text:
User:Egm6321.f10.team3.Sudheesh/Homeworks/Homework3

= Problem 1 - Motion of a Projectile =

Given
Figure shows the Trajectory of a projectile (ex:Rocket):

Find

 * $$\left( 1 \right)$$ Drive Equation of Motion (EOM)
 * $$\left( 2 \right)$$ Particular case $$\displaystyle k = 0 $$ : Verify $$y\left( x \right)$$ is parabolla
 * $$\left( 3 \right)$$ Consider $$k\ne 0$$, $$\displaystyle {{v}_{x}}=0$$,

 * $$ (3.1) $$ Find $${{v}_{y}}\left( t \right)$$, $$\displaystyle y(t)$$ for $$ \displaystyle m$$ = constant
 * $$ (3.2) $$ Find $${{v}_{y}}\left( t \right)$$, $$\displaystyle y(t)$$ if $$m=m\left( t \right)$$

Solution
Part 1 Consider the trajectory of a projectile (ex. Rocket)

 * Various forces acting on the projectile at time 't' are:

 * 1) Weight of the projectile

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

 * $$\displaystyle W=mg$$
 * }
 * }
 * }

 * 2) Inertia force

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

 * $${{F}_{a}}=ma=m\cdot \frac{dv}{dt}$$ for particle with constant mass
 * $${{F}_{a}}=ma=m\cdot \frac{dv}{dt}$$ for particle with constant mass

 * }
 * }

 * 3) Air resistance which is proportional to the velocity of particle

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

 * $$\displaystyle {{F}_{D}}=k{{v}^{n}}$$
 * }
 * }
 * }

Now consider the force equilibrium in both horizontal and vertical direction

 * a) Force Equilibrium in horizontal direction:

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

 * $$\sum{{{F}_{H}}=0}$$
 * }
 * }
 * }

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

 * $$m\cdot \frac{d{{v}_{x}}}{dt}+k{{v}^{n}}\cos \alpha =0$$,
 * $$m\cdot \frac{d{{v}_{x}}}{dt}+k{{v}^{n}}\cos \alpha =0$$,

 * where $$\displaystyle {{v}_{x}}\to $$ horizontal component of velocity

 * }
 * }

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

$$m\cdot \frac{d{{v}_{x}}}{dt}=-k{{v}^{n}}\cos \alpha $$ $$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * $$\displaystyle (Eq. 1.1)
 * }
 * }

 * b) Force Equilibrium in the vertical direction:

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

 * $$\sum{{{F}_{V}}=0}$$
 * }
 * }
 * }

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

$$m\cdot \frac{d{{v}_{y}}}{dt}+k{{v}^{n}}\sin \alpha +mg=0$$,

 * where $$\displaystyle {{v}_{y}}\to $$ vertical component of velocity

 * }
 * }

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

$$m\cdot \frac{d{{v}_{y}}}{dt}=-k{{v}^{n}}\sin \alpha -mg$$ $$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * $$\displaystyle (Eq. 1.2)
 * }
 * }

Part 2

 * Particular case: When $$\displaystyle k=0$$

Eq.1.1 reduces to

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

 * $$m\frac{d{{v}_{x}}}{dt}=0\Rightarrow \frac{d{{v}_{x}}}{dt}=0$$
 * }
 * }
 * }

Integrating the above equation gives:

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

$$
 * $${{v}_{x}}\left( t \right)={{c}_{1}}$$
 * $$\displaystyle (Eq. 1.3)
 * $$\displaystyle (Eq. 1.3)
 * }
 * }

Apply 'initial condition' to determine integration constant,$$\displaystyle{{c}_{1}}$$

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

 * $${{v}_{x}}\left( t=0 \right)={{v}_}={{c}_{1}}$$
 * }
 * }
 * }

Now Eq.1.3 becomes:
 * {| style="width:100%" border="0"

 * $${{v}_{x}}\left( t \right)={{v}_}$$
 * }
 * }
 * }

Integrate the above equation to obtain ,$$\displaystyle x$$

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

 * $$x\left( t \right)={{v}_}t+{{c}_{2}}$$
 * }
 * }
 * }

Then 'Initial condition' is applied to determine ,$$\displaystyle{{c}_{2}}$$

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

 * $$x\left( t=0 \right)={{v}_}\times 0+{{c}_{2}}\Rightarrow {{c}_{2}}={{x}_{0}}$$
 * }
 * }
 * }

Therefore,

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

 * $$x\left( t \right)={{v}_}t+{{x}_{0}}$$
 * }
 * }
 * }

Now $$\displaystyle t$$, can be expressed in terms of $$\displaystyle x$$,

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

$$
 * $$t=\frac{x-{{x}_{0}}}$$
 * $$\displaystyle (Eq. 1.4)
 * $$\displaystyle (Eq. 1.4)
 * }
 * }

Similarly When $$\displaystyle k=0$$, Eq.1.2 ruduces to

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

 * $$m\frac{d{{v}_{y}}}{dt}=-mg\Rightarrow \frac{d{{v}_{y}}}{dt}=-g$$
 * }
 * }
 * }

Integrate the above equation to evaluate, $$\displaystyle {{v}_{y}}$$

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

$$
 * $${{v}_{y}}\left( t \right)=-gt+{{c}_{3}}$$
 * $$\displaystyle (Eq. 1.5)
 * $$\displaystyle (Eq. 1.5)
 * }
 * }

Apply 'initial condition' to obtain $$\displaystyle {{c}_{3}}$$

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

 * $${{v}_{y}}\left( t=0 \right)={{v}_}={{c}_{3}}$$
 * }
 * }
 * }

Now Eq.1.5 becomes,

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

 * $${{v}_{t}}\left( t \right)=-gt+{{v}_}$$
 * }
 * }
 * }

Then integrate the above equation to determine, $$\displaystyle y $$

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

 * $$y\left( t \right)=\frac{-g{{t}^{2}}}{2}+{{v}_}t+{{c}_{4}}$$
 * }
 * }
 * }

 * $${{c}_{4}}$$ is determined using 'initial condition' as:

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

 * $$y\left( t=0 \right)=\frac{-g\times 0}{2}+{{v}_}\times 0+{{c}_{4}}\Rightarrow {{c}_{4}}={{y}_{0}}$$
 * }
 * }
 * }

Therefore,

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

 * $$y\left( t \right)=\frac{-g{{t}^{2}}}{2}+{{v}_}t+{{y}_{0}}$$
 * }
 * }
 * }

Now Substitute Eq.1.4 for $$\displaystyle t$$ in the above equation;

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

$$y\left( x \right)=-\frac{g}{2}{{\left( \frac{x-{{x}_{0}}} \right)}^{2}}+{{v}_}\left( \frac{x-{{x}_{0}}} \right)+{{y}_{0}}$$ $$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * $$\displaystyle (Eq. 1.6)
 * }
 * }

Eq.1.6 is in the form of a parabollic equation. Therefore $$\displaystyle y\left( x \right)$$ is parabolla.

Part 3

= Problem 2- Equation of Motion of Pendulums Connected by a Spring =

Given
Shown in figure is the two Pendulums connected by a spring:

Find

 * $$\left( 1 \right)$$ Derive equation of motion:

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

$$
 * $${{m}_{1}}{{l}^{2}}{{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{1}}=-k{{a}^{2}}\left( {{\theta }_{1}}-{{\theta }_{2}} \right)-{{m}_{1}}gl{{\theta }_{1}}+{{u}_{1}}l$$
 * $$\displaystyle (Eq. 2.1)
 *
$$\displaystyle (Eq. 2.1)

 * }
 * }

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

$$
 * $${{m}_{2}}{{l}^{2}}{{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{2}}=-k{{a}^{2}}\left( {{\theta }_{2}}-\theta 1 \right)-{{m}_{2}}gl{{\theta }_{2}}+{{u}_{2}}l$$
 *
$$\displaystyle (Eq. 2.2)
 *
$$\displaystyle (Eq. 2.2)
 * }
 * }

 * $$\left( 2 \right)$$ Write Eq.2.1 and Eq.2.2 in the form of [[media:2010_09_21_14_56_48.djvu|Mtg 13 (c),page2 ]], of:

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

$$
 * $$\underset{-}{\overset{\centerdot }{\mathop{x}}}\,=\underset{-}{\mathop{A}}\,\left( t \right)\underset{-}{\mathop{x}}\,\left( t \right)+\underset{-}{\mathop{B}}\,\left( t \right)\underset{-}{\mathop{u}}\,\left( t \right)$$
 *
$$\displaystyle (Eq. 2.3)
 *
$$\displaystyle (Eq. 2.3)
 * }
 * }

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

$$\underset{-}{\mathop{x}}\,={{\left[ \begin{matrix} {{\theta }_{1}} & \overset{\centerdot }{\mathop{{{\theta }_{1}}}}\, & {{\theta }_{2}} & \overset{\centerdot }{\mathop{{{\theta }_{2}}}}\, \\ \end{matrix} \right]}^{T}}$$ and $$\underset{-}{\mathop{u}}\,={{\left[ \begin{matrix} {{u}_{1}}l & {{u}_{2}}l \\ \end{matrix} \right]}^{T}}$$
 * Given
 * Given
 * }
 * }

Solution

 * $$\left( 1 \right)$$ Derive equation of motion:
 * (a) Consider Free Body Diagram of left pendulum:

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

 * $$\displaystyle l\sin {{\theta }_{1}}=l{{\theta }_{1}}$$ and $$\displaystyle l\cos {{\theta }_{1}}=l$$
 * }
 * }
 * }

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

 * $$acceleration, {{a}_{1}}=\frac{{{d}^{2}}\left( l{{\theta }_{1}} \right)}{d{{t}^{2}}}=l{{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{1}}$$
 * }
 * }
 * }

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

 * $$ Inertia force ={{m}_{1}}l{{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{1}}$$
 * }
 * }
 * }

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

 * $$\displaystyle Spring force =ka{{\theta }_{1}}-ka{{\theta }_{2}}=ka({{\theta }_{1}}-{{\theta }_{1}})$$
 * }
 * }
 * }

Now using D'Alembert's_principle, sum of the moments about pivot(A)is equal to zero

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

 * $$\sum{{{M}_{A}}=0\Rightarrow }\left( {{m}_{1}}l{{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{1}} \right)\cdot l+\left( ka({{\theta }_{1}}-{{\theta }_{2}}) \right)\cdot a+\left( {{m}_{1}}g \right)\cdot l{{\theta }_{1}}-\left( {{u}_{1}} \right)\cdot l=0$$
 * }
 * }
 * }

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

$$\Rightarrow {{m}_{1}}{{l}^{2}}{{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{1}}=-k{{a}^{2}}({{\theta }_{1}}-{{\theta }_{2}})-{{m}_{1}}gl{{\theta }_{1}}+{{u}_{1}}l$$ $$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 *
$$\displaystyle (Eq. 2.1)
 * }
 * }

 * (b) Consider Free Body Diagram of right pendulum:

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

 * $$\displaystyle l\sin {{\theta }_{2}}=l{{\theta }_{2}}$$ and $$\displaystyle l\cos {{\theta }_{2}}=l$$
 * }
 * }
 * }

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

 * $$acceleration, {{a}_{2}}=\frac{{{d}^{2}}\left( l{{\theta }_{2}} \right)}{d{{t}^{2}}}=l{{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{2}}$$
 * }
 * }
 * }

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

 * $$ Inertia force ={{m}_{2}}l{{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{2}}$$
 * }
 * }
 * }

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

 * $$\displaystyle Spring force =ka{{\theta }_{2}}-ka{{\theta }_{1}}=ka({{\theta }_{2}}-{{\theta }_{1}})$$
 * }
 * }
 * }

Using D'Alembert's_principle, sum of the moments about pivot(B)is equal to zero

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

 * $$\sum{{{M}_{B}}=0\Rightarrow }\left( {{m}_{2}}l{{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{2}} \right)\cdot l+\left( ka({{\theta }_{2}}-{{\theta }_{1}}) \right)\cdot a+\left( {{m}_{2}}g \right)\cdot l{{\theta }_{2}}-\left( {{u}_{2}} \right)\cdot l=0$$
 * }
 * }
 * }

