# 6.6: Hyperbolic Functions

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The hyperbolic functions are a set of functions that have many applications to mathematics, physics, and engineering. Among many other applications, they are used to describe the formation of satellite rings around planets, to describe the shape of a rope hanging from two points, and have application to the theory of special relativity. This section defines the hyperbolic functions and describes many of their properties, especially their usefulness to calculus.

These functions are sometimes referred to as the "hyperbolic trigonometric functions" as there are many, many connections between them and the standard trigonometric functions. Figure (PageIndex{1}) demonstrates one such connection. Just as cosine and sine are used to define points on the circle defined by (x^2+y^2=1), the functions hyperbolic cosine and hyperbolic sine are used to define points on the hyperbola (x^2-y^2=1).

Figure (PageIndex{1}): Using trigonometric functions to define points on a circle and hyperbolic functions to define points on a hyperbola. The area of the shaded regions are included in them.

We begin with their definition.

Definition (PageIndex{1}): Hyperbolic Functions

1. ( cosh x = frac{e^x+e^{-x}}2)
2. ( sinh x = frac{e^x-e^{-x}}2)
3. ( anh x = frac{sinh x}{cosh x})
4. ( ext{sech} x = frac{1}{cosh x})
5. ( ext{csch} x = frac{1}{sinh x})
6. ( coth x = frac{cosh x}{sinh x})

These hyperbolic functions are graphed in Figure (PageIndex{2}). In the graphs of (cosh x) and (sinh x), graphs of (e^x/2) and (e^{-x}/2) are included with dashed lines. As (x) gets "large," (cosh x) and (sinh x) each act like (e^x/2); when (x) is a large negative number, (cosh x) acts like (e^{-x}/2) whereas $sinh x$ acts like (-e^{-x}/2).

Notice the domains of ( anh x) and ( ext{sech} x) are ((-infty,infty)), whereas both (coth x) and ( ext{csch} x) have vertical asymptotes at (x=0). Also note the ranges of these functions, especially ( anh x): as (x oinfty), both (sinh x) and (cosh x) approach (e^{-x}/2), hence ( anh x) approaches (1).

The following example explores some of the properties of these functions that bear remarkable resemblance to the properties of their trigonometric counterparts.

Pronunciation Note: "cosh" rhymes with "gosh," "sinh" rhymes with "pinch," and "tanh" rhymes with "ranch."

Figure (PageIndex{2}): Graphs of the hyperbolic functions.

Example (PageIndex{1}): Exploring properties of hyperbolic functions

Use Definition (PageIndex{1}) to rewrite the following expressions.

1. (cosh^2 x-sinh^2x)
2. ( anh^2 x+ ext{sech}^2 x)
3. (2cosh xsinh x)
4. (frac{d}{dx}ig(cosh xig))
5. (frac{d}{dx}ig(sinh xig))
6. (frac{d}{dx}ig( anh xig))

Solution

1. [egin{align} cosh^2x-sinh^2x &= left(frac{e^x+e^{-x}}2 ight)^2 -left(frac{e^x-e^{-x}}2 ight)^2 &= frac{e^{2x}+2e^xe^{-x} + e^{-2x}}4 - frac{e^{2x}-2e^xe^{-x} + e^{-2x}}4 &= frac44=1.end{align}]So (cosh^2 x-sinh^2x=1).
2. [egin{align} anh^2 x+ ext{sech}^2 x &=frac{sinh^2x}{cosh^2 x} + frac{1}{cosh^2 x} &= frac{sinh^2x+1}{cosh^2 x}qquad ext{Now use identity from #1.} &= frac{cosh^2 x}{cosh^2 x} = 1. end{align}]So ( anh^2 x+ ext{sech}^2 x=1).
3. [egin{align} 2cosh xsinh x &= 2left(frac{e^x+e^{-x}}2 ight)left(frac{e^x-e^{-x}}2 ight) &= 2 cdotfrac{e^{2x} - e^{-2x}}4 &= frac{e^{2x} - e^{-2x}}2 = sinh (2x). end{align}]Thus (2cosh xsinh x = sinh (2x)).
4. [egin{align} frac{d}{dx}ig(cosh xig) &= frac{d}{dx}left(frac{e^x+e^{-x}}2 ight) &= frac{e^x-e^{-x}}2 &= sinh x. end{align}]So (frac{d}{dx}ig(cosh xig) = sinh x.)
5. [egin{align} frac{d}{dx}ig(sinh xig) &= frac{d}{dx}left(frac{e^x-e^{-x}}2 ight) &= frac{e^x+e^{-x}}2 &= cosh x. end{align}]So (frac{d}{dx}ig(sinh xig) = cosh x.)
6. [egin{align} frac{d}{dx}ig( anh xig) &= frac{d}{dx}left(frac{sinh x}{cosh x} ight) &= frac{cosh x cosh x - sinh x sinh x}{cosh^2 x} &= frac{1}{cosh^2 x} &= ext{sech}^2 x. end{align}]So (frac{d}{dx}ig( anh xig) = ext{sech}^2 x.)

