User:Egm6341.s10.Team4.nimaa&m/HW6

= Problem 2: Rate of momentum change for optimal control problem =

Given
Envisage the below figure as free body diagram of an aircraft:

.

likewise consider the shown axes and vectors, for $$\displaystyle v+dv $$ at $$\displaystyle t+dt $$:

.

Find
Show that $$\displaystyle dp_{\bar y}=mvd\gamma $$.

Solution
According to the above figure, we can survey these two cases at $$\displaystyle t $$ and $$\displaystyle t+dt $$, the velocity of the aircraft after $$\displaystyle dt $$ will reach to $$\displaystyle v+dv $$ and the angle between the aircraft and horizontal axis will reach to the $$\displaystyle \gamma+d\gamma $$. Thus, regarding $$\displaystyle d\gamma $$ generated angle between two velocity vectors, we can write:


 * $$\displaystyle

d\gamma\approxeq sin(d\gamma)=\frac{dv_{\bar y}}{v+dv} $$


 * $$\displaystyle

\Rightarrow dv_{\bar y}=v.d\gamma+dv.d\gamma $$

The amount of $$\displaystyle dv.d\gamma $$ can be neglected in front of $$\displaystyle v.d\gamma $$.


 * $$\displaystyle

\Rightarrow dv_{\bar y}=v.d\gamma $$

On the other hand, momentum is defined as $$\displaystyle p=mv $$. So, we have:



dp_{\bar y}=d(m.v_{\bar y})=v_{\bar y}.dm+m.dv_{\bar y} $$

Assuming the amount of $$\displaystyle dm $$ to be negligible in front of changes in velocity;



dp_{\bar y}=d(m.v_{\bar y})=m.dv_{\bar y} $$

Finally, we can summarize the answer as:



dp_{\bar y}=m.dv_{\bar y}=m.v.d\gamma $$


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$$\displaystyle dp_{\bar y}=m.v.d\gamma $$
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Author
Solved and typed by - Egm6341.s10.Team4.nimaa&amp;m 03:08, 3 April 2010 (UTC) .

= Problem 7: Expression for Hermitian interpolation at $$\displaystyle t_{i+\frac{1}{2}} $$ =

Given
Consider the Hermitian interpolation by the following equation (on slide [[media:Egm6341.s10.mtg35.djvu|35-2]]):
 * {| style="width:100%" border="0" align="left"

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle z(s)=\sum_{i=0}^3 c_is^i


 * }
 * }

Find
Show the following expression can be obtained for $$\displaystyle z'_{i+\frac{1}{2}} $$:


 * $$\displaystyle

z_{i+\frac{1}{2}}=z(s=\frac{1}{2})=\frac{1}{2}(z_i+z_{i+1})+\frac{h}{8}(f_i-f_{i+1}) $$

Solution
By differentiating from the equation for $$\displaystyle z(s) $$, we will attain:


 * $$\displaystyle

z'(s)=\sum_{i=1}^3 ic_is^{i-1} $$

Now, we can compute the followings:


 * $$\displaystyle

z_i=z(s=0)=\sum_{i=0}^3 c_i(0)^i=c_0+0+0+0=c_0 $$


 * $$\displaystyle

z_{i+1}=z(s=1)=\sum_{i=0}^3 c_i(1)^i=c_0+c_1+c_2+c_3 $$


 * $$\displaystyle

z'_i=z'(s=0)=\sum_{i=1}^3 ic_i(0)^{i-1}=c_1+0+0=c_1 $$


 * $$\displaystyle

z'_{i+1}=z'(s=1)=\sum_{i=1}^3 ic_i(1)^{i-1}=c_1+2c_2+3c_3 $$


 * $$\displaystyle

\dot{z}=\frac{dz}{dt}=\frac{dz}{ds}\times\frac{ds}{dt}=z'\times\frac{1}{h} $$


 * $$\displaystyle

\Rightarrow h(f_i-f_{i+1})=h(\dot{z}_i-\dot{z}_{i+1})=(z'_i-z'_{i+1}) $$


 * $$\displaystyle

\Rightarrow \frac{h}{8}(f_i-f_{i+1})=\frac{h}{8}(\dot{z}_i-\dot{z}_{i+1})=\frac{1}{8}(z'_i-z'_{i+1}) $$


 * $$\displaystyle

\Rightarrow\begin{cases} z_i+z_{i+1}=2c_0+c_1+c_2+c_3 \\ f_i-f_{i+1}=-2c_2-3c_3 \end{cases} $$


 * $$\displaystyle

\Rightarrow\begin{cases} \frac{1}{2}(z_i+z_{i+1})=c_0+\frac{1}{2}c_1+\frac{1}{2}c_2+\frac{1}{2}c_3 \\ \frac{h}{8}(f_i-f_{i+1})=-\frac{1}{4}c_2-\frac{3}{8}c_3 \end{cases} $$