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

$$\Rightarrow {{m}_{2}}{{l}^{2}}{{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{2}}=-k{{a}^{2}}({{\theta }_{2}}-{{\theta }_{1}})-{{m}_{2}}gl{{\theta }_{2}}+{{u}_{2}}l$$ $$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 *
$$\displaystyle (Eq. 2.2)
 * }
 * }

 * $$\left( 2 \right)$$ Write Eq.2.1 and Eq.2.2 in the form of Eq.2.3(system of coupled equation):

 * Eq.2.1 can be rearranged as,

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

$$
 * $${{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{1}}=\frac{-\left( k{{a}^{2}}+{{m}_{1}}gl \right)}{m{}_{1}{{l}^{2}}}{{\theta }_{1}}+\frac{k{{a}^{2}}}{m{}_{1}{{l}^{2}}}{{\theta }_{2}}+\frac{{{u}_{1}}l}{m{}_{1}{{l}^{2}}}$$
 *
$$\displaystyle (Eq. 2.4)
 *
$$\displaystyle (Eq. 2.4)
 * }
 * }

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

$$
 * $${{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{2}}=\frac{k{{a}^{2}}}{m{}_{2}{{l}^{2}}}{{\theta }_{1}}+\frac{-\left( k{{a}^{2}}+{{m}_{2}}gl \right)}{m{}_{2}{{l}^{2}}}{{\theta }_{2}}+\frac{{{u}_{2}}l}{m{}_{2}{{l}^{2}}}$$
 *
$$\displaystyle (Eq. 2.5)
 *
$$\displaystyle (Eq. 2.5)
 * }
 * Now Eq.2.4 and Eq.2.5 can be put in the form of Eq.2.3 as:
 * Now Eq.2.4 and Eq.2.5 can be put in the form of Eq.2.3 as:

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

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

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

$$
 *
$$\displaystyle (Eq. 2.6)
 * }
 * }

 * Where:

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

0 & 1 & 0 & 0 \\   \frac{-\left( k{{a}^{2}}+{{m}_{1}}gl \right)}{m{}_{1}{{l}^{2}}} & 0 & \frac{k{{a}^{2}}}{m{}_{1}{{l}^{2}}} & 0  \\ 0 & 0 & 0 & 1 \\   \frac{k{{a}^{2}}}{m{}_{2}{{l}^{2}}} & 0 & \frac{-\left( k{{a}^{2}}+{{m}_{2}}gl \right)}{m{}_{2}{{l}^{2}}} & 0  \\ \end{matrix} \right]$$
 * $$\underset{-}{\mathop{A}}\,=\left[ \begin{matrix}
 * $$\underset{-}{\mathop{A}}\,=\left[ \begin{matrix}

 * }
 * }

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

0 & 0 \\   \frac{1}{m{}_{1}{{l}^{2}}} & 0  \\ 0 & 0 \\   0 & \frac{1}{m{}_{2}{{l}^{2}}}  \\ \end{matrix} \right]$$
 * $$\underset{-}{\mathop{B}}\,=\left[ \begin{matrix}
 * $$\underset{-}{\mathop{B}}\,=\left[ \begin{matrix}
 * }
 * }

= Problem 7 - Roll Control of Rocket By Actuating Ailerons =

Given

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

$$
 * $$\overset{\centerdot }{\mathop{\phi }}\,=\omega $$
 *
$$\displaystyle (Eq. 7.1)
 *
$$\displaystyle (Eq. 7.1)
 * }
 * }

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

$$
 * $$\overset{\centerdot }{\mathop{\omega }}\,=\frac{-1}{\tau }\omega +\frac{Q}{\tau }\delta $$
 *
$$\displaystyle (Eq. 7.2)
 *
$$\displaystyle (Eq. 7.2)
 * }
 * }

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

$$
 * $$\overset{\centerdot }{\mathop{\delta }}\,=u$$
 *
$$\displaystyle (Eq. 7.3)
 *
$$\displaystyle (Eq. 7.3)
 * }
 * }

Find
Control behavior in form of

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

$$
 * $$\overset{\centerdot }{\mathop{x\left( t \right)=\underset{-}{\mathop{A}}\,}}\,\underset{-}{\mathop{x}}\,\left( t \right)+\underset{-}{\mathop{B}}\,\underset{-}{\mathop{u}}\,\left( t \right)$$
 *
$$\displaystyle (Eq. 7.4)
 *
$$\displaystyle (Eq. 7.4)

 * }
 * }

Solution

 * Integrating Eq.7.3 gives,

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

$$
 * $$\displaystyle \delta =ut$$
 *
$$\displaystyle (Eq. 7.5)
 *
$$\displaystyle (Eq. 7.5)
 * }
 * Substitute Eq.7.5 for $$\displaystyle \delta $$, in Eq.7.2:
 * Substitute Eq.7.5 for $$\displaystyle \delta $$, in Eq.7.2:

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

$$
 * $$\overset{\centerdot }{\mathop{\omega }}\,=\frac{-1}{\tau }\omega +\frac{Q}{\tau }ut$$
 *
$$\displaystyle (Eq. 7.6)
 *
$$\displaystyle (Eq. 7.6)
 * }
 * }

 * Now integrate Eq.7.6:

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

$$
 * $$\omega =\frac{-1}{\tau }\omega t+\frac{Q}{2\tau }u{{t}^{2}}$$
 *
$$\displaystyle (Eq. 7.7)
 *
$$\displaystyle (Eq. 7.7)
 * }
 * }

 * Similarly integrating Eq.7.1 gives,

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

$$
 * $$\displaystyle \phi =\omega t$$
 *
$$\displaystyle (Eq. 7.8)
 *
$$\displaystyle (Eq. 7.8)
 * }
 * }

 * Substitute Eq.7.8 in Eq.7.7 implies,

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

$$
 * $$\omega =\frac{-1}{\tau }\phi +\frac{Q{{t}^{2}}}{2\tau }u$$
 *
$$\displaystyle (Eq. 7.9)
 *
$$\displaystyle (Eq. 7.9)
 * }
 * }

 * From Eq.7.1 and Eq.7.9,

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

$$
 * $$\overset{\centerdot }{\mathop{\phi }}\,=\frac{-1}{\tau }\phi +\frac{Q{{t}^{2}}}{2\tau }u$$
 *
$$\displaystyle (Eq. 7.10)
 *
$$\displaystyle (Eq. 7.10)
 * }
 * }

 * Now, Eq.7.6 and Eq.7.10 can be written in the required form as,

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

$$\left[ \begin{matrix} \overset{\centerdot }{\mathop{\phi }}\, \\ \overset{\centerdot }{\mathop{\omega }}\, \\ \end{matrix} \right]=\left[ \begin{matrix} \frac{-1}{\tau } & 0 \\ 0 & \frac{-1}{\tau } \\ \end{matrix} \right]\left[ \begin{matrix} \phi  \\ \omega  \\ \end{matrix} \right]+\left[ \begin{matrix} \frac{Q{{t}^{2}}}{2\tau } \\ \frac{Qt}{\tau } \\ \end{matrix} \right]\left[ u \right]$$ $$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 *
$$\displaystyle (Eq. 7.11)
 * }
 * }

 * where,

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

\frac{-1}{\tau } & 0 \\ 0 & \frac{-1}{\tau } \\ \end{matrix} \right]$$
 * $$\underset{-}{\mathop{A}}\,=\left[ \begin{matrix}
 * $$\underset{-}{\mathop{A}}\,=\left[ \begin{matrix}
 * }
 * }

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

\frac{Q{{t}^{2}}}{2\tau } \\ \frac{Qt}{\tau } \\ \end{matrix} \right]$$ |-
 * $$\underset{-}{\mathop{B}}\,=\left[ \begin{matrix}
 * $$\underset{-}{\mathop{B}}\,=\left[ \begin{matrix}
 * }

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

\phi  \\ \omega  \\ \end{matrix} \right]$$ |-
 * $$\underset{-}{\mathop{x}}\,=\left[ \begin{matrix}
 * $$\underset{-}{\mathop{x}}\,=\left[ \begin{matrix}
 * }

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

|-
 * $$\underset{-}{\mathop{u}}\,=\left[ u \right]$$
 * $$\underset{-}{\mathop{u}}\,=\left[ u \right]$$
 * }


text:
User:Egm6321.f10.team03/Hwk3

= Problem 1 - Motion of a Projectile with Air Resistance =

Given
Figure shows the Trajectory of a projectile (ex:Rocket):

Find

 * $$\left( 1 \right)$$ Drive Equation of Motion (EOM)
 * $$\left( 2 \right)$$ Particular case $$\displaystyle k = 0 $$ : Verify $$y\left( x \right)$$ is parabolla
 * $$\left( 3 \right)$$ Consider $$k\ne 0$$, $$\displaystyle {{v}_{x}}=0$$,

 * $$ (3.1) $$ Find $${{v}_{y}}\left( t \right)$$, $$\displaystyle y(t)$$ for $$ \displaystyle m$$ = constant
 * $$ (3.2) $$ Find $${{v}_{y}}\left( t \right)$$, $$\displaystyle y(t)$$ if $$m=m\left( t \right)$$

Solution
Part 1 Consider the trajectory of a projectile (ex. Rocket)

 * Various forces acting on the projectile at time 't' are:

 * 1) Weight of the projectile

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

 * $$\displaystyle W=mg$$
 * }
 * }
 * }

 * 2) Inertia force

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

 * $${{F}_{a}}=ma=m\cdot \frac{dv}{dt}$$ for particle with constant mass
 * $${{F}_{a}}=ma=m\cdot \frac{dv}{dt}$$ for particle with constant mass

 * }
 * }

 * 3) Air resistance which is proportional to the velocity of particle

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

 * $$\displaystyle {{F}_{D}}=k{{v}^{n}}$$
 * }
 * }
 * }

Now consider the force equilibrium in both horizontal and vertical direction

 * a) Force Equilibrium in horizontal direction:

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

 * $$\sum{{{F}_{H}}=0}$$
 * }
 * }
 * }

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

 * $$m\cdot \frac{d{{v}_{x}}}{dt}+k{{v}^{n}}\cos \alpha =0$$,
 * $$m\cdot \frac{d{{v}_{x}}}{dt}+k{{v}^{n}}\cos \alpha =0$$,

 * where $$\displaystyle {{v}_{x}}\to $$ horizontal component of velocity

 * }
 * }

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

$$m\cdot \frac{d{{v}_{x}}}{dt}=-k{{v}^{n}}\cos \alpha $$ $$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * $$\displaystyle (Eq. 1.1)
 * }
 * }

 * b) Force Equilibrium in the vertical direction:

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

 * $$\sum{{{F}_{V}}=0}$$
 * }
 * }
 * }

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

$$m\cdot \frac{d{{v}_{y}}}{dt}+k{{v}^{n}}\sin \alpha +mg=0$$,

 * where $$\displaystyle {{v}_{y}}\to $$ vertical component of velocity

 * }
 * }

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

$$m\cdot \frac{d{{v}_{y}}}{dt}=-k{{v}^{n}}\sin \alpha -mg$$ $$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * $$\displaystyle (Eq. 1.2)
 * }
 * }

Part 2

 * Particular case: When $$\displaystyle k=0$$

Eq.1.1 reduces to

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

 * $$m\frac{d{{v}_{x}}}{dt}=0\Rightarrow \frac{d{{v}_{x}}}{dt}=0$$
 * }
 * }
 * }

Integrating the above equation gives:

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

$$
 * $${{v}_{x}}\left( t \right)={{c}_{1}}$$
 * $$\displaystyle (Eq. 1.3)
 * $$\displaystyle (Eq. 1.3)
 * }
 * }

Apply 'initial condition' to determine integration constant,$$\displaystyle{{c}_{1}}$$

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

 * $${{v}_{x}}\left( t=0 \right)={{v}_}={{c}_{1}}$$
 * }
 * }
 * }

Now Eq.1.3 becomes:
 * {| style="width:100%" border="0"

 * $${{v}_{x}}\left( t \right)={{v}_}$$
 * }
 * }
 * }

Integrate the above equation to obtain ,$$\displaystyle x$$

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

 * $$x\left( t \right)={{v}_}t+{{c}_{2}}$$
 * }
 * }
 * }

Then 'Initial condition' is applied to determine ,$$\displaystyle{{c}_{2}}$$

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

 * $$x\left( t=0 \right)={{v}_}\times 0+{{c}_{2}}\Rightarrow {{c}_{2}}={{x}_{0}}$$
 * }
 * }
 * }

