$$

Rearraing the eq

$$\displaystyle [e^{-At}x(t)]' = \mathbf{B}\mathbf{u}(t)e^{-At}$$3.8.4

Interval is $$\displaystyle t_0< \tau < t$$

The equation is following that

$$\displaystyle \int^t_{t_0} [e^{-A\tau}x(\tau)]'d\tau = \int^t_{t_0}Bu(\tau)e^{-A\tau}d\tau $$

$$\displaystyle e^{-At}x(t)-e^{-At_0}x(t_0) = \int^t_{t_0}e^{-A\tau}Bu(\tau)d\tau$$

Dividing by $$\displaystyle e^{-At}$$

And rearranging for x(t)

$$\begin{align} \displaystyle x(t)&=\frac {\int^t_{t_0}e^{-A\tau}\mathbf{B}\mathbf{u}(\tau)d\tau}{e^{-At}} + e^{A(t-t_0)}x(t_0) \\&={\int^t_{t_0}e^{A(t-\tau)}\mathbf{B}\mathbf{u}(\tau)d\tau} + e^{A(t-t_0)}x(t_0)\end{align} $$ SC-L1-ODE-VC can be based on the SC-L1-ODE-CC.

=R*3.9 - System of Coupled L1-ODE-VC=

Given
The following equation (1),(2),(3)p.15-4

$$ \displaystyle \mathbf{\Phi}(t,t_{0})_{n\times n}=\exp(\mathbf{A}(t-t_{0}))$$ (3.9.1)

The fundamental or state transition matrix $$ \displaystyle \mathbf{\Phi}$$ is related to the integrating factor.

Properties of the state transition matrix $$ \displaystyle \mathbf{\Phi}$$

$$ \displaystyle \frac{d}{dt}\mathbf{\Phi}(t,t_{0})_{n\times n}=\mathbf{A}\mathbf{\Phi}(t,t_{0})_{n\times n}$$ (3.9.2)

$$ \displaystyle \mathbf{\Phi}(t_0,t_0)=\mathbf{I}$$
(3.9.1)

Find
Verify that (3.9.1) satisfies (3.9.2)-(3.9.3)

Solution
$$ \displaystyle \frac{d}{dt}\mathbf{\Phi}(t,t_{0})_{n\times n}=\frac{d}{dt}\exp(\mathbf{A}(t-t_{0}))=\mathbf{A}\exp(\mathbf{A}(t-t_{0}))=\mathbf{A}\mathbf{\Phi}(t,t_{0})_{n\times n} $$

(3.9.4) According to the formula of exponentiation of a matrix

$$ \displaystyle \exp(\mathbf{A})=\sum_{k=0}^{\infty}\mathbf{A}^{k}$$

(3.9.5) and

$$ \displaystyle \exp(\mathbf{A}(t-t_{0}))=\sum_{k=0}^{\infty}\mathbf{A}^{k}(t-t_{0})^{k}$$
(3.9.6)

Rebember $$ \displaystyle (t-t_0)^0=1$$, and expand 3.9.6

$$ \displaystyle \mathbf{\Phi}(t_0,t_0)=\mathbf{I}$$

-- Remark Sophus Lie is the first one who found that the solutions of (3.9.2) form a sort of group, we then call it Lie Group. Lie group equips both geometric characteristic (differentiable manifold) and algebraic characteristic (group). =R3.10 Free Vibration of Coupled Pendulums =

Given
The matrix foam of linear time-variant systems is

The solution of generalized SC-L1-ODE-CC is

Coupled pendulums system is shown as follows:

Pendulums:

{| style="width:100%" border="0" $$\displaystyle a=0.3,l=1,k=0.2 $$
 * style="width:95%" |
 * style="width:95%" |

$$\displaystyle m_{1}g=3,m_{2}g=6,g=10$$

No applied forces:

{| style="width:100%" border="0" $$\displaystyle u_{1}=u_{2}=0$$
 * style="width:95%" |
 * style="width:95%" |

Initial conditions:

{| style="width:100%" border="0" $$\displaystyle \theta_{1}(0)=0,\dot\theta_{1}(0)=-2 $$
 * style="width:95%" |
 * style="width:95%" |

$$\displaystyle \theta_{2}(0)=0,\dot\theta_{2}(0)=1$$

Find
1. Use matlab's ode45 command to integrate the system, (3.10.3) and (3.10.4), in matrix form of (3.10.1) for $$\displaystyle t\in[0,7] $$

2. Use (3.10.2) to find the solution at the same time stations as in Q1.

3. Plot $$\displaystyle \theta_{1}(t) $$ and $$\displaystyle \theta_{2}(t) $$ from Q1 and Q2.

Solution
This problem was solved without referring to past solutions.

(3.10.3) and (3.10.4) are reordered as follows:

Finally we obtain the following matrix:

Mathlab codes and plot from Q1

Mathlab codes and plot from Q2

=R*3.11 Representing the solution of SC-L1-ODE-VC in the terms of State Transition Matrix using the properties of State Transition Matrix =

Find
Express Equation 3.11.1 in terms of [[media:pea1.f11.mtg16.djvu|State Transition Matrix]] $$\displaystyle \mathbf {X}(t)=\mathbf{\Phi}(t,t_0)\mathbf{X}(t_0)+\int^t_{t_0}\mathbf{\Phi}(t,\tau)\,\mathbf{B}(\tau)\,\mathbf{U}(\tau)\,d\tau $$ using properties of State Transition Matrix Equation 3.11.2 and 3.11.3.

Solution
Let,

Equation 3.11.1 becomes, Rearranging Equation 3.11.5 Clearly Equation 3.11.6 satisfies the first exactness condition but doesn't hold good for 2nd exactness condition so in order to make it exact we multiply it by an Integrating Factor $$\displaystyle h(t,{\Phi}) $$.

Equation 3.11.6 becomes Now applying the Second Exactness Condition to Equation 3.11.7. Since $$\displaystyle N(t,\phi)=1 $$.

$$\displaystyle \frac{\partial N}{\partial t}=0 $$ Let $$\displaystyle \frac{\partial h}{\partial \phi}=0 $$

Equation 3.11.9 becomes Since $$\displaystyle M(t,\phi)= -\mathbf {A}(t)\mathbf{\Phi}(t,t_0) $$

$$\displaystyle M_\phi= -\mathbf {A}(t) $$

Solving for integrating factor $$\displaystyle h(t) $$ from Equation 3.11.10.

$$\displaystyle \int \frac{dh}{h}= - \int \mathbf{A(t)}dt $$

Hence $$\displaystyle h(t)=\exp \{-\int \mathbf{A(t)}dt\}$$

Equation 3.11.7 transforms to : Clearly Equation 3.11.11 can be written as: Integrating Equation 3.11.12 within limits $$\displaystyle t_0-t $$ Solving 3.11.13 for $$\displaystyle \mathbf{\Phi}(t,{t_0}) $$ we get: Now from property 1 of State Transition Matrix i.e Equation 3.11.2, (Rearranging and integrating within the limits $$\displaystyle t_0-t $$) we get : Now using property 2 of State Transition Matrix $$\displaystyle\mathbf{\Phi}({t_0},{t_0})= \mathbf I $$ Now  $$\displaystyle \ln \,\mathbf I $$   is a zero matrix.

Therefore Equation 3.11.7 becomes: Similarly, since $$\displaystyle \tau $$ is a dummy integration variable: Now looking at the assumption we did in the beginning i.e Equation 3.11.4 And Using Equation 3.11.14, 3.11.18 , 3.11.19 , 3.11.20 and 3.11.21 we get:

=R*3.12 - Determining the SC-L1-ODE-CC Equation for the Roll Control of a Rocket=

Given

$$\delta$$ = aileron deflection (angle)

$$\phi$$ = roll angle

$$\omega$$ = roll angular velocity

$$Q$$ = aileron effectiveness proportional to inverse of rocket roll inertia

$$\tau$$ = roll time constant

$$\dot \phi = \omega$$ (1)

$$\dot \omega = -\frac{1}{\tau} \omega + \frac{Q}{\tau} \delta$$ (2)

$$\dot \delta = u$$ (3)

Find
Put equations (1)-(3) in the form of a SC-L1-ODE-CC (System of Coupled Linear First Order Differential Equations with constant coefficients) with form:

$$\mathbf{\dot{x}}(t) = \mathbf{A}(t)\,\mathbf{x}(t) + \mathbf{B}(t) \,\mathbf{u}(t)$$ (4)

with $$\mathbf{A}, \mathbf{B}$$ being constant matrices.

Solution
This problem was solved without referring to past solutions.

