User:Egm6341.s2010.Team1/HW7

= Problem 1: Linearization about xmax =

Given
We are given that


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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle f(x)=rx(1-\dfrac{x}{x_{max}})
 * }.
 * }.

and


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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle x=\hat{x}+y
 * }.
 * }.

where


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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \hat{x}=x_{max}
 * }.
 * }.

Find
Need to find the growth equation $$\dfrac{dy}{dt}$$

Solution
Using the above relations we have that


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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle f(x)=r(x_{max}+y)(1-\dfrac{x_{max}+y}{x_{max}})
 * }.
 * }.

canceling out some terms leads to


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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle r(x_{max}+y)(1-1-\dfrac{y}{x_{max}})
 * }.
 * }.

If we continue on algebraic manipulation we have


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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle -r(x_{max}+y)(\dfrac{y}{x_{max}})
 * }.
 * }.


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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle -r(y+\dfrac{y^2}{x_{max}})
 * }.
 * }.


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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle -ry(1+\dfrac{y}{x_{max}})
 * }.
 * }.

Thus we then can approximate


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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \dfrac{dy}{dt}=-ry
 * }.
 * }.

which can then lead to the exponential relation


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

$$ $$ Egm6341.s10.team1.toddmock 23:38, 20 April 2010 (UTC)
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle y=y_0e^{-rt}
 * }.
 * }.

= Problem 2: Verification of differential equation =

Given
We know that the rate of change is given as:


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

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

Find
Need to solve the differential equation to come up with a relation for x(t)

Solution
From separation of variables we can go from the original equation


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

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

to


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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \int_{x_0}^{x}\frac{x_{max}}{x(x_{max}-x)}dx=\int_{t_0}^{t}rdt
 * }.
 * }.

we can separate this into fractions to solve it and we obtain


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

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

from here we must determine the coefficients a and b. To do this we rearrange to get


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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \frac{a-\frac{a}{x_{max}}x+bx}{x(1-\frac{1}{x_{max}}x)}
 * }.
 * }.

combining like terms and setting this equation equal to 1 we get


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

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

right away we can see that


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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle a=1
 * }.
 * }.

and


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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle b-\frac{1}{x_{max}}=0, b=\frac{1}{x_{max}}
 * }.
 * }.

Thus giving the integral


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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \int_{x_0}^{x}\frac{1}{x}dx+\int_{x_0}^{x}\frac{1}{(x_{max}-x)}dx=\int_{t_0}^{t}rdt
 * }.
 * }.

evaluation of the integral leads to


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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle ln(\frac{x}{x_0})-ln(\frac{x_{max}-x}{x_{max}-x_0})= rt
 * }.
 * }.

which then gives upon rearranging


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

$$ $$ Egm6341.s10.team1.toddmock 23:39, 20 April 2010 (UTC)
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle x(t)=\frac{x_{max}x_0e^{rt}}{x_{max}+x_0(e^{rt}-1)}
 * }.
 * }.

= Problem 3: Hermite-Simpson Algorithm to integrate Verhulst equation =

Find
Use Hermite-Simpson rule to integrate the above given Verhulst equation

Given
Refer Lecture slide (1) [[media:Egm6341.s10.mtg38.djvu|p.38-3]]

$$ f(x)=rx (1-\frac {x}{x_{max}}) $$

$$ x_{max}=10\,$$

$$ r=1.2 \,$$

2cases:

$$ x_{0}=2 <\frac {1}{2}x_{max} $$

$$ x_{0}=7 >\frac {1}{2}x_{max} $$

Such that $$ AbsTol= O(10^{-6})\, $$ for $$ t \isin [0,10]$$

Solution
For the first case:

$$ x_{0}=2 \, $$

Matlab Code:

---

For the second case:

$$ x_{0}=7 \, $$

Matlab Code:



Solved by :Egm6341.s10.team1.lei03:52, 23 April 2010 (UTC)

