User:Egm6341.s10.Team4.nimaa&m/HW7

= Problem 2: Analytical equation for $$\displaystyle x(t) $$ in Newton-Raphson method =

Given
The relationship between population growth $$\displaystyle x $$ respect to time $$\displaystyle t $$ can be expressed as follows:


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$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \dot x=\frac{dx}{dt}=rx(1-\frac{x}{x_{max}})


 * }
 * }

Find
Find the analytical equation to show $$\displaystyle x $$ as a function of time $$\displaystyle t $$:

Solution

 * $$\displaystyle

\dot x=\frac{dx}{dt}=rx(1-\frac{x}{x_{max}}) $$


 * $$\displaystyle

\Rightarrow \frac{dx}{rx(1-\frac{x}{x_{max}}})=dt $$

For LHS we have:


 * $$\displaystyle

\frac{1}{x(1-\frac{1}{x_{max}}x)}=\frac{a}{x}+\frac{b}{1-\frac{1}{x_{max}}x}=\frac{a-\frac{a}{x_{max}}x+bx}{x(1-\frac{1}{x_{max}}x)} $$


 * $$\displaystyle

\Rightarrow a-\frac{a}{x_{max}}x+bx=1\Rightarrow x(b-\frac{a}{x_{max}})+a=1 $$


 * $$\displaystyle

\Rightarrow a=1 $$


 * $$\displaystyle

\Rightarrow b-\frac{a=1}{x_{max}}=0\Rightarrow b=\frac{1}{x_{max}} $$


 * $$\displaystyle

\Rightarrow \frac{1}{x(1-\frac{1}{x_{max}}x)}=\frac{1}{x}+\frac{\frac{1}{x_{max}}}{1-\frac{1}{x_{max}}x} $$


 * $$\displaystyle

\Rightarrow \frac{1}{r}\int\limits_{x_0}^{x}\frac{1}{x(1-\frac{1}{x_{max}}x)}\, dx=\frac{1}{r}\bigg[\int\limits_{x_0}^{x}\frac{1}{x}\,dx+\frac{1}{x_{max}}\int\limits_{x_0}^{x}\frac{1}{1-\frac{1}{x_{max}}x}\,dx\bigg]=\int\limits_{t_0}^{t}\,dt $$


 * $$\displaystyle

\Rightarrow \frac{1}{r}\bigg[ln(x)-ln(1-\frac{1}{x_{max}}x)]\Bigg ]^{x}_{x_0}=\frac{1}{r}\bigg[ln(\frac{x}{1-\frac{x}{x-{max}}})-ln(\frac{x_0}{1-\frac{x_0}{x_{max}}})\Bigg]=\int\limits_{t_0}^{t}\,dt $$


 * $$\displaystyle

\Rightarrow \frac{1}{r}ln(\frac{\frac{x}{1-\frac{x}{x_{max}}}}{\frac{x_0}{1-\frac{x_0}{x_{max}}}})=\frac{1}{r}ln(\frac{x(1-\frac{x_0}{x_{max}})}{x_0(1-\frac{x}{x_{max}})})=\int\limits_{t_0}^{t}\, dt=t-t_0 $$

Initial condition:


 * $$\displaystyle

t_0=0\Rightarrow x=x_0. $$


 * $$\displaystyle

\Rightarrow ln(\frac{x(1-\frac{x_0}{x_{max}})}{x_0(1-\frac{x}{x_{max}})})=r(t-t_0) $$


 * $$\displaystyle

\Rightarrow e^{rt}=\frac{x(1-\frac{x_0}{x_{max}})}{x_0(1-\frac{x}{x_{max}})}=\frac{x}{1-\frac{x}{x_{max}}}\times\frac{1-\frac{x_0}{x_{max}}}{x_0}=\frac{x_{max}.x}{x_{max}-x}\times\frac{x_{max}-x_0}{x_{max}.x_0} $$


 * $$\displaystyle

\Rightarrow \frac{x^2_{max}.x-x_{max}.x_0.x}{x^2_{max}.x_0-x_{max}.x_0.x}=e^{rt} $$


 * $$\displaystyle

\Rightarrow (x^2_{max}.x-x_{max}.x_0.x)=(x^2_{max}.x_0-x_{max}.x_0.x)e^{rt} $$


 * $$\displaystyle

\Rightarrow x(x^2_{max}-x_0.x_{max})=(x^2_{max}.x_0)e^{rt}-x_{max}.x_0.e^{rt}.x $$


 * $$\displaystyle

\Rightarrow x(x^2_{max}-x_0.x_{max}+x_{max}.x_0.e^{rt})=x^2_{max}.x_0.e^{rt} $$


 * $$\displaystyle

\Rightarrow x=\frac{x^2_{max}.x_0.e^{rt}}{x^2_{max}-x_0.x_{max}+x_{max}.x_0.e^{rt}}=\frac{x_{max}.x_0.e^{rt}}{x_{max}-x_0+x_0.e^{rt}} $$


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$$\displaystyle \Rightarrow x(t)=\frac{x_0.x_{max}.e^{rt}}{x_{max}+x_0(e^{rt}-1)} $$
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Author
Solved and typed by - Egm6341.s10.Team4.nimaa&amp;m 15:50, 11 April 2010 (UTC) .

= Problem 8: Rate of momentum change for optimal control problem =

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



Find
Show that $$\displaystyle \bar{CD}=\bar{AB} $$ + higher order terms of $$\displaystyle d\gamma $$.

