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#### Hyperbolic Functions

**By V. G. Shervatov, B. I. Argunov, L. A. Skornyakov, V. G. Boltyanskii**

**Dover Publications, Inc.**

**Copyright © 2007 Dover Publications, Inc.**

All rights reserved.

ISBN: 978-0-486-17005-3

All rights reserved.

ISBN: 978-0-486-17005-3

CHAPTER 1

**The Hyperbolic Rotation**

**1. HOMOTHETIC TRANSFORMATIONS**

In solving problems in geometry we often use the following type of transformation. Pick a point *O* in the plane and a positive number *k.* If *A* is any other point, let us transform the point *A* into the point *A*' such that *A*' lies on the half-line *OA* and ** OA'/OA = k** (

**Fig. 1**). We leave the point

*a, b**O*fixed by the transformation. Such a transformation is called a

*homothetic transformation, O*is called the

*homothetic center*(or

*center of similitude*), and

*k*is called the

*homothetic ratio*(or

*ratio of similitude*). Obviously, if

*k*< 1, then

*OA*' <

*OA*(

**Fig. 1**), while if

*a**k*> 1, then

*OA*' >

*OA*(

**Fig. 1**).

*b*Under a homothetic transformation every figure *F* is transformed into a figure *F*', similar to *F* (**Fig. 2**). If *k*< 1, *F*' is smaller than *F,* while if *k* > 1, *F*' is larger than *F.*

A homothetic transformation transforms a straight line into a straight line (**Fig. 3 a**), parallel lines into parallel lines (

**Fig. 3**), and circles into circles (

*b***Fig. 3**).

*c*A homothetic transformation changes the lengths of all line segments in the plane by a constant ratio, the homothetic ratio *k.* It also changes the areas of all figures by the constant ratio *k*2, the square of the homothetic ratio.

To see this, let *F* be a figure in the plane. Consider also a grid of small squares in the plane (**Fig. 4**). The area of *F* is approximately equal to the number of squares which are contained in *F* multiplied by the area of one square; the smaller the squares of the grid, the smaller will be the error. (In fact, if we choose any small positive number ε whatsoever, we can make the error smaller than ε by making the squares sufficiently small.) Under a homothetic transformation the grid of squares is transformed into a new grid of squares and the figure *F* into the figure *F*', which will contain exactly as many squares of the new grid (the squares will be smaller if *k*< 1 and larger if *k* > 1) as there were squares of the first grid inside the figure *F.* The area of *F*' is approximately equal to the number of squares contained in it multiplied by the area of one square.

But the area of each new square is equal to the area of one of the original squares multiplied by *k*2, since the lengths of the sides of the original squares are multiplied by *k.* Therefore, the area of *F*' is equal to the area of *F* multiplied by *k*2.

As an example of the application of homothetic transformations, let us solve the following problem: *Inscribe in a given right triangle ABC a rectangle BDEF so that the lengths of its sides are in a given ratio* (**Fig. 5**).

First let us construct an arbitrary rectangle *BD'E'F'*, with the lengths of its sides in the given ratio and such that the vertices *D*' and *F*' lie respectively on the sides *BC* and *AB* of the given triangle. Let *E* be the point of intersection of the line *BE*' and the side *AC* of the triangle. It is easy to show that the homothetic transformation with *B* as the homothetic center and

k = BE/BE'

as the homothetic ratio will transform the rectangle *BD'E'F'* into the required rectangle*BDEF.* From this we can easily construct, the desired rectangle. When the given triangle *ABC* is not a right triangle, the problem can be solved in an analogous way. We shall not dwell on this here.

**2. STRAINS**

We shall need to use another transformation called a *strain.* Let *l* be a straight line in the plane. Let us transform the point *A* in the plane into the point *A*' such that *A*' lies on the perpendicular *PA* dropped from *A* to *l* and ** PA'/PA = k**, some positive constant (

**Fig. 6**). Every point of the line

*a, b**l*is left fixed by the transformation. The straight line

*l*is called the

*axis of the strain,*and

*k*is called the

*coefficient of the strain.*If

*k*is greater than 1, then

*PA*' >

*PA*(

**Fig. 6**), and the strain is sometimes called an

*b**elongation.*If

*k*is less than 1, then

*PA*' <

*PA*(

**Fig. 6**), and the strain is sometimes called a

*a**compression.*A figure

*F*is transformed by a strain into a new figure

*F*', which is usually

*not*similar to

*F.*In

**Fig. 7**, k = 1/3; that is,

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

Several properties of strains are analogous to the properties of homothetic transformations.