Therefore,

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

 * $$x\left( t \right)={{v}_}t+{{x}_{0}}$$
 * }
 * }
 * }

Now $$\displaystyle t$$, can be expressed in terms of $$\displaystyle x$$,

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

$$
 * $$t=\frac{x-{{x}_{0}}}$$
 * $$\displaystyle (Eq. 1.4)
 * $$\displaystyle (Eq. 1.4)
 * }
 * }

Similarly When $$\displaystyle k=0$$, Eq.1.2 reduces to

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

 * $$m\frac{d{{v}_{y}}}{dt}=-mg\Rightarrow \frac{d{{v}_{y}}}{dt}=-g$$
 * }
 * }
 * }

Integrate the above equation to evaluate, $$\displaystyle {{v}_{y}}$$

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

$$
 * $${{v}_{y}}\left( t \right)=-gt+{{c}_{3}}$$
 * $$\displaystyle (Eq. 1.5)
 * $$\displaystyle (Eq. 1.5)
 * }
 * }

Apply 'initial condition' to obtain $$\displaystyle {{c}_{3}}$$

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

 * $${{v}_{y}}\left( t=0 \right)={{v}_}={{c}_{3}}$$
 * }
 * }
 * }

Now Eq.1.5 becomes,

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

 * $${{v}_{t}}\left( t \right)=-gt+{{v}_}$$
 * }
 * }
 * }

Then integrate the above equation to determine, $$\displaystyle y $$

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

 * $$y\left( t \right)=\frac{-g{{t}^{2}}}{2}+{{v}_}t+{{c}_{4}}$$
 * }
 * }
 * }

$$\displaystyle {{c}_{4}}$$ is determined using 'initial condition' as:

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

 * $$y\left( t=0 \right)=\frac{-g\times 0}{2}+{{v}_}\times 0+{{c}_{4}}\Rightarrow {{c}_{4}}={{y}_{0}}$$
 * }
 * }
 * }

Therefore,

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

 * $$y\left( t \right)=\frac{-g{{t}^{2}}}{2}+{{v}_}t+{{y}_{0}}$$
 * }
 * }
 * }

Now Substitute Eq.1.4 for $$\displaystyle t$$ in the above equation;

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

$$y\left( x \right)=-\frac{g}{2}{{\left( \frac{x-{{x}_{0}}} \right)}^{2}}+{{v}_}\left( \frac{x-{{x}_{0}}} \right)+{{y}_{0}}$$ $$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * $$\displaystyle (Eq. 1.6)
 * }
 * }

Eq.1.6 is in the form of a parabolic equation. Therefore $$\displaystyle y\left( x \right)$$ is parabola.

Part 3

Part 3.1: Constant mass, $$ k \neq0$$ Then
 * {| style="width:100%" border="0"

$$ m \frac{d {v}_{g}}{dt}=-k {{v}_{g}}^{n}-mg$$ || $$\displaystyle (Eq. 1.7.1) $$ Or
 * }
 * }
 * $$m \dot{v}=-k {v}^{n}-mg$$
 * Let
 * {| style="width:100%" border="0"

$$F \left(t,v, \dot{v} \right):=m \dot{v}+k {v}^{n}+mg=0= \frac{d \phi(t,v)}{dt}$$ ||
$$\displaystyle (Eq. 1.7.2) $$ Therefore
 * }
 * }
 * {| style="width:100%" border="0"

$$ \phi \left(t,v \right)=k$$ ||
$$\displaystyle (Eq. 1.7.3) $$ And equation 1.7.2 can be written in the form
 * }
 * }
 * {| style="width:100%" border="0"

$$F= \phi \left(x \right)+ \phi \left( y\right) y'=M+Ny'$$ ||
$$\displaystyle (Eq. 1.7.4) $$ Where
 * }
 * }
 * {| style="width:100%" border="0"

$$ \phi \left(x \right)=k {v}^{n}+mg$$ ||
$$\displaystyle (Eq. 1.7.5) $$ and
 * }
 * }
 * {| style="width:100%" border="0"

$$\phi \left(y \right)=m$$ ||
$$\displaystyle (Eq. 1.7.6) $$ Check the exactness of equation 1.7.4. First condition is satisfied as it is in the form
 * }
 * }
 * {| style="width:100%" border="0"

$$ F= \phi \left(x \right)+ \phi \left( y\right) y'=M+Ny'$$ ||
$$\displaystyle (Eq. 1.7.7) $$ Second condition is satisfied if
 * }
 * }
 * {| style="width:100%" border="0"

$$ {N}_{t}= {m}_{v}$$ ||
$$\displaystyle (Eq. 1.7.8) $$ Differentiating:
 * }
 * }
 * {| style="width:100%" border="0"

$$ {N}_{t}= \frac{d}{dt} \left(m \right)=0$$ ||
$$\displaystyle (Eq. 1.7.9) $$ and
 * }
 * }
 * {| style="width:100%" border="0"

$$ {M}_{v}= \frac{d}{dv} \left( k {v}^{n}+mg\right)=kn {v}^{n-1} \neq0 $$ ||
$$\displaystyle (Eq. 1.7.10) $$ Therefore equation 1.7.4 is not exact and must be made exact by finding an integration factor $$ h \left(t,v \right)$$
 * }
 * }
 * {| style="width:100%" border="0"

$$ h \left(t,v \right) \left[m \dot{v}+k {v}^{n}+mg\right]=0 $$ ||
$$\displaystyle (Eq. 1.7.11) $$ Choose an integrating factor as a function of time only
 * }
 * }
 * {| style="width:100%" border="0"

$$ h \left(t,v \right)=h \left(t \right)$$ ||
$$\displaystyle (Eq. 1.7.12) $$ Therefore $$ {h}_{v}=0$$ and
 * }
 * }
 * {| style="width:100%" border="0"

$$ {h}_{t}N+h \left( {N}_{t}- {M}_{v} \right)=0 $$ ||
$$\displaystyle (Eq. 1.7.13) $$ Rearranging and substituting
 * }
 * }
 * {| style="width:100%" border="0"

$$ \frac{1}{h}dh= -\frac{1}{m} \left[0-kn {v}^{n-1} \right]dt$$ ||
$$\displaystyle (Eq. 1.7.14) $$ Integrate equation 1.7.14 to get the integrating factor, $$h=exp \left( \frac{kn {v}^{n-1}}{m}t\right)$$. Multiply equation 1.7.2 by this gives us the exact form
 * }
 * }
 * {| style="width:100%" border="0"

$$F=exp \left( \frac{kn {v}^{n-1}}{m}t\right)v'+exp \left(  \frac{kn {v}^{n-1}}{m}t\right) \left[  \frac{k}{m} {v}^{n}+g\right]=0$$ Therefore
 *
$$\displaystyle (Eq. 1.7.15) $$
 * }
 * }
 * {| style="width:100%" border="0"

$${v}_{y} \left(t \right)= \frac{1}{exp \left( \frac{kn {v}^{n-1}}{m}t\right)}   \int^{t}-exp \left(  \frac{kn {v}^{n-1}}{m}\right)gds$$ ||
$$\displaystyle (Eq. 1.7.16) $$ To test this solution, assume $$n=1$$. Equation 1.7.16 then becomes
 * }
 * }

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

$$ {v}_{y} \left(t \right)= exp \left(- \frac{k}{m}t \right) \left[- \frac{mg}{k}exp \left( \frac{k}{m}t \right)+c \right]$$ ||
$$\displaystyle (Eq. 1.7.17) $$ Plug in initial conditions of $$ {t}_{0}=0$$, and $$ {v}_{y}={v}_{ {y}_{0}}$$ to obtain $$c= {v}_{ {y}_{0}}+ \frac{mg}{k}$$. Equation 1.7.17 becomes
 * }
 * }
 * {| style="width:100%" border="0"

$$ {v}_{y} \left(t \right)= \left( {v}_{ {y}_{0}}+ \frac{mg}{k} \right)exp \left(- \frac{k}{m}t \right)- \frac{mg}{k}$$ ||
$$\displaystyle (Eq. 1.7.18) $$ Integrate equation 1.7.18 to obtain
 * }
 * }
 * {| style="width:100%" border="0"

$$y \left(t \right)=- \frac{m}{k} \left( {v}_{ {y}_{0}}+ \frac{mg}{k} \right)exp \left(- \frac{k}{m}t \right)- \frac{mg}{k}t+ {c}_{2}$$ ||
$$\displaystyle (Eq. 1.7.19) $$ Plug in initial conditions of $$ {t}_{0}=0$$, and $$ y= {y}_{0}$$ to obtain $$$$. Equation 1.7.19 becomes
 * }
 * }

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

$$ {c}_{2}= {y}_{0}- \frac{mg}{k}+ \frac{m}{k} \left( {v}_{ {y}_{0}}+ \frac{mg}{k}\right) \left[1-exp( \frac{k}{m}t \right] $$ ||
$$\displaystyle (Eq. 1.7.20) $$ Since equations 1.7.18 and 1.7.20 agree with common solutions found in undergraduate mechanics, the general case can be stated after introducing dummy variables $$s$$, and $$w$$
 * }
 * }

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

$${v}_{y} \left(t \right)= \frac{1}{-exp \left( \frac{kn {v}^{n-1}}{m}t\right)}   \int^{t}-exp \left(  \frac{kn {v}^{n-1}}{m}\right) \left(g \right)ds $$ ||
$$\displaystyle (Eq. 1.7.21) $$ and
 * style="width:25%; padding:1px; border:2px solid #0000FF" |
 * style="width:25%; padding:1px; border:2px solid #0000FF" |
 * style= |
 * }
 * {| style="width:100%" border="0"

$${y} \left(t \right)= \int_{ }^{t} {-exp \left( \frac{kn {v}^{n-1}}{m}w\right)}   \int^{w}-exp \left(  \frac{kn {v}^{n-1}}{m}s\right) \left(g \right)dsdw $$ ||
$$\displaystyle (Eq. 1.7.22) $$ Part 3.2 Find $$ {v}_{y} \left(t \right) $$ and $${y} \left(t \right)$$ for $$m=m\left(t \right)$$ To solve we first apply a force balance on the rocket
 * style="width:25%; padding:1px; border:2px solid #0000FF" |
 * style="width:25%; padding:1px; border:2px solid #0000FF" |
 * style= |
 * }
 * {| style="width:100%" border="0"

$$ \frac{d \left(m {v}_{y} \right)}{dt}= -k{{v}_{y}}^{n}-mg$$ ||
$$\displaystyle (Eq. 1.8.1) $$ Use the conservation of linear momentum
 * }
 * }

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

$$d \left(m {v}_{y} \right)= md{v}_{y}+{v}_{y}dm$$ ||
$$\displaystyle (Eq. 1.8.2) $$ Plugging equation 1.8.2 into 1.8.1
 * }
 * }
 * {| style="width:100%" border="0"

$$m \frac{d{v}_{y}}{dt}+ {v}_{y} \frac{dm}{dt}=-k{{v}_{y}}^{n}-mg$$ ||
$$\displaystyle (Eq. 1.8.3) $$ Since the mass is reduced at a constant rate while the fuel is burned for $${t}_{0} \rightarrow {t}_{1}$$ the rate in equation 1.8.3 becomes a constant
 * }
 * }
 * {| style="width:100%" border="0"

$$\frac{dm}{dt}= \frac{m- {m}_{0}}{ {t}_{1}}:=- \alpha $$ ||
$$\displaystyle (Eq. 1.8.4) $$ And equation 1.8.3 becomes
 * }
 * }
 * {| style="width:100%" border="0"

$$ m \frac{d{v}_{y}}{dt}- {v}_{y} \alpha=-k{{v}_{y}}^{n}-mg $$ ||
$$\displaystyle (Eq. 1.8.5) $$ Rearranging gives us
 * }
 * }
 * {| style="width:100%" border="0"

$$ F \left( \cdot \right)= \dot{ {v}_{y}}- \frac{ \alpha}{m}{v}_{y}+ \frac{k}{m} {{v}_{y}}^{n}+g $$ ||
$$\displaystyle (Eq. 1.8.6) $$ Check for exactness. Equation 1.8.6 is in the form to satisfy condition 1. Checking condition 2
 * }
 * }
 * {| style="width:100%" border="0"

$$ {N}_{t}= \frac{d}{dt} \dot{ {v}_{y}}=0 $$ ||
 * }
 * {| style="width:100%" border="0"
 * {| style="width:100%" border="0"

$$ {M}_{{v}_{y}}= \frac{d}{d {v}_{y}} \left( \frac{k}{m} {{v}_{y}}^{n}- \frac{ \alpha}{m} {v}_{y}+g \right)= \frac{kn}{m} {{v}_{y}}^{n-1}- \frac{ \alpha}{m} \neq0$$ || Second condition is not satisfied. Make equation 1.8.6 exact using an integrating factor
 * }
 * }
 * {| style="width:100%" border="0"

$$hF \left( \cdot \right)=hM \left(t, {v}_{y} \right)+hN \left(t, {v}_{y} \right) \dot{ {v}_{y}}$$ ||
$$\displaystyle (Eq. 1.8.7) $$ Choose an integrating factor as a function of time only, such that $${h}_{v}=0 $$ and therefore
 * }
 * }