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, for two generated triangles $$\displaystyle \triangle OAB $$ and $$\displaystyle \triangle ODC $$ following relationships can be written:


 * $$\displaystyle

tan(d\gamma)=\frac{\bar {AB}}{\bar {OA}}=\frac{\bar {AB}}{v} $$

Likewise we have below relationship in $$\displaystyle \triangle ODC $$:


 * $$\displaystyle

sin(d\gamma)=\frac{\bar {CD}}{\bar {OD}}=\frac{\bar {CD}}{v+dv} $$

Hence, by applying Taylor series for these two cases:


 * $$\displaystyle

tan(d\gamma)=d\gamma+\frac{d\gamma^3}{3}+\frac{2d\gamma^5}{15}+...=\frac{\bar{AB}}{v} $$


 * $$\displaystyle

\Rightarrow \bar{AB}=v(d\gamma+\frac{d\gamma^3}{3}+...)=v.d\gamma+v\frac{d\gamma^3}{3}+v\frac{2d\gamma^5}{15}+... $$


 * $$\displaystyle

sin(d\gamma)=d\gamma-\frac{d\gamma^3}{3!}+\frac{d\gamma^5}{5!}+...=\frac{\bar{CD}}{v+dv} $$


 * $$\displaystyle

\Rightarrow \bar{CD}=(v+dv)(d\gamma-\frac{d\gamma^3}{3!}+\frac{d\gamma^5}{5!}...)=v.d\gamma-v\frac{d\gamma^3}{3!}+v\frac{d\gamma^5}{5!}+... $$


 * $$\displaystyle

\Rightarrow \bar{CD}-\bar{AB}=-\frac{1}{2}vd\gamma^3-\frac{1}{8}vd\gamma^5... $$


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$$\displaystyle \Rightarrow \bar{CD}=\bar{AB}+v.f(d\gamma^3)+v.f(d\gamma^5)+... $$
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Author
Solved and typed by - Egm6341.s10.Team4.nimaa&amp;m 22:19, 22 April 2010 (UTC) .

= Problem 14: Obtaining $$\displaystyle a_k $$ in cosine fourier transform =

Given
The Cosine series as Fourier transform for function $$\displaystyle f(x) $$ is expressed as follows:


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$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle f(cos\theta)=\frac{a_0}{2}+\sum_{k=1} a_k.cos(k\theta)


 * }
 * }

Find
Demonstrate the following expression for $$\displaystyle a_k $$:


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$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle a_k=\frac{2}{\pi}\int\limits_{0}^{\pi} f(cos\theta).cos(k\theta)\, d\theta


 * }
 * }

Solution
Considering $$\displaystyle x=cos(\theta) $$, cosine Fourier series for $$\displaystyle f(x) $$ is written as:


 * $$\displaystyle

f(x=cos(\theta))=\frac{a_0}{2}+\sum_{k=1} a_k.cos(k\theta) $$


 * $$\displaystyle

\xrightarrow{\times cos(k\theta)}f(x).cos(k\theta)=\frac{a_0}{2}.cos(k\theta)+\sum_{k=1} a_k.cos^2(k\theta) $$


 * $$\displaystyle

\xrightarrow{\int\limits_{0}^{\pi}}\int\limits_{0}^{\pi}f(x).cos(k\theta)\, d\theta=\frac{a_0}{2}\int\limits_{0}^{\pi}cos(k\theta)\, d\theta+\sum_{k=1} a_k\int\limits_{0}^{\pi}cos^2(k\theta)\, d\theta $$


 * $$\displaystyle

\frac{a_0}{2}\times\underbrace{\frac{1}{k}[sin(k\theta)]_{0}^{\pi}}_{=0}+\sum_{k=1}a_k\int\limits_{0}^{\pi}cos^2(k\theta)\, d\theta=\sum_{k=1}a_k\int\limits_{0}^{\pi}cos^2(k\theta)\, d\theta $$


 * $$\displaystyle

cos(2\theta)=cos^2(\theta)-sin^2(\theta)=2cos^2(\theta)-1\Rightarrow cos^2(\theta)=\frac{cos(2\theta)+1}{2} $$


 * $$\displaystyle

\Rightarrow \int\limits_{0}^{\pi}f(x).cos(k\theta)\, d\theta=\sum_{k=1} \lbrack a_k\times\frac{1}{2}\int\limits_{0}^{\pi}(cos(2\theta)+1)\, d\theta\rbrack=\frac{1}{2}\sum_{k=1}a_k\lbrack\int\limits_{0}^{\pi}cos(2\theta)\, d\theta+\int\limits_{0}^{\pi}\, d\theta]=\frac{1}{2}\sum_{k=1}a_k\lbrack[\underbrace{\frac{1}{2}sin(2\theta)]_{0}^{\pi}}_{=0}+\underbrace{[\theta]_{0}^{\pi}}_{\pi}\rbrack=\frac{\pi}{2}\sum_{k=1}a_k $$


 * $$\displaystyle

\Rightarrow a_k=\frac{2}{\pi}\int\limits_{0}^{\pi}f(x).cos(k\theta)\, d\theta $$


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$$\displaystyle a_k=\frac{2}{\pi}\int\limits_{0}^{\pi}f(x).cos(k\theta)\, d\theta $$
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Author
Solved and typed by - Egm6341.s10.Team4.nimaa&amp;m 19:15, 18 April 2010 (UTC) .