The following Key Idea summarizes many of the important identities relating to hyperbolic functions. Each can be verified by referring back to Definition (PageIndex{1}).

Key Idea 16: Useful Hyperbolic Function Properties

Basic Identities

1. (cosh^2x-sinh^2x=1)
2. ( anh^2x+ ext{sech}^2x=1)
3. (coth^2x- ext{csch}^2x = 1)
4. (cosh 2x=cosh^2x+sinh^2x)
5. (sinh 2x = 2sinh xcosh x)
6. (cosh^2x = frac{cosh 2x+1}{2})
7. (sinh^2x=frac{cosh 2x-1}{2})

Derivatives

1. (frac{d}{dx}ig(cosh xig) = sinh x)
2. (frac{d}{dx}ig(sinh xig) = cosh x)
3. (frac{d}{dx}ig( anh xig) = ext{sech}^2 x)
4. (frac{d}{dx}ig( ext{sech} xig) = - ext{sech} x anh x)
5. (frac{d}{dx}ig( ext{csch} xig) = - ext{csch} xcoth x)
6. (frac{d}{dx}ig(coth xig) = - ext{csch}^2x)

Integrals

1. (int cosh x dx = sinh x+C)
2. (int sinh x dx = cosh x+C)
3. (int anh x dx = ln(cosh x) +C)
4. (int coth x dx = ln|sinh x,|+C)

We practice using Key Idea 16

Example (PageIndex{2}): Derivatives and integrals of hyperbolic functions

Evaluate the following derivatives and integrals.

1. (frac{d}{dx}ig(cosh 2xig))
2. (int ext{sech}^2(7t-3) dt)
3. ( int_0^{ln 2} cosh x dx)

Solution

1. Using the Chain Rule directly, we have (frac{d}{dx} ig(cosh 2xig) = 2sinh 2x).
Just to demonstrate that it works, let's also use the Basic Identity found in Key Idea 16: (cosh 2x = cosh^2x+sinh^2x).
[egin{align}frac{d}{dx}ig(cosh 2xig) = frac{d}{dx}ig(cosh^2x+sinh^2xig) &= 2cosh xsinh x+ 2sinh xcosh x &= 4cosh xsinh x.end{align}]Using another Basic Identity, we can see that (4cosh xsinh x = 2sinh 2x). We get the same answer either way.
2. We employ substitution, with (u = 7t-3) and (du = 7dt). Applying Key Ideas 10 and 16 we have:
$$int ext{sech}^2 (7t-3) dt = frac17 anh (7t-3) + C.$$
3. $$int_0^{ln 2} cosh x dx = sinh xBig|_0^{ln 2} = sinh (ln 2) - sinh 0 = sinh(ln 2).$$
We can simplify this last expression as (sinh x) is based on exponentials:
$$sinh(ln 2) = frac{e^{ln 2}-e^{-ln 2}}2 = frac{2-1/2}{2} = frac34.$$

## Inverse Hyperbolic Functions

Just as the inverse trigonometric functions are useful in certain integrations, the inverse hyperbolic functions are useful with others. Figure 16 shows the restrictions on the domains to make each function one-to-one and the resulting domains and ranges of their inverse functions. Their graphs are shown in Figure (PageIndex{3})

Because the hyperbolic functions are defined in terms of exponential functions, their inverses can be expressed in terms of logarithms as shown in Key Idea 17. It is often more convenient to refer to (sinh^{-1}x) than to (lnig(x+sqrt{x^2+1}ig)), especially when one is working on theory and does not need to compute actual values. On the other hand, when computations are needed, technology is often helpful but many hand-held calculators lack a extit{convenient} (sinh^{-1}x) button. (Often it can be accessed under a menu system, but not conveniently.) In such a situation, the logarithmic representation is useful. The reader is not encouraged to memorize these, but rather know they exist and know how to use them when needed.

Table (PageIndex{1}): Graphs of (cosh x), (sinh x) and their inverses.

Figure (PageIndex{3}): Graphs of the hyperbolic functions and their inverses.

The following Key Ideas give the derivatives and integrals relating to the inverse hyperbolic functions. In Key Idea 19, both the inverse hyperbolic and logarithmic function representations of the antiderivative are given, based on Key Idea 17. Again, these latter functions are often more useful than the former. Note how inverse hyperbolic functions can be used to solve integrals we used Trigonometric Substitution to solve in Section 6.4.

Key IDea 17: Logarithmic definitions of the inverse hyperbolic functions.

