User:Egm6321.f10.team5/hw3

= Problem 1 - Derive EOM (SC-N1-ODE) = From Meeting 13, p. 13-1 ~ p. 13-2

Given
A projectile of mass (m) is being shot through the air at an angle (α) to the horizontal with velocity (v). The projectile experiences a drag force due to air resistance, $$F_{air} = k v^n \,$$ Where $$k \,$$ is the coefficient of air resistance and is a constant in this problem. The force equations are therefore given as such, $$m(\frac {dv_x} {dt}) = -k v^{n} {\cos {\alpha}}$$ $$\displaystyle (Eq. 1.1) $$ $$m(\frac {dv_y} {dt}) = -k v^{n} {\sin {\alpha}} - mg$$ $$\displaystyle (Eq. 1.2) $$

Find
For the given situation: $$m(\frac {dv_y} {dt}) = -k v^{n} {\sin {\alpha}} - mg$$
 * 1) Derive the equations of motion for the projectile
 * 2) For the particular case $$k = 0 \,$$ verify that $$y(x)\,$$ is a parabola
 * 3) Consider $$k \ne 0 \,$$ and $$v_{xo}=0 \,$$, therefore

becomes

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


 * 3.1 Find $$v_y (t)\,$$ and $$y(t) \,$$ for $$m = constant \,$$
 * 3.2 Find $$v_y (t)\,$$ and $$y(t) \,$$ for $$m = m(t) \,$$

Part 1
The projectile is given an initial velocity $$ v $$ in a direction that is at an angle $$ \alpha $$ to the horizontal. The Equations of motion (Eq. 1.1 and Eq. 1.2) given in the problem statement are acquired through summing the forces on the projectile and decomposing those forces into their vector quantities acting on the projectile. We sum the forces by looking at which ones are having an effect on the motion of the object.

1) inertia of the projectile after being fired from its initial position:

$$F_{inertia} = m(\frac {dv} {dt}) $$

which must be decomposed into its x and y components:

$$F_{x-inertia} = m(\frac {dv_x} {dt}) $$

$$F_{y-inertia} = m(\frac {dv_y} {dt}) $$

2) The effects of gravity on the y-component of the projectile's inertia:

$$F_{g} = -m g \,$$

$$g =  \,$$ acceleration  due  to  gravity

3) The force of air resistance on the projectile's motion decomposed into its x and y directional components:

$$F_{x-air} = -k v^{n} {\cos {\alpha}} \,$$

$$F_{y-air} = -k v^{n} {\sin {\alpha}} \,$$

Now if we sum the forces in the x and y directions and consider the force equilibrium in those directions we obtain the Equations of Motion given in the problem statement:

$$\sum {F_x} = m(\frac {dv_x} {dt}) = -k v^{n} {\cos {\alpha}}$$

$$\sum {F_y} = m(\frac {dv_y} {dt}) = -k v^{n} {\sin {\alpha}} - mg$$

Part 2
For the particular case where $$ k = 0 $$ equations 1.1 and 1.2 reduce to:

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

which reduces to

$$ (\frac {dv_x} {dt}) = 0$$$$\displaystyle (Eq. 1.3) $$

and

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

which reduces further to

$$(\frac {dv_y} {dt}) = -g $$$$\displaystyle (Eq. 1.4) $$

Now we can integrate Eqs. 1.3 and 1.4 with respect to time in order to get expressions for velocity and position in the x and y directions. We will start with the x-direction equations:

x-component:

$$ \frac {dv_x} {dt} = 0$$

Integrating with respect to time we obtain,

$$ v_x(t) = c_1 \,$$$$\displaystyle (Eq. 1.5) $$

Using the initial condition $$ v_x(t=0) = v_{xo} \,$$ the constant of integration $$ c_1 \,$$ can be found,

$$ c_1 = v_{xo} \,$$

So Eq. 1.5 becomes,

$$ v_x(t) = v_{xo} \,$$ $$\displaystyle (Eq. 1.6) $$

Now we repeat the same procedure for Eq. 1.4 to obtain the y component of the velocity vector. Once we integrate Eq. 1.4 with respect to time we get,

