It is a good idea to enjoy a cup of coffee before starting a busy day.

Suppose the coffee fresh out of the pot with temperature is too hot, we can immediately add cream to reduce the temperature by instantly, then wait for the coffee to cool down naturally to before sipping it comfortably. We can also wait until the temperature of the coffee drops to first, then add the cream to further reduce it instantly to .

Typically, , and .

If we are in a hurry and want to wait the shortest possible time, should the cream be added right after the coffee is made, or should we wait for a while before adding the cream?

The heat flow from the hot water to the surrounding air obeys Newton’s cooling and heating law, described by the following ordinary differential equation:

where , a function of time , is the temperature of the water, is the temperature of its surroundings, and is a constant depends on the heat transfer mechanism, the contact are with the surroundings, and the thermal properties of the water.

Fig. 1 a place where Newton’s law breaks down

Under normal circumstances, we have

Based on Newton’s law, the mathematical model of coffee cooling is:

Fig. 2

Solving (2), an initial-value problem (see Fig. 2) gives

Therefore,

If cream is added immediately (see Fig. 3),

Fig. 3 : cream first

by (3),

Otherwise (see Fig. 4),

Fig. 4: cream last

And so,

Fig. 5

Since

implies

from (4) , we see that

i.e.,

Hence,

If we are in a hurry and want to wait the shortest possible time, we should wait for a while before adding the cream!

A turkey is taken from the refrigerator at and placed in an oven preheated to and kept at that temperature; after minutes the internal temperature of the turkey has risen to . The fowl is ready to be taken out when its internal temperature reaches .

Typically, .

Determine the cooking time required.

According to Newton’s law of heating and cooling (see “Convective heat transfer“), the rate of heat gain or loss of an object is directly proportional to the difference in the temperatures between the object and its surroundings. This law is best described by the following ODE (Ordinary Differential Equation):

where are the temperatures of the object and its surroundings respectively. is the constant of proportionality.

Fig. 1

We see that (1) has a critical point . Fig. 1 illustrates the fact that depending on its initial temperature, an object either heats up or cools down, trending towards in both cases.

We formulate the problem as a system of differential-algebraic equations:

To find the required cooking time, we solve (2) for (see Fig. 2).

Fig. 2

Using Omega CAS Explorer, the typical cooking time is found to be approximately hours ()

Luise Lange of Woodrow Wilson Junior College once wrote (see ” A Century of Calculus, Part I”, p. 50):

“In many calculus texts problems are formulated too one-sidedly in terms of particular, numerical data rather than in general terms. While pedagogically it may be wise to begin a new type of problem with some numerical examples, it is only the general formulation, and the interpretation of the answer in general terms, which can give insight into the functional relation between the given and the derived data.”

I agree with her wholeheartedly! On encountering a mathematical modeling problem stated with numerical values, I prefer to re-state it using symbols first. Then solve the problem symbolically. The numerical values are substituted for the symbols at the very end.

This post is a case in point, as the problem is re-formulated from page 1005 of Jan Gullberg’s “Mathematics From the Birth of Numbers”:

Exercise-1 Solving (2) without using a CAS.

Exercise-2 Given , show that

Exercise-3 Given , verify that from

Exercise-4 A slice is cut from a loaf of bread fresh from the oven at and placed in a room with a constant temperature of . After 1 minute, the temperature of the slice is . When has the slice of bread cooled to ?

Question: What is the shape of a flexible rope hanging from nails at each end and sagging under the gravity?

Answer:

First, observe that no matter how the rope hangs, it will have a lowest point (see Fig. 1)

Fig. 1

It follows that the hanging rope can be placed in a coordinate system whose origin coincides with the lowest point and the tangent to the rope at is horizontal:

Fig. 2

At , the rope to its left exerts a horizontal force. This force (or tension), denoted by , is a constant:

Fig. 3

Shown in Fig. 3 also is an arbitrary point with coordinates on the rope. The tension at , denoted by , is along the tangent to the rope curve. is the angle makes with the horizontal.

Since the section of the rope from to is stationary, the net force acting on it must be zero. Namely, the sum of the horizontal force, and the sum of the vertical force, must each be zero:

where is the hanging rope’s mass density and its length from to .

