User:EGM6321.F10.TEAM1.HW3

Given
For $$ k \ne \ 0\ $$ and $$ v_{x,0}=0\ $$, Eq 1.2 becomes

Find
1) Derive the Equations of motion, (1.1) and (1.2) 2) Verify $$ y(x)\ $$ is a parabola, for the particular case of $$ k=0\ $$ 3.1) Utilizing eq1.3, find $$ v_y(t)\ $$ and $$ y(t)\ $$ for m=constant 3.2) Utilizing eq1.3, find $$ m=m(t)\ $$

Solution
1) Derive the Equations of motion, (1.1) and (1.2) Newton's 1st Law of motion says that a change in a particle's momentum with respect to time is due directly to the net (resultant) force acting on that object. If we assume that the object is moving in an inertial reference frame then we may express Newton's 1st Law mathematically as

What we mean by inertial reference frame is that the coordinate system in which we are observing the motion of the particle is not accelerating; i.e. is moving at the same exact speed as the coordinate system in which we are calculating the motion of the particle. Momentum is defined by the product of both the mass of the particle, and the velocity at which the particle is traveling.

Therefore we may rewrite Newton's 1st Law as If we look closely we see that by stating the change in momentum with respect time is due to the net force, and not just to any force it is implied that a summation of multiple forces acting on the particle is required. Also we must note that because force is a vector quantity, we will be summing vectors which have both a magnitude and direction (this will be important when calculating a force's action on the path and velocity a particle travels). We will assume that the only forces acting on the particle while in motion are gravity and some retarding force from the viscosity of the medium the particle is traveling through. Their summation is expressed as

In cases where we are observing the motion of an object that does not lose a measurable amount of mass during its travel (measurable is a relative term, meaning measurable by only techniques used with in the scale of the experiment), we can assume m to be a constant and pull it out of the time derivative.

Let's consider a case such as a baseball being thrown between two people. Imagine that player A (we pick arbitrarly) throws a baseball to player B. As the baseball flies through the air it will experience the force of gravity acting straight down on it, as well as interactions with the molecules that make up the air the baseball is flying through. This interaction with the air molecules is very complex and therefore we need to make an assumption about it that will allow us to mathematically describe it in a simple way. It turns out that if the speed of the baseball is not changing very quickly during its flight between player A and B, then we can use whats known in physics and engineering as the Power Law. We say that the force opposing the motion of the baseball due to the air molecules colliding with it is directly proportional to the velocity to some n power, and in the opposite direction of its velocity. Using k as a constant to represent the strength of the retarding force we may express it as

Therefore we arrive at the following vector equation to describe the flight of the baseball between player A and B expressed as

From experience we know that a baseball flying between two people will travel both horizontally (or else how would it move from player A to B?) and vertically forming an arc-like path between player A and B. Therefore, we will choose to express this motion in 2 dimensional cartesian coordinates (x,y). Due to the nature of a vector quantity and the fact that both force and velocity are vectors, we must come up with a way of describing the variation of the angels between the force vectors (gravity and retarding) acting on the object and the objects velocity vector during its flight. Due to gravity always acting in a down position we will say that a relationship between the retarding force and velocity will suffice; i.e. gravity has no influence on the x-component of velocity. Therefore, we can say ɑ is the angle between the horizontal direction and the velocity vector of the baseball at any point in time during its flight between player A and B. Using trigonometric arguments to determine the x and y components of the vectors, we arrive at the follow equations describing the flight of the baseball in both the x and y directions.

2) Verify $$ y(x)\ $$ is a parabola, for the particular case of $$ k=0\ $$ First let's attack the x-direction of motion. Setting k = 0 we arrive at the following expression.

Integrating once over x with t gives

Now lets stop and notice something about this expression. By integration we have found the the velocity in the x-direction is constant, i.e. in does not change. From start to finish, the baseball's velocity strictly in the x-direction will not change, and we can see this prediction from the above equation. Now integrating again over the x-direction with time will give us an expression for the instanteous x position of the baseball at any moment during its flight in the form of 2 constants.