$$ v_y(t) = -gt + c_2 \,$$$$\displaystyle (Eq. 1.7) $$

Then using the initial condition $$ v_y(t=0) = v_{yo} \,$$ we can determine that $$ c_2 = v_{yo} \,$$ so then Eq. 1.7 becomes,

$$ v_y(t) = -gt + v_{yo} \,$$$$\displaystyle (Eq. 1.8) $$

Now we integrate Eq. 1.7 and 1.8 with respect to time once more to obtain the equations for position as a function of time in the x and y directions. After integrating with respect to time Eq. 1.7 and 1.8 become,

$$ x(t) = v_{xo}t + c_3 \,$$$$\displaystyle (Eq. 1.9) $$

and

$$ y(t) = - \frac 1 2 gt^2 + v_{yo}t + c_4 \,$$$$\displaystyle (Eq. 1.10) $$

Using initial conditions for position $$ x(t=0) = x_{o}\,$$ and $$ y(t=0) = y_{o}\,$$ then $$c_3\,$$ and $$c_4\,$$ can be solved for,

$$ c_3 = x_{o}\,$$

and

$$ c_4 = y_{o}\,$$

Plugging these values back into Eq. 1.9 and 1.10 yields the position equations for the x and y directions,

$$ x(t) = v_{xo}t + x_o \,$$$$\displaystyle (Eq. 1.11) $$

$$ y(t) = - \frac 1 2 gt^2 + v_{yo}t + y_o \,$$$$\displaystyle (Eq. 1.12) $$

To express y as a function of x we will manipulate eq. 1.11 to get t in terms of x. This yeilds the expression,

$$ t = \frac {x-x_o} {v_{xo}} \,$$$$\displaystyle (Eq. 1.13) $$

We then plug this value in for t in Eq. 1.12 to get,

$$ y(x) = - \frac 1 2 g(\frac {x-x_o} {v_{xo}})^2 + v_{yo}(\frac {x-x_o} {v_{xo}}) + y_o \,$$$$\displaystyle (Eq. 1.14) $$

to confirm that Eq. 1.14 is a parabola we can simply compare the general form of a parabolic function to Eq. 1.14,

$$ y(x) = ax^2 + bx + c \,$$

$$ y(x) = - \frac 1 2 g(\frac {x-x_o} {v_{xo}})^2 + v_{yo}(\frac {x-x_o} {v_{xo}}) + y_o \,$$

where $$a$$, $$b$$ and $$c$$ are all constants. It is apparent that Eq. 1.14 fits the form of a parabolic function so it can be concluded that our expression for $$y(x)$$ is a parabola.

Part 3
For the case where $$v_{xo}=0 \,$$, and $$k \ne 0$$ we can see from Eq. 1.1 that as long as $$v^n \ne 0 \,$$ then $$\alpha = 90^\circ \,$$ so that there are no net forces acting on the projectile in the x direction which leaves Eq. 1.2 as the net force acting on the projectile. However, since $$\alpha = 90^\circ \,$$ Eq. 1.2 can be re-written as,

$$m(\frac {dv_y} {dt}) = -k v_y^{n}- mg \,$$ $$\displaystyle (Eq. 1.15) $$

Part 3.1
Solving Eq. 1.15 for $$v_y(t)$$ and $$y(t)$$ when $$m =$$ constant. Start with the equation of motion (Eq. 1.15),

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

In order to solve this we must first show exactness. So we first put Eq 1.15 into a form that shows the first condition of exactness is met

$$m(\frac {dv_y} {dt})+ k v_y^{n}+ mg = 0\,$$ $$\displaystyle (Eq. 1.16) $$

So that we can see that the two parts, $$M(v_y,t)$$ and $$N(v_y,t)$$ are,

$$M(v_y,t)=k v_y^{n}+ mg $$

$$N(v_y,t)=m \,$$

Now that the first condition of exactness has been shown we must see if the second condition of exactness is met,

$$ M(v_y,t)_{v_y} = N(v_y,t)_t$$<p style="text-align:right;">$$\displaystyle (Eq. 1.17) $$

