# 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 ${\displaystyle \displaystyle v+dv}$ at ${\displaystyle \displaystyle t+dt}$:

.

## Find

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

## Solution

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

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

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

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

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

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

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

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

Finally, we can summarize the answer as:

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

## Author

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

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

## Given

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

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

## Find

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

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

## Solution

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

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

Now, we can compute the followings:

${\displaystyle \displaystyle z_{i}=z(s=0)=\sum _{i=0}^{3}c_{i}(0)^{i}=c_{0}+0+0+0=c_{0}}$
${\displaystyle \displaystyle z_{i+1}=z(s=1)=\sum _{i=0}^{3}c_{i}(1)^{i}=c_{0}+c_{1}+c_{2}+c_{3}}$
${\displaystyle \displaystyle z'_{i}=z'(s=0)=\sum _{i=1}^{3}ic_{i}(0)^{i-1}=c_{1}+0+0=c_{1}}$
${\displaystyle \displaystyle z'_{i+1}=z'(s=1)=\sum _{i=1}^{3}ic_{i}(1)^{i-1}=c_{1}+2c_{2}+3c_{3}}$
${\displaystyle \displaystyle {\dot {z}}={\frac {dz}{dt}}={\frac {dz}{ds}}\times {\frac {ds}{dt}}=z'\times {\frac {1}{h}}}$
${\displaystyle \displaystyle \Rightarrow h(f_{i}-f_{i+1})=h({\dot {z}}_{i}-{\dot {z}}_{i+1})=(z'_{i}-z'_{i+1})}$
${\displaystyle \displaystyle \Rightarrow {\frac {h}{8}}(f_{i}-f_{i+1})={\frac {h}{8}}({\dot {z}}_{i}-{\dot {z}}_{i+1})={\frac {1}{8}}(z'_{i}-z'_{i+1})}$
${\displaystyle \displaystyle \Rightarrow {\begin{cases}z_{i}+z_{i+1}=2c_{0}+c_{1}+c_{2}+c_{3}\\f_{i}-f_{i+1}=-2c_{2}-3c_{3}\end{cases}}}$
${\displaystyle \displaystyle \Rightarrow {\begin{cases}{\frac {1}{2}}(z_{i}+z_{i+1})=c_{0}+{\frac {1}{2}}c_{1}+{\frac {1}{2}}c_{2}+{\frac {1}{2}}c_{3}\\{\frac {h}{8}}(f_{i}-f_{i+1})=-{\frac {1}{4}}c_{2}-{\frac {3}{8}}c_{3}\end{cases}}}$
${\displaystyle \displaystyle \Rightarrow {\frac {1}{2}}(z_{i}+z_{i+1})+{\frac {h}{8}}(f_{i}-f_{i+1})=c_{0}+{\frac {1}{2}}c_{1}+{\frac {1}{2}}c_{2}+{\frac {1}{2}}c_{3}-{\frac {1}{4}}c_{2}-{\frac {3}{8}}c_{3}=c_{0}+{\frac {1}{2}}c_{1}+{\frac {1}{4}}c_{2}+{\frac {1}{8}}c_{3}}$

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

${\displaystyle \displaystyle LHS=z_{i+{\frac {1}{2}}}=z(s={\frac {1}{2}})=\sum _{i=0}^{3}c_{i}({\frac {1}{2}})^{i}=c_{0}\times 1+c_{1}\times {\frac {1}{2}}+c_{2}\times {\frac {1}{4}}+c_{3}\times {\frac {1}{8}}=c_{0}+{\frac {1}{2}}c_{1}+{\frac {1}{4}}c_{2}+{\frac {1}{8}}c_{3}}$
${\displaystyle \displaystyle \Rightarrow LHS=RHS}$
 ${\displaystyle \displaystyle z_{i+{\frac {1}{2}}}=z(s={\frac {1}{2}})={\frac {1}{2}}(z_{i}+z_{i+1})+{\frac {h}{8}}(f_{i}-f_{i+1})}$
.

## Author

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

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

## Given

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

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

## Find

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

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

## Solution

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

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

Now, we can compute the followings:

${\displaystyle \displaystyle z_{i}=z(s=0)=\sum _{i=0}^{3}c_{i}(0)^{i}=c_{0}+0+0+0=c_{0}}$
${\displaystyle \displaystyle z_{i+1}=z(s=1)=\sum _{i=0}^{3}c_{i}(1)^{i}=c_{0}+c_{1}+c_{2}+c_{3}}$
${\displaystyle \displaystyle z'_{i}=z'(s=0)=\sum _{i=1}^{3}ic_{i}(0)^{i-1}=c_{1}+0+0=c_{1}}$
${\displaystyle \displaystyle z'_{i+1}=z'(s=1)=\sum _{i=1}^{3}ic_{i}(1)^{i-1}=c_{1}+2c_{2}+3c_{3}}$
${\displaystyle \displaystyle \Rightarrow {\begin{cases}z_{i}-z_{i+1}=c_{0}-(c_{0}+c_{1}+c_{2}+c_{3})=-c_{1}-c_{2}-c_{3}\\z'_{i}+z'_{i+1}=2c_{1}+2c_{2}+3c_{3}\end{cases}}}$
${\displaystyle \displaystyle \Rightarrow {\begin{cases}-{\frac {3}{2}}(z_{i}-z_{i+1})={\frac {3}{2}}c_{1}+{\frac {3}{2}}c_{2}+{\frac {3}{2}}c_{3}\\-{\frac {1}{4}}(z'_{i}+z'_{i+1})=-{\frac {1}{2}}c_{1}-{\frac {1}{2}}c_{2}-{\frac {3}{4}}c_{3}\end{cases}}}$
${\displaystyle \displaystyle \Rightarrow -{\frac {3}{2}}(z_{i}-z_{i+1})-{\frac {1}{4}}(z'_{i}+z'_{i+1})=({\frac {3}{2}}-{\frac {1}{2}})c_{1}+({\frac {3}{2}}-{\frac {1}{2}})c_{2}+({\frac {3}{2}}-{\frac {3}{4}})c_{3}=c_{1}+c_{2}+{\frac {3}{4}}c_{3}}$

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

${\displaystyle \displaystyle LHS=z'_{i+{\frac {1}{2}}}=z'(s={\frac {1}{2}})=\sum _{i=1}^{3}ic_{i}({\frac {1}{2}})^{i-1}=1\times c_{1}\times ({\frac {1}{2}})^{1-1}+2\times c_{2}\times ({\frac {1}{2}})^{2-1}+3\times c_{3}\times ({\frac {1}{2}})^{3-1}=c_{1}+c_{2}+{\frac {3}{4}}c_{3}}$

${\displaystyle \displaystyle \Rightarrow LHS=RHS}$
 ${\displaystyle \displaystyle z'_{i+{\frac {1}{2}}}=z'(s={\frac {1}{2}})=-{\frac {3}{2}}(z_{i}-z_{i+1})-{\frac {1}{4}}(z'_{i}+z'_{i+1})}$
.

## Author

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