Equations (1)-(3) constitute a system of 3 coupled linear equations. Therefore

$$\mathbf{\dot x}:= \left \lfloor \ \dot \phi \ \dot \omega \ \dot \delta \ \right \rfloor ^T \in \mathbb R^{3 \times 1}$$

$$\mathbf{x}:= \left \lfloor \ \phi \ \omega \ \delta \ \right \rfloor ^T \in \mathbb R^{3 \times 1}$$

Since the aileron's angular velocity is the control,

$$\mathbf{u} := \dot \delta = u$$

Thus, placing equations (1)-(3) in the form of equation (4) yields

$$\begin{Bmatrix} \dot \phi \\ \dot \omega \\ \dot \delta \end{Bmatrix} = \begin{Bmatrix} 0 \ 1 \ 0 \\ 0 \ \frac{-1}{\tau} \ \frac{Q}{\tau} \\ 0 \ 0 \ 0\end{Bmatrix} \ \begin{Bmatrix} \phi \\ \omega \\ \delta\end{Bmatrix} + \begin{Bmatrix} 0 \\ 0 \\ 1\end{Bmatrix} \ u$$

= R*3.13 - Derivation of the 2nd Exactness Conditions for N2-ODEs =

Given
If an N2-ODE is exact, then it must be expressible in the form

$$ \begin{align} \frac{d\phi}{dx} &= \phi_x + \phi_y\, p + \phi_p\, p_x \\ &= g + f\, p_x \end{align} $$

where

$$ \begin{align} p &:= \frac{dy}{dx} \\ g &:= \phi_x + \phi_y\, p \\ f &:= \phi_p \end{align} $$

Furthermore, since we assume that $$\phi$$ is sufficiently smooth,

$$ \begin{align} \phi_{xy} &= \phi_{yx} \\ \phi_{yp} &= \phi_{py} \\ \phi_{xp} &= \phi_{px} \end{align} $$

Finally, we are given a particular equation,

Find
Derive the 2nd Exactness Conditions, given below

Also, show that 3.13.1 satisfies these conditions.

Solution
This problem was solved without referring to past solutions, though reference was made to the lecture notes from Meeting 17 for the second part.

3.13.3 is much easier to derive, so let's begin there

$$ \begin{align} g &= \phi_x + \phi_y\, p \\ g_p &= \phi_{xp} + \phi_{yp}\, p + \phi_y \\ &= f_x + f_y\, p + \phi_y \\ g_{pp} &= f_{xp} + p\, f_{yp} + f_y + \phi_{yp} \\ &= f_{xp} + p\, f_{yp} + 2 f_y \end{align} $$

In deriving 3.13.2, we will make much use of the continuity of the second derivatives of $$\phi$$. Much of this material is a restatement of that contained in the lecture notes from Meeting 17.

$$ \begin{align} \phi_{px} &= \phi_{xp} \\ f_x &= [ g - \phi_y\, p ]_p \\ f_x &= g_p - \phi_{yp}\, p - \phi_y \\ f_x &= g_p - f_y\, p - \phi_y \end{align} $$

where use was made of the definition of g in the second step to obtain an expression for $$\phi_x$$. We can rearrange this last expression to arrive at an expression for $$\phi_y$$

This expression will be useful later on.

$$ \begin{align} \phi_{py} &= \phi_{yp} \\ f_y &= [ \frac{1}{p} (g - \phi_x) ]_p \\ f_y &= - \frac{1}{p^2} (g - \phi_x) + \frac{1}{p} (g_p - \phi_{xp}) \end{align} $$

where again the definition of g was used in the second step, this time to obtain an expression for $$\phi_y$$. Solving this expression for $$\phi_x$$ yields

Finally, we make use of $$\phi_{xy} = \phi_{yx}$$. We first differentiate 3.13.4 with respect to x

$$ \begin{align} \phi_{y} &= g_p - f_y\, p - f_x \\ \phi_{yx} &= g_{px} - f_{yx}\, p - f_y\, p_x - f_{xx} \\ \phi_{yx} &= g_{xp} - f_{xy}\, p - f_y\, p_x - f_{xx} \end{align} $$

We now differentiate 3.13.5 with respect to y, noting that $$p_y = \partial^2 y\, / \partial x\, \partial y = \partial^2 y\, / \partial y\, \partial x = \partial\, 1 / \partial x = 0$$

$$ \begin{align} \phi_x = g - p (g_p - f_x) + p^2\,f_y \\ \phi_{xy} = g_y - p (g_{py} - f_{xy}) + p^2\, f_{xy} \\ \phi_{xy} = g_y - p (g_{yp} - f_{xy}) + p^2\, f_{xy} \end{align} $$

Equating these two expressions and collecting like terms yields

$$\displaystyle f_{xx} + 2 p f_{xy} + p^2 f_{yy} + f_y p_x = g_{xp} + p g_{yp} - g_y$$

Note that there is an extra term on the left hand side of this expression compared to the desired result.

Applying the definition of g and f to 3.13.1 yields

$$ \begin{align} g &= x (y\,')^2 + y y\,' \\ &= x p^2 + y p \\ f &= xy \end{align} $$

Performing the necessary derivatives is simple

$$ \begin{align} g_p &= 2xp + y \\ g_{pp} &= 2x \\ g_y &= p \\ g_{yp} &= 1 \\ g_{xp} &= 2p \\ \\ f_{xx} &= f_{yy} = f_{xp} = f_{yp} = 0 \\ f_{xy} &= 1 \\ f_y &= x \end{align} $$

The exactness conditions are

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

=R*3.14 - Find solution for h(x,y) =

Given
$$\displaystyle g=\cancel{xp^2}+\cancel{yp} = \phi_x + \phi_y p = \underbrace{(h_x + \cancel{yp})}_{\phi_x}+\underbrace{(h_y+\cancel{xp})}_{\phi_y}p $$

$$\displaystyle p(x)= y'(x) $$

Find
$$\displaystyle h_x + h_y p = 0$$
3.14.1

Find h(x,y).

Solution
$$\displaystyle h_x + h_y p = 0 $$

where

$$\displaystyle h_x = \frac {\partial h}{\partial x}$$

$$\displaystyle h_y = \frac {\partial h}{\partial y}$$

$$\displaystyle p = \frac {dy}{dx} $$

The eq 3.14.1 is following that

$$ \displaystyle \frac {\partial h(x,y)}{\partial x}+\frac {\partial h(x,y)}{\partial y} \frac {dy}{dx} = \frac {d}{dx} h(x,y) = 0$$

Since $$\displaystyle \frac {d}{dx} h(x,y)$$ is 0, h(x,y) is only the function of y.

$$\displaystyle h(x,y) = f(y) $$ It shows that

$$\displaystyle h_x = 0 $$ Thus, the eq 3.14.1 is $$\displaystyle h_y \cdot p = 0 $$
3.14.2

In order to satisfy the eq 3.14.2,

1.$$\displaystyle h_y = 0 $$ or

2.$$\displaystyle p = \frac {dy}{dx} = 0 $$

If the second one is possible, the entire equation will be zero.It is trivial.

Thus, the possible solution is the first one.

$$\displaystyle h_x = 0, h_y = 0$$

Therefore, we can conclude that

h = constant = k

=R*3.15 - Second Exact condition=

Given
Given the following equations (1)p.18-3

$$ \displaystyle (15p^{4}\cos x^{2})y''+(6xy^{2})y'+[-6xy^{5}\sin x^{2}+2y^{3}]=0$$
(3.15.1)

$$ \displaystyle f_{xx}+2pf_{xy}+p^{2}f_{yy}=g_{xp}+pg_{yp}-g_{y}$$

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

(3.15.3)

Find
Verify whether (3.15.1) satisfies the 2nd exactness condition (3.15.2) and (3.15.3).

Solution
According to Prof. Loc Vu-Quoc's lecture note, we know that

$$ \displaystyle \phi_{x}=3p^{5}(-\sin x^{2})(2x)+2y^{3}$$

(3.15.4)

$$ \displaystyle \phi_{y}=6xy^{2}$$

(3.15.5)

$$ \displaystyle f:=\phi_{p}=15p^{4}\cos x^{2}$$
(3.15.6)

$$ \displaystyle g:=\phi_{y}y'+\phi_{x}=(6xy^{2})y'+[-6xp^{5}\sin x^{2}+2y^{3}]$$
(3.15.7)

Hence

$$ \displaystyle f_{xx}=(\phi_{p})_{xx}=-30p^{4}\sin x^{2}-60x^{2}p^{4}\cos x^{2}$$
(3.15.8)

$$ \displaystyle f_{xp}=-120xp^{3}\sin x^{2},\, f_{xy}=0,\, f_{yy}=0,\, f_{y}=0,\, f_{yp}=0$$
(3.15.9) $$ \displaystyle g_{y}=(12xy)y'+6y^{2},\, g_{yp}=0$$
(3.15.10) $$ \displaystyle g_{xp}=-30p^{4}\sin x^{2}-60x^{2}p^{4}\cos x^{2},\,g_{pp}=-120xp^{3}\sin x^{2}$$
(3.15.11)

Substitute (3.15.8)-(3.15.11) to (3.15.2)

$$ \displaystyle LHS=-30p^{4}\sin x^{2}-60x^{2}p^{4}\cos x^{2}+2p\times0+p^{2}\times0$$

$$ \displaystyle LHS=-30p^{4}\sin x^{2}-60x^{2}p^{4}\cos x^{2}$$
(3.15.12) $$ \displaystyle RHS=-30p^{4}\sin x^{2}-60x^{2}p^{4}\cos x^{2}+p\times0-((12xy)y'+6y^{2})$$

$$ \displaystyle RHS=-30p^{4}\sin x^{2}-60x^{2}p^{4}\cos x^{2}-((12xy)y'+6y^{2})$$
(3.15.13)

Thus $$ \displaystyle LHS\neq RHS$$, then (3.15.11) does not satisfy (3.15.2), but we can easily find that (3.15.1) does satisfy (3.15.2)

=R*3.16- Completion of Second Choice=

Given
From (3.16.1) and (3.16.2), we get

For the first choice of the solution, assume that

The process to find $$\displaystyle \phi(x,y,p)$$ is as follows:

Find
Finish the second choice assuming that

Solution
This problem was solved without referring to past solutions.

