User:EGM6341.s10.Team1.toddmock/HW7

= Problem 1: Linearization about xmax =

Given
We are given that


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

and


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

where


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

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

Solution
Using the above relations we have that


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

canceling out some terms leads to


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

If we continue on algebraic manipulation we have


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


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


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

Thus we then can approximate


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

which can then lead to the exponential relation


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$$ $$ 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:


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$$ $$
 * $$\displaystyle
 * $$\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


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

to


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$$ $$
 * $$\displaystyle
 * $$\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


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$$ $$
 * $$\displaystyle
 * $$\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


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$$ $$
 * $$\displaystyle
 * $$\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


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

right away we can see that


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

and


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

Thus giving the integral


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$$ $$
 * $$\displaystyle
 * $$\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


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

which then gives upon rearranging


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$$ $$ 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 10: Verification of elliptical equation =

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


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


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

Find
We are tasked with showing that


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$$ $$
 * $$\displaystyle
 * $$\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


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$$ $$
 * $$\displaystyle
 * $$\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


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

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


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$$ $$
 * $$\displaystyle
 * $$\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


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$$ $$ 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


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

and the Eccentricity is


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$$ $$
 * $$\displaystyle
 * $$\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


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

which then gives


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$$ $$
 * $$\displaystyle
 * $$\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:


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$$ $$
 * $$\displaystyle
 * $$\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:


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$$ $$
 * $$\displaystyle
 * $$\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


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$$ $$
 * $$\displaystyle
 * $$\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}
 * }.
 * }.