= Problem 4: Reproduce the Chaotic System of Population Dynamics =

Problem Statement
Reproduce Figure 15.6 and 15.7 in "Differential equations : linear, nonlinear, ordinary, partial " on page 455-456

Ref: Lecture Notes [[media:Egm6341.s10.mtg40.djvu|p.40-1]]  

Solution
(1) Figure 15.6 Copy the code directly from the book (2) Figure 15.7 Change initial conditions from random to 0.1 and 0.1+10e-16 at r=4



- EGM6341.s10.Team1.Ya-Chiao Chang 18:08, 21 April 2010 (UTC)

=Problem 5: Determine the limit on the step size in the Hermite-Simpson method, Forward Euler, Backward Euler=

Statement
Hermite-Simpson algorithm was used to find $$\hat{h}=0.125$$ to give an error of order $$10^{-6}$$ from the true value where $$\hat{h}$$ is the step size between nodes in the numerical solution. The function to be integrated was the population logistic equation,


 * $$\dot{x}=rx\left ( 1-\frac{x}{x_{max}} \right )$$

The numerical integration using the Hermite-Simpson algorithm was compared to the true value for x,


 * $$x\left ( t \right )=\frac{x_0x_{max}e^{rt}}{x_{max}+x_0\left ( e^{rt}-1 \right )}$$

For this problem, the step size is increased at a rate of $$h_{new}=2^k\hat{h}$$ where $$k=1,2,3,...$$

Solution
The differential equation is first integrated using ODE45 (at MATLAB command).

Next the Hermite-Simpson algorithm is run using increasing values of step size. The max step size is limited by the size of the interval $$t\in [0,10]$$. The step size maxes out at 8. A step size of 16 is not allowed. Therefore this algorithm is stable at integrating this function.

Next the Forward Euler integration is used. The step size maxes at 2 at which point it wildly oscillates about $$x_{max}=10$$

Finally, backward Euler integration is used. The step size is allowed to go to 8 so it is stable like the Hermite-Simpson algorithm. Notice, though, that the answer is less accurate for the same number of steps compared to the Hermite-Simpson method

EGM6341.s10.Team1.Kumanchik 20:15, 23 April 2010 (UTC)

=Problem 6: Interceptor Simulation using Hermite-Simpson algorithm=

Statement
Use the Hermite-Simpson algorithm to determine the trajectory of an interceptor given the following inputs.


 * $$T(t)=\left\{\begin{matrix}

6000 N, & 0\leq t< 27, & 33\leq t\leq 40\\ 1000 N, & 27\leq t< 33 & \end{matrix}\right.$$


 * $$\alpha(t)=\left\{\begin{matrix}

0.03 rad & 0\leq t< 21\\ \frac{0.13-0.09}{27-21}(t-21)+0.09 rad & 21\leq t< 27\\ \frac{-0.2+0.13}{33-27}(t-27)-0.13 rad & 27\leq t< 33\\ \frac{-0.13+0.2}{40-33}(t-33)-0.2 rad & 33\leq t\leq 40 \end{matrix}\right.$$


 * $$\mathbf{f}=

\begin{bmatrix} \dot{\gamma}\\ \dot{v}\\ \dot{x}\\ \dot{y} \end{bmatrix}=\begin{bmatrix} \frac{T-D}{mv}sin(\alpha)+\frac{L}{mv}cos(\alpha)-\frac{g}{v}cos(\gamma)\\ \frac{T-D}{m}cos(\alpha)+\frac{L}{m}sin(\alpha)-gsin(\gamma)\\ vcos(\gamma)\\ vsin(\gamma) \end{bmatrix}$$


 * $$\mathbf{z}=\begin{bmatrix}

\gamma\\ v\\ x\\ y \end{bmatrix}$$


 * $$DL=\left\{\begin{matrix}

D(\alpha,v,y)=0.5C_d\rho v^2S_{ref}\\ L(\alpha,v,y)=0.5C_c\rho v^2S_{ref} \end{matrix}\right. $$