Which is clearly not met if $$m$$ is a constant

$$M(v_y,t)_{v_y} = knv_y^{n-1} \ne N(v_y,t)_{t} = \frac{dm(t)}{dt}$$<p style="text-align:right;">$$\displaystyle (Eq. 1.18) $$

Therefore, the equation is non-exact.

However, the equation can be made exact through the integrating factor method

Then expressing as a total derivative,

Keeping in mind that Eq. 1.20 is a sum of the forces so we can introduce the $$\phi$$ notation that will simplify the equation a bit.

$$ \sum F = \frac {\partial \phi} {\partial t} dt + \frac {\partial \phi} {\partial v_y} dv_y =0 $$

Therefore the $$\phi$$ derivatives are as follows,

$$\phi_t = h(v_y,t)(kv_y^n+mg)$$

$$\phi_{v_y} = h(v_y,t)m)$$

$$\phi_{t,v_y} = \phi_{v_y,t}$$

Now, we test for exactness. In order for Eq. 1.20 to be exact it must meet the condition,

$$h_{v_y}N+hN_{v_y}=h_tM+hM_t \,$$<p style="text-align:right;">$$\displaystyle (Eq. 1.21) $$

So we substitute the problem parameters into Eq. 1.21 and get,

$$\frac{\partial h(v_y,t)}{\partial v_y}(kv_y^n+mg)+h(v_y,t)\frac{\partial(kv_y^n+mg)}{\partial v_y} = \frac{\partial h(v_y,t)}{\partial t}m+h(v_y,t)\frac{\partial m}{\partial t}$$<p style="text-align:right;">$$\displaystyle (Eq. 1.22) $$

Letting $$\partial h / \partial v_y = 0$$ and since $$m$$ is a constant $$\partial m / \partial t = 0$$. So Eq. 1.22 reduces to,

$$h(v_y,t)\frac{\partial(kv_y^n+mg)}{\partial v_y} = \frac{\partial h(v_y,t)}{\partial t}m$$<p style="text-align:right;">$$\displaystyle (Eq. 1.23) $$

Now, we manipulate Eq 1.23 in order to obtain an expression for $$h(v_y,t)$$,

$$h(v_y,t)\frac{\partial(kv_y^n+mg)}{\partial v_y} = \frac{\partial h(v_y,t)}{\partial t}m$$<p style="text-align:right;">$$\displaystyle (Eq. 1.24) $$

$$h(v_y,t)\frac{nkv_y^n}{m} = \frac{\partial h(v_y,t)}{\partial t}$$<p style="text-align:right;">$$\displaystyle (Eq. 1.25) $$

$$h(v_y,t) = \exp(\frac{kv_y^n}{m}t)$$<p style="text-align:right;">$$\displaystyle (Eq. 1.26) $$

Therefore the $$\phi$$ derivative equations can be solved to get $$\phi$$,

$$ \phi_t = \exp(\frac{kv_y^n}{m}t)(kv_y^n+mg) $$ <p style="text-align:right;">$$\displaystyle (Eq. 1.27) $$

$$ \phi = \int{\phi_t dt} = \int{ \exp(\frac{kv_y^n}{m}t)(kv_y^n+mg)dt} $$ <p style="text-align:right;">$$\displaystyle (Eq. 1.28) $$

$$ \phi =(kv_y^n+mg)\frac{m}{kv_y^n}\exp(\frac{kv_y^n}{m}t)+f(v_y) = c_5 $$ <p style="text-align:right;">$$\displaystyle (Eq. 1.29) $$

$$ \phi_v = m\exp(\frac{kv_y^n}{m}t) $$ <p style="text-align:right;">$$\displaystyle (Eq. 1.30) $$

$$ \phi = \int{\phi_{v_y} dv_y} = \int{ m\exp(\frac{kv_y^n}{m}t)dv_y} = c_5 $$ <p style="text-align:right;">$$\displaystyle (Eq. 1.31) $$

$$ \phi = m\int{\exp(\frac{kv_y^n}{m}t)dv_y} + f(t) = c_5 $$ <p style="text-align:right;">$$\displaystyle (Eq. 1.32) $$