1. (cosh^{-1}x=lnig(x+sqrt{x^2-1}ig); xgeq1)
2. ( anh^{-1}x = frac12lnleft(frac{1+x}{1-x} ight); |x|<1)
3. ( ext{sech}^{-1}x = lnleft(frac{1+sqrt{1-x^2}}x ight); 0
4. (sinh^{-1}x = lnig(x+sqrt{x^2+1}ig))
5. (coth^{-1}x = frac12lnleft(frac{x+1}{x-1} ight); |x|>1)
6. ( ext{csch}^{-1}x = lnleft(frac1x+frac{sqrt{1+x^2}}{|x|} ight); x eq0)

Key Idea 18: Derivatives Involving Inverse Hyperbolic Functions

1. (frac{d}{dx}ig(cosh^{-1} xig) = frac{1}{sqrt{x^2-1}}; x>1)
2. (frac{d}{dx}ig(sinh^{-1} xig) = frac{1}{sqrt{x^2+1}})
3. (frac{d}{dx}ig( anh^{-1} xig) = frac{1}{1-x^2}; |x|<1)
4. (frac{d}{dx}ig( ext{sech}^{-1} xig) = frac{-1}{xsqrt{1-x^2}}; 0
5. (frac{d}{dx}ig( ext{csch}^{-1} xig) = frac{-1}{|x|sqrt{1+x^2}}; x eq0)
6. (frac{d}{dx}ig(coth^{-1} xig) = frac{1}{1-x^2}; |x|>1)

Key Idea 19: Integrals Involving Inverse Hyperbolic Functions

1. (int frac{1}{sqrt{x^2-a^2}} dx) (=qquad cosh^{-1}left(frac xa ight)+C; 0
2. (int frac{1}{sqrt{x^2+a^2}} dx) (=qquad sinh^{-1}left(frac xa ight)+C; a>0) (qquad=lnBig|x+sqrt{x^2+a^2}Big|+C)
3. (int frac{1}{a^2-x^2} dx) (=qquad left{egin{array}{ccc} frac1a anh^{-1}left(frac xa ight)+C & & x^2
4. (int frac{1}{xsqrt{a^2-x^2}} dx ) (=qquad -frac1a ext{sech}^{-1}left(frac xa ight)+C; 0
5. (int frac{1}{xsqrt{x^2+a^2}} dx) (=qquad -frac1a ext{csch}^{-1}left|frac xa ight| + C; x eq 0, a>0) (quad= frac1a lnleft|frac{x}{a+sqrt{a^2+x^2}} ight|+C)

We practice using the derivative and integral formulas in the following example.

Example (PageIndex{3}): Derivatives and integrals involving inverse hyperbolic functions

Evaluate the following.

1. ( frac{d}{dx}left[cosh^{-1}left(frac{3x-2}{5} ight) ight])
2. ( intfrac{1}{x^2-1} dx)
3. ( int frac{1}{sqrt{9x^2+10}} dx)

Solution

1. Applying Key Idea 18 with the Chain Rule gives:
$$frac{d}{dx}left[cosh^{-1}left(frac{3x-2}5 ight) ight] = frac{1}{sqrt{left(frac{3x-2}5 ight)^2-1}}cdotfrac35.$$
2. Multiplying the numerator and denominator by ((-1)) gives: ( int frac{1}{x^2-1} dx = int frac{-1}{1-x^2} dx). The second integral can be solved with a direct application of item #3 from Key Idea 19, with (a=1). Thus [ egin{align} int frac{1}{x^2-1} dx &= -int frac{1}{1-x^2} dx &= left{egin{array}{ccc} - anh^{-1}left(x ight)+C & & x^2<1 -coth^{-1}left(x ight)+C & & 1We should note that this exact problem was solved at the beginning of Section 6.5. In that example the answer was given as (frac12ln|x-1|-frac12ln|x+1|+C.) Note that this is equivalent to the answer given in Equation (PageIndex{29}), as (ln(a/b) = ln a - ln b).

3. This requires a substitution, then item #2 of Key Idea 19 can be applied.

Let (u = 3x), hence (du = 3dx). We have
[int frac{1}{sqrt{9x^2+10}} dx = frac13intfrac{1}{sqrt{u^2+10}} du. ]
Note (a^2=10), hence (a = sqrt{10}.) Now apply the integral rule.
[egin{align} &= frac13 sinh^{-1}left(frac{3x}{sqrt{10}} ight) + C &= frac13 ln Big|3x+sqrt{9x^2+10}Big|+C. end{align}]

This section covers a lot of ground. New functions were introduced, along with some of their fundamental identities, their derivatives and antiderivatives, their inverses, and the derivatives and antiderivatives of these inverses. Four Key Ideas were presented, each including quite a bit of information.

Do not view this section as containing a source of information to be memorized, but rather as a reference for future problem solving. Key Idea 19 contains perhaps the most useful information. Know the integration forms it helps evaluate and understand how to use the inverse hyperbolic answer and the logarithmic answer.