 * $$\left\{\begin{matrix}

C_d(\alpha)=A_1\alpha^2+A_2\alpha+A_3\\ C_l(\alpha)=B_1\alpha+B_2\\ \rho(y)=C_1y^2+C_2y+C_3 \end{matrix}\right.$$


 * $$\left\{\begin{matrix}

m=1005 kg\\ g=9.81 \frac{m}{s^2}\\ S_{ref}=0.3376 m^2\\ A_1=-1.9431\\ A_2=-0.1499\\ A_3=0.2359\\ B_1=21.9\\ B_2=0\\ C_1=3.312e^{-9} \frac{kg}{m^5}\\ C_2=-1.142e^{-4} \frac{kg}{m^4}\\ C_3=1.224 \frac{kg}{m^3} \end{matrix}\right.$$


 * $$t\in [0,40]$$ seconds
 * $$h=1$$ seconds

Solution
The control inputs are the thrust, T, and the angle of attack, $$\alpha$$. They are plotted below.

The Hermite-Simpson algorithm takes the solution to a differential equation and subdivides it into nodes. Each node is related to the next by,


 * $$z_{i+1}=z_{i}+\frac{h}{6}\left [ f_i+4f_{i+\frac{1}{2}}+f_{i+1} \right ]$$
 * $$z_{i+\frac{1}{2}}=\frac{1}{2}\left [ z_i+z_{i+1} \right ]+\frac{h}{8}\left [ f_i-f_{i+1} \right ]$$

However, this equation is dependent on the $$z_{i+1} \,$$ node which appears in the $$f_{i+1} \,$$ and $$f_{i+\frac{1}{2}}$$ terms. Therefore, the equation is solved for by the Newton Raphson method,


 * $$F=-z_{i+1}+z_{i}+\frac{h}{6}\left [ f_i+4f_{i+\frac{1}{2}}+f_{i+1} \right ]=0$$

Intelligently choosing values for $$z_{i+1} \,$$ allows $$F \,$$ to approach 0. The intelligent algorithm is,


 * $$z^{(k+1)}=z^{(k)}-\frac{dF(z^{(k)})}{dz}^{-1}F(z^{(k)})$$

where $$k $$ is the previous guess and $$k+1$$ is the next guess. This algorithm can search for $$z_{i+1}$$ by starting with $$z^{k}=z_i$$. Using $$F$$ as a metric, the algorithm can stop when $$F\approx0$$ or for this problem we choose $$F$$ of $$O(10^{-9})$$.

This solution can be extended to this interceptor problem which solves a system of differential equations. Placing the differential equations in matrix form, the solution remains identical. Keep in mind that the differential of a matrix is the Jacobian. The result of the interceptor problem is plotted below. The time step is 1 second between 0 and 40 seconds.



The source code of the interceptor algorithm

Thrust and angle functions (control variables)

EGM6341.s10.Team1.Kumanchik 22:55, 23 April 2010 (UTC) =Problem 7: Integrate the logistic equation using an inconsistent trapezoidal-Simpson algorithm=

Statement
Integrate the logistic equation,
 * $$\dot{x}=rx\left ( 1-\frac{x}{x_{max}} \right )$$

Using the Hermite-Simpson algorithm and the inconsistent Trapezoidal-Simpson algorithm, compare their errors for the same step size, $$h=0.125$$. Both methods require the Newton-Raphson method so use a error for each node of $$O(10^{-9})$$. Compare the integrations to the true integral of the logistic equation,
 * $$x\left ( t \right )=\frac{x_0x_{max}e^{rt}}{x_{max}+x_0\left ( e^{rt}-1 \right )}$$

Solution
The Hermite-Simpson algorithm was adjusted to create the inconsistent trapezoidal-Simpson algorithm. Both solutions were subtracted from the true solution. The error in the Hermite-Simpson algorithm was designed from problem 5 as less than $$O(10^{-6})$$. Using the same parameters on the inconsistent trapezoidal-Simpson algorithm yields worse error by about 3 orders of magnitude. The errors from the true value are plotted below. The line on the bottom is the Hermite-Simpson error. It is much smaller than the inconsistent trapezoidal-Simpson error.

