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):

$\displaystyle$

$\displaystyle z(s)=\sum _{i=0}^{3}c_{i}s^{i}$

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:

$\displaystyle z'(s)=\sum _{i=1}^{3}ic_{i}s^{i-1}$

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):

$\displaystyle$

$\displaystyle z(s)=\sum _{i=0}^{3}c_{i}s^{i}$

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:

$\displaystyle z'(s)=\sum _{i=1}^{3}ic_{i}s^{i-1}$

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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