We can now combine these $$\phi$$ equations to come up with an expression of the sum of the forces that is exact,

$$ \sum F =\frac {\partial \phi} {\partial t} dt + \frac {\partial \phi} {\partial v_y} dv_y =\phi = m\int{\exp(\frac{kv_y^n}{m}t)dv_y} + (kv_y^n+mg)\frac{m}{kv_y^n}\exp(\frac{kv_y^n}{m}t) = c_5 $$ <p style="text-align:right;">$$\displaystyle (Eq. 1.33) $$

This resulting equation is not integrable and therefore no explicit general form of $$v_y$$ was found. However, through the integrating factor method an exact non-linear expression was found. Numerical solutions can be found given a value of n, and initial conditions.

Part 3.2
For the condition where $$m=m(t) \ne 0$$, the integrating factor becomes even more complicated because in Eq 1.22 the initial simplification that occurs when $$ \frac {\partial m} {\partial t} = 0$$ does not happen. This will result in another implicit solution for $$v_y$$ when the integrating factor method is applied.

Author and Proof-reader
[Author] Rob Carroll

[Proof-reader] Scott Elliott

= Problem 2 - Derive EOM (SC-L1-ODE) = From Meeting 13, p. 13-3

Given
A system of pendulum.

Part 1
Derive the following equations.

Part 2
Write the above equations in the form of $$\dot x(t) = A(t) x(t) + B(t) u(t)$$. Find A, B, and u.

Part 1
We use Newtons's second law for Rotation which is $$\Sigma \tau = I \alpha$$ for the derivation. We note that there are three torques applied on the particle 1; gravitational torque, controlling torque, and spring torque. Gravitational torque is Controlling torque is Spring torque is Moment of inertia for the point mass is Angular acceleration for the point mass is We set up the equation of motion for the first particle. Apply the small-angle approximation which states that $$\sin x \approx x$$ and $$\cos x \approx 1$$ to simplify equation as We use the same techniques to derive the equation of motion for second particle. Gravitational torque is Controlling torque is Spring torque is Moment of inertia for the point mass is Angular acceleration for the point mass is We set up the equation of motion for the second particle. Apply the small-angle approximation which states that $$\sin x \approx x$$ and $$\cos x \approx 1$$ to simplify equation as

Part 2
We write $$\dot x(t) = A(t) x(t) + B(t) u(t)$$ in matrix form. We divide the both side of $$m_1 l^2 \ddot \theta_1\ = - k a^2 (\theta_1 - \theta_2) - l \theta_1 m_1 g + l u_1$$ by $$m_1 l^2$$ to get Similarly, we divide the both side of $$m_2 l^2 \ddot \theta_2\ = - k a^2 (\theta_2 - \theta_1) - l \theta_2 m_2 g + l u_2$$ by $$m_2 l^2$$ to get We can easily determine that the all the components on row 1 and row 3 of matrix A and matrix B are null except for $$A_{12}$$ and $$A_{34}$$, so we write that Now we plug in the expressions of $$\ddot \theta_1$$ and $$\ddot \theta_2$$ into the matrix.

Author and Proof-reader
[Author] Egm6321.f10.team5.abc

[Proof-reader] Mike Faraone / Oh, Sang min

= Problem 3 - Derive (L1-ODE-CC) = From Meeting 14, p. 14-1

Given
An L1-ODE-CC equation.

Find
Solve $$\dot x(t) = a(t) x(t) + b(t) u(t)$$ for the case where $$a(t)$$ and $$b(t)$$ are constant coefficients while u(t) is prescribed.

