where molecule of combine reversibly to form and, are the reaction rates.

If are the concentrations of respectively, then according to the Law of Mass Action, the reaction is governed by

Without solving this initial-value problem quantitatively, the future state of system can be predicted through qualitatively analyzing how the value of changes over the course of time.

We will study a simple chemical reaction described by

where two molecules of are combined reversibly to form and, are the reaction rates.

If is the concentration of , of , then according to the Law of Mass Action,

or equivalently,

We seek first the equilibrium points that represent the steady state of the system. They are the constant solutions where and , simultaneously.

From and , it is apparent that

is an equilibrium point.

To find the value of , we solve for from (0-1),

Substitute it in (0-2),

,

i.e.,

This is a 2nd order nonlinear differential equaion. Since it has no direct dependence on , we can reduce its order by appropriate substitution of its first order derivative.

Let

,

we have

so that (0-6) is reduced to

a 1st order differential eqution. It follows that either or .

The second case gives

Integrate it with respect to ,

Hence, the equilibrium points of (0-1) and (0-2) can be obtained by solving a quadratic equation

Notice in order to have as a solution, must be non-negative .

Fig. 1

The valid solution is

Fig. 2

By (0-4),

and so, the equilibrium point is

Next, we turn our attentions to the phase-plan trajectories that describe the paths traced out by the pairs over the course of time, depending on the initial values.

For . Dividing (0-2) by (0-1) yields

i.e.,

Integrating it with respect to ,

By (0-3),

Therefore,

Moreover, by (0-5)

As a result,

Substitute in (1-1), we have

This is the trajectory of the system. Clearly, all trajectories are monotonically decreaseing lines.

At last, let us examine how the system behaves in the long run.

If then (see Fig. 2) and will increase. As a result, will decrease. Similarly, if ensures that will decrease. Consequently, will increase.

Fig. 3 Trajectories and Equilibriums

It is evident that as time advances, on the trajectory approaches the equilibrium point

A phase portrait of the system is illustrated in Fig. 4.

Fig. 4

It shows that the system is asymptotically stable.

In this appendix to my previous post “From Dancing Planet to Kepler’s Laws“, we derive the polar form for an ellipse that has a rectangular coordinate system’s origin as one of its foci.

Fig. 1

We start with the ellipse shown in Fig. 1. Namely,

Clearly,

After shifting the origin to the right by , the ellipse has the new origin as one of its foci (Fig. 2).

Fig. 2

Since , the ellipse in is

Substituting in (2) by

yields equation

Replacing by respectively, the equation becomes

Fig. 3

Solving (3) for (see Fig. 3) gives

or .

The first solution

.

Let

,

we have

.

The second solution is not valid since it suggests that :

“This most beautiful system of the sun, planets, and comets, could only proceed from the counsel and dominion of an intelligent powerful Being” – Sir. Issac Newton

When I was seven years old, I had the notion that all planets dance around the sun along a wavy orbit (see Fig. 1).

Fig. 1

Many years later, I took on a challenge to show mathematically the orbit of my ‘dancing planet’ . This post is a long overdue report of my journey.

Shown in Fig. 2 is the sun and a planet in a x-y-z coordinate system. The sun is at the origin. The moving planet’s position is being described by .

Fig. 2

According to Newton’s theory, the gravitational force sun exerts on the planet is

where is the gravitational constant, the mass of the sun and planet respectively. .

By Newton’s second law of motion,

(0-3) (0-2) yields

.

Since

,

it must be true that

.

i.e.

where is a constant.

Similarly,

where are constants.

Consequently,

,

,

.

Hence

If then by the following well known theorem in Analytic Geometry:

“If A, B, C and D are constants and A, B, and C are not all zero, then the graph of the equation Ax+By+Cz+D=0 is a plane“,

(0-7) represents a plane in the x-y-z coordinate system.

For , we have

.

It means

where is a constant. Simply put,

.

Hence, (0-7) still represents a plane in the x-y-z coordinate system (see Fig. 3(a)).

Fig. 3

The implication is that the planet moves around the sun on a plane (see Fig. 4).

Fig. 4

By rotating the axes so that the orbit of the planet is on the x-y plane where (see Fig. 3), we simplify the equations (0-1)-(0-3) to

It follows that

.

i.e.,

.

Integrate with respect to ,

where is a constant.

We can also re-write (0-6) as

where is a constant.

Using polar coordinates

Fig. 5

we obtain from (1-2) and (1-3) (see Fig. 5):

If the speed of planet at time is then from Fig. 6,

Fig. 6

gives

Suppose at , the planet is at the greatest distance from the sun with and speed . Then the fact that attains maximum at implies . Therefore, by (1-4) and (1-5),

i.e.,

We can now express (1-4) and (1-5) as:

Let

then

By chain rule,

.

Thus,

.

That is,

Since

,

we let

.

Notice that .