(3.16.10) can be rewritten as follows:

And then the both sides of (3.16.11) are integrated as follows:

The above integration results in

The result is exactly same with (3.16.5).

Thus, the next process and resulted $$\displaystyle \phi(x,y,p)$$ would be same with the above given functions from (3.16.6) to (3.16.9).

=R*3.17- Application of Reynold's Transport Theorem =

Given
Considering: in [[media:pea1.f11.mtg19-20.djvu|Reynold's Transport Theorem]] Where:

$$\displaystyle \mathbf u = \mathbf u (x_i,t):= u_i(x_i,t) $$

Find
Show: Using Equations 3.17.4 and 3.17.5: Conservation of Mass:

Solution
Putting the value of $$\displaystyle f(x,t)$$ i.e Equation 3.17.1 in Reynold's Transport Theorem, Equation 3.17.2 we get: Partially differentiating within the integral on the right hand side we get: Rearranging terms: Since: Taking the Material Time Derivative of Equation 3.17.9 : Term $$\displaystyle \frac{\partial x_i}{\partial t}= u_i $$ in Equation 3.17.10.

We see that the first bracket of of right hand side in Equation 3.17.8 is nothing but $$\displaystyle \frac {D\mathbf u}{Dt} $$ of Equation 3.17.10 and second bracket is zero by Conservation of Mass (Equation 3.17.5).

Hence Equation 3.17.8 reduces to:

=R*3.18 - Deriving the 1-Dimensional Case from Reynolds Transport Theorem=

Given
The alternate version of Reynolds Transport Theorem is:

$$\frac{D}{Dt} \int_{\mathcal B_t}f(x,t) \ d \mathcal B_t = \int_{\mathcal B_t} \frac{\partial f(x,t)}{\partial t} \ d \mathcal B_T + \int_{\partial \mathcal B_t} \mathbf n \cdot (f(x,t) \mathbf u(x,t)) \ d (\partial \mathcal B_t)$$ (1)

$$\mathcal B_t$$ = region of integration $$\displaystyle x$$ = a point in the region $$\mathbf u(x,t)$$ = the velocity of x at time t $$\mathbf n$$ = outward unit normal to the boudary

A particular case of equation (1) for 1-Dimension is:

$$\frac{d}{dt} \int^{s=b(t)}_{s=a(t)} f(s,t) \ ds = \int^{s=b(t)}_{s=a(t)} \frac{\partial f(s,t)}{\partial t} \ ds + f(b(t),t) \ \frac{db(t)}{dt} - f(a(t),t) \ \frac{da(t)}{dt}$$ (2)

Find
Obtain particular case of equation (2) from the general case of equation (1)

Solution
This problem was solved without referring to past solutions.

Various terms simplify in the 1-dimensional case, since all measurements are now 1-dimensional:

$$\mathcal B_t = s$$ $$\displaystyle f(x,t) = f(s,t)$$ $$\mathbf u(x,t) = u(s,t)$$ $$\mathbf n = 1$$

$$\displaystyle \mathbf n$$ becomes 1 because it is a unit vector. Since a unit vector must have a magnitude of 1, and it only has 1 dimension, $$\displaystyle \mathbf n$$ must equal 1. The dot product also becomes a multiplication in the 1-dimensional case. Substituting these new values into equation (1) we obtain,

$$\frac{d}{dt} \int_{s}f(s,t) \ ds = \int_{s} \frac{\partial f(s,t)}{\partial t} \ ds + \int_{\partial s} 1 \times (f(s,t) u(s,t)) \ d (\partial s)$$ (3)

Where s is defined over the interval $$ \left [ a(t),b(t) \right ]$$ making equation (3):

$$\frac{d}{dt} \int^{s=b(t)}_{s=a(t)}f(s,t) \ ds = \int^{s=b(t)}_{s=a(t)} \frac{\partial f(s,t)}{\partial t} \ ds + \int_{\partial s} f(s,t) u(s,t) \ d (\partial s)$$ (4)

$$\displaystyle \partial s$$ is the boundary of s. The boundary of s is only defined at two points, a(t) and b(t). Because of this the last term in equation (4) becomes:

$$\int_{\partial s} f(s,t) u(s,t) \ d (\partial s) = f(b(t),t)u(b(t),t) - f(a(t),t)u(a(t),t)$$ (5)

Since u(b(t),t) is the velocity of b(t) at time t

$$u(b(t),t) = \frac{db(t)}{dt}$$

Substituting for the last term we obtain equation (2).

$$\frac{d}{dt} \int^{s=b(t)}_{s=a(t)}f(s,t) \ ds = \int^{s=b(t)}_{s=a(t)} \frac{\partial f(s,t)}{\partial t} \ ds + f(b(t),t) \frac{db(t)}{dt} - f(a(t),t) \frac{da(t)}{dt}$$ (6)

=References=

1. Malvern 1969, Introduction to the mechanics of a continuous medium, p.211

=Contributing Members(need updating)=

Egm6321.f11.team2.rho I solved R*3.4, R*3.10, and R*3.16 and read the proof of R*3.2, and  R*3.3. 20:12, 5 October 2011 (UTC) Egm6321.f11.team2.Xia I solved R*3.3, R*3.9, and R*3.15 and read the proof of R*3.1, R*3.2. and R*3.5   14:21, 3 October 2011 (UTC) Egm6321.f11.team2.kim I solved R*3.2, R*3.8, and R*3.14 and read the proof of R*3.1 and R*3.18. 02:12, 4 October 2011 (UTC) Egm6321.f11.team2.epps I solved R3.1, R3.7, and R3.13, reviewed R3.18 and R3.17, and made a small correction to the A matrix in R3.6.) Egm6321.f11.team2.epps 01:53, 5 October 2011 (UTC) Egm6321.f11.team2.johri.y 04:53, 5 October 2011 (UTC) I solved R*3.5, R*3.11 and R*3.17 and read the proof of R 3.1, R*3.4, R*3.6, R*3.7, R*3.12, R*3.13, R*3.14, R*3.16 and R*3.18 20:29, 5 October 2011 (UTC) Egm6321.f11.team2.wise 22:00, 5 October 2011 (UTC) I solved R*3.6, R*3.12,R*3.18; and read the proof of R*3.16 and R*3.17

title: User:Egm6321.f11.team2/HW3


text:
User:Egm6321.f11.team2/HW4

=R4.1 - A Physical Derivation of the Reynolds Transport Theorem=

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

Find
A direct, physical derivation of the given equation, given by Malvern, 1969, p. 211.

Solution
Consider a region of space that moves with a given quantity of matter, such that it always contains exactly the same particles.

Let f(x, t) be any physical property of the material per unit mass.

As Malvern states, the rate of increase of the total amount of f within the region of space is equal to the rate of increase of the amount of f possessed by the material instantaneously inside the region minus the net rate of outward flux of f carried by mass transport through the control surface. .

Translated into mathematics, this states

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

which, upon a slight rearrangement of the terms, is the sought after equation!

=R*4.2 - Verify the exactness for L2-ODE-VC =

Given
$$\displaystyle \sqrt{x} \, y'' + 2xy' + 3y = 0 $$  (4.2.1)

Find
Verify the exactness of eq (4.2.1)

Solution
Let $$\displaystyle g(x,y,p) = 2xy' + 3y, \, \, f(x,y,p) = \sqrt{x} $$ $$\displaystyle p := y' $$

$$\displaystyle \underbrace {\sqrt{x}}_{f(x,y,p)}\, y'' + \underbrace {2xp + 3y}_{g(x,y,p)} = 0 $$ It shows that above equation satisfy the first exactness. In order to satisfy 2nd exacntess condition, $$\displaystyle f_{xx} + 2pf_{xy}+p^2f_{yy} = g_{xp} + pg_{yp} - g_{y} $$ (4.2.2)

$$\displaystyle f_{xp} + pf_{yp} + 2f_y = g_{pp}$$ (4.2.3) From the eq (4.2.1),

$$\displaystyle f_{xx} = -\frac {1}{4}x^{-\frac{3}{2}}, \, \, f_{xy} = 0, \, \, f_{yy} = 0 $$ $$\displaystyle g_{xp} = 2, \, \, g_{yp} = 0, \, \, g_{y} = 3, \, \, f_{xp} = 0, \, \, f_{yp} = 0, \, \, f_{y} = 0, \, \, g_{pp} = 0 $$ From the above values, eq (4.2.2) and eq (4.2.3) is that $$\displaystyle f_{xx} = g_{xp} - g_{y}$$  (4.2.4) $$\displaystyle f_{xx} = \frac {1}{4}x^{-\frac{3}{2}}, \, \, \, g_{xp} - g_{y} = -1 $$ $$\displaystyle f_{xp} + pf_{yp} + 2f_{y} = g_{pp} = 0 $$ (4.2.5)

This shows that although the eq (4.2.5) satisfies the 2nd exactness condition, eq (4.2.4) does not satisfy the 2nd exactness condition. Thus, the eq (4.2.1) is not exact.