The next section takes a brief break from demonstrating new integration techniques. It instead demonstrates a technique of evaluating limits that return indeterminate forms. This technique will be useful in Section 6.8, where limits will arise in the evaluation of certain definite integrals.

## 6.6: Hyperbolic Functions

If the second order partial differential equations are classified with the help of its characteristics, an hyperbolic equation in a two-dimensional domain will have two real characteristics whereas a parabolic has one and elliptic equation has only complex characteristics. Since at any point of the domain these two characteristics are orthogonal in direction (one is called the left characteristic and the other is called the right characteristic) these curves can be used as a mesh to find solutions of hyperbolic equations and the scheme is called method of characteristics.

Since a first order partial differential equation has also a real characteristic exist at each point of the domain similar to second order hyperbolic equation, these equations are also called as hyperbolic equations. Hence first order PDE also comes under hyperbolic type and is important to understand the method of characteristics and application of finite differnce methods to these first order equations.

In the present section the numerical methods and the method of characteristics to first order PDE is considered first and then these methods are extended to second order PDE in the consequent sections.

Consider the first order PDE

. (6.6.1)

where a, b and c are functions of x, y and u but not of and so that the equation (6.6.1) is called quasi-linear. (It is linear if they are functions of x and y alone and are non-linear if they are also functions of and ).

An equivalent form of (6.6.1) is

. (6.6.2)

where p= and q=.

If we consider a curve C in the xy-plane other than the curve at which the initial conditions of u have been specified, then in directions tangential to C from the points on C. We have

. (6.6.3)

By eliminating p from (6.4.2) and (6.4.3) we get,

. (6.6.4)

Now if we choose the curve C such that

. (6.6.5)

on C then

. (6.6.6)

The curve C is called the characteristic curve and along this curve we have from (6.6.5) and (6.6.6)

. (6.6.7)

or

. (6.6.8)

u can be obtained at any C by integrating (6.6.8).

Example 6.6.1: Find the solution u satisfying and the initial condition.

Solution: The family of characteristics from equation (6.6.5) is

(where k is a constant.)

Now the values of kalong the two characteristics passing through the end points (0,0) and (0,1) of the initial curve are k=0 and k=1respectively. For the characteristic through (0, yc). We have k=yc, hence this characterisic can be written as

The solution along this characteristic is ( from equation (6.6.8))

gives u = 5x+ A

where A is a constant.

If u=uc at (0, yc) then A = ucand along the characteristics . We have the solution

since u is known only on the line segment , the solution is also defined only in this region with bounding characteristics and .

In this region the solution is unique and is not defined outside the region . If the initial curve is given along any of the characteristics then the initial values can' t be fixed arbitrarily since the solution has to satisfy the characteristic equation

.

However the initial curve differs from any characteristic then the solution along this curve can be fixed arbitrarily. Also if the initial conditions are given along any characteristic say along then the equation

is also a solution at any point of the domain. That is in this case the solution is not unique at any point not on the chosen characteristic .

## 6.9 Calculus of the Hyperbolic Functions

We were introduced to hyperbolic functions in Introduction to Functions and Graphs, along with some of their basic properties. In this section, we look at differentiation and integration formulas for the hyperbolic functions and their inverses.

### Derivatives and Integrals of the Hyperbolic Functions

Recall that the hyperbolic sine and hyperbolic cosine are defined as

It is easy to develop differentiation formulas for the hyperbolic functions. For example, looking at sinh x sinh x we have

These differentiation formulas for the hyperbolic functions lead directly to the following integral formulas.

### Example 6.47

#### Differentiating Hyperbolic Functions

Evaluate the following derivatives:

#### Solution

Using the formulas in Table 6.2 and the chain rule, we get

Evaluate the following derivatives:

### Example 6.48

#### Integrals Involving Hyperbolic Functions

Evaluate the following integrals:

#### Solution

We can use u-substitution in both cases.

Evaluate the following integrals:

### Calculus of Inverse Hyperbolic Functions

Looking at the graphs of the hyperbolic functions, we see that with appropriate range restrictions, they all have inverses. Most of the necessary range restrictions can be discerned by close examination of the graphs. The domains and ranges of the inverse hyperbolic functions are summarized in the following table.

The graphs of the inverse hyperbolic functions are shown in the following figure.

To find the derivatives of the inverse functions, we use implicit differentiation. We have

We can derive differentiation formulas for the other inverse hyperbolic functions in a similar fashion. These differentiation formulas are summarized in the following table.