Solve
We state that $$\displaystyle a(t) = a$$ and $$\displaystyle b(t) = b$$ for this case. So we have the equation in the form of $$\displaystyle hx' + h'x = hk$$.

Note that we cannot use the equation of $$\displaystyle y(x) = \frac 1 {h(x)} \int^x h(s) d(s) ds$$

since the lower limit is bounded as $$\displaystyle t_0$$.

Deriving an equation for the lower bound

Apply the modified equation given above to our problem. It is possible to simplify the equation further as

Author and Proof-reader
[Author] Egm6321.f10.team5.abc

[Proof-reader] Mike Faraone / Oh, Sang min

= Problem 4 - Expand Taylor series(exponential and exponential matrix) = From Meeting 14, p. 14-2

Given

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$$\displaystyle {e}^{x} = 1 + \frac{x}{1!} + \frac{x^2}{2!} + ... $$ $$\displaystyle \ = \sum_{k=0}^{\infty} \frac{x^k}{k!} $$ <p style="text-align:right;">$$\displaystyle (4-1) $$
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$$\displaystyle \underbrace{{e}^{\underline{A}}}_{\underline{A} \ : \ n \times n \ Matrix} = \underbrace{\underline{\mathit I }}_{n \times n \ Unit \ mat.} + \frac{\underline{A}}{1!} + \frac{\underline{A}^2}{2!} + ... $$ $$\displaystyle \ = \sum_{k=0}^{\infty} \frac{\underline{A}^k}{k!} $$ <p style="text-align:right;">$$\displaystyle (4-2) $$
 * }
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Find
1) Derive (4-1) 

2) Derive (4-2) 

Solve
Solution of 1)

Using Taylor series ,

$$f(a)+\frac {f'(a)}{1!} (x-a)+ \frac{f''(a)}{2!} (x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+ \cdots. $$

which can be written in the more compact sigma notation as


 * $$ \sum_{n=0} ^ {\infin } \frac {f^{(n)}(a)}{n!} \, (x-a)^{n} $$

In the particular case where a = 0, the series is also called a Maclaurin series
 * {| style="width:100%" border="0" align="left"
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$$f(x) = f(0)+f'(0)x + \frac{f''(0)}{2!}x^2+\frac{f^{(3)}(0)}{3!}x^3+ \cdots .$$ $$ \ = \sum_{n=0} ^ {\infin } \frac {f^{(n)}(0)}{n!} \, (x)^{n}$$ <p style="text-align:right;">$$\displaystyle (4-3) $$
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$$\displaystyle f(x) := {e}^{x} $$ <p style="text-align:right;">$$\displaystyle (4-4) $$
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$$\displaystyle f(0) = 1, \ f'(x) = {e}^{x}, \ f'(0) = 1, \ ... $$ <p style="text-align:right;">$$\displaystyle (4-5) $$
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$$\displaystyle f(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} \ ... $$ <p style="text-align:right;">$$\displaystyle (4-6) $$
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$$\therefore {e}^{x} = \sum_{k=0}^{\infty} \frac{x^k}{k!} $$
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Solution of 2)

Using Taylor series and Maclaurin series


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$$\displaystyle f(\underline{x}) :=  \underbrace{{e}^{\underline{A}}}_{\underline{A} \ : \ n \times n \ Matrix} $$ <p style="text-align:right;">$$\displaystyle (4-7) $$
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<Background Knowledge> - Exponential Matrix, Identity Matrix


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$$\displaystyle f(\underline{0}) = \underbrace{\underline{\mathit  I }}_{n \times n \ Unit \ mat.}, \ f'(\underline{x}) = {e}^{\underline{A}}, \ f'(\underline{0}) = \underline{\mathit  I }, \ ... $$ <p style="text-align:right;">$$\displaystyle (4-8) $$
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$$\displaystyle f(\underline{x}) = 1 + \underline{A} + \frac{\underline{A}^2}{2!} + \frac{\underline{A}^3}{3!} \ ... $$ <p style="text-align:right;">$$\displaystyle (4-9) $$
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$$\therefore {e}^{\underline{A}} = \sum_{k=0}^{\infty} \frac{\underline{A}^k}{k!} $$
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Author and Proof-reader
[Author] Oh, Sang Min

