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Calculus Volume 2

2.9 Calculus of the Hyperbolic Functions

Calculus Volume 22.9 Calculus of the Hyperbolic Functions
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  1. Preface
  2. 1 Integration
    1. Introduction
    2. 1.1 Approximating Areas
    3. 1.2 The Definite Integral
    4. 1.3 The Fundamental Theorem of Calculus
    5. 1.4 Integration Formulas and the Net Change Theorem
    6. 1.5 Substitution
    7. 1.6 Integrals Involving Exponential and Logarithmic Functions
    8. 1.7 Integrals Resulting in Inverse Trigonometric Functions
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Chapter Review Exercises
  3. 2 Applications of Integration
    1. Introduction
    2. 2.1 Areas between Curves
    3. 2.2 Determining Volumes by Slicing
    4. 2.3 Volumes of Revolution: Cylindrical Shells
    5. 2.4 Arc Length of a Curve and Surface Area
    6. 2.5 Physical Applications
    7. 2.6 Moments and Centers of Mass
    8. 2.7 Integrals, Exponential Functions, and Logarithms
    9. 2.8 Exponential Growth and Decay
    10. 2.9 Calculus of the Hyperbolic Functions
    11. Key Terms
    12. Key Equations
    13. Key Concepts
    14. Chapter Review Exercises
  4. 3 Techniques of Integration
    1. Introduction
    2. 3.1 Integration by Parts
    3. 3.2 Trigonometric Integrals
    4. 3.3 Trigonometric Substitution
    5. 3.4 Partial Fractions
    6. 3.5 Other Strategies for Integration
    7. 3.6 Numerical Integration
    8. 3.7 Improper Integrals
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Chapter Review Exercises
  5. 4 Introduction to Differential Equations
    1. Introduction
    2. 4.1 Basics of Differential Equations
    3. 4.2 Direction Fields and Numerical Methods
    4. 4.3 Separable Equations
    5. 4.4 The Logistic Equation
    6. 4.5 First-order Linear Equations
    7. Key Terms
    8. Key Equations
    9. Key Concepts
    10. Chapter Review Exercises
  6. 5 Sequences and Series
    1. Introduction
    2. 5.1 Sequences
    3. 5.2 Infinite Series
    4. 5.3 The Divergence and Integral Tests
    5. 5.4 Comparison Tests
    6. 5.5 Alternating Series
    7. 5.6 Ratio and Root Tests
    8. Key Terms
    9. Key Equations
    10. Key Concepts
    11. Chapter Review Exercises
  7. 6 Power Series
    1. Introduction
    2. 6.1 Power Series and Functions
    3. 6.2 Properties of Power Series
    4. 6.3 Taylor and Maclaurin Series
    5. 6.4 Working with Taylor Series
    6. Key Terms
    7. Key Equations
    8. Key Concepts
    9. Chapter Review Exercises
  8. 7 Parametric Equations and Polar Coordinates
    1. Introduction
    2. 7.1 Parametric Equations
    3. 7.2 Calculus of Parametric Curves
    4. 7.3 Polar Coordinates
    5. 7.4 Area and Arc Length in Polar Coordinates
    6. 7.5 Conic Sections
    7. Key Terms
    8. Key Equations
    9. Key Concepts
    10. Chapter Review Exercises
  9. A | Table of Integrals
  10. B | Table of Derivatives
  11. C | Review of Pre-Calculus
  12. Answer Key
    1. Chapter 1
    2. Chapter 2
    3. Chapter 3
    4. Chapter 4
    5. Chapter 5
    6. Chapter 6
    7. Chapter 7
  13. Index

Learning Objectives

  • 2.9.1. Apply the formulas for derivatives and integrals of the hyperbolic functions.
  • 2.9.2. Apply the formulas for the derivatives of the inverse hyperbolic functions and their associated integrals.
  • 2.9.3. Describe the common applied conditions of a catenary curve.

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

sinhx=exex2andcoshx=ex+ex2.sinhx=exex2andcoshx=ex+ex2.

The other hyperbolic functions are then defined in terms of sinhxsinhx and coshx.coshx. The graphs of the hyperbolic functions are shown in the following figure.