(a) *A straight line is transformed by a strain into a straight line.*

Suppose that a strain has axis *l* and coefficient *k;* let *m* be any straight line. If *m* is parallel to *l* and at a distance *d* from *l,* then *m* is transformed into a straight line *m*' parallel to *l* and at a distance *kd* from *l* (**Fig. 8 a**). On the other hand, suppose that

*m*and

*l*intersect, say at the point

*O*(

**Fig. 8**). Under the strain the point

*b**O*remains fixed. Let

*A*be any point different from

*O*on the line

*m,*and let

*A*' be the point into which

*A*is transformed by the strain. Then

*PA*' =

*k*·

*PA.*Let

*B*be another point on the straight line

*m.*Let

*B*' be the point where the perpendicular

*BQ*from

*B*to

*l*intersects the straight line

*OA*'. The triangles

*OQB*and

*OPA*are similar, and the triangles

*OQB*' and

*OPA*' are similar; hence,

QB'/QB = PA'/PA = k,

that is, *QB*' = *k* · *QB.* From this last equality we see that the strain transforms the point *B* into the point *B*'. Since *B* was an arbitrary point of the line *m,* this line is transformed by the strain into the line *OA*', which we shall call *m*'.

(b) *Parallel lines are transformed by a strain into parallel lines.*

Let the lines *m* and *n* be parallel; then they have no point of intersection. In that case the lines *m*' and *n*' into which *m* and *n* are transformed also have no point of intersection (since a point common to *m*' and *n*' could only result from a point common to the lines *m* and *n*); this shows that the lines *m*' and *n*' are also parallel (**Fig. 9**). Note that if ψ and ψ' are the angles formed by the original line *m* and the image line *m*' with the axis of the strain *l,* then from **Fig. 8 b** it is easy to see that

Hence, it also follows from this that parallel lines (intersecting *l* with the same angle ψ) are transformed into parallel lines (intersecting *l* with the same angle ψ').

(c) *The ratios of lengths of line segments lying on the same straight line are unchanged by a strain.*

Indeed, ** AB/BC = A'B'/B'C'** (

**Fig. 10**).

(d) *A strain changes the areas of all figures by the constant ratio k, the coeficient of strain.*

To see this, let *F* be a figure in the plane. Consider also a grid of small squares whose sides are parallel and

perpendicular to the axis of the strain. The area of *F* is approximately equal to the number of squares which are contained in *F* multiplied by the area of one square (**Fig. 11**). The squares are transformed by the strain into a grid of rectangles, the area of each of which is the area of one of the original squares multiplied by *k* (one of the dimensions of the square is unchanged, while the other is multiplied by *k*). The figure *F*' contains exactly the same number of rectangles as the figure *F* did of squares, and the area of *F*' is approximately equal to the number of rectangles contained in it multiplied by the area of one rectangle. Hence, the area of *F*' is equal to the area of *F* multiplied by *k.*

In order to illustrate the use of strains, let us solve the following problem (compare this with the problem on **p. 3**): *In a given right triangle ABC inscribe a rectangle BDEF of given area T* (**Fig. 12**). For the solution, apply a strain with axis *BC* and coefficient

k = BC/BA

to the triangle *ABC.* This strain transforms the triangle *ABC* into an isosceles right triangle *A*'*BC.* Since

BA' = k · BA = BC/BA · BA = BC,

the area of the triangle *A'BC* is *kS,* where *S* is the area of the triangle *ABC.* By property (d), any rectangle *BDEF* of area *T* inscribed in *ABC* would be transformed into a rectangle *BDE'F'* with area *kT.* Therefore, in the isosceles right triangle *A'BC* let us inscribe a rectangle *BDE'F* of area *kT.*

It is easy to do this, for

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII];

therefore, *S*Δ*E'DC* + *S*Δ*A'F'E'* = *S*Δ*A'BC* - *SBDE'F*' = *kS - kT.*

On the other hand,

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

(Here we make use of the fact that the triangle *A'BC,* and consequently the similar triangles *E'DC* and *A'F'E'*, are isosceles.) Thus, we obtain

1/2 BE'2 = kS - kT.

Now that we know the length of the segment *BE*', we can at once find the point *E*'. After that we can construct the rectangle *BDE'F'*' inscribed in the triangle *A'BC,* and finally the rectangle *BDEF* inscribed in the triangle *ABC.*

Depending on the value of *T,* the problem may have two, one, or no solutions. When the given triangle *ABC* is not a right triangle, the problem can be solved in an analogous way, but we shall not consider this here. A geometric solution of this problem without using strains is unknown.

As contrasted with homothetic transformations, strains do not transform circles into circles. A circle is transformed by a strain into a curve called an *ellipse* (**Fig. 13**). With properties (a) - (d), we could use strains to study a number of geometric properties of the ellipse, but such a study would be beyond the scope of this booklet.

**3. THE HYPERBOLA**

In the following sections an important part will be played by the graph of the curve whose equation is

y = a/x or xy = a (a ≠ 0).

This curve is called a *hyperbola* (**Fig. 14**).