EGM6341.s10.Team1.Kumanchik 20:15, 23 April 2010 (UTC)

= Problem 8: Solve Logistic Equation by Using Inconsistent Trap-Simpson Algorithm =

Problem Statement
Compare with Hermite-Simpson results using same value of h Ref: Lecture Notes [[media:Egm6341.s10.mtg41.djvu|p.41-3]] [[media:Egm6341.s10.mtg39.djvu|p.39-1]] 

Solution
Inconsistent Trap-Simpson Algorithm Matlab code change from $$ x_{\frac{1}{2}} = \!\,\frac{1}{2}*[x(i)+x(i+1)]+\frac{h}{8}*(f_1-f_2)$$ to $$ x_{\frac{1}{2}} = \!\,\frac{1}{2}*[x(i)+x(i+1)]$$

(1) initial condition is $$x_0 = \!\, 2 $$



- (2) initial condition is $$x_0 = \!\, 7 $$

II Compare with Problem 3-Hermite-Simpson Algorithm to integrate Verhulst equation the reslt is pretty much the same EGM6341.s10.Team1.Ya-Chiao Chang 19:13, 23 April 2010 (UTC)

=Problem 9: Show that CD = AB plus H.O.T. for a small change in gamma=

Find
Verify that $$ \bar{CD} = \bar{AB} + H.O.T. \,$$ for small $$ d\gamma \,$$ and determine what the higher-order terms are.

Solution
$$ tan(d\gamma)=\frac{\bar{AB}}{V} \Rightarrow \bar{AB}=Vtan(d\gamma) \,$$ $$ sin(d\gamma)=\frac{\bar{CD}}{V+dV} \Rightarrow \bar{CD}=(V+dv)sin(d\gamma) \,$$ $$ \bar{CD} = \bar{AB} + (V+dv)sin(d\gamma) - Vtan(d\gamma) \,$$ Both $$ sin(x)\,$$ ≈ $$ x \,$$ and $$ tan(x) \,$$ ≈ $$ x \,$$ when x is small, thus $$ \bar{CD} \,$$ ≈ $$ \bar{AB} + (V+dV)d\gamma - V{d\gamma} \,$$

To determine what the higher-order terms are, use Taylor series expansions. $$ sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + ... \,$$ $$ tan(x) = x + \frac{x^3}{3} + \frac{2x^5}{15} + \frac{17x^7}{315} + ... \,$$

Egm6341.s10.team1.andrewdugan 07:42, 23 April 2010 (UTC)

= Problem 10: Verification of elliptical equation =

Given
From [[media:Egm6341.s10.mtg42.djvu|42-1]] we are given that


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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle x=aCos(t)
 * }.
 * }.


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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle y=bSin(t)
 * }.
 * }.

Find
We are tasked with showing that


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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \qquad\frac{x^2}{a^2}+ \frac{y^2}{b^2}=1
 * }.
 * }.

Solution
if we square both the x and y terms from above we obtain


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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle x^2=a^2Cos^2(t)
 * }.
 * }.


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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle y^2=b^2Sin^2(t)
 * }.
 * }.

Using the trigonometric relation


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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle Cos^2(t)+Sin^2(t)=1
 * }.
 * }.

we can see that by cancelation of the a and b terms in


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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \dfrac{a^2Cos^2(t)}{a^2}+\dfrac{b^2Sin^2(t)}{b^2}
 * }.
 * }.

will indeed yield the result that we are looking for


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

$$ $$ Egm6341.s10.team1.toddmock 23:39, 20 April 2010 (UTC)
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \qquad\frac{x^2}{a^2}+ \frac{y^2}{b^2}=1
 * }.
 * }.

= Problem 11: Eccentricity and differential arc length =

Given
From [[media:Egm6341.s10.mtg42.djvu|42-2]] we are given that


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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle dl= \sqrt{dx^2+dy^2}
 * }.
 * }.

and the Eccentricity is


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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle e=[1-\dfrac{b^2}{a^2}]^{1/2}
 * }.
 * }.