### Example 6.49

#### Differentiating Inverse Hyperbolic Functions

Evaluate the following derivatives:

#### Solution

Using the formulas in Table 6.4 and the chain rule, we obtain the following results:

Evaluate the following derivatives:

### Example 6.50

#### Integrals Involving Inverse Hyperbolic Functions

Evaluate the following integrals:

#### Solution

Evaluate the following integrals:

### Applications

One physical application of hyperbolic functions involves hanging cables . If a cable of uniform density is suspended between two supports without any load other than its own weight, the cable forms a curve called a catenary . High-voltage power lines, chains hanging between two posts, and strands of a spider’s web all form catenaries. The following figure shows chains hanging from a row of posts.

Hyperbolic functions can be used to model catenaries. Specifically, functions of the form y = a cosh ( x / a ) y = a cosh ( x / a ) are catenaries. Figure 6.84 shows the graph of y = 2 cosh ( x / 2 ) . y = 2 cosh ( x / 2 ) .

### Example 6.51

#### Solution

Now recall that 1 + sinh 2 x = cosh 2 x , 1 + sinh 2 x = cosh 2 x , so we have

### Section 6.9 Exercises

Use the quotient rule to verify that tanh ( x ) ′ = sech 2 ( x ) . tanh ( x ) ′ = sech 2 ( x ) .

Take the derivative of the previous expression to find an expression for sinh ( 2 x ) . sinh ( 2 x ) .

Take the derivative of the previous expression to find an expression for cosh ( x + y ) . cosh ( x + y ) .

For the following exercises, find the derivatives of the given functions and graph along with the function to ensure your answer is correct.

[T] cosh ( 3 x + 1 ) cosh ( 3 x + 1 )

[T] sinh ( x 2 ) sinh ( x 2 )

[T] 1 cosh ( x ) 1 cosh ( x )

[T] sinh ( ln ( x ) ) sinh ( ln ( x ) )

[T] cosh 2 ( x ) + sinh 2 ( x ) cosh 2 ( x ) + sinh 2 ( x )

[T] cosh 2 ( x ) − sinh 2 ( x ) cosh 2 ( x ) − sinh 2 ( x )

[T] tanh ( x 2 + 1 ) tanh ( x 2 + 1 )

[T] 1 + tanh ( x ) 1 − tanh ( x ) 1 + tanh ( x ) 1 − tanh ( x )

[T] sinh 6 ( x ) sinh 6 ( x )

[T] ln ( sech ( x ) + tanh ( x ) ) ln ( sech ( x ) + tanh ( x ) )

For the following exercises, find the antiderivatives for the given functions.

For the following exercises, find the derivatives for the functions.

For the following exercises, find the antiderivatives for the functions.

For the following exercises, use the fact that a falling body with friction equal to velocity squared obeys the equation d v / d t = g − v 2 . d v / d t = g − v 2 .

Derive the previous expression for v ( t ) v ( t ) by integrating d v g − v 2 = d t . d v g − v 2 = d t .

[T] Estimate how far a body has fallen in 12 12 seconds by finding the area underneath the curve of v ( t ) . v ( t ) .

For the following exercises, use this scenario: A cable hanging under its own weight has a slope S = d y / d x S = d y / d x that satisfies d S / d x = c 1 + S 2 . d S / d x = c 1 + S 2 . The constant c c is the ratio of cable density to tension.

Sketch the cable and determine how far down it sags at x = 0 . x = 0 .

For the following exercises, solve each problem.

[T] A chain hangs from two posts four meters apart to form a catenary described by the equation y = 4 cosh ( x / 4 ) − 3 . y = 4 cosh ( x / 4 ) − 3 . Find the total length of the catenary (arc length).

[T] A high-voltage power line is a catenary described by y = 10 cosh ( x / 10 ) . y = 10 cosh ( x / 10 ) . Find the ratio of the area under the catenary to its arc length. What do you notice?

Prove the formula for the derivative of y = cosh −1 ( x ) y = cosh −1 ( x ) by differentiating x = cosh ( y ) . x = cosh ( y ) .

(Hint: Use hyperbolic trigonometric identities.)

Prove that ( cosh ( x ) + sinh ( x ) ) n = cosh ( n x ) + sinh ( n x ) . ( cosh ( x ) + sinh ( x ) ) n = cosh ( n x ) + sinh ( n x ) .

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## STPM Further Mathematics T

The hyperbolic functions, of which there are six, are so named because they are related to the parametric equations for a hyperbola.

The 2 main hyperbolic functions are sinh x and cosh x (and so now you know what the ‘hyp’ button on your calculator is for). The hyperbolic functions are actually functions of the natural exponents e x through the following equations:

We now relate the hyperbolic functions with the hyperbola. The equation for the hyperbola is

We let
x = a cosh u
y = b sinh u

We find that cosh 2 u – sinh 2 u = 1, which is true (This can be proven by substituting the e x into the equation). Now that we have 2 hyperbolic functions, we use it to further derive a few other functions following a similar convention which the trigonometric function uses:

All these 6 hyperbolic functions have their special pronunciation. sinh is read as ‘shine’, cosh as ‘cosh’, tanh as ‘than’, sech as ‘sheck’, csch as ‘co-sheck’ and coth as ‘cough’.