Now we must use 2 conditions in which we know about the motion of a the baseball traveling between player A and B to solve for these 2 constants. First we know that at time = 0, the baseball is in the starting position. Therefore, let us take the starting position of the baseball to be the x = 0 position on our x,y plane.

We also know that at the final time of travel T, the baseball has to transverse the entire path between player A and B. Let's call this path length L.

Therefore our final expression for the x-direction of motion is Because we are proving that y(x) is a parabola we know that we must eliminate t between x(t) and y(t). Here is a very good opportunity due to the simplicity of the x(t) expression to use the inverse of x(t) in order to get t(x).

Following the same logic as we did before with the x-direction we start with the following 2nd order differential equation. Integrating over the y-direction with respect to t gives Now, because the velocity in the y-direction is not constant, we must apply a initial condition to it. We state that at t = 0, the velocity in the y-direction will have some generic value of $${{v}_{0}}$$.

Integrate once again over y with respect to t

Now again, to keep the solution to this problem as general as possible, we would like to give the option of inputing different starting heights for the projectile, i.e. in our case allowing for player A to stand on a hill and throw the baseball to player B who is at the bottom of the hill. This is done by specifying the following initial condition.

Now plugging in for the fourth constant we have the equation of motion for the y-direction as the following

Now plugging in x(t) into the above y(t) equation we arrive at the follow expression.

We see that y(x) is of the following form

3.1) Utilizing eq1.3, find $$ v_y(t)\ $$ and $$ y(t)\ $$ for m=constant

Due to the velocity being a function of time, if we take the integrating factor as only a function of time and leave it in its integral form, then technically we have not violated any rules. We can not integrate the integrating factor because we do not have the velocity in an explicit form. Therefore we have the following. Which then will simplify to the following ordinary differential equation. Integrating the left side and leaving the right side in integral form we arrive at the following result for v(t).

Integrating again over time we arrive at the following expression for y(t).

3.2) Utilizing eq1.3, find $$ m=m(t)\ $$

EGM6321.F10.TEAM1.Lang70.185.108.64 07:45, 6 October 2010 (UTC)

Author EGM6321.Langpm

Find
1) Derive eq2.1 and eq 2.2 2.1) Write eq2.1 and eq 2.2 in the form of: 2.2) Find $$ \underline{A}\ $$, $$ \underline{B}\ $$ and $$ \underline{u}\ $$,

Solution
1) Derive equation (2.1) and (2.2)

Equation (2.1) can be derived by linearizing (2.10) for small argument.

Equation (2.2) can be derived by linearizing (2.19) for small argument.

2.1) Write eq2.1 and eq 2.2 in the form of:

2.2) Find $$ \underline{A}\ $$, $$ \underline{B}\ $$ and $$ \underline{u}\ $$,

Author EGM6321.Taeho Kim

Find
Use the integrating factor method to obtain the result of eq3.1 directly

Solution
The given solution is one of a first order linear ODE. The derivative of such an equation has the form:

Where $$\ \tau\, $$ is the constant of integration. We will assume the case of constant coefficients, so this equation becomes:

Equation 3.3 can be rearranged as:

We desire a solution for $$\ x(\tau\,) $$, which may be found using the integration by factors method. First, the function $$\ h(\tau\,) $$ must be found:

Multiplying both sides of equation 3.4 by $$\ h(\tau\,) $$ yields:

This equation may be simplified by recalling that $$\ h(\tau\,)a(\tau\,) = \dot{h}(\tau\,) $$:

Substituting equation 3.5 into equation 3.7 for $$h(\tau\,)$$ yields:

It is now possible to find an expression for x(t) by integrating both sides of this equation with respect to $$\ \tau\, $$:

Performing this integration yields:

By performing some algebra on this equation, x(t) is found:

Author: EGM6321.f10.team1.russo 07:15, 6 October 2010 (UTC)

Find
Expand Eq4.1 in a taylor series

Solution
One definition of the exponential function is:

In addition, the Taylor series of a function about some real or complex point, a, is defined as:

For the specific case of a=0, the series is also called the Maclaurin series which is defined as:

Substituting equation 4.2 into the first part of equation 4.4 yields:

Since the derivative of $${{e}^{x}}$$ is simply $${{e}^{x}}$$ and anything raised to the zeroth power is 1, equation 4.5 can be simplified to.