Find
We are to show both of these relations are true

Solution
For the differential arc length shown below


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


 * [[Image:elly.png]]
 * }.
 * }.
 * }.

we can use simple geometric relations specifically the Pythagorean


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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle dl^2= {dx^2+dy^2}
 * }.
 * }.

which then gives


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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle dl= \sqrt{dx^2+dy^2}
 * }.
 * }.

For the eccentricity consider the next figure


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


 * [[Image:elly2.png]]
 * }.
 * }.
 * }.

The eccentricity is defined as the fraction of the distance c (the center to the focus) to the major axis a:


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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle e=\dfrac{c}{a}
 * }.
 * }.

It is known that the sum of the distance from one focus to the edge of the ellipse and from that same edge to the other focus is equal to 2a (Stewart, James. Calculus . Pacific Grove : Brooks/Cole publishing company, 1999. 0534359493):


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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle |(f1) b| + |(f2) b| = 2a
 * }.
 * }.

Since the two lines were chosen to intersect at the y axis at point (b,0) then we know for sure that the line segments are equal to each other i.e. both are equal to major axis a. Therefore the determination of c becomes a simple geometry problem in which:


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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle c^2=a^2-b^2
 * }.
 * }.

of


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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle c=\sqrt{a^2-b^2}
 * }.
 * }.

Therefore using the above definition of eccentricity


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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle e=\dfrac{\sqrt{a^2-b^2}}{a}
 * }.
 * }.

which is the same as


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

$$ $$ Egm6341.s10.team1.toddmock 23:39, 20 April 2010 (UTC)
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle e=[1-\dfrac{b^2}{a^2}]^{1/2}
 * }.
 * }.

=Problem 12: Verify the expression for the circumference of an ellipse=

Given
The arc length of an ellipse is $$ dl = [dx^2+dy^2]^\frac{1}{2} \,$$. $$ x = acost \,$$ $$ y = bsint \,$$ The eccentricity $$ e \,$$ is defined as $$ e = [1-\frac{b^2}{a^2}]^\frac{1}{2} \,$$.

Find
Verify that $$ C = \int{dl}=a\int_{t=0}^{2\pi}[1-e^2cos^2t]^\frac{1}{2}dt \,$$.

Solution
$$ dx = \frac{d}{dt}[acost]=-asint \,$$ $$ dy = \frac{d}{dt}[bsint]=bcost \,$$ $$ C = \int{dl}=\int_{t=0}^{2\pi}[(-asint)^2+(bcost)^2]^\frac{1}{2}dt=\int_{t=0}^{2\pi}[a^2sin^2t+b^2cos^2t]^\frac{1}{2}dt \,$$ $$ C = a\int_{t=0}^{2\pi}[sin^2+\frac{b^2}{a^2}cos^2t]^\frac{1}{2}dt \,$$, and since $$ sin^2t+cos^2t=1 \,$$, $$ C = a\int_{t=0}^{2\pi}[1-cos^2t+\frac{b^2}{a^2}cos^2t]^\frac{1}{2}dt=a\int_{t=0}^{2\pi}[1-(1-\frac{b^2}{a^2})cos^2t]^\frac{1}{2}dt \,$$

Egm6341.s10.team1.andrewdugan 05:23, 23 April 2010 (UTC)

= Problem 13: Use $$\alpha$$ As A Differential Variable to Calculate the Circumference of Ellipse =

Problem Statement
We already have $$C= \!\,\int dl \!\,= \!\,a\int_{t=0}^{2 \pi}[1-e^2cos^2t]^\frac{1}{2}dt$$ What if we change the differential variable from t to $$\alpha$$ Ref: Lecture Notes [[media:Egm6341.s10.mtg42.djvu|p.42-2]]  

Solution

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

C = \!\,\int_{o}^{\pi/2}dl = 4*\!\,[dx^2+dy^2]^\frac{1}{2} $$
 * $$ \displaystyle
 * $$ \displaystyle
 * }