=R*4.3 - Integrating Factor Methods - Power-Form=

Given
Giving the following N2-ODE $$\displaystyle \sqrt{x}y''+2xy'+3y=0$$ (4.3.1)

Find
find $$ \displaystyle m,n\in\mathbb{R} $$ such that 4.3.2 is exact: $$ \displaystyle (x^{m}y^{n})\left[\sqrt{x}y''+2xy'+3y\right]=0 $$ (4.3.2) show that the first integral is a L1-ODE-VC $$ \displaystyle \phi(x,y,p)=xp+(2x^{\frac{3}{2}}-1)y=k $$ (4.3.3) with $$ \displaystyle p(x):=y'(x) $$ Solve 4.3.3 for $$ \displaystyle y $$.

Solution
The $$ 2^{nd} $$ exactness condition is , $$\displaystyle f_{xx} + 2pf_{xy}+p^2f_{yy} = g_{xp} + pg_{yp} - g_{y}$$  (4.3.4) $$\displaystyle f_{xp} + pf_{yp} + 2f_y = g_{pp} $$  (4.3.5) From 4.3.2, we have $$ \displaystyle \underbrace{(x^{m+\frac{1}{2}}y^{n})}_{f(x,y,p)}y''+\underbrace{2(x^{m+1}y^{n})p+3x^{m}y^{n+1}}_{g(x,y,p)}=0 $$  (4.3.5) Then $$\displaystyle f_{x}=(m+\frac{1}{2})x^{m-\frac{1}{2}}y^{n} $$ $$\displaystyle f_{xx}=(m+\frac{1}{2})(m-\frac{1}{2})x^{m-\frac{3}{2}}y^{n} $$ $$\displaystyle f_{xy}=n(m+\frac{1}{2})x^{m-\frac{1}{2}}y^{n-1} $$ $$\displaystyle f_{yy}=n(n-1)x^{m-\frac{1}{2}}y^{n-2} $$ $$\displaystyle f_{xp}=0,\qquad f_{yp}=0 $$ $$\displaystyle g_{x}=2(m+1)x^{m}y^{n}p+3mx^{m-1}y^{n+1} $$ $$\displaystyle g_{y}=2nx^{m+1}y^{n-1}p+3(n+1)x^{m}y^{n} $$ $$\displaystyle g_{xp}=2(m+1)x^{m}y^{n}+3mx^{m-1}y^{n+1} $$ $$\displaystyle g_{yp}=2nx^{m+1}y^{n-1}+3(n+1)x^{m}y^{n} $$ $$\displaystyle g_{pp}=0 $$

Substitute the above equation to 4.3.4 and 4.3.5, we have $$\displaystyle (m+\frac{1}{2})(m-\frac{1}{2})x^{m-\frac{3}{2}}y^{n}+2pn(m+\frac{1}{2})x^{m-\frac{1}{2}}y^{n-1}+p^{2}n(n-1)x^{m-\frac{1}{2}}y^{n-2}=2(m+1)x^{m}y^{n} $$ $$\displaystyle +3mx^{m-1}y^{n+1}+p\left(2nx^{m+1}y^{n-1}+3(n+1)x^{m}y^{n}\right)-2nx^{m+1}y^{n-1}p-3(n+1)x^{m}y^{n} $$  (4.3.6) $$\displaystyle 2nx^{m+\frac{1}{2}}y^{n-1}=0 $$  (4.3.7) From 4.3.7 we know $$\displaystyle n=0 $$ , then examine 4.3.6 we have $$\displaystyle m= \frac{1}{2} $$. Thus $$\displaystyle f(x,y,p)=\underbrace{x}_{\phi_{p}},\quad g(x,y,p)=\underbrace{3x^{\frac{1}{2}}y}_{\phi_{x}}+\underbrace{2x^{\frac{3}{2}}}_{\phi_{y}}p $$  (4.3.8) Integrating 4.3.8, we have $$\displaystyle \begin{array}{l} \phi=xp+a(x,y)\\ \phi=2x^{\frac{3}{2}}y+b(y,p)\\ \phi=2x^{\frac{3}{2}}y+c(x,p)\end{array} $$ Thus $$ \displaystyle \phi(x,y,p)=xp+(2x^{\frac{3}{2}}-1)y=k $$  (4.3.9) From 4.3.9, we have $$\displaystyle \underbrace{1}_{a_{1}(x)}\cdot y'+\underbrace{(2x^{\frac{1}{2}}-\frac{1}{x})}_{a_{0}(x)}y=\underbrace{\frac{k}{x}}_{b(x)} $$ So, using Maple, we have $$\displaystyle y\left( x\right) =x\,{e}^{-\frac{4\,{x}^{\frac{3}{2}}}{3}}\,k_2-\frac{8\,k\,\Pi\,x\,{e}^{-\frac{4\,{x}^{\frac{3}{2}}}{3}}}{\sqrt{3}\,{36}^{\frac{1}{3}}\,\Gamma \left( \frac{2}{3}\right) }+\frac{4\,\Gamma \left( \frac{1}{3},-\frac{4\,{x}^{\frac{3}{2}}}{3}\right) \,k\,x\,{e}^{-\frac{4\,{x}^{\frac{3}{2}}}{3}}}{{36}^{\frac{1}{3}}}-k $$

=R*4.4 - Derivation of a class of exact L2-ODE-VC=

Given
The form of general N2-ODE is

The (4.4.1) is considered as follows:

A generated class of exact L2-ODE-VC is

Find
Show that (4.4.2) and (4.4.3) leads to (4.4.4).

Solution
From (4.4.1) and (4.4.2), we can obtain

Integrating the both sides of (4.4.5), the following is obtained.

Using (4.4.6),

whrere

$$ \displaystyle \begin{align} &p = y'(x) \\ &\phi_x = P'(x)p + \frac{\partial k(x,y)}{\partial x} \\ &\phi_y = \frac{\partial k(x,y)}{\partial y} \end{align} $$

Integrating the term of R(x)y in (4.4.7),

With (4.4.8), (4.4.6) can be rewritten as follows:

To find k(y), $$\displaystyle \phi_y $$ is found and compared to Q(x) because $$\displaystyle \phi_y = Q(x) $$ in (4.4.7).

Frrom (4.4.10), we can find the following:

Therefore, k(y) is verified with constant k and (4.4.9) is resulted in

$$ \displaystyle \phi(x,y,p) = P(x)p + T(x)y + k $$

=R*4.5 - To show an L2-ODE-VC to be exact and solve for first integral and final solution=

Find
a) Is F exact?

b) Find $$\phi$$

c) Solve for $$y$$

Solution
Part a

In order for F to be exact, it should follow the two exactness condition. It is already in the particular form but it has to satisfy the two equations of second exactness condition. From equation 4.5.1 :

The partial derivatives are as follows

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

$$\displaystyle g_y=2x $$
 * }
 * }

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

$$\displaystyle g_{yp}=0 $$
 * }
 * }

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

$$\displaystyle g_x=2y+(2x-cox(x))p $$
 * }
 * }

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

$$\displaystyle g_{xp}=2x-cos(x) $$
 * }
 * }

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

$$\displaystyle f_x=-sin(x) $$
 * }
 * }

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

$$\displaystyle f_{xp}=0 $$
 * }
 * }

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

$$\displaystyle f_{xy}=0 $$
 * }
 * }

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

$$\displaystyle f_{xx}=-cos(x) $$
 * }
 * }

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

$$\displaystyle f_y=0 $$
 * }
 * }

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

$$\displaystyle g_{pp}=0 $$
 * }
 * }

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

$$\displaystyle f_{yp}=0 $$
 * }
 * }

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$$\displaystyle f_{yy}=0 $$
 * }
 * }

If we take the above differentials and replace them in Equation 4.5.2 and 4.5.3. Both conditions are satisfied.