Now we shall see the graphs of the 6 hyperbolic functions. Note that they are all derived from the exponential function:

cosh x sinh x tanh x

/>
sech x csch x coth x

Their domain and ranges are as follows:

Now that you know the basic information of these functions, it’s time to memorize formulas. But before you start, I need to introduce a special rule which makes the memorizing easier.

The Osborne’s Rule states that to change a standard ordinary trigonometric identities into the equivalent standard hyperbolic identity, change the sign of the term which is the product of two sines, and substitute the corresponding hyperbolic functions. This means that if you remember all the trigonometric identities, you can remember the hyperbolic identities. Please note that all the trigonometric formulas which have the periodic characteristics (for example, the R formula and the phase shifts) do not apply to hyperbolic functions, as they are not periodic.

For each case, you should be able to derive them. Proving them is simple, just plug in the e x relation into it and you are sure to get it.

The formulas and identities are as follows:

Double-Angle Formula

Besides all these formulas, you should also know the relations between hyperbolic functions and trigonometric functions. Use the following to derive those for tanh x, sech x, csch x and coth x too. Bear in mind that i × i = 𔂿.

To prove this, you will need to wait till you learn Power Series in Chapter 7.

Easy to memorize, isn’t it? Exam questions normally focus on proving stuff, and probably sketching graphs too. Be very careful not to make mistakes. ☺

## 6.6 Moments and Centers of Mass

In this section, we consider centers of mass (also called centroids, under certain conditions) and moments. The basic idea of the center of mass is the notion of a balancing point. Many of us have seen performers who spin plates on the ends of sticks. The performers try to keep several of them spinning without allowing any of them to drop. If we look at a single plate (without spinning it), there is a sweet spot on the plate where it balances perfectly on the stick. If we put the stick anywhere other than that sweet spot, the plate does not balance and it falls to the ground. (That is why performers spin the plates the spin helps keep the plates from falling even if the stick is not exactly in the right place.) Mathematically, that sweet spot is called the center of mass of the plate.

In this section, we first examine these concepts in a one-dimensional context, then expand our development to consider centers of mass of two-dimensional regions and symmetry. Last, we use centroids to find the volume of certain solids by applying the theorem of Pappus.

### Center of Mass and Moments

Let’s begin by looking at the center of mass in a one-dimensional context. Consider a long, thin wire or rod of negligible mass resting on a fulcrum, as shown in Figure 6.62(a). Now suppose we place objects having masses m 1 m 1 and m 2 m 2 at distances d 1 d 1 and d 2 d 2 from the fulcrum, respectively, as shown in Figure 6.62(b).

The most common real-life example of a system like this is a playground seesaw, or teeter-totter, with children of different weights sitting at different distances from the center. On a seesaw, if one child sits at each end, the heavier child sinks down and the lighter child is lifted into the air. If the heavier child slides in toward the center, though, the seesaw balances. Applying this concept to the masses on the rod, we note that the masses balance each other if and only if m 1 d 1 = m 2 d 2 . m 1 d 1 = m 2 d 2 .

In the seesaw example, we balanced the system by moving the masses (children) with respect to the fulcrum. However, we are really interested in systems in which the masses are not allowed to move, and instead we balance the system by moving the fulcrum. Suppose we have two point masses, m 1 m 1 and m 2 , m 2 , located on a number line at points x 1 x 1 and x 2 , x 2 , respectively (Figure 6.63). The center of mass, x – , x – , is the point where the fulcrum should be placed to make the system balance.

#### Center of Mass of Objects on a Line

and the center of mass of the system is given by

We apply this theorem in the following example.

### Example 6.29

#### Finding the Center of Mass of Objects along a Line

Suppose four point masses are placed on a number line as follows:

Find the moment of the system with respect to the origin and find the center of mass of the system.

#### Solution

First, we need to calculate the moment of the system:

Now, to find the center of mass, we need the total mass of the system:

The center of mass is located 1/2 m to the left of the origin.

Suppose four point masses are placed on a number line as follows:

Find the moment of the system with respect to the origin and find the center of mass of the system.

If we have several point masses in the xy-plane, we can use the moments with respect to the x- and y-axes to calculate the x- and y-coordinates of the center of mass of the system.

#### Center of Mass of Objects in a Plane

Also, the coordinates of the center of mass ( x – , y – ) ( x – , y – ) of the system are

The next example demonstrates how to apply this theorem.

### Example 6.30

#### Finding the Center of Mass of Objects in a Plane

Suppose three point masses are placed in the xy-plane as follows (assume coordinates are given in meters):

Find the center of mass of the system.