Finally, equation 4.2 can be used to reintroduce the original equation.

Author Egm6321.f10.team1.allison 13:24, 28 September 2010 (UTC)

Find
Generalize Eq 5.1 to the case of SC_L1_ODE_VC.

Solution
For a L1_ODE_CC we have

and for a L1_ODE_VC we have

Similarly, with

Author

Egm6321.f10.team1.allen 04:30, 6 October 2010 (UTC)

Find
Use Eq6.1, Eq6.2 and Eq6.3 to obtain Eq6.4:

Solution
From Bryson & Ho, 1975, pgs.449-450 Relating Eq(6.3) to Eq (A4.5) from Bryson & Ho, 1975, p.450

The initial condition of the output is from Bryson and Ho Substitute 6.11 into 6.13 and

Author 173.18.203.75 05:36, 6 October 2010 (UTC)EGM6321,F10.TEAM1.WILKS

Find
Place Eq7.1, Eq7.2 and Eq7.3 in the form of Eq7.4

Solution
Author EGM6321.Taeho Kim

Given
if: then 8.1 is satisfied

Find
Are there any other solutions, without assuming solution given by eq 8.2? Discuss the search for the solution of 8.1.

Solution
Author EGM6321.XXXXXXXXXXXXXXXXXXXX

Given
Exactness condition 1:

Find
Check 2nd condition of exactness using the following:

Solution
First, the various derivatives used must be found:

Now, substituting equations 9.6 and 9.10-9.14 into equation 9.4.

When the terms on the right hand side are canceled with each other and the zero terms dropped, the equation simplifies to:

which can easily be verified by the reflexive property.

Similarly, equations 9.7-9.9 and 9.15 are substituted into 9.5 to yield:

Once again, the zero terms must be dropped and the results is proven by the reflexive property

$$\therefore \begin{align} f_{xp}+pf_{yp}+2f_y=g_{pp} \end{align}$$

Author Egm6321.f10.team1.allison 14:37, 5 October 2010 (UTC)

Given
with cancelled terms removed The first choice for solution assumes thus and since: now 10.4 becomes: solving for $$ \phi\ (x,y,p) $$ moving $$ k_1\ $$ to the other side of the equation and using the equivalency $$ k_2-k_1=k_3\ $$, the solution for choice 1 bcomes:

Find
Finish the argument assuming the following

Solution
Now and using 10.6 and 10.7 and solving for $$ h(x,y)\ $$

solving for $$ \phi\ (x,y,p) $$ moving $$ k_1\ $$ to the other side of the equation and using the equivalency $$ k_2-k_1=k_3\ $$, the solution for choice 2 is:

Author EGM6321.F10.TEAM1.WILKS 03:12, 4 October 2010 (UTC)EGM6321,F10.TEAM1.WILKS

Signatures
Solved problem 1 -- EGM6321.F10.TEAM1.HW3 03:50, 22 September 2010 (UTC)

Solved problem 2 -- EGM6321.F10.TEAM1.HW3 03:50, 22 September 2010 (UTC)

Solved problem 3 -- EGM6321.f10.team1.russo 07:14, 6 October 2010 (UTC)

Solved problem 4 -- Egm6321.f10.team1.allison 13:27, 28 September 2010 (UTC)

Solved problem 5 -- EGM6321.F10.TEAM1.HW3 03:50, 22 September 2010 (UTC)

Solved problem 6 -- EGM6321.F10.TEAM1.WILKS 06:00, 6 October 2010 (UTC)EGM6321,F10.TEAM1.WILKS

Solved problem 7 -- EGM6321.F10.TEAM1.HW3 03:50, 22 September 2010 (UTC)

Solved problem 8 -- EGM6321.F10.TEAM1.HW3 03:50, 22 September 2010 (UTC)

Solved problem 9 -- Egm6321.f10.team1.allison 14:36, 5 October 2010 (UTC)

Solved problem 10 -- EGM6321.F10.TEAM1.WILKS 03:13, 4 October 2010 (UTC)EGM6321,F10.TEAM1.WILKS