Part b

Equation 4.5.1 can be written as: $$\displaystyle {\phi}_x + {{\phi}_y}y' + {{\phi}_p}y'' $$ Where : $$\displaystyle {\phi}_x + {{\phi}_y}y' = 2xy+(x^2-sin x)p $$ And: From Equation 4.5.6 Taking partial derivatives w.r.t 'x' and 'y' for equation 4.5.7 we get: And Now: Putting the values of $$\displaystyle {\phi}_x, {\phi}_y $$ from Equation 4.5.8 and 4.5.9 in Equation 4.5.10 we get: From Equation 4.5.11 Partially differentiating Equation 4.5.13 Comparing Equation 4.5.14 and 4.5.12
 * {| style="width:100%" border="0" align="left"

$$\displaystyle k'(y)=0$$
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle k(y)=k$$ (constant). Hence Equation 4.5.13 becomes And the first integral i.e Equation 4.5.7 becomes:
 * }
 * }

Part c Hence Let $$\displaystyle k_2=0 $$ Equation 4.5.17 becomes Integrating both sides: We get:

=R*4.6 - Equivalence of Two Forms of the Second Exactness Conditions for N2-ODEs, Method 1=

Given
The second exactness condition for a Nn-ODE to be exact is
 * $$ \displaystyle \sum^n_{j=0} (-1)^j g^{(j)}_j = 0$$(4.6.1)

Where,
 * $$\displaystyle g^{(j)}_j = \frac{d^j}{dx^j} \frac{\partial G}{\partial y^{(j)}}$$

 * $$\displaystyle G(x,y',y'') = \frac{d}{dx}\phi (x,y,y') = 0$$

Find
Show that the first and second forms of the 2nd exactness condition are equivalent, where the first form is,
 * $$\displaystyle \phi_{xy} = \phi_{yx}, \ \ \phi_{yp} = \phi_{py}, \ \ \phi_{px} = \phi_{xp}$$

where,
 * $$\displaystyle p = y'$$

or equivalently,
 * $$\displaystyle f_{xx} + 2pf_{xy} + p^2 f_{yy} = g_{xp} + p g_{yp} - g_y$$(4.6.2)

 * $$\displaystyle f_{xp} + p f_{yp} + 2 f_y = g_{pp}$$(4.6.3)

And the second form is
 * $$\displaystyle g_o - \frac{d g_1}{dx} + \frac{d^2 g_2}{dx^2} = 0$$(4.6.4)

Solution
First we should derive the expressions for the terms of equation (4.6.4).
 * $$\displaystyle g_0 := \frac{\partial G}{\partial y} = \frac{\partial}{\partial y} (\frac{d \phi}{dx})$$

 * $$\displaystyle \frac{d}{dx} g_1 = \frac{d}{dx} \frac{\partial G}{\partial y'} = \frac{d}{dx} [ \phi_{xy'} + \phi_{yy'} y' + \phi_y + \phi_{y'y'} y'' ]$$

 * $$\displaystyle \frac{d^2}{dx^2} g_2 = \frac{d}{dx} \frac{d}{dx} \frac{\partial G}{\partial y} = \frac{d}{dx} [ \phi_{y'x} + \phi_{y'y} y' + \phi_{y'y'} y]$$

Replacing the terms in equation (4.6.4) with their derived expressions we obtain
 * $$\frac{\partial}{\partial y} (\frac{d \phi}{dx}) - \frac{d}{dx} [ \phi_{xy'} + \phi_{yy'} y' + \phi_y + \phi_{y'y'} y ] + \frac{d}{dx} [ \phi_{y'x} + \phi_{y'y} y' + \phi_{y'y'} y] = 0$$

It can be seen that by separating out like terms we can form the first form of the second exactness condition. $$\displaystyle \frac{\partial}{\partial y} (\frac{d \phi}{dx}) = \frac{d}{dx} \phi_y$$ $$\displaystyle \frac{d}{dx} \phi_{xy'} = \frac{d}{dx} \phi_{y'x}$$ $$\displaystyle \frac{d}{dx} \phi_{yy'} y' = \frac{d}{dx} \phi_{y'y} y'$$ By simplifying the equations and replacing $$\displaystyle y'$$ with $$\displaystyle p$$ we obtain $$\displaystyle \phi_{yx} = \phi_{xy}$$ $$\displaystyle \phi_{xp} = \phi_{px}$$ $$\displaystyle \phi_{yp} = \phi_{py}$$ Which is the first form of the 2nd exactness condition.

=R*4.7 - Equivalence of Two Forms of the Second Exactness Conditions for N2-ODEs, Method 2=

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

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

where

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

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

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

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

Then,

Plugging 4.7.1, 4.7.2, and 4.7.3 into the given form of the second exactness condition yields

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

By equating in turn the coefficients of q0 and q1 to zero, the desired relations are obtained.

=References=

=Contributing Members=

Egm6321.f11.team2.epps I solved R4.1 and R*4.7 and proofread all the other problems. 00:41, 19 October 2011 (UTC)

Egm6321.f11.team2.johri.y 00:50, 19 October 2011 (UTC) I solved R*4.5 and proofread all other problems.

Egm6321.f11.team2.wise 01:37, 19 October 2011 (UTC) I solved R*4.6 and proofread problems R4.4 and R4.5.

Egm6321.f11.team2.kim 14:23, 19 October 2011 (UTC) I solved R*4.2 and proofread problems R*4.3, R*4.6 and R*4.7

Egm6321.f11.team2.Xia 23:50, 19 October 2011 (UTC) I solved R*4.3 and proofread R*4.1,R*4.2 and R*4.7.

Egm6321.f11.team2.rho15:53, 19 October 2011 (UTC) I solved R*4.4 and proofread R*4.1, R*4.3, and R*4.5.

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

title: User:Egm6321.f11.team2/HW4


text:
User:Kathleenmccaskie

Hi, my name is Kathleen and I'm studying a Bachelors in Coaching and Exercise Science and Sports Management at the university of Canberra.

 * /Sport, Sponsorship and Social Responsibility/

title: User:Kathleenmccaskie


text:
User talk:Kathleenmccaskie

 Hello Kathleenmccaskie, 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! --Abd 19:44, 11 September 2011 (UTC)

Thanks for meeting today.

Separate page
I will check out the add page function!

title: User talk:Kathleenmccaskie


text:
User:SarahProudman

Hi, my name is Sarah and I'm studying a Bachelor in Coaching and Exercise Science and Sports Management at the University of Canberra

 * /Price to play? Australia’s sports Injuries Epidemic/

title: User:SarahProudman


text:
User:EGM6341.s11.team1.Chiu/PEA1 F09 Mtg32

Mtg 32: Thu, 05 Nov 09

[[media: Egm6321.f09.mtg32.djvu| page32-1]]

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

 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

HW: Verify that $$ P_{0}, \cdots, P_{4} \ $$ [[media: Egm6321.f09.mtg31.djvu| (1)-(5) page31-3]] are solutions of Legendre equation [[media: Egm6321.f09.mtg14.djvu| (1) page14-2]], [[media: Egm6321.f09.mtg30.djvu| page30-4]]

 * }

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

 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

HW:(See [[media: Egm6321.f09.mtg31.djvu| page31-1]]) Obtain the separate equations for Laplace equation on circle, cylindrical coordinates and identify the Bessel differential equation.

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

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

K.p.84 $$ x^{2}y''+xy'-(x^{2}- \nu^{2})y \underset{ \color{red} \underset{(1)}{\uparrow}}=0 \ $$

 * }

 * }

Heat Problem continued [[media: Egm6321.f09.mtg31.djvu| page31-3]]

Superposition:

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

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

$$ \psi(r, \theta) \underset{ \color{red} \underset{(2)}{\uparrow}}= \sum_{n=0} \left [ A_{n}r^{n}+B_{n}r^{-(n+1)} \right ] P_{n}(\mu) \ { \color{blue} \mu= \sin \theta= \cos \bar{ \theta}}$$

 * }

As $$ r \to 0, r^{-(n+1)} \to \infty ; \ $$ so for $$ \psi \ $$ to be physical meaningful, set

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

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

$$ B_{n}=0, \ \forall n=0,1,2, \cdots \ $$

(3)
 *
 * }

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

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

Boundary Condition $$ \psi(1, \theta)= \underbrace{f( \theta)= \sum_{n}A_{n} P_{n}( \sin \theta)}_{ \color{blue} Fourier-Legendre \ series} \ $$

(4)
 *
 * }

[[media: Egm6321.f09.mtg32.djvu| page32-2]]

Recall Fourier series:

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

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

$$ f( \theta)= { \color{red}1 \cdot}a_{0}+ \sum a_{n} \cos n \omega \theta+ \sum b_{n} \sin n \omega \theta \ $$

(1)
 *
 * }

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

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

$$ \left \{ 1, \cos n \omega \theta, \sin n \omega \theta \right \}, \ n=1,2, \cdots \ \ $$   linear independent

(2)
 *
 * }

End Recall Fourier series

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

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

Recall: $$ \left \{ \underbrace{ \mathbf{b}_{1}, \mathbf{b}_{2}, \cdots, \mathbf{b}_{n}}_{ \color{blue} linear \ independent} \right \} \in \mathbb{R}^{n} \ $$

(3)
 *
 * }

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

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

Consider $$ \mathbf{v} \in \mathbb{R}^{n}, \ \mathbf{v}= \sum_{i=1}^{n}a_{i} \mathbf{b}_{i} \ $$

(4)
 *
 * }

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

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

Find $$ a_{i}, \ i=1, \cdots, n \ $$

 * }

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

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

$$ \mathbf{v} \cdot \mathbf{b}_{j}= \sum_{i} a_{i} \mathbf{b}_{i} \cdot \mathbf{b}{j}, \ j=1, \cdots, n \ $$

 * }

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

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

$$ \Rightarrow \underbrace{ \left [ ( \mathbf{b}_{i} \cdot \mathbf{b}_{j}) \right ]_{ \color{red} n \times n}}_{ \color{blue} Gram \ matrix \ \Gamma} \left \{ a_{i} \right \}_{ \color{red} n \times 1}= \left \{ ( \mathbf{v} \cdot \mathbf{b}_{j}) \right \}_{ \color{red} n \times 1} \ $$

 * }

Theorem: $$ \left \{ \mathbf{b}_{1}, \cdots, \mathbf{b}_{n} \right \} \ $$ linear independent iff det $$ \Gamma \neq 0 \ $$ End Theorem

[[media: Egm6321.f09.mtg32.djvu| page32-3]]

[[media: Egm6321.f09.mtg32.djvu| (4) page32-1:]] functions

Consider inner (scalar) products:

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

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

$$g,h \ $$ 2 functions on $$ \left [ a,b \right ] \ $$

 * }

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

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

$$ = \int_{a}^{b}g(x)h(x)dx \ $$

(1)
 *
 * }

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

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

$$ f( \theta)= \sum_{n=0}^{ \infty} A_{n} P_{n} ( \sin \theta) \ $$

(2)
 *
 * }

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

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

$$ = \sum_{n { \color{red}=0}}^{ \color{red} \infty} A_{n} < P_{n}, P_{m}> \ $$

(3)
 *
 * }

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

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

$$ \underbrace{ \left [ < P_{n}, P_{m}> \right ]_{ \color{red} \infty \times \infty}}_{ \color{blue} \Gamma ( P_{0}, \cdots)} \left \{ A_{n} \right \}_{ \color{red} \infty \times 1}= \left \{  \right \}_{ \color{red} \infty \times 1} \ $$

(4)
 *
 * }


text:
Engineering Projects/Engines

Project Engines
The goal is to create model engines with moving parts made out of easily found materials or 3D printer parts. Stirling engines can run off a candle flame. Steam, Rotary, and Quasiturbine engines can run off compressed air. Gas (Otto), Diesel and Jet engines are best turned with an electric motor. Single piston engines like a lawnmower can be taken apart and displayed. Odd motors like the ballbearing motor are dangerous and need more work on the safety precautions than the actual motor.

Engine conversion is illegal, but some are advocating flex-fuel conversions that would enable a competition among different types of fuel to power our cars.

Tutorials
old documentation

title: Engineering Projects/Engines


text:
User:WatsonS

I am a 2nd year student studing Sports Management at the University of Canberra.

/Women in Sport/

title: User:WatsonS


text:
User:WatsonS/Women in Sport

Presentation slides

Introduction
Women in Sport in an ongoing topic of interest, and over the past decade we have witness women being a part of our elite sporting teams as coaches, trainers, managers and physiotherapist. However we are now finding women taking control and leading many of our countries government sporting businesses (Stell, 1991). This journal piece will explore many aspects within the Women in Sport status. It will address and give detail to the history of women in sport. We will show the development and the current success of Women in Australia and international sport, not only on field athletes but those in successful organisations. Finally it will elucidate two of the political issues involved in women in sport.

History
Women’s participation in sport has a long history. It is a history marked by division and discrimination but also one filled with major accomplishments by female athletes and important advances for gender equality and the empowerment of women and girls (Affairs, 2007).

Between the 18th and 19th century women’s sport was practically non- existent (Rostkowska, 2007). The only form of physical activity that women in this time experienced was those identified as leisure. Such activities as marbles, skipping and puzzles were the most common recreational activities that women were involved in (Stell, 1991). In the mid 18 hundreds swimming soon became popular and women were draw to this new and exciting idea, and in those days swimming pools were known as baths. In every ‘bath’ around the country women and men were allocated different swimming times and as expected women did not receive equal access. In addition to this, women were still bound by the fashion of the day, which stipulated neck to knee woollen costumes (Commission A. S.). As time passed more and more women learnt to swim, and in 1881 swimming races were organised in the hope that society would accept the idea of sporting events for women (Stell, 1991). In 1912 women’s swimming had a major boost when Fanny Durack and Mina Wylie took out the 100m freestyle gold and silver medals respectively at the Stockholm Olympics. This was the first ever swimming event open to women at an Olympic Games (Commission A. S.). It was shown that in those days society’s belief was that a female participating in athletic behaviour was both ‘unladylike’ and potentially hazardous to their reproductive lives (Bower, 2007).

In the 1930 women and sport lobby groups began to spring up around the country. High on their agenda was the need for more women’s sports ground. Fanned by a new wave of confident and empowered women fresh from university, women’s sports began a new era, played, administered and promoted by women for women (Rostkowska, 2007).

The sports between 1950 and 1970 that grew for women were tennis, golf and squash. These were seen as the most ‘ladylike’ sports on offer. This was the era that saw the rise of innovative administrators like Heather McKay (Squash), Gertrude McLeod (golf) and Margaret Court (tennis) (Commission A. S.). A memorial moment came in 1968 when Enriquetta Basilio became the first female to ever light the Olympic cauldron. This was part of the 19th Olympic Games held in Mexico City, and was identified as a moment that changed history for ever (Rostkowska, 2007).

A new ear began in the 1980’s, especially for Australian athletes, because in 1984 the Australian Institute of Sport opened in Canberra. This enabled both male and female athletes to get the expert training needed to gain a podium finish at international events. It was the chance for women to further develop their skills alongside men (Commission A. S.). Since the opening of the AIS Australia has developed some of the greatest and most inspiring women athletes in the world.

Leadership for Women
Women have made tremendous strides, entering the world of professional sports as both participants and front office leaders. However, that does not mean women are equally represented in these front office staff. Over the years women have been participating in sport in greater numbers and have a greater interest in sport. These increased opportunities to participate will hopefully translate into increased opportunities to lead professional sport organizations (Bower, 2007).

Do you recognise these names – Anita DeFrantz, Babe Didrickson, Billie Jean King and Pat Summit? Well these women have emerged as not only great sporting athletes but as advocates, coaches and top executives in sport throughout the United States. Although women such as these are now at the forefront of the sport industry, this has not always been the case. Historically in all countries women had been excluded from educational opportunities, sport opportunities, and career opportunities. In Germany for example 34% of leadership roles are held by women and in Norway less than 20% of leading positions are held by women (Bower, 2007). Through the years many laws and the general acceptance from society have provided a growth for the equal opportunities for women in all areas in the sporting industry. Although the playing field in not yet level, significant progress has been made. More female athletes compete today than ever before. Women are accepting leadership roles in all facets of the sport industry as athletic directors, coaches, CEO’s, general managers, athletics trainers, and journalists. Although the number of women holding top leadership positions in sport is underrepresented, substantial and noteworthy progress is in the works. Furthermore, the progress made by women in sport leadership creates visible role models, enables women’s voices, and opens up opportunities for women to have a greater impact and influence in the sporting industry as well as in society (Bower, 2007).

So where do female leaders come from? Are they born or made? One train of thought is that leaders are made, and one path people may use to home their leadership skills is through sport. Teamwork, winning and losing, social skills and responsibilities are lessons learned by playing sport (Stewart, 2004). There are obviously many qualifications that you can gain from educational experiences however research has shown that 85% of the qualifications needed to become a successful leader come from participating in sport. Such skills include, creating an understanding of teamwork and loyalty, builds confidence, teaches you how to fail and recover, builds competitive spirit and builds courage (Bower, 2007). These skills are found in both male and female athletes, so why do males still have the upper edge in our sporting industry?

A number of initiatives are being implemented to give women the chance to be leaders, improve their confidence, increase their self-awareness and strengthen their capacities in terms of decision-making, critical thinking and negotiating. One such project is from the Australian Sports Commission who have set up a program title ‘the sport leadership grants’. These grants were set up to provide financial support for women who endeavour to get the extra skills and education needed to gain a leading role in the sporting industry (Commission). They are offered to women, who have been, or still are involved in sport and who wish to push through and start a career in the industry (Commission A. S.).

Political Issues
BODY IMAGE –

Television, magazines, movies, newspapers, billboards, and the internet are all powerful creators of the 'desirable' woman. They reinforce and perpetuate stereotypes with which we compare the reality of our own body. Similarities between the 'ideal' and our own bodies are commended and differences are considered unattractive (Commision, 2008/2009). Visual images of the 'ideal' woman are used to sell everything from cars to ice cream and sport is no different. This commoditisation of the female body can lead women to identify their physical appearance as a type of currency – personal worth or value as measured by body type (Commision, 2008/2009). When we see female athletes in the media, almost 45% of them are portrayed in a sexual way (Commision, 2008/2009). Why? For the past decade female athletes both domestically and internationally have been portrayed in the media through there looks and body image, of cause the main focus of the newspaper articles and interviews are on their true athletic ability and success, however it is very rare that the media only focuses on this part (Commision, 2008/2009). Let use Stephanie Rice as an example. We all know she is an incredible swimmer for Australia and having won 3 Olympic gold medals she has found herself surrounded and confronted by media all the time. However all the promotion and media attention she has done has all been subjected around her looks and personal appearance (Stewart, 2004). You wonder if Steph was less attractive would she have gotten the amount of attention that she did. This is an issue evident in sport as a whole industry, however more women than men are likely to be targeted for this type of promotion. Women are offered large amounts of money to ‘bare all’ and promote a particular brand. This idea of marketing women in a sexual way is becoming part of the future and is almost expected to happen with all ‘attractive’ female athletes (Affairs, 2007).

MEDIA –

Consistent and positive media coverage is one of the prime goals of all sports administrators and players. The reason for this is simple. Sport and the media are two of the most powerful influences that affect how society works. They are also intertwined in that both affect how people think and shape ideas and emotions. Consistent media coverage can benefit a sport in a number of ways. It can provide a visual profile, create positive role models and, by increasing spectator appeal, help attract lucrative sponsorship opportunities for the sport. How the media portrays a particular sport or athlete can also impact on both the sport and/or the athlete’s credibility.

