A guided missile is launched to destroy a fighter jet (Fig. 1).

Fig. 1

We introduce a coordinate axes such that at the missile is at origin and the jet at . The jet flies parallel to the x-axis with constant speed . The missile has locked onto the jet so it is always pointing at the jet as it moves. Its speed is

Find the time and position the missile strikes its target.

“The missile has locked onto the jet so it is always pointing at the jet as it moves” means that the tangent to the missile’s path at any point will pass through the position of the jet. The equation of the tangent is

In addition,

Notice are functions of time

Differentiate (1) with respecte to

By (1), substituting for ,

Let we express (2) as

Suppose , which implies that (see Exercise-4). Solving for gives

Submitting (4) into (3),

Let

(5) becomes

We solve this non-linear differential equation as follows:

Illustrated in Fig. 2 are two circular hoops of unit radius, centered on a common x-axis and a distance apart. There is also a soap films extends between the two hoops, taking the form of a surface of revolution about the x-axis. If gravity is negligible, the film takes up a state of stable, equilibrium in which its surface area is a minimum.

Fig. 2

Our problem is to find the function , satisfying the boundary conditions

This is the differential equation solved in “A Relentless Pursuit” whose solution is

We must then find and subject to the boundary condition (1), i.e.,

The fact that is an even function implies either

or

While (3) is clearly false since it claims for all , (4) gives

And so, the solution to boundary-value problem

is

To determine , we deduce the following equation from the boundary conditions that at

This is a transcendental equation for that can not be solved explicitly. Nonetheless, we can examine it qualitatively.

Let

and express (7) as

Fig. 4

A plot of (8)’s two sides in Fig. 4 shows that for sufficient small , the curves and will intersect. However, as increases, , a line whose slope is rotates clockwise towards -axis. The curves will not intersect if is too large. The critical case is when , the curves touch at a single point, so that

and is the tangent line of i.e.,

Dividing (9) by (10) yields

What the mathematical model (5) predicts then is, as we gradually move the rings apart, the soap film breaks when the distance between the two rings reaches , and for , there is no more soap film surface connects the two rings. This is confirmed by an experiment (see Fig. 1).

We compute the value of , the maximum value of that supports a minimum area soap film surface as follows.

Fig. 5

Solving (11) for numerically (see Fig. 5), we obtain

By (10), the corresponding value of

.

We also compute the surface area of the soap film from (2) and (6) (see Fig. 6). Namely,

Fig. 6

Exercise-1 Given , solve (7) numerically for

Exercise-2 Without using a CAS, find the surface area of the soap film from (2) and (6).

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