This figure has six graphs. The first graph labeled “a” is of the function y=sinh(x). It is an increasing function from the 3rd quadrant, through the origin to the first quadrant. The second graph is labeled “b” and is of the function y=cosh(x). It decreases in the second quadrant to the intercept y=1, then becomes an increasing function. The third graph labeled “c” is of the function y=tanh(x). It is an increasing function from the third quadrant, through the origin, to the first quadrant. The fourth graph is labeled “d” and is of the function y=coth(x). It has two pieces, one in the third quadrant and one in the first quadrant with a vertical asymptote at the y-axis. The fifth graph is labeled “e” and is of the function y=sech(x). It is a curve above the x-axis, increasing in the second quadrant, to the y-axis at y=1 and then decreases. The sixth graph is labeled “f” and is of the function y=csch(x). It has two pieces, one in the third quadrant and one in the first quadrant with a vertical asymptote at the y-axis.
Figure 2.81 Graphs of the hyperbolic functions.

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

ddx(sinhx)=ddx(exex2)=12[ddx(ex)ddx(ex)]=12[ex+ex]=coshx.ddx(sinhx)=ddx(exex2)=12[ddx(ex)ddx(ex)]=12[ex+ex]=coshx.

Similarly, (d/dx)coshx=sinhx.(d/dx)coshx=sinhx. We summarize the differentiation formulas for the hyperbolic functions in the following table.

f(x)f(x) ddxf(x)ddxf(x)
sinhxsinhx coshxcoshx
coshxcoshx sinhxsinhx
tanhxtanhx sech2xsech2x
cothxcothx csch2xcsch2x
sechxsechx sechxtanhxsechxtanhx
cschxcschx cschxcothxcschxcothx
Table 2.2 Derivatives of the Hyperbolic Functions

Let’s take a moment to compare the derivatives of the hyperbolic functions with the derivatives of the standard trigonometric functions. There are a lot of similarities, but differences as well. For example, the derivatives of the sine functions match: (d/dx)sinx=cosx(d/dx)sinx=cosx and (d/dx)sinhx=coshx.(d/dx)sinhx=coshx. The derivatives of the cosine functions, however, differ in sign: (d/dx)cosx=sinx,(d/dx)cosx=sinx, but (d/dx)coshx=sinhx.(d/dx)coshx=sinhx. As we continue our examination of the hyperbolic functions, we must be mindful of their similarities and differences to the standard trigonometric functions.

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

sinhudu=coshu+Ccsch2udu=cothu+Ccoshudu=sinhu+Csechutanhudu=sechu+Csech2udu=tanhu+Ccschucothudu=cschu+Csinhudu=coshu+Ccsch2udu=cothu+Ccoshudu=sinhu+Csechutanhudu=sechu+Csech2udu=tanhu+Ccschucothudu=cschu+C

Example 2.47

Differentiating Hyperbolic Functions

Evaluate the following derivatives:

  1. ddx(sinh(x2))ddx(sinh(x2))
  2. ddx(coshx)2ddx(coshx)2

Solution

Using the formulas in Table 2.2 and the chain rule, we get

  1. ddx(sinh(x2))=cosh(x2)·2xddx(sinh(x2))=cosh(x2)·2x
  2. ddx(coshx)2=2coshxsinhxddx(coshx)2=2coshxsinhx
Checkpoint 2.47

Evaluate the following derivatives:

  1. ddx(tanh(x2+3x))ddx(tanh(x2+3x))
  2. ddx(1(sinhx)2)ddx(1(sinhx)2)

Example 2.48

Integrals Involving Hyperbolic Functions

Evaluate the following integrals:

  1. xcosh(x2)dxxcosh(x2)dx
  2. tanhxdxtanhxdx

Solution

We can use u-substitution in both cases.

  1. Let u=x2.u=x2. Then, du=2xdxdu=2xdx and
    xcosh(x2)dx=12coshudu=12sinhu+C=12sinh(x2)+C.xcosh(x2)dx=12coshudu=12sinhu+C=12sinh(x2)+C.
  2. Let u=coshx.u=coshx. Then, du=sinhxdxdu=sinhxdx and
    tanhxdx=sinhxcoshxdx=1udu=ln|u|+C=ln|coshx|+C.tanhxdx=sinhxcoshxdx=1udu=ln|u|+C=ln|coshx|+C.

    Note that coshx>0coshx>0 for all x,x, so we can eliminate the absolute value signs and obtain
    tanhxdx=ln(coshx)+C.tanhxdx=ln(coshx)+C.
Checkpoint 2.48

Evaluate the following integrals:

  1. sinh3xcoshxdxsinh3xcoshxdx
  2. sech2(3x)dxsech2(3x)dx

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.