#### Solution

First we calculate the total mass of the system:

Next we find the moments with respect to the x- and y-axes:

The center of mass of the system is ( 1 , 1 / 3 ) , ( 1 , 1 / 3 ) , in meters.

Suppose three point masses are placed on a number line as follows (assume coordinates are given in meters):

Find the center of mass of the system.

### Center of Mass of Thin Plates

So far we have looked at systems of point masses on a line and in a plane. Now, instead of having the mass of a system concentrated at discrete points, we want to look at systems in which the mass of the system is distributed continuously across a thin sheet of material. For our purposes, we assume the sheet is thin enough that it can be treated as if it is two-dimensional. Such a sheet is called a lamina . Next we develop techniques to find the center of mass of a lamina. In this section, we also assume the density of the lamina is constant.

Laminas are often represented by a two-dimensional region in a plane. The geometric center of such a region is called its centroid . Since we have assumed the density of the lamina is constant, the center of mass of the lamina depends only on the shape of the corresponding region in the plane it does not depend on the density. In this case, the center of mass of the lamina corresponds to the centroid of the delineated region in the plane. As with systems of point masses, we need to find the total mass of the lamina, as well as the moments of the lamina with respect to the x- and y-axes.

We first consider a lamina in the shape of a rectangle. Recall that the center of mass of a lamina is the point where the lamina balances. For a rectangle, that point is both the horizontal and vertical center of the rectangle. Based on this understanding, it is clear that the center of mass of a rectangular lamina is the point where the diagonals intersect, which is a result of the symmetry principle , and it is stated here without proof.

#### The Symmetry Principle

If a region R is symmetric about a line l, then the centroid of R lies on l.

Let’s turn to more general laminas. Suppose we have a lamina bounded above by the graph of a continuous function f ( x ) , f ( x ) , below by the x-axis, and on the left and right by the lines x = a x = a and x = b , x = b , respectively, as shown in the following figure.

As with systems of point masses, to find the center of mass of the lamina, we need to find the total mass of the lamina, as well as the moments of the lamina with respect to the x- and y-axes. As we have done many times before, we approximate these quantities by partitioning the interval [ a , b ] [ a , b ] and constructing rectangles.

To get the approximate mass of the lamina, we add the masses of all the rectangles to get

This is a Riemann sum. Taking the limit as n → ∞ n → ∞ gives the exact mass of the lamina:

We derive the moment with respect to the y-axis similarly, noting that the distance from the center of mass of the rectangle to the y-axis is x i * . x i * . Then the moment of the lamina with respect to the y-axis is given by

We find the coordinates of the center of mass by dividing the moments by the total mass to give x – = M y / m and y – = M x / m . x – = M y / m and y – = M x / m . If we look closely at the expressions for M x , M y , and m , M x , M y , and m , we notice that the constant ρ ρ cancels out when x – x – and y – y – are calculated.

## Hyperbolic Functions

The hyperbolic functions enjoy properties similar to the trigonometric functions their definitions, though, are much more straightforward:

Here are their graphs: the (pronounce: "kosh") is pictured in red, the function (rhymes with the "Grinch") is depicted in blue.

As their trigonometric counterparts, the function is even, while the function is odd.

Their most important property is their version of the Pythagorean Theorem.

The verification is straightforward:

While , , parametrizes the unit circle, the hyperbolic functions , , parametrize the standard hyperbola , x >1.

In the picture below, the standard hyperbola is depicted in red, while the point for various values of the parameter t is pictured in blue.

The other hyperbolic functions are defined the same way, the rest of the trigonometric functions is defined:

For every formula for the trigonometric functions, there is a similar (not necessary identical) formula for the hyperbolic functions:

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## Feature Details

### 64-bit Integer Atomic Operations

Shader Model 6.6 will introduce the ability to perform atomic arithmetic, bitwise, and exchange/store operations on 64-bit values.

All the following atomic intrinsic functions and methods will take 64-bit values when used on RWByteAddressBuffer and RWStructuredBuffer types in all shader stages:

Where RWByteAddressBuffer methods are concerned, each of these will have a *64 suffix to indicate the expected type.

Shader Model 6.6 will include optional support for other resource and variable types. Typed resources, including writeable typed buffers and textures, will be supported where AtomicInt64OnTypedResourceSupported option is set. Shared memory groupshared variables will be supported where AtomicInt64OnGroupSharedSupported is set.

### Integer Atomics on Float-Typed Resources

Shader Model 6.6 will introduce support for using floating point values in the existing integer compare and exchange intrinsic functions. The functions that use compares use bitwise compares and not true floating point compares:

InterlockedExchange was an existing intrinsic function extended to include floats since it involved no compare, no new suffix was needed. The ByteAddressBuffer version was given a *Float suffix to indicate the intended type.