Dividing (2) by (1), we have

Since

, the slope of the curve at ,

and

,

we rewrite (3) as

and so,

i.e.,

To solve (4), let

.

We have

Integrate (5) with respect to gives

i.e.,

Solving (6) for yields

Integrate (7) with respect to ,

.

Hence,

.

Essentially, it is the hyperbolic cosine function that describes the shape of a hanging rope.

“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

The stages of a two stage rocket have initial masses and respectively and carry a payload of mass . Both stages have equal structure factors and equal relative exhaust speeds. If the rocket mass, , is fixed, show that the condition for maximal final speed is

.

Find the optimal ratio when .

According to multi-stage rocket’s flight equation (see “Viva Rocketry! Part 2“), the final speed of a two stage rocket is

Let , we have

and,

Differentiate with respect to gives

It follows that implies

.

That is, . i.e.,

It is the condition for an extreme value of . Specifically, the condition to attain a maximum (see Exercise-2)

When , solving

yields two pairs:

and

Only (3) is valid (see Exercise-1)

Hence

The entire process is captured in Fig. 2.

Fig. 2

Exercise-1 Given , prove:

Exercise-2 From (1), prove the extreme value attained under (2) is a maximum.

The stages of a two-stage rocket have initial masses and respectively and carry a payload of mass . Both stages have equal structure factors and equal relative exhaust speed . The rocket mass, is fixed and .

According to multi-stage rocket’s flight equation (see “Viva Rocketry! Part 2“), the final speed of a two-stage rocket is

Let , it becomes

where . We will maximize with an appropriate choice of .

That is, given

where . Maximize with an appropriate value of .

The above optimization problem is solved using calculus (see “Viva Rocketry! Part 2“). However, there is an alternative that requires only high school mathematics with the help of a Computer Algebra System (CAS). This non-calculus approach places more emphasis on problem solving through mathematical thinking, as all symbolic calculations are carried out by the CAS (e.g., see Fig. 2). It also makes a range of interesting problems readily tackled with minimum mathematical prerequisites.

The fact that

is a monotonic increasing function

where

or

(1) can be written as

where

,

.

Since means

.

That is

.

Solve

for gives if .

Hence, (1) is a quadratic equation. For it to have solution, its discriminant must be nonnegative, i.e.,

Consider

If , (3) is a quadratic equation.

Solving (3) yields two solutions

,

.

Since ,

(4) implies

and, the solution to (2) is

or

i.e.,

or

We prove that (4) is true by showing (5) is false:

Consider :

where

.

It can be written as

where

,

,

.

Since (see Exercise 1) and,

solve (7) for yields

.

It follows that for .

Consequently, is a negative quantity. i.e.,

which tells that (5) is false.

Hence, when , the global maximum is .

Solving for :

,

we have

.

Therefore,

attains maximum at .

In fact, attains maxima at even when , as shown below:

Solving for , we have

or .

Only is valid (see Exercise-2),

When ,

where

Solve quadratic equation for yields

.

The coefficient of in is , a negative quantity (see Exercise-3).

The implication is that is a negative quantity when .

A rocket with stages is a composition of single stage rocket (see Fig. 1) Each stage has its own casing, instruments and fuel. The th stage houses the payload.

Fig. 2

The model is illustrated in Fig. 2, the stage having initial total mass and containing fuel . The exhaust speed of the stage is .

The flight of multi-stage rocket starts with the stage fires its engine and the rocket is lifted. When all the fuel in the stage has been burnt, the stage’s casing and instruments are detached. The remaining stages of the rocket continue the flight with stage’s engine ignited.

Generally, the rocket starts its stage of flight with final velocity achieved at the end of previous stage of flight. The entire rocket is propelled by the fuel in the casing of the rocket. When all the fuel for this stage has been burnt, the casing is separated from the rest of the stages. The flight of the rocket is completed if . Otherwise, it enters the next stage of flight.

Let the velocity of rocket at the end of stage of flight.

Since , (2) becomes

i.e.,

For a single stage rocket (), (3) is

In my previous post “Viva Rocketry! Part 1“, it shows that given and , (4) yields , a value far below , the required speed of an earth orbiting satellite.