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

$$ {h}_{t}N+h \left( {N}_{x}- {M}_{y} \right)=0$$ ||
$$\displaystyle (Eq. 1.8.8) $$ Plugging in
 * }
 * }
 * {| style="width:100%" border="0"

$$ \frac{{h}_{t}}{h}= \frac{-1}{1} \left[0- \frac{kn{ {v}_{y}}^{n-1}}{m \left(t \right)}+ \frac{ \alpha}{m \left(t \right)} \right]$$ ||
$$\displaystyle (Eq. 1.8.9) $$ Integrating, using a dummy variable $$s$$
 * }
 * }
 * {| style="width:100%" border="0"

$$ \int_{}^{}\frac{1}{h}dh= \int_{}^{t} \left( \frac{kn{ {v}_{y} \left(s \right)}^{n-1}}{m \left(t \right)}+ \frac{ \alpha}{m \left(t \right)}\right)ds$$ ||
$$\displaystyle (Eq. 1.8.10) $$ Therefore
 * }
 * }
 * {| style="width:100%" border="0"

$$ h= exp \left[ \int_{}^{t} \left( \frac{kn{ {v}_{y} \left(s \right)}^{n-1}}{m \left(s \right)}+ \frac{ \alpha}{m \left(s \right)}\right)ds\right]$$ ||
$$\displaystyle (Eq. 1.8.11) $$ Using a dummy variable $$w$$, a solution can be written as
 * }
 * }
 * {| style="width:100%" border="0"

$$ {v}_{y} \left(t \right)= exp \left[ -\int_{}^{t} \left( \frac{kn{ {v}_{y} \left(s \right)}^{n-1}}{m \left(s \right)}+ \frac{ \alpha}{m \left(s \right)}\right)ds\right] \int_{}^{t} exp \left[ \int_{}^{s} \left( \frac{kn{ {v}_{y} \left(w \right)}^{n-1}}{m \left(w \right)}+ \frac{ \alpha}{m \left(w \right)}\right)dw\right]gds$$ ||
$$\displaystyle (Eq. 1.8.12) $$ and
 * style="width:25%; padding:1px; border:2px solid #0000FF" |
 * style="width:25%; padding:1px; border:2px solid #0000FF" |
 * style= |
 * }
 * {| style="width:100%" border="0"

$$ y \left(t \right)= \int_{}^{t} {v}_{y} \left(s \right)ds$$ ||
$$\displaystyle (Eq. 1.8.) $$ With varying mass and air resistance, the equations are typically solved by numerical methods.
 * style="width:25%; padding:1px; border:2px solid #0000FF" |
 * style="width:25%; padding:1px; border:2px solid #0000FF" |
 * style= |
 * }

Contributing Members

 * Solved and posted part 1 & part 2 by Egm6321.f10.team3.Sudheesh 15:36, 4 October 2010 (UTC)
 * Proofread part 1 & 2 and posted part 3 by James Roark 19:48, 5 October 2010 (UTC)
 * Solved part 3 by Chris Cook --Egm6321.f10.team3.cook 22:10, 5 October 2010 (UTC)

= Problem 2- Equation of Motion of Pendulums Connected by a Spring =

Given
Shown in figure are the two Pendulums connected by a spring:

Find

 * $$\left( 1 \right)$$ Derive equation of motion:

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

$$
 * $${{m}_{1}}{{l}^{2}}{{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{1}}=-k{{a}^{2}}\left( {{\theta }_{1}}-{{\theta }_{2}} \right)-{{m}_{1}}gl{{\theta }_{1}}+{{u}_{1}}l$$
 *
$$\displaystyle (Eq. 2.1)
 *
$$\displaystyle (Eq. 2.1)

 * }
 * }

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

$$
 * $${{m}_{2}}{{l}^{2}}{{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{2}}=-k{{a}^{2}}\left( {{\theta }_{2}}-{{\theta }_{1}}\right)-{{m}_{2}}gl{{\theta }_{2}}+{{u}_{2}}l$$
 *
$$\displaystyle (Eq. 2.2)
 *
$$\displaystyle (Eq. 2.2)
 * }
 * }

 * $$\left( 2 \right)$$ Write Eq.2.1 and Eq.2.2 in the form of [[media:2010_09_21_14_56_48.djvu|Mtg 13 (c),page2 ]], of:

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

$$
 * $$\underset{-}{\overset{\centerdot }{\mathop{x}}}\,=\underset{-}{\mathop{A}}\,\left( t \right)\underset{-}{\mathop{x}}\,\left( t \right)+\underset{-}{\mathop{B}}\,\left( t \right)\underset{-}{\mathop{u}}\,\left( t \right)$$
 *
$$\displaystyle (Eq. 2.3)
 *
$$\displaystyle (Eq. 2.3)
 * }
 * }

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

$$\underset{-}{\mathop{x}}\,={{\left[ \begin{matrix} {{\theta }_{1}} & \overset{\centerdot }{\mathop{{{\theta }_{1}}}}\, & {{\theta }_{2}} & \overset{\centerdot }{\mathop{{{\theta }_{2}}}}\, \\ \end{matrix} \right]}^{T}}$$ and $$\underset{-}{\mathop{u}}\,={{\left[ \begin{matrix} {{u}_{1}}l & {{u}_{2}}l \\ \end{matrix} \right]}^{T}}$$
 * Given
 * Given
 * }
 * }

Solution

 * $$\left( 1 \right)$$ Derive equation of motion:
 * (a) Consider Free Body Diagram of left pendulum:

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

 * $$\displaystyle l\sin {{\theta }_{1}}=l{{\theta }_{1}}$$ and $$\displaystyle l\cos {{\theta }_{1}}=l$$
 * }
 * }
 * }

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

 * $$acceleration, {{a}_{1}}=\frac{{{d}^{2}}\left( l{{\theta }_{1}} \right)}{d{{t}^{2}}}=l{{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{1}}$$
 * }
 * }
 * }

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

 * $$ Inertia force ={{m}_{1}}l{{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{1}}$$
 * }
 * }
 * }

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

 * $$\displaystyle Spring force =ka{{\theta }_{1}}-ka{{\theta }_{2}}=ka({{\theta }_{1}}-{{\theta }_{1}})$$
 * }
 * }
 * }

Now using D'Alembert's_principle, sum of the moments about pivot(A)is equal to zero

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

 * $$\sum{{{M}_{A}}=0\Rightarrow }\left( {{m}_{1}}l{{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{1}} \right)\cdot l+\left( ka({{\theta }_{1}}-{{\theta }_{2}}) \right)\cdot a+\left( {{m}_{1}}g \right)\cdot l{{\theta }_{1}}-\left( {{u}_{1}} \right)\cdot l=0$$
 * }
 * }
 * }

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

$$\Rightarrow {{m}_{1}}{{l}^{2}}{{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{1}}=-k{{a}^{2}}({{\theta }_{1}}-{{\theta }_{2}})-{{m}_{1}}gl{{\theta }_{1}}+{{u}_{1}}l$$ $$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 *
$$\displaystyle (Eq. 2.1)
 * }
 * }

 * (b) Consider Free Body Diagram of right pendulum:

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

 * $$\displaystyle l\sin {{\theta }_{2}}=l{{\theta }_{2}}$$ and $$\displaystyle l\cos {{\theta }_{2}}=l$$
 * }
 * }
 * }

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

 * $$acceleration, {{a}_{2}}=\frac{{{d}^{2}}\left( l{{\theta }_{2}} \right)}{d{{t}^{2}}}=l{{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{2}}$$
 * }
 * }
 * }

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

 * $$ Inertia force ={{m}_{2}}l{{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{2}}$$
 * }
 * }
 * }

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

 * $$\displaystyle Spring force =ka{{\theta }_{2}}-ka{{\theta }_{1}}=ka({{\theta }_{2}}-{{\theta }_{1}})$$
 * }
 * }
 * }

Using D'Alembert's_principle, sum of the moments about pivot(B)is equal to zero

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

 * $$\sum{{{M}_{B}}=0\Rightarrow }\left( {{m}_{2}}l{{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{2}} \right)\cdot l+\left( ka({{\theta }_{2}}-{{\theta }_{1}}) \right)\cdot a+\left( {{m}_{2}}g \right)\cdot l{{\theta }_{2}}-\left( {{u}_{2}} \right)\cdot l=0$$
 * }
 * }
 * }

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

$$\Rightarrow {{m}_{2}}{{l}^{2}}{{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{2}}=-k{{a}^{2}}({{\theta }_{2}}-{{\theta }_{1}})-{{m}_{2}}gl{{\theta }_{2}}+{{u}_{2}}l$$ $$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 *
$$\displaystyle (Eq. 2.2)
 * }
 * }

 * $$\left( 2 \right)$$ Write Eq.2.1 and Eq.2.2 in the form of Eq.2.3(system of coupled equation):

 * Eq.2.1 can be rearranged as,

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

$$
 * $${{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{1}}=\frac{-\left( k{{a}^{2}}+{{m}_{1}}gl \right)}{m{}_{1}{{l}^{2}}}{{\theta }_{1}}+\frac{k{{a}^{2}}}{m{}_{1}{{l}^{2}}}{{\theta }_{2}}+\frac{{{u}_{1}}l}{m{}_{1}{{l}^{2}}}$$
 *
$$\displaystyle (Eq. 2.4)
 *
$$\displaystyle (Eq. 2.4)
 * }
 * }

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

$$
 * $${{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{2}}=\frac{k{{a}^{2}}}{m{}_{2}{{l}^{2}}}{{\theta }_{1}}+\frac{-\left( k{{a}^{2}}+{{m}_{2}}gl \right)}{m{}_{2}{{l}^{2}}}{{\theta }_{2}}+\frac{{{u}_{2}}l}{m{}_{2}{{l}^{2}}}$$
 *
$$\displaystyle (Eq. 2.5)
 *
$$\displaystyle (Eq. 2.5)
 * }
 * Now Eq.2.4 and Eq.2.5 can be put in the form of Eq.2.3 as:
 * Now Eq.2.4 and Eq.2.5 can be put in the form of Eq.2.3 as:

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

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

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

$$
 *
$$\displaystyle (Eq. 2.6)
 * }
 * }

 * Where:

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

0 & 1 & 0 & 0 \\   \frac{-\left( k{{a}^{2}}+{{m}_{1}}gl \right)}{m{}_{1}{{l}^{2}}} & 0 & \frac{k{{a}^{2}}}{m{}_{1}{{l}^{2}}} & 0  \\ 0 & 0 & 0 & 1 \\   \frac{k{{a}^{2}}}{m{}_{2}{{l}^{2}}} & 0 & \frac{-\left( k{{a}^{2}}+{{m}_{2}}gl \right)}{m{}_{2}{{l}^{2}}} & 0  \\ \end{matrix} \right]$$
 * $$\underset{-}{\mathop{A}}\,=\left[ \begin{matrix}
 * $$\underset{-}{\mathop{A}}\,=\left[ \begin{matrix}

 * }
 * }

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

0 & 0 \\   \frac{1}{m{}_{1}{{l}^{2}}} & 0  \\ 0 & 0 \\   0 & \frac{1}{m{}_{2}{{l}^{2}}}  \\ \end{matrix} \right]$$
 * $$\underset{-}{\mathop{B}}\,=\left[ \begin{matrix}
 * $$\underset{-}{\mathop{B}}\,=\left[ \begin{matrix}
 * }
 * }

Contributing Members
Solved and posted by Egm6321.f10.team3.Sudheesh 15:39, 4 October 2010 (UTC)

Proofed by Kurt Schulze

Proofed by Egm6321.f10.team3.franklin 06:04, 6 October 2010 (UTC)

= Problem 3- Solution of State Equation (Control Engineering) =

Given

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

$$
 * $$ x'(t)=a(t)\cdot x(t)+b(t)\cdot u(t)$$
 *
$$\displaystyle
 *
$$\displaystyle
 * }
 * }

Find
Solve given equation by using Integrating Factor Method.