[Proof-reader] Mike Faraone

= Problem 5 - Generalized to SC-L1-ODE-VC = From Meeting 14, p. 14-2

Given
Dimension of matrix

$$\displaystyle \underline{x} \ : \ n \times \ 1 $$

$$\displaystyle \underline{A} \ : \ n \times \ n $$

$$\displaystyle \underline{B} \ : \ n \times \ m $$

$$\displaystyle \underline{U} \ : \ m \times \ 1 $$

Find
Generalized (5-3) to SC-L1-ODE-VC

Solve
 SC-L1-ODE-CC can be generalized to SC-L1-ODE-CC as same as L1-ODE-CC is generalized to L1-ODE-VC 

Using (5-1) ~ (5-3)

 SC-L1-ODE-VC 


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$$ \underline{x}(t)= {e}^{\int_{{t}_{0}}^{t}{e}^{\underline{A} (\tau)}d\tau}\cdot \underline{x}({t}_{0}) + \int_{{t}_{0}}^{t}{e}^{\int_{\tau}^{t}{e}^{\underline{A} (s)}ds} \cdot \underline{B}\cdot \underline{U}(\tau)d\tau $$
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Dimension of matrix

$$\displaystyle \underline{x} \ : \ n \times \ 1 $$

$$\displaystyle \underline{A} \ : \ n \times \ n $$

$$\displaystyle \underline{B} \ : \ n \times \ m $$

$$\displaystyle \underline{U} \ : \ m \times \ 1 $$

Author and Proof-reader
[Author] Oh, Sang Min

[Proof-reader]

= Problem 6 - Obtaining SC-L1-ODE-CC with int. factor method = From Meeting 15, p. 15-1

Given
We are given a system of coupled linear first order ODEs with constant coefficients,
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A part of the solution to this system is the state transition matrix, $$\underline{\Phi}$$, which has the properties
 * style="width:95%" | $$\displaystyle \underline{\dot{x}}(t)=\underline{A}\underline{x}(t)+\underline{B}\underline{u}(t)$$
 * <p style="text-align:right">(6.A)
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and
 * style="width:95%" | $$\displaystyle \frac{d}{dt}\underline{\phi}(t,t_0)=\underline{A}\underline{\phi}(t,t_0)$$
 * <p style="text-align:right">(6.B)
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 * }
 * {| style="width:100%" border="0"

where $$I$$ is the identity matrix.
 * style="width:95%" | $$\displaystyle \underline{\phi}(t_0,t_0)=I,$$
 * <p style="text-align:right">(6.C)
 * }
 * }

Find
Use equations (6.A), (6.B), and (6.C), to obtain the following equation:
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$$\displaystyle \underline{x}(t)=\underbrace{\left[{\rm exp}\underline{A}(t-t_0)\right]}_{\underline{\phi}(t,t_0)}\underline{x}(t_0)+\int\limits_{t_0}^{t}{\underbrace{\left[{\rm exp}\underline{A}(t-\tau)\right]}_{\underline{\phi}(t,\tau)}\underline{B}\underline{u}(\tau)\,d\tau} $$
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 * style="width:95%" |
 * <p style="text-align:right">(6.D)
 * }

Solution
We can derive from $$\frac d {dt} \underline \phi (t,t_0) = \underline A \underline \phi (t,t_0)$$ that $$\underline A = \frac {\frac d {dt} \underline \phi (t,t_0)} {\underline \phi (t,t_0)}$$. Because $$\phi (t_0,t_0) = \underline 1$$ which is an identity matrix.

