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:

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

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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:

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

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

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Author

Solved and typed by - Egm6341.s10.Team4.nimaa&m 03:08, 3 April 2010 (UTC) .

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

Given

Consider the Hermitian interpolation by the following equation (on slide 35-2):

$\displaystyle$

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

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:

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

$\displaystyle \Rightarrow LHS=RHS$

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

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Author

Solved and typed by - Egm6341.s10.Team4.nimaa&m 04:15, 3 April 2010 (UTC) .

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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 35-2):

$\displaystyle$

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

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:

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

$\displaystyle \Rightarrow LHS=RHS$

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

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Author

Solved and typed by - Egm6341.s10.Team4.nimaa&m 04:23, 3 April 2010 (UTC) .

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