Solution
Let a(t) = a  &  b(t) = b

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

$$
 * $$ x'(t)=a\cdot x(t)+b\cdot u(t)$$
 *
$$\displaystyle
 *
$$\displaystyle
 * }
 * {| style="width:100%" border="0"
 * {| style="width:100%" border="0"

$$
 * $$ x'(t)-a\cdot x(t)=b\cdot u(t)$$
 *
$$\displaystyle (Eq. 3.1)
 *
$$\displaystyle (Eq. 3.1)
 * }
 * }

We determine integrating factor from the coefficient of x(t) in Eq(3.1)

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

$$
 * $$ h(t) = e^{\int {-a \cdot dt}} = e^{-at} $$
 *
$$\displaystyle
 *
$$\displaystyle
 * }
 * }

Apply obtained integrating factor to Eq(3.1)

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

$$
 * $$ [x'(t)-a\cdot x(t)]\cdot e^{-at} = [b\cdot u(t)]\cdot e^{-at}$$
 *
$$\displaystyle
 *
$$\displaystyle
 * }
 * }

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

$$
 * $$\underbrace{{x}'(t)\cdot {{e}^{-at}}-a\cdot x(t)\cdot {{e}^{-at}}}_{\because f(x)\cdot {g}'(x)+{f}'(x)\cdot g(x)={{\left( f(x)\cdot g(x) \right)}^{'}}}=b\cdot u(t)\cdot {{e}^{-at}}={{\left( f(x)\cdot g(x) \right)}^{'}}$$
 *
$$\displaystyle
 *
$$\displaystyle
 * }
 * }

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

$$ $$
 * $$ [e^{-at}\cdot x(t)]' = b\cdot u(t)\cdot e^{-at}
 * $$ [e^{-at}\cdot x(t)]' = b\cdot u(t)\cdot e^{-at}
 *
$$\displaystyle
 * }
 * }

Integrate between $$\displaystyle t_{0} $$ < $$\displaystyle \tau $$ < $$\displaystyle t $$

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

$$ $$
 * $$ \int\limits_^{t}{[e^{-a\tau}\cdot x(\tau)]'}d\tau = \int\limits_^{t} {b \cdot u(\tau)\cdot e^{-a\tau}}d\tau
 * $$ \int\limits_^{t}{[e^{-a\tau}\cdot x(\tau)]'}d\tau = \int\limits_^{t} {b \cdot u(\tau)\cdot e^{-a\tau}}d\tau
 *
$$\displaystyle
 * }
 * }

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

$$ $$
 * $$ e^{-at}\cdot x(t) - e^{-at_0}\cdot x(t_0) = \int\limits_^{t} {b \cdot u(\tau)\cdot e^{-a\tau}}d\tau
 * $$ e^{-at}\cdot x(t) - e^{-at_0}\cdot x(t_0) = \int\limits_^{t} {b \cdot u(\tau)\cdot e^{-a\tau}}d\tau
 *
$$\displaystyle
 * }
 * }

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

$$ $$
 * $$ e^{-at}\cdot x(t) = e^{-at_0}\cdot x(t_0) + \int\limits_^{t} {b \cdot u(\tau)\cdot e^{-a\tau}}d\tau
 * $$ e^{-at}\cdot x(t) = e^{-at_0}\cdot x(t_0) + \int\limits_^{t} {b \cdot u(\tau)\cdot e^{-a\tau}}d\tau
 *
$$\displaystyle
 * }
 * }

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

$$ $$
 * $$ x(t) = \frac {e^{-at_0}}{e^{-at}} \cdot x(t_0) + \frac {\int\limits_^{t} {b \cdot u(\tau)\cdot e^{-a\tau}}d\tau}{e^{-at}}
 * $$ x(t) = \frac {e^{-at_0}}{e^{-at}} \cdot x(t_0) + \frac {\int\limits_^{t} {b \cdot u(\tau)\cdot e^{-a\tau}}d\tau}{e^{-at}}
 *
$$\displaystyle
 * }
 * }

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

$$x(t) = e^{a(t-t_0)}\cdot x(t_0) + \int\limits_^{t} {e^{a(t-\tau)}}\cdot b \cdot u(\tau)d\tau$$ $$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 *
$$\displaystyle (Eq. 3.2)
 * }
 * }

Contributing Members

 * Solved and posted by-User:Egm6321.f10.team3.Hong SJ 21:18, 03 October 2010 (UTC)
 * Proof read by- Egm6321.f10.team3.Sudheesh 00:11, 5 October 2010 (UTC)
 * Proof read by-Egm6321.f10.team03.sigillo 13:57, 5 October 2010 (UTC)

=Problem 4: Expansion of a Taylor Series=

Given
Given an expression of $$\exp \left( x \right)$$

Find
Find the Taylor Series Expansion

Solution
In general, the Taylor Series can be defined as:

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

$$f\left( x \right)=\sum\limits_{n=0}^{\infty }{\frac{{{f}^{\left( n \right)}}\left( a \right)}{n!}}{{\left( x-a \right)}^{n}}$$
 *
$$\,(Eq.4.1)$$
 * }

Where $$\displaystyle f^{\left( n \right)}$$ is the $$\displaystyle n$$th derivative of $$\displaystyle  f$$ evaluated at $$\displaystyle a$$. Expanding the series around point $$\displaystyle a=0$$ gives each $$\displaystyle  n$$th derivative of $$ \displaystyle  f$$ equal to zero, or $$\displaystyle f^{\left( n \right)}\left( 0 \right)=1$$. Thus, the Taylor series in this case can be defined as:

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

$$f\left( x \right)=\sum\limits_{n=0}^{\infty }{\frac{n!}}$$
 *
$$\,(Eq.4.2)$$
 * }

Hence, for $$\displaystyle f\left( x \right)=\exp \left( x \right)$$, Taylor's series expansion is

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

$$\exp \left( x \right)=1+\frac{x}{1!}+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\frac{x^{4}}{4!}+...+\frac{x^{n}}{n!}$$
 * style="width:25%; padding:1px; border:2px solid #0000FF" |
 * style="width:25%; padding:1px; border:2px solid #0000FF" |
 *
$$\,(Eq.4.3)$$
 * style= |
 * }

Contributing Members

 * Solved and posted by James Roark 23:02, 3 October 2010 (UTC)
 * Proof read by Egm6321.f10.team3.Sudheesh 00:36, 5 October 2010 (UTC)
 * Proof read byEgm6321.f10.team03.sigillo 14:05, 5 October 2010 (UTC)
 * Proof read byEgm6321.f10.team03.Hong SJ 11:24, 5 October 2010 (UTC)
 * Proof read by Egm6321.f10.team3.franklin 05:59, 6 October 2010 (UTC)

= Problem 5 - Generalization of a SC-L1-ODE-CC to SC-L1-ODE-VC =

Given
Given is the solution for $$\underline{x}(t)$$ of a first order linear ODE for a coupled system with constant coefficients (SC-L1-ODE-CC) out of the lecture notes in [[media:2010_09_23_13_57_17.djvu|Mtg 14 page 1-2, Eq.4]]:

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

$$\underline{x}(t)={{e}^{\underline{A}\cdot \left( t-{{t}_{0}} \right)}}\cdot \underline {x}({{t}_{0}})+\int\limits_^{t}{{{e}^{\underline{A}\cdot \left( t-\tau \right)}}\cdot \underline{B}\cdot \underline{u}\left( \tau  \right)d\tau }$$
 *
$$\,(Eq. 5.1) $$
 * style= |
 * }

Where $$\underline{x}$$ is a n x 1 vector, $$\underline{u}$$ a m x 1 vector, $$\underline{A}$$ is a n x n matrix and $$\underline{B}$$ is a n x m matrix.

Find
The general form for SC-L1-ODEs with varying coefficient (SC-L1-ODE-VC) has to be found based on the given SC-L1-ODE-CC

Solution
Following the approach of the integrating factor method and comparing the general form for L1-ODE-VC given in [[media:2010_09_23_13_57_17.djvu|Eq.3,page 1, Mtg 14 ]], it can be seen (and calculated through the intergration factor method - refer to Problem 3-) that the generalized form for (Eq. 5.1) can be achieved by integrating the argument of the exp(.)-function in the limits $$\,{{t}_{0}}$$ and $$\,t$$ for the first term, $$\,\tau $$ and $$\,t$$ for the second term:
 * {| style="width:100%" border="0"

$$\underline{x}(t)={{e}^{\int\limits_^{t}{\underline{A}(\tau )d\tau }}}\cdot \underline{x}({{t}_{0}})+\int\limits_^{t}{{{e}^{\int\limits_{\tau }^{t}{\underline{A}(s)ds}}}\cdot \underline{B}\cdot \underline{u}\left( \tau \right)d\tau }$$ Note: The integral of a matrix is calculated by integrating each element that composes the matrix.
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 *
$$\,(Eq. 5.2) $$
 * style= |
 * }

Contributing Members

 * Solved by Michele Sigilló
 * Posted by Michele Sigilló
 * Proof read by Egm6321.f10.team3.Sudheesh 18:20, 5 October 2010 (UTC)

= Problem 6 - General SC-L1-ODE-CC with state transition matrix $$\underline{\Phi} $$ =

Given
The state equation for a coupled system like it's usual for a openloop control system is given as:
 * {| style="width:100%" border="0"

$$\underline{\dot{x}}(t)=\underline{A}\underline{x}(t)+\underline{B}\underline{u}(t)$$ The state transition matrix $$\underline{\Phi} $$ has the following properties: and where I is the identity matrix.
 *
$$\,(Eq. 6.1) $$
 * style= |
 * }

Find
With (Eq. 6.1) through (Eq. 6.3), Eq. (4) on pp. 14-1 of the lecture notes, has to be found:
 * {| style="width:100%" border="0"

$$\underline{x}(t)={{e}^{\underline{A}(t-{{t}_{0}})}}\cdot \underline{x}({{t}_{0}})+\int\limits_^{t}{{{e}^{\underline{A}(t-\tau )}}\underline{B}\underline{u}(\tau )\,d\tau }$$
 *
$$\,(Eq. 6.4) $$
 * style= |
 * }

Solution
In (Eq. 6.2) the homogeneous differential equation for $$\underline{\Phi} (t,{{t}_{0}})$$ is given. Thus replacing $$\underline{x}(t)$$ with $$\underline{\Phi} (t,{{t}_{0}})$$ in (Eq. 6.1) gives the inhomogeneous equation for $$\underline{\Phi} (t,{{t}_{0}})$$:
 * {| style="width:100%" border="0"

$$\dot{\underline{\phi} }(t,{{t}_{0}})=\underline{A}\underline{\phi} (t,{{t}_{0}})+\underline{B}\underline{u}(t)$$
 *
$$\,(Eq. 6.5) $$
 * style= |
 * }