By doing so, (1-14) can be expressed as

.

Take the first case,

.

Integrate it with respect to gives

where is a constant.

When ,

or .

And so,

or .

For ,

.

By (1-11), it is

Fig. 7

Solving (1-15) for yields

.

i.e.,

Studies in Analytic Geometry show that for an orbit expressed by (1-16), there are four cases to consider depend on the value of :

We can rule out parabolic and hyperbolic orbit immediately for they are not periodic. Given the fact that a circle is a special case of an ellipse, it is fair to say:

The orbit of a planet is an ellipse with the Sun at one of the two foci.

In fact, this is what Kepler stated as his first law of planetary motion.

Fig. 8

For ,

from which we obtain

This is an ellipse. Namely, the result of rotating (1-16) by hundred eighty degrees or assuming attains its minimum at .

The second case

can be written as

.

Integrate it with respect to yields

from which we can obtain (1-16) and (1-17) again.

Fig. 9

Over the time duration , the area a line joining the sun and a planet sweeps an area (see Fig. 9).

.

It means

or that

is a constant. Therefore,

A line joining the Sun and a planet sweeps out equal areas during equal intervals of time.

This is Kepler’s second law. It suggests that the speed of the planet increases as it nears the sun and decreases as it recedes from the sun (see Fig. 10).

Fig. 10

Furthermore, over the interval , the period of the planet’s revolution around the sun, the line joining the sun and the planet sweeps the enire interior of the planet’s elliptical orbit with semi-major axis and semi-minor axis . Since the area enlosed by such orbit is (see “Evaluate a Definite Integral without FTC“), setting in (2-1) to gives

Research on rocket flight performance has shown that typical single-stage rockets cannot serve as the carrier vehicle for launching satellite into orbit. Instead, multi-stage rockets are used in practice with two-stage rockets being the most common. The jettisoning of stages allows decreasing the mass of the remaining rocket in order for it to accelerate rapidly till reaching its desired velocity and height.

Optimizing flight performance is a non-trivial problem in the field of rocketry. This post examines a two-stage rocket flight performance through rigorous mathematical analysis. A Computer Algebra System (CAS) is employed to carry out the symbolic computations in the process. CAS has been proven to be an efficient tool in carry out laborious mathematical calculations for decades. This post reports on the process and the results of using Omega CAS explorer, a Maxima based CAS to solve this complex problem.

A two-stage rocket consists of a payload propelled by two stages of masses (first stage) and (second stage), both with structure factor . The exhaust speed of the first stage is , and of second stage . The initial total mass, is fixed. The ratio is small.

Based on Tsiolkovsky’s equation, we derived the multi-stage rocket flight equation . For a two-stage rocket, the final velocity can be calculated from the following:

Let , so that (1) becomes

where .

We seek an appropriate value of that maximizes .

Consider as a function of , its derivative is computed (see Fig. 1)

Fig. 1

We have .

That is, .

Fig. 2

As shown in Fig. 2, can be expressed as

Notice that .

Solving for gives two solutions (see Fig. 3)

Fig. 3

We rewrite the expression under the square root in and as a quadratic function of : and compute (see Fig. 4)

Fig. 4

If , . It implies that is positive since . When , where is still positive since as a result of , the zero point of function is .

The expression under the square root is positive means both and are real-valued and (see Fig. 5), i.e., .

Fig. 5

From (3) where , we deduce the following:

For all , if then

For all , if then

For all , if then

Fig. 6

Moreover, from Fig. 6,

.

Since the expression in the numerator of , namely

,

It follows that

The implication is that has at least one zero point between and .

However, if both and , the two known zero points of are between and , by () and (), must be positive, which contradicts (4). Therefore, must have only one zero point between and .

We will proceed to show that the only zero lies between and is .

There are two cases to consider.

Case 1 () since and . But this contradicts (). Therefore, must not be positive.

Case 2 () The denominator of is negative since . However, , the terms not under the square root in the numerator of can be expressed as . This is a positive expression since implies that . Therefore, .

The fact that only lies between and , together with () and () proves that is where the global maximum of occurs.

can be simplified to a Taylor series expansion (see Fig. 7)

Fig. 7

The result produced by CAS can be written as . However, it is incorrect as would suggest that is a negative quantity when is small.

To obtain a correct Taylor series expansion for , we rewrite as first where

Its first order Taylor series is then computed (see Fig. 8)

Fig. 8

The first term of the result can be written as . Bring the value of into the result, we have:

To compute from (6) , we substitute for in (2) and compute its Taylor series expansion about (see Fig. 9 )

Fig. 9

Writing its first term as and substituting the value yields:

It is positive when is small.

We have shown the time-saving, error-reduction advantages of using CAS to aid manipulation of complex mathematical expressions. On the other hand, we also caution that just as is the cases with any software system, CAS may contain software bugs that need to be detected and weeded out with a well- trained mathematical mind.