It is obvious that the greater the absolute value of *x,* the smaller will be the absolute value of *y,* and conversely. Symbolically, if *x* -> ∞, then *y* -> 0, and if *y* -> ∞, then *x* -> 0. Geometrically, this means that the hyperbola comes arbitrarily close to the coordinate axes but does not actually intersect them, for it follows from the equation *xy* = *a* that neither *x* nor *y* can be zero.

A straight line which a curve approaches nearer and nearer without touching it is called an *asymptote* of that curve. Thus, the coordinate axes are asymptotes of the hyperbola.

The hyperbola is composed of two *branches,* which for *a* > 0 lie in the fist quadrant of the coordinate system (*x* and *y* both positive) and in the third quadrant (*x* and *y* both negative).

The algebraic equation *xy* = *a* has a simple geometric interpretation with reference to the hyperbola. If *M* is an arbitrary point of the hyperbola, then the area of the rectangle *MQOP,* which is bounded by the coordinate axes and the two straight lines through *M* and parallel to the axes (**Fig. 14**), is equal to *a.* In other words, the area of such a rectangle is independent of our choice of the point *M.* That is, since *OP* = *x* and *PM* = *y,* we have

SMQOP = OP · PM = x · y = a.

Calling the rectangle *MQOP* the *coordinate rectangle of the point M,* we can say that *the hyperbola is the locus of those points lying in the first and third quadrants of the coordinate system whose coordinate rectangles have a given area.*

The hyperbola has a *center of symmetry:* the two branches of the hyperbola are symmetric with respect to the origin *O* of the coordinate system. (The coordinate rectangles *MQOP* and *M'Q'OP'* symmetric with respect to *O* (**Fig. 14**) have equal areas.) The hyperbola also has two *axes of symmetry,* the lines *aa* and *bb* which bisect angles between the coordinate axes (**Fig. 15**). In fact, the coordinate rectangles *MQOP* and *M*1*Q*1*OP*1 (symmetric with respect to *aa*) have equal areas; similarly, the coordinate rectangles *MQOP* and *M*2*Q*2*OP*2 (symmetric with respect to *bb*) have equal areas. The center of symmetry *O* and the axes of symmetry *aa* and *bb* are usually simply called the *center* and the *axes* of the hyperbola; the points *A* and *B* in which the hyperbola intersects the axis *aa* are called the *vertices* of the hyperbola.

**4. THE HYPERBOLIC ROTATION**

Consider the hyperbola *xy* = *a,* and apply a strain with axis the *x*-axis and with coefficient *k.* Under this transformation the hyperbola *xy* = *a* is transformed into the hyperbola *xy* = *ka:* the abscissa *x* of every point remains unchanged, while the ordinate *y* is changed to *ky* (**Fig. 16**). Now let us apply another strain—this time with axis the *y*-axis and with coefficient 1/*k*. Under this transformation the hyperbola *xy* = *ka* is transformed into the hyperbola *xy = ka/k,* that is, *xy* = *a;* for, this strain leaves the ordinate *y* of each point unchanged, while it changes the abscissa *x* to *x/k.* In this way we see that *two successive strains of the plane with respective axes the x-axis and the y-axis and with respective coefficients k and 1/k transform the hyperbola xy = a into itself.* The successive application of these two strains results in a transformation which is known as a *hyperbolic rotation.* The term "hyperbolic rotation" is related to the fact that under such a transformation all the points of the hyperbola "slide" along the curve; thus, in **Fig. 16** the point *M* is first transformed into the point *M*1, and then *M*1 is transformed into *M*'; that is, the effect of the hyperbolic rotation is to transform the point *M* of the hyperbola into the point *M*' of the same hyperbola. Such a transformation is analogous to a rotation—we might say that the hyperbola "rotates."

Note the following properties of a hyperbolic rotation:

(a) *A hyperbolic rotation transforms a straight line into a straight line.* (This follows from property (a), **section 2**.)

(b) *A hyperbolic rotation transforms the coordinate axes* (*the asymptotes of the hyperbola*) *into themselves* (since each of the two strains which constitute the hyperbolic rotation transforms each coordinate axis into itself).

(c) *A hyperbolic rotation transforms parallel lines into parallel lines.* (This follows from property (b), **section 2**.)

(d) *A hyperbolic rotation does not change the ratio of the lengths of line segments lying on the same straight line.* (This follows from property (c), **section 2**.)

(e) *A hyperbolic rotation does not change the areas of figures in the plane.* (For, as a result of the first strain, all figures have their areas multiplied by *k,* and as a result of the second they are divided by *k.* This follows from property (d), **section 2**.)

*(Continues...)*

Excerpted fromHyperbolic FunctionsbyV. G. Shervatov, B. I. Argunov, L. A. Skornyakov, V. G. Boltyanskii. Copyright © 2007 Dover Publications, Inc.. Excerpted by permission of Dover Publications, Inc..

All rights reserved. No part of this excerpt may be reproduced or reprinted without permission in writing from the publisher.

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