Regrouping the equation above results in:
 * {| style="width:100%" border="0"

$$\underbrace{-\left( \underline{A}\underline{\phi} (t,{{t}_{0}})+\underline{B}\underline{u}(t) \right)}_{\underline{M}(\underline{\phi} ,\underline{u})}+\underbrace{1}_{\underline{N}(\underline{\phi} ,\underline{u})}\cdot \dot{\underline{\phi} }(t,{{t}_{0}})=0$$ (Eq. 6.6) meets the first condition of exactness, but the second condition $${{\underline{M}}_{\underline{\phi} }}={{\underline{N}}_{\underline{u}}}$$ needs to be checked:
 *
$$\,(Eq. 6.6) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$${{\underline{M}}_{\underline{\phi} }}=-\underline{A}\ne 0={{\underline{N}}_{\underline{u}}}$$ The second condition is not met, thus a function $$h(t)$$ to make (Eq. 6.6) exact needs to be found. Therefore (Eq. 6.6) will be rewritten as:
 *
$$\,(Eq. 6.7) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$-\underline{A}\underline{\phi} (t,{{t}_{0}})+1\cdot \dot{\underline{\phi} }(t,{{t}_{0}})=\underline{B}\underline{u}(t)$$ The integrating factor $$h(t)$$ is then given as:
 *
$$\,(Eq. 6.8) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$h(t)={{e}^{\int\limits_^{t}{-\underline{A}\,ds}}}={{e}^{-\underline{A}(t-{{t}_{0}})}}$$ Multiplying $$h(t)$$ throughout (Eq. 6.8) gives:
 *
$$\,(Eq. 6.9) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$-\underline{A}\underline{\phi} (t,{{t}_{0}})\cdot {{e}^{-\underline{A}(t-{{t}_{0}})}}+{{e}^{-\underline{A}(t-{{t}_{0}})}}\cdot \dot{\underline{\phi} }(t,{{t}_{0}})=\underline{B}\underline{u}(t)\cdot {{e}^{-\underline{A}(t-{{t}_{0}})}}$$ which is equal to
 *
$$\,(Eq. 6.10) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\frac{d\left( {{e}^{-\underline{A}(t-{{t}_{0}})}}\cdot \underline{\phi} (t,{{t}_{0}}) \right)}{dx}=\underline{B}\underline{u}(t)\cdot {{e}^{-\underline{A}(t-{{t}_{0}})}}$$ Integrating (Eq. 6.11) on both sides between $${{t}_{0}}$$ and $$t$$ gives:
 *
$$\,(Eq. 6.11) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$${{e}^{-\underline{A}(t-{{t}_{0}})}}\cdot \underline{\phi} (t,{{t}_{0}})-{{e}^{-\underline{A}({{t}_{0}}-{{t}_{0}})}}\cdot \underline{\phi} ({{t}_{0}},{{t}_{0}})={{e}^{\underline{A}{{t}_{0}}}}\cdot \int\limits_^{t}\cdot \underline{B}\underline{u}(t)\,d\tau $$ that is equal to
 *
$$\,(Eq. 6.12) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\underline{\phi} (t,{{t}_{0}})={{e}^{\underline{A}(t-{{t}_{0}})}}\cdot \underline{\phi} ({{t}_{0}},{{t}_{0}})+\int\limits_^{t}{{{e}^{\underline{A}(t-\tau )}}\cdot }\underline{B}\underline{u}(t)\,d\tau $$ Replacing $$\underline{\Phi} (t,{{t}_{0}})$$ with $$\underline{x}(t)$$ gives:
 *
$$\,(Eq. 6.13) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\underline{x}(t)={{e}^{\underline{A}(t-{{t}_{0}})}}\cdot \underline{x}({{t}_{0}})+\int\limits_^{t}{{{e}^{\underline{A}(t-\tau )}}\underline{B}\underline{u}(\tau )\,d\tau }$$ which is the same as the given solution in (Eq. 6.4)!
 * style="width:2%; padding:1px; border:2px solid #0000FF" |
 * style="width:2%; padding:1px; border:2px solid #0000FF" |
 *
$$\,(Eq. 6.13) $$
 * style= |
 * }

Contributing Members

 * Solved by Michele Sigilló
 * Posted by Michele Sigilló
 * Proof read by Egm6321.f10.team3.franklin 06:05, 6 October 2010 (UTC)
 * Proof read by James Roark 12:18, 6 October 2010 (UTC)

= Problem 7 - Roll Control of Rocket By Actuating Ailerons =

Given

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

$$
 * $$\overset{\centerdot }{\mathop{\phi }}\,=\omega $$
 *
$$\displaystyle (Eq. 7.1)
 *
$$\displaystyle (Eq. 7.1)
 * }
 * }

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

$$
 * $$\overset{\centerdot }{\mathop{\omega }}\,=\frac{-1}{\tau }\omega +\frac{Q}{\tau }\delta $$
 *
$$\displaystyle (Eq. 7.2)
 *
$$\displaystyle (Eq. 7.2)
 * }
 * }

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

$$
 * $$\overset{\centerdot }{\mathop{\delta }}\,=u$$
 *
$$\displaystyle (Eq. 7.3)
 *
$$\displaystyle (Eq. 7.3)
 * }
 * }

Find
Control behavior in form of

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

$$
 * $$\overset{\centerdot}{\mathop{x\left( t \right)=\underset{-}{\mathop{A}}\,}}\,\underset{-}{\mathop{x}}\,\left( t \right)+\underset{-}{\mathop{B}}\,\underset{-}{\mathop{u}}\,\left( t \right)$$
 *
$$\displaystyle (Eq. 7.4)
 *
$$\displaystyle (Eq. 7.4)

 * }
 * }

Solution

 * Integrating Eq.7.3 gives,

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

$$
 * $$\displaystyle \delta =ut$$
 *
$$\displaystyle (Eq. 7.5)
 *
$$\displaystyle (Eq. 7.5)
 * }
 * Substitute Eq.7.5 for $$\displaystyle \delta $$, in Eq.7.2:
 * Substitute Eq.7.5 for $$\displaystyle \delta $$, in Eq.7.2:

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

$$
 * $$\overset{\centerdot }{\mathop{\omega }}\,=\frac{-1}{\tau }\omega +\frac{Q}{\tau }ut$$
 *
$$\displaystyle (Eq. 7.6)
 *
$$\displaystyle (Eq. 7.6)
 * }
 * }

 * Now integrate Eq.7.6:

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

$$
 * $$\omega =\frac{-1}{\tau }\omega t+\frac{Q}{2\tau }u{{t}^{2}}$$
 *
$$\displaystyle (Eq. 7.7)
 *
$$\displaystyle (Eq. 7.7)
 * }
 * }

 * Similarly integrating Eq.7.1 gives,

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

$$
 * $$\displaystyle \phi =\omega t$$
 *
$$\displaystyle (Eq. 7.8)
 *
$$\displaystyle (Eq. 7.8)
 * }
 * }

 * Substitute Eq.7.8 in Eq.7.7 implies,

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

$$
 * $$\omega =\frac{-1}{\tau }\phi +\frac{Q{{t}^{2}}}{2\tau }u$$
 *
$$\displaystyle (Eq. 7.9)
 *
$$\displaystyle (Eq. 7.9)
 * }
 * }

 * From Eq.7.1 and Eq.7.9,

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

$$
 * $$\overset{\centerdot }{\mathop{\phi }}\,=\frac{-1}{\tau }\phi +\frac{Q{{t}^{2}}}{2\tau }u$$
 *
$$\displaystyle (Eq. 7.10)
 *
$$\displaystyle (Eq. 7.10)
 * }
 * }

 * Now, Eq.7.3, Eq.7.6 and Eq.7.10 can be written in the required form as,

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

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

$$\left[ \begin{matrix} \overset{\centerdot }{\mathop{\phi }}\, \\ \overset{\centerdot }{\mathop{\omega }}\, \\ \overset{\centerdot }{\mathop{\delta }}\, \\ \end{matrix} \right]=\left[ \begin{matrix} \frac{-1}{\tau } & 0 & 0 \\ 0 & \frac{-1}{\tau } & 0 \\ 0 & 0 & 0 \\ \end{matrix} \right]\left[ \begin{matrix} \phi  \\ \omega  \\ \delta  \\ \end{matrix} \right]+\left[ \begin{matrix} \frac{Q{{t}^{2}}}{2\tau } \\ \frac{Qt}{\tau } \\ 1 \\ \end{matrix} \right]\left[ u \right]$$ $$
 *
$$\displaystyle (Eq. 7.11)
 * }
 * }

 * where,

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

\frac{-1}{\tau } & 0 & 0 \\ 0 & \frac{-1}{\tau } & 0 \\ 0 & 0 & 0 \\ \end{matrix} \right]$$
 * $$\underset{-}{\mathop{A}}\,=\left[ \begin{matrix}
 * $$\underset{-}{\mathop{A}}\,=\left[ \begin{matrix}

 * }
 * }

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

\frac{Q{{t}^{2}}}{2\tau } \\ \frac{Qt}{\tau } \\ 1 \\ \end{matrix} \right]$$
 * $$\underset{-}{\mathop{B}}\,=\left[ \begin{matrix}
 * $$\underset{-}{\mathop{B}}\,=\left[ \begin{matrix}

|-
 * }

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

\phi  \\ \omega  \\ \delta  \\ \end{matrix} \right]$$ |-
 * $$\underset{-}{\mathop{x}}\,=\left[ \begin{matrix}
 * $$\underset{-}{\mathop{x}}\,=\left[ \begin{matrix}
 * }

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

|-
 * $$\underset{-}{\mathop{u}}\,=\left[ u \right]$$
 * $$\underset{-}{\mathop{u}}\,=\left[ u \right]$$
 * }

Contributing Members

 * Solved by: Egm6321.f10.team03 19:29, 4 October 2010 (UTC).Kurt Schulze
 * Posted by: Egm6321.f10.team3.Sudheesh 18:59, 4 October 2010 (UTC)
 * Proof read by Egm6321.f10.team03.sigillo 14:30, 5 October 2010 (UTC)
 * Proof read by Egm6321.f10.team3.franklin 05:58, 6 October 2010 (UTC)

= Problem 8- Discuss the search for h =

Given
From the lecture notes of Mtg [|Mtg 17]

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

$$\displaystyle {h}_{x} + {h}_{y} {P} = 0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
$$\displaystyle (1) p. 17-1
 * }

Find
Without assuming a priory that $$ h = const$$, discuss the search for the solution of $$ (1) p. 17-1$$

Solution
To find $$ h(x,y)$$ we begin by rearranging $$ (1) p. 17-1 $$ to find

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

$$\displaystyle {h}_{x} = - {h}_{y}P $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
$$\displaystyle Eq. (2)
 * }

Next we will use the property that

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

$$\displaystyle {h}_{xy} = {h}_{yx} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
$$\displaystyle Eq. (3)
 * }

and see if it holds. First we take evaluate $$ {h}_{xy}$$

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

$$ {h}_{x}= - {h}_{y} \underbrace_{:= {y}^{'}}
 * style="width:95%" |
 * style="width:95%" |

\Rightarrow

{h}_{xy} = - {h}_{yy}{y}^{'} - {h}_{y} {y}^{''} $$ $$
 *
$$\displaystyle Eq. (4)
 * }

By inserting $$ Eq. (4) $$ into $$ Eq. (3)$$ we see that this property does not hold

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

$$ {h}_{xy} = - {h}_{yy}{y}^{'} - {h}_{y} {y}^{''} \neq {h}_{yx} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
$$\displaystyle Eq. (5)
 * }

So for $$ (1) p. 17-1 $$ to be true, $$ {h}_{x} = {h}_{y} = 0$$.