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

x = \!\,asin(\alpha), y = \!\,bcos(\alpha) $$
 * $$ \displaystyle
 * $$ \displaystyle
 * }


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

dx = \!\,acos(\alpha)d\alpha, dy = \!\,-bsin(\alpha)d\alpha $$
 * $$ \displaystyle
 * $$ \displaystyle
 * }


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

dx^2 = \!\,a^2cos^2(\alpha)(d\alpha)^2, dy^2 = \!\,b^2sin^2(\alpha)(d\alpha)^2 $$
 * $$ \displaystyle
 * $$ \displaystyle
 * }


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

C = \!\, \int_{o}^{\pi/2}dl = \!\,4[a^2cos^2(\alpha)+b^2sin^2(\alpha)]^\frac{1}{2}d\alpha $$
 * $$ \displaystyle
 * $$ \displaystyle
 * }


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

C = \!\, \int_{o}^{\pi/2}dl = \!\,4[a^2cos^2(\alpha)+a^2sin^2(\alpha)-a^2sin^2(\alpha)+b^2sin^2(\alpha)]^\frac{1}{2}d\alpha $$
 * $$ \displaystyle
 * $$ \displaystyle
 * }


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

C = \!\, \int_{o}^{\pi/2}dl = \!\,4[\underbrace{(a^2cos^2(\alpha)+a^2sin^2(\alpha))}_{a^2}-a^2sin^2(\alpha)+b^2sin^2(\alpha)]^\frac{1}{2}d\alpha $$
 * $$ \displaystyle
 * $$ \displaystyle
 * }


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

C = \!\,\int_{o}^{\pi/2}dl = \!\,4*a*[1-\underbrace{(1-(\frac{b}{a})^2)}_{e^2}sin^2(\alpha)]^\frac{1}{2}d\alpha $$
 * $$ \displaystyle
 * $$ \displaystyle
 * }


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

C = \!\,\int_{o}^{\pi/2}dl = \!\,4a[1-e^2sin^2(\alpha)]^\frac{1}{2}d\alpha $$ - $$C = \!\,\int_{o}^{\pi/2}dl = \!\,4a[1-e^2sin^2(\alpha)]^\frac{1}{2}d\alpha$$ - EGM6341.s10.Team1.Ya-Chiao Chang 18:09, 21 April 2010 (UTC)
 * $$ \displaystyle
 * $$ \displaystyle
 * }

=Problem 14: Verify the change of variables expression for I in Clenshaw-Curtis quadrature=

Given
Clenshaw-Curtis quadrature utilizes the change of variables of $$ x=cos\theta \,$$. Before the change of variables, $$ I = \int_{-1}^{+1}f(x)dx \,$$.

Find
Verify that $$ I = \int_{0}^{\pi}f(cos\theta)sin{\theta}d\theta \,$$.

Solution
$$ I = \int_{-1}^{+1}f(x)dx \,$$ $$ f(x)=f(cos\theta) \,$$ $$ dx = \frac{d}{dx}(cos\theta) = -sin{\theta}d\theta \,$$ $$ sin^{-1}(-1)=\pi \,$$ thus, lower limit $$ -1 \Rightarrow \pi \,$$ $$ sin^{-1}(1)=0 \,$$ thus, lower limit $$ 1 \Rightarrow 0 \,$$ $$ I = \int_{\pi}^{0}f(cos\theta)(-sin{\theta}d\theta) \,$$

Egm6341.s10.team1.andrewdugan 05:25, 23 April 2010 (UTC)