As women what do we want from the media? It is simple, a level playing field and equal opportunities. In 2008 the coverage of women’s sport was only 9% of all sport coverage in Australia. While male sports took out 81% of all coverage. In the same study it was shown that the only sport that women received equal media attention was tennis (Commision, 2008/2009). Let’s take an international look from the 2008 Olympics, in rowing, US women won two medals to the one for US men, yet received less than one third as much media coverage. Also in the US the NBC broadcasting program aired nearly 24 minutes of men’s weightlifting, although no US men gained a medal. By comparison, two US women weightlifters won a gold and a bronze medal, but their success warranted no prime time coverage (Commision, 2008/2009). A big reason for the lack of women’s sport in the media is money. The financial funding that sports get from the government go towards training equipment, uniforms, tours and many other aspect. To get the media to promote and broadcast their sport would result in losing out on other essential needs of the sport (Stewart, 2004). Another reason is the popularity; if a television program air’s a female sporting game/event will it get the enough hits to warrant that time slot? Free to air and foxtel programs don’t want to experiment on viewing unknown women’s sports, as this would result in loss of money and time. So how can we help, or is anything already being down? A program called womensport is a non government organisation, which are representing women and helping to improve the media coverage of women. They have initiated research that also helps further build the picture of media coverage of women’s sport. Their focus is not only aimed at National and International events, they also want to push for local and state competitive events to get the attention they deserve. This is a very slow and extensive process, however to see campaigns and programs like this who are out there giving women the opportunities they deserve is the real key (Affairs, 2007).

Conclusion
As we have shown, Women in sport is a very broad and in depth topic within the business politics and sport subject. There are many areas across the subject that can be addressed. The particular sections that have been spoken about throughout this essay are that of the history, the development of women in leadership roles and we also addressed two of the most identified political issues surrounding women in sport, body image and the media. The lesson we have learnt through this essay is that the right for equal opportunities for women is still an uphill battle and even tho there are programs, campaigns and support out there for women, the gap between male and females in the sporting industry will always be there. The important fact is that we are aware of the issue and knowing this is a great start to help aid and improve the women the in sport status.

title: User:WatsonS/Women in Sport


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Motivation and emotion/Book/2011/Volunteerism

Overview
The answer to the question of what motivates people to volunteer their time and expertise in the community is a simple one – they want to help, make a difference, find happiness and fulfilment, and make others happy. What motivates volunteerism is that simple and so much more complex. This book chapter explores the theories, motivations, implications and some of the practical applications of volunteerism. It is also intended to assist readers to identify what personally motivates them to volunteer, or make the decision to volunteer and to consider whether volunteerism is likely to promote happiness.

http://www.youtube.com/watch?v=9hbVuiU9Zzw/3000 Dedicated Volunteers | Wesley Mission, Sydney - YouTube] (Enrich and empower another life, meet new and interesting people, learn new skills, explore career development and discover training opportunities through volunteering, 6:19mins)

Volunteerism defined
Definitions of volunteerism tend to be multifaceted, focussing on the long-term helping of others without material rewards and of free will, without coercion or financial payment, in a formal setting (Cnaan, Handy & Wadsworth, 1996; Fiorillo, 2011; Haski-Leventhal, 2009; Volunteering Australia). Rather than being a spontaneous act, such as in the immediate response to an emergency, Clary et al. (1998) describes volunteerism as being a planned form of helping. In the ‘planned helping’ context, those interested in volunteering proactively search for opportunities to help others; they deliberate about the extent of their involvement in terms of time, costs, opportunity and which activities will be of benefit to the volunteer’s personal needs and those of the community.

What is a volunteer?

It is difficult to define what a volunteer comprises because volunteers are not a homogenous group. Volunteers can operate in a variety of organisations and roles, be of all ages, backgrounds, male or female and possess a wide range of interests, experiences and skills (Bussell & Forbes, 2002). A study by Cnaan, Handy and Wadsworth (1996) found that people tend to define volunteers in four dimensions:

 * Free Will (free to choose to be a volunteer);
 * Remuneration (none at all);
 * Structure (formal); and
 * Intended Beneficiaries (benefit or help others or strangers).

The study also found that public perception of ‘what is a volunteer’ is frequently viewed in terms of the concept of ‘net cost’. That is, the higher the amount of work undertaken in terms of the four dimensions (cost), the more likely the person will be perceived as a volunteer. Those who are perceived as receiving rewards (benefits) are less likely to be thought of as a volunteer.

Demographics and motivations
Many studies have labelled the ‘typical volunteer’ as being female, middle-aged to older, well educated and with a higher income (Bussell & Forbes, 2002; Holmes, 2009; Shye, 2010; Warburton & Stirling, 2007). Warburton and Stirling (2007) indicated that these factors appear to be predictive of volunteering, explained by the fact that higher levels of education and income can produce greater opportunities to contribute. Gender differences in volunteering tend to differ, with some research indicating that more females than males volunteer (Bussell & Forbes, 2002), and others reporting no significant gender differences (Choi, 2003; Warburton & Stirling, 2007).

A study by Dolnicar and Randle (2007) studied volunteers’ demographic profiles alongside motivational patterns and identified six ‘segments’ of volunteering. The six segments were classified as follows: “classic volunteers” are older, very active volunteers who are involved to gain satisfaction and help others; “dedicated volunteers” who volunteer the most hours, over a number of volunteering organisations; “personally involved volunteers” who volunteer temporarily, for example in school activities as parental support; “Volunteers for personal satisfaction” are the least distinct segment; “altruists” who befriend and listen to people; and “niche volunteers” are young, highly educated, and enter volunteering through obligation, to gain work experience or honour religious beliefs. The identification of the six segments indicates that the generalised labelling of volunteers into specific ‘typical’ groups such as female, well educated with a high income is quite constrictive and may not present volunteers’ true demographics and motivations. Studies such as undertaken by Dolnicar and Randle (2007), reflect researchers’ and voluntary organisations’ considerable interest in obtaining a more complete picture of what volunteers ‘look like’ and what motivates them to volunteer. Such research has the potential to assist, for example, in effectively targeted marketing in the recruitment and retention of volunteers. Understanding volunteerism can also help individuals to more effectively identify the type of prosocial activities which will suit their tastes and give satisfaction. The following section considers motivation in more detail by discussing a few of the theories on volunteerism that have arisen over the past few decades. Before reading further, however, you may wish to reflect on your own demographic profile and what has motivated you to volunteer – or what could motivate you to become a volunteer.

For the latest facts and figures on volunteering in Australia, please click on the following link : http://www.volunteeringaustralia.org/html/s02_article/article_view.asp?id=4333&nav_cat_id=222&nav_top_id=50

 Reflect on your own demographic profile. What is your demographic profile for volunteering? Try writing a list for yourself, or use the following checklist, based on the Dolnicar & Randle, 2007 study

 * Classic Volunteer – aged 45 years or older, probably working reduced hours or retired. Wants to gain satisfaction and help others
 * Dedicated Volunteer – have time and knowledge or expertise available to volunteer very regularly, for a number of different organisations
 * Personally Involved Volunteer – available for short-term volunteering in a specific area of interest or need, for example helping out with your child’s school activities or your sporting or social group
 * Volunteer for Personal Satisfaction – your motivation to volunteer could include any of these six categories, or a combination
 * Altruistic Volunteer – helping others, for example through friendship, ‘being there’ for others, being a good listener, with no regard to personal gain
 * Niche Volunteer – a young motivated volunteer who wishes to gain experience in their chosen career, build on knowledge gained through education. Also includes mandatory volunteerism, for example as part of an educational course.

The above checklist is by no means a comprehensive or complete list of demographics and motivations for volunteerism. However, it is hoped that by reflecting at this point in the chapter, you will begin to form a framework of who you are or could be as a volunteer. The following section on theories will further assist in developing that framework.

Theories and contexts - What motivates people to volunteer?
The 1970s saw the emergence of an increasing number of studies into volunteerism and motivation (Okun, Barr, & Herzog, 1998), along with corresponding theories on the subject. Generally, these theories have revolved around ‘models’ which attribute one or more ‘factors’ to volunteerism and motivation. For example, volunteerism has been described as a single factor model where individuals are motivated to volunteer by a single element which is meaningful for them, such as a rewarding experience (Okun, et al., 1998). The ‘rewarding experience’ may consist of a number of motives which in combination make up this single factor.

In the 1980s several researchers found that people volunteer for more than one reason. For example, Frisch and Gerrard (1981) undertook a study involving Red Cross volunteers which reinforced the concept of a two-factor model of volunteerism. They found that people were motivated to volunteer by altruistic motives, such as a concern for others or egoistic motives involving the desire to satisfy their own needs. In this study, altruistic and egoistic motives were thought to be distinct elements of volunteer motivation. There has, however, been some debate about these factors (eg. Dolnicar & Randle, 2007; Haski-Leventhal, 2009) in the context of more recent theoretical models. Although much of the early research into volunteerism was not derived within a theoretical framework, it did cast a new perspective on motivations and by the beginning of 1990, a more complex model was evolving. Omoto and Snyder (1990) began by studying social and personality aspects of HIV/AIDS volunteers. The multifactor model of volunteerism began to develop when it was observed that there were several distinct motives for volunteering (Okun, et al., 1998).