Therefore $$ h = const $$

Contributing Members
Solved and Posted By Chris Cook --Egm6321.f10.team3.cook 22:08, 5 October 2010 (UTC)

= Problem 9  -  Check Exactness of Differential Equation =

Given

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

$$ $$
 * $$ \underbrace{\phi_p}_{f}\cdot y'' + \underbrace{\phi_y \cdot y' + \phi_x}_{g} = 0
 * $$ \underbrace{\phi_p}_{f}\cdot y'' + \underbrace{\phi_y \cdot y' + \phi_x}_{g} = 0
 *
$$\displaystyle
 * }
 * }

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

$$ $$
 * $$ \underbrace{(15p^4\cdot cos (x^2))}_{\phi_p}\cdot y'' + \underbrace{(6xy^2)}_{\phi_y} \cdot y' + \underbrace{(-6xp^5\cdot sin(x^2) + 2y^3)}_{\phi_x} = 0
 * $$ \underbrace{(15p^4\cdot cos (x^2))}_{\phi_p}\cdot y'' + \underbrace{(6xy^2)}_{\phi_y} \cdot y' + \underbrace{(-6xp^5\cdot sin(x^2) + 2y^3)}_{\phi_x} = 0
 *
$$\displaystyle
 * }
 * }

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

$$ $$
 * $$ \therefore f = \phi_p = 15p^4\cdot cos (x^2)
 * $$ \therefore f = \phi_p = 15p^4\cdot cos (x^2)
 *
$$\displaystyle
 * }
 * }

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

$$
 * $$ \therefore g = \phi_y \cdot y' + \phi_x = (6xy^2)\cdot y' + (-6xyp^5\cdot sin(x^2) + 2y^3)$$
 *
$$\displaystyle
 *
$$\displaystyle
 * }
 * }

Find

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

$$ $$
 * $$ (1) f_{xx} + 2p\cdot f_{xy} + p^2\cdot f_{yy} = g_{xp} + p\cdot g_{yp}- g_y
 * $$ (1) f_{xx} + 2p\cdot f_{xy} + p^2\cdot f_{yy} = g_{xp} + p\cdot g_{yp}- g_y
 *
$$\displaystyle Eq(9.1)
 * }
 * }

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

$$ $$
 * $$ (2) f_{xp} + p\cdot f_{yp} + 2\cdot f_{y} = g_{pp}
 * $$ (2) f_{xp} + p\cdot f_{yp} + 2\cdot f_{y} = g_{pp}
 *
$$\displaystyle Eq(9.2)
 * }
 * }

Solution

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

$$
 * $$\displaystyle for Eq(9.1)$$
 *
$$\displaystyle
 *
$$\displaystyle
 * }
 * }

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

$$
 * $$ f_{xx} = \frac{\partial }{\partial x} (\frac{\partial }{\partial x} (15p^4\cdot cos(x^2)))$$
 *
$$\displaystyle
 *
$$\displaystyle
 * }
 * }

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

$$ $$
 * $$\displaystyle = \frac{\partial }{\partial x} (-2x\cdot sin(x^2)\cdot 15p^4)
 * $$\displaystyle = \frac{\partial }{\partial x} (-2x\cdot sin(x^2)\cdot 15p^4)
 *
$$\displaystyle
 * }
 * }

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

$$ $$
 * $$\displaystyle = \frac{\partial }{\partial x} (-30p^4\cdot x\cdot sin(x^2))
 * $$\displaystyle = \frac{\partial }{\partial x} (-30p^4\cdot x\cdot sin(x^2))
 *
$$\displaystyle
 * }
 * }

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

$$ $$
 * $$\displaystyle  = -30p^4\cdot x \cdot 2x \cdot cos(x^2) + (-30p^4\cdot sin(x^2))
 * $$\displaystyle  = -30p^4\cdot x \cdot 2x \cdot cos(x^2) + (-30p^4\cdot sin(x^2))
 *
$$\displaystyle
 * }
 * }

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

$$ $$
 * $$\displaystyle  = -60p^4\cdot x^2 \cdot cos(x^2) - 30p^4\cdot sin(x^2)
 * $$\displaystyle  = -60p^4\cdot x^2 \cdot cos(x^2) - 30p^4\cdot sin(x^2)
 *
$$\displaystyle
 * }
 * }

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

$$
 * $$\displaystyle f_{xy} = \frac{\partial }{\partial y} (\frac{\partial }{\partial x} (15p^4\cdot cos(x^2)))$$
 *
$$\displaystyle
 *
$$\displaystyle
 * }
 * }

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

$$ $$
 * $$\displaystyle = \frac{\partial }{\partial y} (-2x\cdot sin(x^2)\cdot 15p^4)
 * $$\displaystyle = \frac{\partial }{\partial y} (-2x\cdot sin(x^2)\cdot 15p^4)
 *
$$\displaystyle
 * }
 * }

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

$$ $$
 * $$\displaystyle = \frac{\partial }{\partial y} (-30p^4\cdot x\cdot sin(x^2))
 * $$\displaystyle = \frac{\partial }{\partial y} (-30p^4\cdot x\cdot sin(x^2))
 *
$$\displaystyle
 * }
 * }

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

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

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

$$
 * $$\displaystyle f_{yy} = \frac{\partial }{\partial y} (\frac{\partial }{\partial y} (15p^4\cdot cos(x^2)))$$
 *
$$\displaystyle
 *
$$\displaystyle
 * }
 * }

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

$$
 * $$\displaystyle  = \frac{\partial }{\partial y} (0)$$
 *
$$\displaystyle
 *
$$\displaystyle
 * }
 * }

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

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

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

$$
 * $$\displaystyle g_{xp} = \frac{\partial }{\partial p} (\frac{\partial }{\partial x} (6xy^2p - 6xp^5\cdot sin(x^2) + 2y^3))$$
 *
$$\displaystyle
 *
$$\displaystyle
 * }
 * }

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

$$
 * $$\displaystyle = \frac{\partial }{\partial p} (6y^2p-(6p^5x\cdot 2x\cdot cos(x^2) + 6p^5\cdot sin(x^2)))$$
 *
$$\displaystyle
 *
$$\displaystyle
 * }
 * }

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

$$
 * $$\displaystyle = 6y^2 - 60x^2\cdot cos(x^2)\cdot p^4 - 30sin(x^2)\cdot p^4$$
 *
$$\displaystyle
 *
$$\displaystyle
 * }
 * }

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

$$
 * $$\displaystyle g_{yp} = \frac{\partial }{\partial p} (\frac{\partial }{\partial y} (6xy^2p - 6xp^5\cdot sin(x^2) + 2y^3))$$
 *
$$\displaystyle
 *
$$\displaystyle
 * }
 * }

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

$$
 * $$\displaystyle = \frac{\partial }{\partial p} (12xyp - 0 +6y^2)$$
 *
$$\displaystyle
 *
$$\displaystyle
 * }
 * }

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

$$
 * $$\displaystyle = 12xy $$
 *
$$\displaystyle
 *
$$\displaystyle
 * }
 * }

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

$$
 * $$\displaystyle g_{y} = \frac{\partial }{\partial y} (6xy^2p - 6xp^5\cdot sin(x^2) + 2y^3))$$
 *
$$\displaystyle
 *
$$\displaystyle
 * }
 * }

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

$$
 * $$\displaystyle = 12xyp + 6y^2$$
 *
$$\displaystyle
 *
$$\displaystyle
 * }
 * }

Apply obtained factors to Eq(9.1)

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

$$ $$
 * $$\displaystyle L.H.S = -60p^4x^2\cdot cos(x^2) - 30p^4\cdot sin(x^2)
 * $$\displaystyle L.H.S = -60p^4x^2\cdot cos(x^2) - 30p^4\cdot sin(x^2)
 *
$$\displaystyle
 * }
 * }

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

$$ $$
 * $$\displaystyle R.H.S = (6y^2 - 60x^2\cdot cos(x^2)\cdot p^4 - 30sin(x^2)\cdot p^4)+ p(12xy) - (12xyp + 6y^2)
 * $$\displaystyle R.H.S = (6y^2 - 60x^2\cdot cos(x^2)\cdot p^4 - 30sin(x^2)\cdot p^4)+ p(12xy) - (12xyp + 6y^2)
 *
$$\displaystyle
 * }
 * }

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

$$
 * $$\displaystyle = -60p^4x^2\cdot cos(x^2) - 30p^4\cdot sin(x^2)$$
 *
$$\displaystyle
 *
$$\displaystyle
 * }
 * }

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

$$\displaystyle \therefore L.H.S = R.H.S$$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 *
$$$$
 * style= |
 * }

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

$$
 * $$\displaystyle for Eq(9.2)$$
 *
$$\displaystyle
 *
$$\displaystyle
 * }
 * }

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

$$
 * $$\displaystyle f_{xp} = \frac{\partial }{\partial p} (\frac{\partial }{\partial x} (15p^4\cdot cos(x^2)))$$
 *
$$\displaystyle
 *
$$\displaystyle
 * }
 * }

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

$$ $$
 * $$\displaystyle = \frac{\partial }{\partial p} (-2x\cdot sin(x^2)\cdot 15p^4)
 * $$\displaystyle = \frac{\partial }{\partial p} (-2x\cdot sin(x^2)\cdot 15p^4)
 *
$$\displaystyle
 * }
 * }

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

$$ $$
 * $$\displaystyle = \frac{\partial }{\partial p} (-30p^4\cdot x\cdot sin(x^2))
 * $$\displaystyle = \frac{\partial }{\partial p} (-30p^4\cdot x\cdot sin(x^2))
 *
$$\displaystyle
 * }
 * }

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

$$ $$
 * $$\displaystyle = -120x\cdot sin(x^2)\cdot p^3
 * $$\displaystyle = -120x\cdot sin(x^2)\cdot p^3
 *
$$\displaystyle
 * }
 * }

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

$$
 * $$\displaystyle f_{yp} = \frac{\partial }{\partial p} (\frac{\partial }{\partial y} (15p^4\cdot cos(x^2)))$$
 *
$$\displaystyle
 *
$$\displaystyle
 * }
 * }

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

$$
 * $$\displaystyle = \frac{\partial }{\partial p} (0)$$
 *
$$\displaystyle
 *
$$\displaystyle
 * }
 * }

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

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

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

$$
 * $$\displaystyle f_{y} = \frac{\partial }{\partial y} (15p^4\cdot cos(x^2)))$$
 *
$$\displaystyle
 *
$$\displaystyle
 * }
 * }

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

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

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

$$
 * $$\displaystyle g_{pp} = \frac{\partial }{\partial p} (\frac{\partial }{\partial p} (6xy^2p - 6xp^5\cdot sin(x^2) + 2y^3))$$
 *
$$\displaystyle
 *
$$\displaystyle
 * }
 * }

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

$$
 * $$\displaystyle g_{pp} = \frac{\partial }{\partial p} (6xy^2 - 30x/cdot sin(x^2)\cdot p^4)$$
 *
$$\displaystyle
 *
$$\displaystyle
 * }
 * }

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

$$
 * $$\displaystyle -120x\cdot sin(x^2)\cdot p^3$$
 *
$$\displaystyle
 *
$$\displaystyle
 * }
 * }

Apply obtained factors to Eq(9.2)

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

$$ $$
 * $$\displaystyle L.H.S = -120x\cdot sin(x^2)\cdot p^3
 * $$\displaystyle L.H.S = -120x\cdot sin(x^2)\cdot p^3
 *
$$\displaystyle
 * }
 * }

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

$$ $$
 * $$\displaystyle R.H.S = -120x\cdot sin(x^2)\cdot p^3
 * $$\displaystyle R.H.S = -120x\cdot sin(x^2)\cdot p^3
 *
$$\displaystyle
 * }
 * }

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

$$\displaystyle \therefore L.H.S = R.H.S$$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 *
$$$$
 * style= |
 * }

Contributing Members

 * Solved and posted by-User:Egm6321.f10.team3.Hong SJ 17:09, 04 October 2010 (UTC)
 * Proof read by Kurt Schulze
 * Proof read by Egm6321.f10.team3.Sudheesh 18:27, 6 October 2010 (UTC)

= Problem 10 - Generate exact N2-ODE's - Reverse Engineering=

Given
Given is the following function from meeting 17
 * {| style="width:100%" border="0"

$$\phi \left( x,y,p \right)=3{{p}^{5}}\cos \left( {{x}^{2}} \right)+2x{{y}^{3}}=k$$ and Eq. (2) from meeting 17
 *
$$\,(Eq. 10.1) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\left( 6x{{y}^{2}} \right)\cdot {y}'+2{{y}^{3}}=\left( {{h}_{y}}+0 \right){y}'+{{h}_{x}}$$ on page 17-1 and 17-3 respectively in the lecture notes.
 *
$$\,(Eq. 10.2) $$
 * style= |
 * }

Find
Using (Eq. 10.2) the function $$\,h\left( x,y \right)$$ has to be found in order to get again the given function $$\,\phi $$ (reverse engineering) assuming $${{h}_{y}}\cdot {y}'=2{{y}^{3}}$$.