= Problem 16: Find the Cosine Series coeffient ak =

Find
Find

$$ a_{k}=\frac {2}{\pi}\int_{0}^{\pi}f(\cos \theta) \cos (k \theta)d \theta $$

Given
We know Fourier Cosine Series:

$$ f(\cos\theta)=\frac{a_{0}}{2}+\sum _{k=1}^{\infin} a_{k}\cos (k\theta) $$

Solution
$$ a_{0}=\frac{1}{\pi}\int _{-\pi}^{\pi} f(\cos\theta)d\theta $$

$$ =\frac{2}{\pi}\int _{0}^{\pi} f(\cos\theta)d\theta $$

$$ a_{n}=\frac{1}{\pi}\int _{-\pi}^{\pi} f(\cos\theta)\cos(k\theta) d\theta $$

$$ =\frac{2}{\pi}\int _{0}^{\pi} f(\cos\theta) cos(k\theta)d\theta $$

Solved by :Egm6341.s10.team1.lei03:54, 23 April 2010 (UTC)

=Problem 17: Find int if cosine series is know =

Find
[[media:Egm6341.s10.mtg42.djvu|p.42-3]]

$$ I=a_{0}+\sum_{j=1}^{\infin} \frac{2a_{2j}}{1-(2j)^2} $$

Given
[[media:Egm6341.s10.mtg42.djvu|p.42-2]]

$$ I= \int_{-1}^{+1}f(x)dx= \int_{0}^{\pi} f(\cos\theta) \sin\theta d\theta\, $$

[[media:Egm6341.s10.mtg42.djvu|p.42-3]]

$$ f(\cos\theta)=\frac{a_{0}}{2}+\sum _{k=1}^{\infin} a_{k}\cos (k\theta) $$

Solution
$$ I=\int_{0}^{\pi}(\frac{a_{0}}{2}+\sum _{j=1}^{\infin}a_{2j}\cos(2j\theta))\sin\theta d\theta$$

$$ =\int_{0}^{\pi}\frac {a_{0}}{2}\sin\theta d\theta +\int_{0}^{\pi}\sum _{j=1}^{\infin}a_{2j}\cos(2j\theta)\sin\theta d\theta$$

$$ =-\frac{a_{0}}{2} \cos\theta |_{0}^{\pi}+ \sum _{j=1}^{\infin}\int_{0}^{\pi}a_{2j}\cos(2j\theta)\sin\theta d\theta$$

$$ =a_{0}+\sum _{j=1}^{\infin}a_{2j} \int_{0}^{\pi}\cos(2j\theta)\sin\theta d\theta$$

$$ =a_{0}+\sum _{j=1}^{\infin}a_{2j} \int_{0}^{\pi}-[\sin((2j+1)\theta)-\sin((2j-1)\theta)]d\theta $$

$$ =a_{0}+\sum _{j=1}^{\infin}a_{2j}[\int_{0}^{\pi}-\sin((2j+1)\theta)d\theta +\int_{0}^{\pi}\sin((2j-1)\theta)d\theta] $$

$$ =a_{0}+\sum _{j=1}^{\infin}a_{2j}[\frac{1}{2j+1} \cos(2j+1)\theta|_{0}^{\pi}-\frac{1}{2j-1} \cos(2j-1)\theta|_{0}^{\pi}]$$

$$ =a_{0}+\sum _{j=1}^{\infin}a_{2j}\frac{2}{1-(2j)^{2}} $$


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


 * style="width:50%; padding:10px; border:2px solid #8888aa" |
 * $$ \displaystyle
 * $$ \displaystyle

\Rightarrow I=a_{0}+\sum _{j=1}^{\infin}\frac{2a_{2j}}{1-(2j)^{2}} $$ $$
 * $$\displaystyle
 * }
 * }

Solved by :Egm6341.s10.team1.lei03:54, 23 April 2010 (UTC)

=Contributing Members= Egm6341.s10.team1.toddmock 23:47, 20 April 2010 (UTC) EGM6341.s10.Team1.Ya-Chiao Chang 18:10, 21 April 2010 (UTC) Egm6341.s10.team1.lei 03:55, 23 April 2010 (UTC) Author of Problems 3, 16 and 17 Egm6341.s10.team1.andrewdugan 07:44, 23 April 2010 (UTC) Author of Problems 9, 12 and 14 EGM6341.s10.Team1.Kumanchik 20:15, 23 April 2010 (UTC) Author of problems 5, 6, and 7