But is there a value of that will enable the single stage rocket to produce the speed a satellite needs?

Let’s find out.

Differentiate (4) with respect to gives

since are positive quantities and .

It means is a monotonically decreasing function of .

Moreover,

Given , (5) yields approximately

Fig. 3

This upper limit implies that for the given and , no value of will produce a speed beyond (see Fig. 4)

Let’s now turn to a two stage rocket ()

From (3), we have

If and , then

.

Consequently,

When and ,

Fig. 5

This is a considerable improvement over the single stage rocket (). Nevertheless, it is still short of producing the orbiting speed a satellite needs.

In fact,

indicates that is a monotonically decreasing function of .

In addition,

.

Therefore, there is an upper limit to the speed a two stage rocket can produce. When , the limit is approximately

Fig. 6

In the value used above, we have taken equal stage masses, . i.e., the ratio of .

Is there a better choice for the ratio of such that a better can be obtained?

Therefore, to maximize the final speed given to the satellite, we must choose the ratio

.

With , the optimum ratio , showing that the first stage must be about ten times large than the second.

Using this ratio and keep as before, (10) now gives

,

a value very close to the required one.

Fig. 8

Setting , we reach the goal:

Fig. 9

Fig. 10

At last, it is shown mathematically that provided the stage mass ratios ( and )are suitably chosen, a two stage rocket can indeed launch satellites into earth orbit.

Exercise 1. Show that and .

Exercise 2. Using the optimum and , solving (10) numerically for such that .

A rocket is programmed to burn and ejects its propellant at the variable rate , where and are positive constants. The rocket is launched vertically from rest. Neglecting all external forces except gravity, show that the final speed given to the payload, of mass , when all the fuel has been burnt is

.

Here is the speed of the propellant relative to the rocket, the initial rocket mass, excluding the payload. The initial fuel mass is .

Exercise 2. Before firing, a single stage rocket has total mass , which comprises the casing, instruments etc, with mass , and the fuel. The fuel is programmed to burn and to be ejected at a variable rate such that the total mass of the rocket at any time , during which the fuel is being burnt, is given by

where is a constant.

The rocket is launched vertically from rest. Neglect all external forces except gravity, show that the height attained at the instant the fuel is fully consumed is

A single rocket expels its propellant at a constant rate .

Assuming constant gravity is the only external force, show that the equation of motion is

where is the rocket’s speed, the speed of the propellant relative to the rocket, the payload mass, and the initial rocket mass.

If the rocket burn is continuous, show that the burn time is and deduce that the final speed given to the payload is

where is the structural factor of the rocket.

Estimate the percentage reduction in the predicted final speed due to the inclusion of the gravity term if

, and .

Find an expression for the height reached by the rocket during the burn and estimate its value using the data above.

Let’s recall the governing equation of rocket’s flight derived in “Viva Rocketry! Part 1“, namely,

.

In the present context, . It implies that

and,

.

With , we have

,

i.e.,

or

.

The structural factor indicates the amount of fuel is . Since the fuel is burnt at a constant rate , it must be true that at burnt out time ,

.

Therefore,

.

The solution to initial-value problem

tells the speed of the rocket during its flight while fuel is burnt (see Fig. 1):

Fig. 1

Evaluate (1) at burnt out time gives the final speed of the payload:

Notice the first term of (2) is the burnt out velocity without gravity (see “Viva Rocketry! Part 1“)

It follows that the percentage reduction in the predicted final speed due to the inclusion of gravity is

Using the given values which are typical, the estimated value of (3) (see Fig. 2) is

.

Fig. 2

This shows the results obtained without taking gravity into consideration can be regarded as a reasonable approximation and the characteristics of rocket flight indicated in “Viva Rocketry! Part 1” are valid.

Since , (1) can be written as

To find the distance travelled while the fuel is burnt, we solve yet another initial-value problem:

Fig. 3

The solution (see Fig. 3) is

.

Hence, the height reached at the burnt out time is

.

Using the given values, we estimate that (see Fig. 4)

Fig. 4

Exercise 1: Find the distance the rocket travelled while the fuel is burnt by solving the following initial-value problem: