Let’s have a cow first:

There are more:

For example:

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Let’s have a cow first:

There are more:

For example:

The tangent line of a circle can be defined as a line that intersects the circle at one point only.

Put a circle in the rectangular coordinate system.

Let be a point on a circle. The tangent line at is a line intersects the circle at only.

Let’s first find a function that represents the line.

From circle’s equation , we have

Since the line intersects the circle at only,

has only one solution.

That means

has only one solution. i.e., its discriminant

By the definition of slope,

.

It follows that

Substitute (2) into (1) and solve for gives

The slope of line connecting and where is .

Since , the tangent line is perpendicular to the line connecting and .

Substitute (3) into , we have

.

The fact that the line intersects the circle at means

or

.

Hence,

.

It follows that by (4),

(5) is derived under the assumption that . However, by letting in (5), we obtain two tangent lines that can not be expressed in the form of :

The tangent line of a quadratic function at is a line that intersects at only.

The presence of function immediately excludes the vertical line which also intersects at only (see Fig. 1).

Fig. 1

Let’s find .

Line intersects at only means quadratic equation

has only one solution. That is, the discriminant of is zero:

Fig. 2

And, by the definition of slope (see Fig. 2),

.

It follows that

Substituting (2) into (1), we have

.

Solve it for gives

.

Fig. 3

Now, solve .

By inspection, .

Is this the only solution?

Visually, it is difficult to tell (see Fig. 1 and Fig. 2)

Fig. 1

Fig. 2

However, we can prove that is the only solution:

Fig. 3

If then from Fig. 3, we have

Area of* triangle *OAB < Area of* circular sector *OAB.

That is,

.

Hence,

If then

Put (1) and (2) together, we have

If then

i.e.,

Therefore by (3) and (4), we conclude that

only when .

Solve for .

By mere inspection, we have .

Visually, it appears that is the only solution (see Fig.1 or Fig. 2)

Fig. 1

Fig. 2

To show that is the only solution of *analytically, *let

is a solution of means

Suppose there is a solution

then,

Since is an function continuous and differentiable on ,

by Lagrange’s Mean-Value Theorem (see “A Sprint to FTC“), there such that

.

We know

.

From (5), we have

.

i.e.,

.

This is not possible since .

A simpler alternative without direct applying Lagrange’s Mean-Value Theorem is:

is a strictly decreasing function.

Since and .

Therefore, is the only solution of . i.e.,

is the only solution of .

Finding the value of for is equivalent to seeking a positive number whose square yields : . In other words, the solution to .

We can find by solving . It is by mere inspection. can also be obtained by solving , again, by mere inspection. But ‘mere inspection’ can only go so far. For Example, what is ?

Method for finding the square root of a positive real number date back to the acient Greek and Babylonian eras. *Heron’s algorithm*, also known as the *Babylonian method*, is an algorithm named after the – century Greek mathematician Hero of Alexandria, It proceeds as follows:

- Begin with an arbitrary positive starting value .
- Let be the average of and
- Repeat step 2 until the desired accuracy is achieved.

This post offers an analysis of Heron’s algorithm. Our aim is a better understanding of the algorithm through mathematics.

Let us begin by arbitrarily choosing a number . If , then is , and we have guessed the exact value of the square root. Otherwise, we are in one of the following cases:

Case 1:

Case 2:

Both cases indicate that lies somewhere between and .

Let us define as the relative error of approximating by :

The closer is to , the better is as an approximation of .

Since ,

By (1),

Let be the mid-point of and :

and, the relative error of approximating ,

We have

By (5),

i.e., lies between and .

We can generate more values in stages by

Clearly,

.

Let

.

We have

and

Consequently, we can prove that

by induction:

When .

Assume when ,

When .

Since (11) implies

(10) and (13) together implies

It follows that

and

(15) indicates that the error is cut at least in half at each stage after the first.

Therefore, we will reach a stage that

It can be shown that

by induction:

When ,

Assume when k = p,

When ,

From (17) and (18), we see that eventually, the error decreases at an exponential rate.

Illustrated below is a script that implements the Heron’s Algorithm:

/* a - the positive number whose square root we are seeking x0 - the initial guess eps - the desired accuracy */ square_root(a, x0, eps) := block( [xk], numer:true, xk:x0, loop, if abs(xk^2-a) < eps then return(xk), xk:(xk+a/xk)/2, go(loop) )$

Suppose we want to find and start with . Fig. 2 shows the Heron’s algorithm in action.

Fig. 1

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.