Multifactor model - Functional theory of motivation
A key component of the function theory of motivation is that people will engage in the same type of volunteer work for different reasons. The functional theory is the result of analyses of empirical studies on the functions served by volunteerism, from which Clary et al. (1998) identified six personal and social functions that relate to volunteering.

The Six Functions of Volunteerism

Clary and Snyder (1999) found that individuals’ multiple motives for volunteerism arise from their individual experiences, needs and goals, and that motives may change over time. This may help explain observations by Chacon, Vecina and Davila (2007) that the functions contained in the functional theory are most useful in the early stages of volunteering, prior to the development of an individual’s identity as a volunteer and while commitment is still relatively low.

Role Identity Theory
The role identity model helps to explain what motivates sustained volunteering (Chacon, Vecina & Davila, 2007). Role identity theory suggests that an individual’s self-concept is organised into a series of role identities that are consistent with roles undertaken in that individual’s social structure, eg. spouse, activist, employee, volunteer (Charng, Piliavin & Callero, 1988). These multiple role identities develop through on-going external social interactions and the perceived expectations of others, for example, parental or societal expectations may provide the impetus to volunteer (Finkelstein, Penner & Brannick, 2005).

According to Finkelstein and Brannick (2007), the more an individual participates in a role, such as that of a volunteer, the more that role is adopted as part of the self, or ‘role identity’ and the greater the motivation to sustain that role. An example of role identity theory was found in a study by Charng, Piliavin and Callero (1988) where blood donors began to change their attitudes toward donation from a ‘habit’ to a ‘career’. Blood donors began to label themselves as “regular blood donors” after three or four donations and became sensitive to any implication that their donation was due to social pressure, rather than a voluntary act. Interestingly, research has also shown that, in addition to motivating sustained volunteering, there is a correlation between the strength of a person’s role identity and the quantity of voluntary work undertaken (Grube & Piliavin, 2000).

Social capital theory
The social capital theory hypothesises that people are motivated to volunteer because it creates social capital and social networks and thus brings economic, resource and social benefits to a community (Holmes, 2009; Warburton & Stirling, 2007). According to Holmes (2009), volunteering creates good citizens because it builds trust in other people and organisations; enables people to socialise and become involved in community issues and allows people to learn about local social problems and needs. Social capital is frequently given a central role in volunteerism because without the social networks and trusting collaboration between individuals and groups, volunteering would be a solitary individual activity (Clary & Snyder, 2002; Warburton & Stirling, 2007). Factors associated with social capital theory include:

 * Religious affiliation – people with a religious affiliation are motivated to volunteer because of the community and social capital aspects of membership (Evans & Kelley, 2000);
 * People from culturally and linguistically diverse backgrounds may prefer to volunteer informally within their communities, assisting with social integration and the building of trust within their community, as well as bringing with them their own cultural network connections (Warburton & Stirling, 2007).

According to Holmes (2009) evidence suggests that many community volunteer programs attract like-minded groups of people, which leads to the fulfilment of the primary reason given for being a volunteer in Australia ‘helping others and/or the community’ (ABS 2010).

Systematic quality of life theory
The Systematic Quality of Life (SQOL) theory adopts a new approach to volunteerism. The SQOL theory offers a set of possible benefits and rewards for volunteering by inquiring into how volunteering enhances the volunteer’s own quality of life (Shye, 2010). The approach taken by Shye (2010) differs considerably from the majority of other theoretical models which tend to be primarily altruistically or intrinsically oriented. The study reverses the usual order of investigation into volunteer motivations, asking what needs are perceived to be satisfied by volunteering, rather than what motivates individuals to volunteer.

The Shye (2010) study which asked volunteers and non-volunteers what are the benefits of volunteering found that, for the general population, the opportunity to develop friendships, gaining a sense of belonging in the community and expressing one’s personality and beliefs were the primary motivations for volunteering. The very recent nature of the SQOL theory means that there are few studies available to directly refute or support its primarily egoistic view of volunteering. A recent study by Finkelstein (2010), did however, confirm that individualists and collectivists differ in why they choose to volunteer. Finkelstein suggested that, in-line with the Shye (2010) findings, individualists tend to be more self-focussed and use volunteering as a means to meet career aspirations and fulfilment of other personal needs, whereas collectivism is associated with other-oriented motives. A study exploring extrinsic and intrinsic motivational orientation (Finkelstein, 2009) also linked extrinsic orientation with ‘external’ motives such as fulfilling career aspirations, and intrinsic orientation with ‘internal’ motives which were satisfied by the volunteer activity itself.

In defence of the SQOL theory, Shye suggests that one’s self-interest is an intrinsic element of being entangled in a social and cultural network. This suggestion is supported by the findings in the Finkelstein (2009; 2010) studies and is the essence of all the theories presented in this chapter on volunteerism because it acknowledges the multi-facted and complex composition of volunteerism and its motivations.

It is interesting to note the progression from the early ‘single and two-factor’ models of volunteerism, which tended to make the assumption that volunteering was principally motivated by altruistic motives. More recent theories have become bolder by exploring egoistic motives as well as the consideration of personality orientation, social and cultural factors. The knowledge gained from these theoretical models also helps to facilitate mutually satisfying relationships between volunteers and volunteer organisations.

Volunteerism from an organisational perspective
It is widely acknowledged that volunteer work on a sustained basis is fundamental to the effective functioning of many organisations (Finkelstein et al., 2005). In fact, it is also recognized that formal volunteering makes a considerable social and economic contribution to society. The processes that motivate and sustain volunteerism are therefore of great interest to social psychologists (Finkelstein et al., 2005) and ultimately to organisations who utilise volunteers as part of their formal workforce. Research has found that elements of volunteerism such as initial motives and self-concept change and evolve as volunteering experience increases. Volunteers who feel their efforts are nurtured, accepted and valued will be motivated to remain with an organisation long-term (Omoto & Snyder, 1995). Chacon, Vecina and Davila (2007) reported that a combination of organisational dedication, volunteers’ satisfaction with the way in which their motives are fulfilled by the experience, and role identity are predictive of both short and long-term volunteer service. Overall, Finkelstein et al. (2005) suggests that volunteers’ sense of commitment is mediated by role identity, with longevity in volunteering being linked to the strength of role identity.

From the personal fulfilment perspective, it is equally important that prospective volunteers understand what qualities, goals and skills they wish to offer an organisation or volunteer group, as well as rewards they may hope to gain from the experience. On the basis of theoretical studies into the motivational aspects of volunteerism, it cannot be assumed that people volunteer for purely altruistic reasons. It is clear that volunteers expect various rewards from their volunteer work, such as the opportunity to learn, gain experience in a chosen career, building social capital in their community, develop friendships and a sense of belonging. The receipt of these rewards, along with a strong self-concept is important to a volunteer’s decision to remain in their volunteer roles for long periods of time.

Quiz
Altruism versus Egoism

Indicate which of the following describe your motives for volunteerism. Are your motivations altruistic, egoistic, or a mix of both? (There are no right or wrong answers)

{Altruistic? + I volunteer out of my concern for others - I am happy to listen to others - I use my skills and experience to benefit those who need my help
 * type=""}

{Egoistic? + I volunteer because it makes me feel like I belong to a community - I volunteer to gain experience for my new career - Volunteering makes me feel good about myself
 * type=""}

What is happiness?
According to Harris (2007), happiness comprises two very different definitions. Happiness can be linked to feelings or emotions such as “a sense of pleasure, gladness or gratification” (Harris, 2007, p. 15), as well as being “characterised by experiences of contentment, satisfaction, love or joy” (Grant & Leigh, 2011, p. 4). The other part of happiness is “a rich, full and meaningful life” (Harris, 2007, p. 15). Harris goes on to describe living a meaningful life as offering a “profound sense of a life well lived” – which is not fleeting but long lasting. The meaningful life is about feeling that you are “purposefully involved in life activities such as work, relationships and recreational activities” (Grant & Leigh, 2011, p. 5).

If you consider “What is happiness?” against each of the theories of volunteerism discussed in this chapter, you could probably come up with an extensive list which highlights the joyful and meaningful aspects of volunteering. Here are just a few examples of how volunteerism may promote happiness:
 * Gain satisfaction, joy, pleasure from helping others;
 * Volunteering provides a sense of pride and identity;
 * The enjoyment of being able to help kids in their school and other activities;
 * Experience joy in the ability to share knowledge and skills with others;
 * Develop a sense of purpose through voluntary contributions to civic life ie. volunteering builds a better community;
 * Volunteering enables individuals to learn more about social problems and issues;
 * Volunteering helps build career skills and experience;
 * Being a volunteer can help take focus off your own troubles and focus on others;
 * Volunteering can nurture long-last friendships and relationships

You may also wish to reflect on your responses to this chapter’s Motivation Exercises, to help build your own framework of who you are or could be as a volunteer. In many situations the mutual benefits of volunteerism are most easily explained by those at the ‘coal face’. Follow this link http://www.youtube.com/watch?v=ONcpcMFOsjc to hear a touching story of volunteerism – told by Australian volunteers caring for young abandoned Chinese children, through International China Concern (ICC).

Does volunteerism promote happiness? There is a considerable amount of research which suggests that volunteering has the potential to promote happiness in ourselves and others. The ultimate decision is a personal one.