 * $$\displaystyle

\Rightarrow\frac{1}{2}(z_i+z_{i+1})+\frac{h}{8}(f_i-f_{i+1})=c_0+\frac{1}{2}c_1+\frac{1}{2}c_2+\frac{1}{2}c_3-\frac{1}{4}c_2-\frac{3}{8}c_3=c_0+\frac{1}{2}c_1+\frac{1}{4}c_2+\frac{1}{8}c_3 $$

The acquired foregoing equation is equal to RHS of the expression. Now, we can compute the LHS of it as:


 * $$\displaystyle

LHS=z_{i+\frac{1}{2}}=z(s=\frac{1}{2})=\sum_{i=0}^3 c_i(\frac{1}{2})^i=c_0\times 1+c_1\times\frac{1}{2}+c_2\times\frac{1}{4}+c_3\times\frac{1}{8}=c_0+\frac{1}{2}c_1+\frac{1}{4}c_2+\frac{1}{8}c_3 $$


 * $$\displaystyle

\Rightarrow LHS=RHS $$


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

$$\displaystyle z_{i+\frac{1}{2}}=z(s=\frac{1}{2})=\frac{1}{2}(z_i+z_{i+1})+\frac{h}{8}(f_i-f_{i+1}) $$
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Author
Solved and typed by - Egm6341.s10.Team4.nimaa&amp;m 04:15, 3 April 2010 (UTC) .

= Problem 8: Expression for derivative of Hermitian interpolation at $$\displaystyle t_{i+\frac{1}{2}} $$ =

Given
Consider the Hermitian interpolation by the following equation (on slide [[media:Egm6341.s10.mtg35.djvu|35-2]]):
 * {| style="width:100%" border="0" align="left"

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle z(s)=\sum_{i=0}^3 c_is^i


 * }
 * }

Find
Show the following expression can be obtained for $$\displaystyle z'_{i+\frac{1}{2}} $$:


 * $$\displaystyle

z'_{i+\frac{1}{2}}=z'(s=\frac{1}{2})=-\frac{3}{2}(z_i-z_{i+1})-\frac{1}{4}(z'_i+z'_{i+1}) $$

Solution
By differentiating from the equation for $$\displaystyle z(s) $$, we will attain:


 * $$\displaystyle

z'(s)=\sum_{i=1}^3 ic_is^{i-1} $$

Now, we can compute the followings:


 * $$\displaystyle

z_i=z(s=0)=\sum_{i=0}^3 c_i(0)^i=c_0+0+0+0=c_0 $$


 * $$\displaystyle

z_{i+1}=z(s=1)=\sum_{i=0}^3 c_i(1)^i=c_0+c_1+c_2+c_3 $$


 * $$\displaystyle

z'_i=z'(s=0)=\sum_{i=1}^3 ic_i(0)^{i-1}=c_1+0+0=c_1 $$


 * $$\displaystyle

z'_{i+1}=z'(s=1)=\sum_{i=1}^3 ic_i(1)^{i-1}=c_1+2c_2+3c_3 $$


 * $$\displaystyle

\Rightarrow\begin{cases} z_i-z_{i+1}=c_0-(c_0+c_1+c_2+c_3)=-c_1-c_2-c_3 \\ z'_i+z'_{i+1}=2c_1+2c_2+3c_3 \end{cases} $$


 * $$\displaystyle

\Rightarrow\begin{cases} -\frac{3}{2}(z_i-z_{i+1})=\frac{3}{2}c_1+\frac{3}{2}c_2+\frac{3}{2}c_3\\ -\frac{1}{4}(z'_i+z'_{i+1})=-\frac{1}{2}c_1-\frac{1}{2}c_2-\frac{3}{4}c_3 \end{cases} $$


 * $$\displaystyle

\Rightarrow -\frac{3}{2}(z_i-z_{i+1})-\frac{1}{4}(z'_i+z'_{i+1})=(\frac{3}{2}-\frac{1}{2})c_1+(\frac{3}{2}-\frac{1}{2})c_2+(\frac{3}{2}-\frac{3}{4})c_3=c_1+c_2+\frac{3}{4}c_3 $$

The acquired foregoing equation is equal to RHS of the expression. Now, we can compute the LHS of it as:


 * $$\displaystyle

LHS=z'_{i+\frac{1}{2}}=z'(s=\frac{1}{2})=\sum_{i=1}^3 ic_i(\frac{1}{2})^{i-1}=1\times c_1\times (\frac{1}{2})^{1-1}+2\times c_2\times (\frac{1}{2})^{2-1}+3\times c_3\times (\frac{1}{2})^{3-1}=c_1+c_2+\frac {3}{4}c_3 $$


 * $$\displaystyle

\Rightarrow LHS=RHS $$


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

$$\displaystyle z'_{i+\frac{1}{2}}=z'(s=\frac{1}{2})=-\frac{3}{2}(z_i-z_{i+1})-\frac{1}{4}(z'_i+z'_{i+1}) $$
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
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Author
Solved and typed by - Egm6341.s10.Team4.nimaa&amp;m 04:23, 3 April 2010 (UTC) .