Author and Proof-reader
[Author] Egm6321.f10.team5.abc

Egm6321.f2010.team5.riveros 17:18, 6 October 2010 (UTC)

[Proof-reader] Michael Steinberg

= Problem 7 - Application SC-L1-ODE-CC about rolling control of rocket = From Meeting 15, p. 15-1

Given

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

$$\displaystyle \phi $$ =   roll angle

$$\displaystyle \omega $$ =  roll angular velocity

$$\displaystyle Q $$ =   aileron efflectiveness

$$\displaystyle \tau $$ = roll time constant


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Find
Put (7-1) ~ (7-3) in form of (7-4)

Solve

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$$
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\begin{pmatrix} \dot\phi\\ \dot\omega\\ \dot\delta \end{pmatrix} = \begin{pmatrix} 0 & 1 & 0 \\ 0 & \frac{-1}{\tau} & \frac{Q}{\tau}\\ 0 & 0 & 0 \end{pmatrix}

\begin{pmatrix} \phi\\ \omega\\ \delta \end{pmatrix} + \begin{pmatrix} 0\\ 0\\ 1 \end{pmatrix} u $$

$$ \underline{A}= \begin{pmatrix} 0 & 1 & 0 \\ 0 & \frac{-1}{\tau} & \frac{Q}{\tau}\\ 0 & 0 & 0 \end{pmatrix}

, \ \underline{B}= \begin{pmatrix} 0\\ 0\\ 1 \end{pmatrix}

$$
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Author and Proof-reader
[Author] Oh, Sang Min

[Proof-reader] Mike Faraone

= Problem 8 - Searching for solution without assuming = From Meeting 17, p. 17-1

Given
A second order nonlinear ordinary differential equation (N2-ODE) may be represented as follows (First Cond. of Exactness):

$$ F(x,y,y',y) = g(x,y,p) + f(x,y,p)y = 0 \! $$

where $$ p = y' \! $$

If both the first and second conditions of exactness (see Meeting 15, pp. 15-2, 15-3) are satisfied,

Find
Define the method for determining $$ h(x,y) \! $$ for the case when $$ h(x,y) \not= const \! $$

Solve
Using Eq. (8.3), the following partial derivatives can be obtained

Recalling the definition of $$ g \! $$ given in Eq. (8.1) and substituting (8.4) and (8.5) into (8.1) yields

Because we already know $$ g \! $$ from the N2-ODE given, we can set like terms equal in order to solve for $$ h(x,y) \! $$.

For example, should (8.6) simplify into the following form,

$$ ay' + b = h_x + h_y y' \! $$

then we may set $$ h_x = b \! $$ and $$ h_y = a \! $$ and solve for $$ h(x,y) \! $$.

Author and Proof-reader
[Author] Michael Steinberg

[Proof-reader]

=Problem 9 - Show the given equation satisfies the second condition for exactness= From Meeting 17, p. 17-2

Given
We are given a nonlinear second order ODE,
 * $$ \displaystyle

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

f = \phi_p = 15p^4cos (x^2) $$ and,
 * $$ \displaystyle

g = \phi_yy' + \phi_x = (6xy^2)y' + (-6xp^5sin(x^2) + 2y^3) $$

Find
The second condition of exactness for nonlinear second order ODEs states that the ODE must satisfy the following two relations:
 * 1) $$ \displaystyle f_{xx}+2pf_{xy}+p^2f_{yy} = g_{xp}+pg_{yp}-g_{y}$$
 * 2) $$ \displaystyle f_{xp}+pf_{yp}+2f_{y}=g_{pp}$$

Show that the given equation satisfies the second condition of exactness.