Solution
Applying the 'separation of variables'-method to the given assumption, $$\,h\left( x,y \right)$$ can be calculated:
 * {| style="width:100%" border="0"

$${{h}_{y}}\cdot dy=2{{y}^{3}}\cdot dx$$
 *
$$\,(Eq. 10.3) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\int{{{h}_{y}}\cdot dy=\int{2{{y}^{3}}\cdot dx}}$$
 *
$$\,(Eq. 10.4) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$h\left( x,y \right)=2{{y}^{3}}x+{{k}_{1}}(y)$$ Take derivative of $$\,h\left( x,y \right)$$ with respect to $$\,x$$ gives:
 *
$$\,(Eq. 10.5) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\,{{h}_{x}}=2{{y}^{3}}$$ At the same time it can be seen from (Eq. 10.2) that $$\,{{h}_{x}}$$ is equal to $$\left( 6x{{y}^{2}} \right)\cdot {y}'$$, so that the following is valid:
 *
$$\,(Eq. 10.6) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$${{h}_{x}}=2{{y}^{3}}=6x{{y}^{2}}\cdot {y}'$$ What implies that:
 *
$$\,(Eq. 10.7) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$${y}'=\frac{y}{3x}$$ and
 *
$$\,(Eq. 10.8) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\,{{k}_{1}}(y)=0$$
 *
$$\,(Eq. 10.9) $$
 * style= |
 * }

We know from Eq. (1) of page 17-3 in meeting 17

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

 * $$\phi =h\left( x,y \right)+\int{f(x,y,p)dp}$$
 *
$$\,(Eq. 10.10) $$
 * style= |
 * }
 * style= |
 * }

For the given problem,

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

 * $$\int{f(x,y,p)dp}=3{{p}^{5}}\cos {{x}^{2}}$$ from page 17-3 meeting 17
 * style= |
 * }
 * }

Now, inserting the found experssion in (Eq. 10.5) with $$\,{{k}_{1}}(y)=0$$ into Eq. 10.10 gives:
 * {| style="width:100%" border="0"

$$\phi \left( x,y,p \right)=3{{p}^{5}}\cos \left( {{x}^{2}} \right)+2x{{y}^{3}}=k$$ Which is the same as the given function $$\,\phi \left( x,y,p \right)$$
 * style="width:2%; padding:1px; border:2px solid #0000FF" |
 * style="width:2%; padding:1px; border:2px solid #0000FF" |
 *
$$\,(Eq. 10.11) $$
 * style= |
 * }

Contributing Members

 * Solved by Michele Sigilló
 * Posted by Michele Sigilló
 * Proof read by Egm6321.f10.team3.Sudheesh 18:14, 5 October 2010 (UTC)

=Contributing Team Members=
 * Kurt Schulze, solved 7, proofed 2 and 9.
 * Michele Sigilló, solved 5,6,10, proofed 3,4,7.
 * Sung Joo, Hong, solved 3, 9, proofed 4.
 * Sudheesh Thiyyakkandi, Solved and posted 1(part1& part2), 2, Posted 7, proofed 3,4,5,9,10.
 * Adam Franklin, proofed 2, 4, 6, and 7.

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

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Motivation and emotion/Textbook/Emotion/Antisocial personality disorder/Draft

Chapter Outline (draft)

 * 1) What is anti social personality disorder? (DSM-IV-TR Criteria DSM-IV])
 * 2) Overview of ASPD - characteristics
 * 3) causes
 * 4) treatment
 * 5) prevalence
 * 6) Brief overview of emotion
 * 7) Psychodynamic Theory- Absence of parental love could lead to a lack of trust (Absence of emotion)
 * 8) Behavioural Theory- antisocial symptoms are learned through modelling. This behaviour is reinforced by parents rewarding (i.e. giving in to) aggressive behaviour. Needs to be linked to emotion
 * 9) Biological Theory
 * 10) Inheritance/Genetic influence
 * 11) Grey's Model of brain functioning as applied to ASPD
 * 12) Underarousal Hypothesis- low levels of cortical arousal leads to constant boredom (low stimulation)
 * 13) Fearlessness hypothesis- risk taking behaviour caused by a lack of fear
 * 14) ASPD and anxiety
 * 15) ASPD and stress- stress of combat may be a cause of ASPD
 * 16) Conclusion

What is Antisocial Personality Disorder?
Antisocial personality disorder Antisocial personality disorder, as defined by the Diagnostic and Statistical Manual, Fourth Edition, Text Revision (DSM-IV) is an ongoing disregard for the rights and feelings of others. This pattern of behaviour usually begins in childhood or adolescence and continues to develop in adulthood. For a diagnosis of antisocial personality disorder a person must be 18 years old and have displayed symptoms of conduct disorder (Conduct disorder) before the age of 15.

Characteristics of ASPD (Antisocial personality disorder)
The main characteristic of this disorder is the complete disregard for the wishes, rights or feelings of others. ASPD may also be known as psychopathy, sociopathy or dyssocial personality disorder. Traits typical of this disorder are: For a diangosis of ASPD an individual must display more than three of the above symptoms over a prolonged period of time, and they must significantly impair on an individuals daily functioning. If you have a friend who has been in troble with the law on one or two occasions, or know someone who impulsive, these characteristics alone do not necessarily warrant a diagnosis of ASPD. {| cellpadding="10" cellspacing="5" style="float: right; width: 50%; background-color: inherit; margin-left: auto; margin-right: auto"
 * Repeated violation of the rights of others
 * Repeatedly performing criminal acts such as destroying property, harassing others, stealing or pursuing illegal occupations.
 * Deceitful or manipulative behaviour used for personal gain or pleasure
 * Lying
 * Impulsivity, and an inability to plan ahead
 * Easily irritated and aggressive, and as a consequence often get into physical fights or physically abuse others (e.g. spouses or children)
 * Recklessness and a disregard for personal safety. This can be displayed in driving behaviours, such as speeding or driving whilst intoxicated. They may also engage in unsafe sexual relationships or use substances at risky levels.
 * Irresponsible- this often leads to an inability to hold a steady job, and difficulty in handling finances.
 * Show no remorse for their behaviour. Often they will rationalise their actions by blaming the victims of their crimes (e.g. "They deserved it", "If I didn't take advantage of them someone else would have")
 * Lack of empathy and a tendency to be callous, cynical and comtemptuous of the feelings, rights and suffering of others.
 * Inflated sense of self and arrogant demeanor
 * Superficial charm
 * style="width: 10%; background-color:Khaki; border: 1px solid #777777; vertical-align: top; -moz-border-radius-topleft: 8px; -moz-border-radius-bottomleft: 8px; -moz-border-radius-topright: 8px; -moz-border-radius-bottomright: 8px; height: 10px;" |

ASPD at the movies
Writers and directors in Hollywood have had an interest in mental illnesses for years and many successful movies have been made surrounding characters with various disorders. The main character, Sebastian, in the 1999 film Cruel Intentions displays some of the common symptoms of ASPD. He is easily able to manipulate those around him, is arrogant and has an inflated sense of self, and shows a complete disregard for the feelings of others. He frequently uses others for sexual gratification and shows no remorse for his actions. He is also very charming, which is essential for manipulation. However movies are not always accurate, as shown when Sebastion falls in love with a female character, and undergoes a change of heart. In reality, it is very unlikely for a person with ASPD to change their behaviour so quickly and drastically. But thats Hollywood for you. GET ON MY LEVEL. |}
 * }


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User talk:Theerakup

My name is Theerakup Thammaneewong ID 5180712 This is my video that uploaded to YOUTUBE http://www.youtube.com/watch?v=LltV02v8Ys0

I pad CPU & speed: 1 GHz apple A4 chip and 1 GHz clock speed Bus speed: very high bus speeds, according to Anandtech Cache size: 640 KB L2 cache Storage: type and amount: 16 GB – 64 GB flash storage 256 MB RAM but the RAM has higher clocked MH2 Black Berry CPU & speed 520 MHz Processor Bus speed 533 MHz Cache size 1MBL2 cache Storage ; type and amount : Internal memory 256 MB external 16 GB well, from the information above, i think I pad is much faster because ipad has a bigger CPu and speed because it has 1 GHZ, if we compare to Black berry, it contains just 520 MHz processor.

title: User talk:Theerakup


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User:Piraya~enwikiversity

http://www.youtube.com/watch?v=U0wCEy1hbYw

The Specification on iPad and iPhone3D

iPad:

CPU and speed:   1GHz A4 (Cortex A8)

bus speed:     -

cache size:    -

storage type:  internal storage(flash) amount:16GB

iPhone3D:

CPU and speed:  420MHz

bus speed:      100 MHz

cache size:     25KB or less

storage type:   11788 MB amount:16GB

To compare these two devices, we think that Ipad is more efficient than iphone because it has bigger screen which can later on connected to keyborad and work as a computer. It also very thin. Also, the cpu of ipad is faster that it can process quicker when using it.

Group members: Nantiya Arunrungruangkit and Piriya Chintanamankhong

title: User:Piraya~enwikiversity


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User:Iman95

Mariam Ahmed currently teaching at Sultan Qaboos University in Muscat, Oman. Graduated with Masters in Special Education, Masters in Teaching of English as a Second Langauge from the University of illinois, Champaign-Urbana (UIUC). Got Masters and M.phil in Linguistics and Phonetics respectively at Osmania University in Hyderabad, India. Undergraduation at St. Francis College in Hyderabad, India.

Reasearch interests are in latest trends in the teaching of English. Dynamism, Chaos and paradigm shifts are of special interest as these are now around which the discussion of classroom teaching and learning is focussed. Interested in works done by Canagarajah, Kumaravedivelu, Kachru, Prabhu, Finch, Larsen-Freeman to name a few.

title: User:Iman95


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User talk:Praewploy/Youtube Link

The following link is for homework2. This is for introducing myself and welcoming people to MUIC.

http://www.youtube.com/watch?v=kPtLXCLAAzg

title: User talk:Praewploy/Youtube Link


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User talk:Iman95

 Hello Iman95, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random project.
 * Browse Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Ottava Rima (talk) 12:55, 2 October 2010 (UTC)

title: User talk:Iman95


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User talk:Piraya~enwikiversity

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

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random project.
 * Browse Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Ottava Rima (talk) 12:56, 2 October 2010 (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 Piraya. To make sure that both of you can use all Wikimedia projects in future, we have reserved the name Piraya~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 23:41, 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) 06:18, 19 April 2015 (UTC)

title: User talk:Piraya~enwikiversity


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User talk:Fgrosshans

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

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random project.
 * Browse Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Ottava Rima (talk) 12:56, 2 October 2010 (UTC)

title: User talk:Fgrosshans


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User:Exfilia/What the Bible Says/Psalms/Psalm3

CUSTODIANS: Please don't move this yet. I want to build up a substantial volume of work before I go "live;" I don't want this project to be one of those with nine million red resources and one or two blue ones.

Directions
Read Psalm 3 in your preferred translation and in the King James Version. What differences do you find? How do they affect the passage's meaning?

If there are words in the passage that you're not familiar with, look them up now in Wiktionary or your dictionary of choice.

The Quiz
To test your understanding, complete the following texts based on the KJV version of Psalm 3. Yes, this is an open Book quiz.

{ Despair Lord, how are they increased that { trouble } me! many are they that { rise up against } me. Many there be which say of my soul, { There is no help for him in God. } Selah.
 * type="{}"}

{ Salvation But thou, O LORD, art { a shield for me }; { my glory }, and { the lifter up of mine head }. I laid me down and slept; I awaked; for the LORD { sustained } me. { Salvation } belongeth unto the LORD: thy { blessing } is upon thy { people }.
 * type="{}"}

Discussion questions

 * 1) Who do you think is the speaker in this psalm, as opposed to the author?
 * 2) In the psalm, what circumstances might the speaker and his audience be facing?
 * 3) Whom exactly do you think the speaker means to address?  How about the author?
 * 4) Cite examples of repetition in Psalm 3.  Why do you think the author repeated himself in these particular places?

Optional Activities
Won't change much from passage to passage
 * 1) Research the use of repetition as a characteristic of Hebrew poetry.
 * 2) Research techniques of translation of poetry.  How do translators handle conventions that exist in the source language but not the target language?
 * 3) What other passages of scripture reference this one?  What passages does this one reference?
 * 4) Research the effect this passage has had on the thought of one or more theologians.
 * 5) Research ways in which this passage has been used in art, music and literature.
 * 6) Research the ways in which this passage has been or might be used in your congregation's worship.
 * 7) Create an original work of art based on this passage.
 * 8) Design a way to use this passage in your personal devotions.
 * 9) Design a way to teach this passage to preliterate children: a skit, a feltboard presentation,a game, a craft activity, whatever.
 * 10) Design a way to use this passage to address the needs of some specific population: nursing home residents, grief support group, etc.
 * 11) Memory verse: choose the verse from this passage that seems most significant to you, and commit it to memory together with its chapter and verse reference.  For example, "Romans 13:1--Let every soul be subject unto the higher powers. For there is no power but of God: the powers that be are ordained of God"