### Dynamic Resource Binding

Shader Model 6.6 will introduce the ability to create resources from descriptors by directly indexing into the CBV_SRV_UAV heap or the Sampler heap. This resource creation method eliminates the need for root signature descriptor table mapping but requires new global root signature flags to indicate the use of each heap.

The feature is exposed as two new builtin global indexable objects: ResourceDescriptorHeap and SamplerDescriptorHeap . Indexing these global objects returns an internal handle object. This object can be assigned to temporary resource or sampler objects without requiring resource binding locations or mapping through root signature descriptor tables.

The assigned variable must match the heap type of the indexed array.

### Compute Shader Derivatives and Samples

Shader Model 6.6 will introduce derivative and sample intrinsic functions to compute shaders. Previous shader models restricted these functions to pixel shaders.

Derivative operations depend on 2ࡨ quads. Compute shaders don’t have quads. So in order to map these functions to a compute shader which views data as a serial sequence, we’ve defined the quads these functions operate on according to the compute shader lane index. One quad consists of the first four elements in the land index sequence in left-to-right and then top-to-bottom order. Another quad similarly consists of the next four and so on. This gives the 2ࡨ quads that the following intrinsic functions operate on.

The derivative functions added:

The sample functions added:

These operations will be optionally available for Amplification and Mesh shader stages where the D erivativesInMeshAndAmplificationShadersSupported capability bit is set.

### Packed 8-Bit Operations

Shader Model 6.6 will add a new set of intrinsic functions for processing packed 8-bit data. These are useful to reduce bandwidth usage where lower precision calculations are acceptable.

These are the new data types representing a vector of packed 8-bit values:

These new types can be cast to and from uint32_t values without a change in the bitwise representation.

The pack intrinsic functions allow packing a vector of 4 signed or unsigned values into a packed 32-bit value represented by the new packed data types. One version performs a datatype clamp and the other simply drops the unused bits.

To unpack 32-bit values representing 4 8-bit values into a vector of 16 bit or 32 bit signed or unsigned values:

### Wave Size

Shader Model 6.6 will introduce a new compute shader attribute that allows the shader author to specify a wave size that the compute shader is compatible with.

This feature allows the application to guarantee that a shader will be run at the required wave size. With this attribute, DirectX 12 runtime validation will fail if shaders in a pipeline state object have a required wave size that is not in the range reported by the driver. Because use of this feature limits shader flexibility, we only recommended it for shaders compatible with only one wave size.

The required wave size is specified by an attribute before the entry function. The allowed wave sizes that an HLSL shader may specify are the powers of 2 between 4 and 128, inclusive. In other words, the set: [4, 8, 16, 32, 64, 128] .

<numLanes> must be an immediate integer value of an allowed wave size.

## Worked example 12: Sketching a hyperbola

Use horizontal and vertical shifts to sketch the graph of (f(x) = frac<1> + 3).

### Examine the equation of the form (y = frac + q)

We notice that (a > 0), therefore the graph will lie in the first and third quadrants.

### Sketch the standard hyperbola (y = frac)

Start with a sketch of the standard hyperbola (g(x) = frac<1>).

The vertical asymptote is (x = 0) and the horizontal asymptote is (y = 0).

### Determine the vertical shift

From the equation we see that (q = 3), which means (g(x)) must shifted ( ext<3>) units up.

The horizontal asymptote is also shifted ( ext<3>) units up to (y = 3) .

### Determine the horizontal shift

From the equation we see that (p = -2), which means (g(x)) must shifted ( ext<2>) units to the right.

The vertical asymptote is also shifted ( ext<2>) units to the right.

### Determine the (y)-intercept

The (y)-intercept is obtained by letting (x = 0): egin y &= frac<1> <0 - 2>+ 3 &= 2frac<1> <2>end This gives the point ((02frac<1><2>)).

### Determine the (x)-intercept

The (x)-intercept is obtained by letting (y = 0): egin 0 &= frac<1> + 3 -3 &= frac<1> -3(x - 2) &= 1 -3x + 6 &= 1 -3x &= -5 x &= frac<5> <3>end This gives the point ((frac<5><3>0)).

### Determine the domain and range

Johnivan Johnivan took Further Mathematics T as his 5th subject in STPM 2009. With limited resources, and without a teacher, he worked really hard in order to score well in Further Mathematics T. In the end, he was one of the 2 who passed the paper in 2009, in which he obtained an A-.

Johnivan studied Physics (specialized in Astrophysics) in the National University of Singapore. He then pursued his PhD in University College London, specializing in the field of cosmology. Currently he is a senior lecturer in Universiti Sains Malaysia.

He created this blog in 2010 to assist and help those who would like to take this subject. Sad to say that the syllabus of the subject was changed in 2012, and later eliminated from STPM in 2014, so now this blog will act as an archive for the old syllabus. Nevertheless, the contents will still be helpful to high school and college students.