First relation
The terms on the left hand side are found,
 * $$\displaystyle

f_{xx} = -60\,{p}^{4} \ {x}^{2} \ cos({x}^{2})-30\,{p}^{4} \ sin({x}^{2}) $$


 * $$\displaystyle

f_{xy} = 0 $$


 * $$\displaystyle

f_{yy} = 0 $$ Thus, the left hand side is,
 * $$\displaystyle

f_{xx}+2pf_{xy}+p^2f_{yy} = -60\,{p}^{4} \ {x}^{2} \ cos({x}^{2})-30\,{p}^{4}\sin({x}^{2}) $$ The individual terms on the right hand side are found,
 * $$\displaystyle

g_{xp} = 6\,{y}^{2}-30\,{p}^{4}\sin({x}^{2})-60\,{p}^{4}\ {x}^{2} \cos({x}^{2}) $$


 * $$\displaystyle

g_{yp} = 12\,xy $$


 * $$\displaystyle

g_{y} = 12\,xyp+6\,{y}^{2} $$ Thus, the right hand side is,
 * $$\displaystyle

g_{xp}+pg_{yp}-g_{y} = -60\,{p}^{4} \ {x}^{2} \ cos({x}^{2})-30\,{p}^{4}\sin({x}^{2}) $$ We then check for equality,
 * $$\displaystyle f_{xx}+2pf_{xy}+p^2f_{yy} =g_{xp}+pg_{yp}-g_{y} \implies -60\,{p}^{4} \ {x}^{2} \ cos({x}^{2})-30\,{p}^{4}\sin({x}^{2})=-60\,{p}^{4} \ {x}^{2} \ cos({x}^{2})-30\,{p}^{4}\sin({x}^{2})$$

Which is true.

Second relation
The individual terms on the left hand side are found,
 * $$\displaystyle

f_{xp} = -120\,{p}^{3} \ x \ sin({x}^{2}) $$


 * $$\displaystyle

f_{yp} = 0 $$


 * $$\displaystyle

f_{y} = 0 $$ Thus, the left hand side is,
 * $$\displaystyle

f_{xp}+pf_{yp}+2f_{y}=-120\,{p}^{3} \ x \ sin({x}^{2}) $$ The right hand side is then found,
 * $$\displaystyle

g_{pp}=-120\,{p}^{3} \ x \ sin({x}^{2}) $$

We check for equality,
 * $$\displaystyle f_{xp}+pf_{yp}+2f_{y}=g_{pp} \implies -120\,{p}^{3} \ x \ sin({x}^{2})=-120\,{p}^{3} \ x \ sin({x}^{2})$$

Which is also true; therefore, the given equation satisfies the second condition of exactness.

Author and Proof-reader
[Author] Egm6321.f2010.team5.riveros 06:20, 5 October 2010 (UTC)

[Proof-reader] Michael Steinberg / Oh, Sang min

= Problem 10 - Finish story = From Meeting 18, p. 18-3

Given
A nonlinear second order ODE.

Find
Assume $$\displaystyle h_y y' = 2 y^3$$ and solve $$\displaystyle (6 x y^2) y' + 2 y^3 = h_y y' + h_x$$.

Solve
We derive the following procedures according to the given assumption. Note from the equation of $$\displaystyle \ (6 x y^2) y' + 2 y^3 = h_y y' + h_x$$ that $$\displaystyle  \ h_x = (6 x y^2) y'$$. Take a derivative of $$\displaystyle \ h(x,y)$$ with respect to $$\displaystyle \ y$$ to find that Multiply by $$\displaystyle \ y'$$ to determine that $$\displaystyle \ k_1(y)$$ is a constant. Since $$\displaystyle \ {k_1}'(y) = 0$$, then $$\displaystyle \ k_1 (y) = k_1$$.

Author and Proof-reader
[Author] Egm6321.f10.team5.abc [Proof-reader] Oh, Sang min

= References =

= Contributing Members =

Egm6321.f10.team5.oh Solved and typed problems 4, 5 and 7 / proof-read problems 2, 3, 9 and 10

Michael Faraone Solved problem 2. Helped with part1 problem 1. proof-read and checked problems 2,3,4,7

Egm6321.f10.team5.abc Authored problem 2, problem 3, problem 6, problem 10.

Michael Steinberg Authored problem 8, assisted with problem 6, and proof-read problem 6, problem 9.