Function Domain Range
sinh−1xsinh−1x (,)(,) (,)(,)
cosh−1xcosh−1x (1,)(1,) [0,)[0,)
tanh−1xtanh−1x (−1,1)(−1,1) (,)(,)
coth−1xcoth−1x (,−1)(1,)(,−1)(1,) (,0)(0,)(,0)(0,)
sech−1xsech−1x (0, 1)(0, 1) [0,)[0,)
csch−1xcsch−1x (,0)(0,)(,0)(0,) (,0)(0,)(,0)(0,)
Table 2.3 Domains and Ranges of the Inverse Hyperbolic Functions

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

This figure has six graphs. The first graph labeled “a” is of the function y=sinh^-1(x). It is an increasing function from the 3rd quadrant, through the origin to the first quadrant. The second graph is labeled “b” and is of the function y=cosh^-1(x). It is in the first quadrant, beginning on the x-axis at 2 and increasing. The third graph labeled “c” is of the function y=tanh^-1(x). It is an increasing function from the third quadrant, through the origin, to the first quadrant. The fourth graph is labeled “d” and is of the function y=coth^-1(x). It has two pieces, one in the third quadrant and one in the first quadrant with a vertical asymptote at the y-axis. The fifth graph is labeled “e” and is of the function y=sech^-1(x). It is a curve decreasing in the first quadrant and stopping on the x-axis at x=1. The sixth graph is labeled “f” and is of the function y=csch^-1(x). It has two pieces, one in the third quadrant and one in the first quadrant with a vertical asymptote at the y-axis.
Figure 2.82 Graphs of the inverse hyperbolic functions.

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

y=sinh−1xsinhy=xddxsinhy=ddxxcoshydydx=1.y=sinh−1xsinhy=xddxsinhy=ddxxcoshydydx=1.

Recall that cosh2ysinh2y=1,cosh2ysinh2y=1, so coshy=1+sinh2y.coshy=1+sinh2y. Then,

dydx=1coshy=11+sinh2y=11+x2.dydx=1coshy=11+sinh2y=11+x2.

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

f(x)f(x) ddxf(x)ddxf(x)
sinh−1xsinh−1x 11+x211+x2
cosh−1xcosh−1x 1x211x21
tanh−1xtanh−1x 11x211x2
coth−1xcoth−1x 11x211x2
sech−1xsech−1x −1x1x2−1x1x2
csch−1xcsch−1x −1|x|1+x2−1|x|1+x2
Table 2.4 Derivatives of the Inverse Hyperbolic Functions

Note that the derivatives of tanh−1xtanh−1x and coth−1xcoth−1x are the same. Thus, when we integrate 1/(1x2),1/(1x2), we need to select the proper antiderivative based on the domain of the functions and the values of x.x. Integration formulas involving the inverse hyperbolic functions are summarized as follows.

11+u2du=sinh−1u+C1u1u2du=sech−1|u|+C1u21du=cosh−1u+C1u1+u2du=csch−1|u|+C11u2du={tanh−1u+Cif|u|<1coth−1u+Cif|u|>111+u2du=sinh−1u+C1u1u2du=sech−1|u|+C1u21du=cosh−1u+C1u1+u2du=csch−1|u|+C11u2du={tanh−1u+Cif|u|<1coth−1u+Cif|u|>1

Example 2.49

Differentiating Inverse Hyperbolic Functions

Evaluate the following derivatives:

  1. ddx(sinh−1(x3))ddx(sinh−1(x3))
  2. ddx(tanh−1x)2ddx(tanh−1x)2

Solution

Using the formulas in Table 2.4 and the chain rule, we obtain the following results:

  1. ddx(sinh−1(x3))=131+x29=19+x2ddx(sinh−1(x3))=131+x29=19+x2
  2. ddx(tanh−1x)2=2(tanh−1x)1x2ddx(tanh−1x)2=2(tanh−1x)1x2
Checkpoint 2.49

Evaluate the following derivatives:

  1. ddx(cosh−1(3x))ddx(cosh−1(3x))
  2. ddx(coth−1x)3ddx(coth−1x)3

Example 2.50

Integrals Involving Inverse Hyperbolic Functions

Evaluate the following integrals:

  1. 14x21dx14x21dx
  2. 12x19x2dx12x19x2dx

Solution

We can use u-substitutionu-substitution in both cases.

  1. Let u=2x.u=2x. Then, du=2dxdu=2dx and we have
    14x21dx=12u21du=12cosh−1u+C=12cosh−1(2x)+C.14x21dx=12u21du=12cosh−1u+C=12cosh−1(2x)+C.
  2. Let u=3x.u=3x. Then, du=3dxdu=3dx and we obtain
    12x19x2dx=121u1u2du=12sech−1|u|+C=12sech−1|3x|+C.12x19x2dx=121u1u2du=12sech−1|u|+C=12sech−1|3x|+C.
Checkpoint 2.50

Evaluate the following integrals:

  1. 1x24dx,x>21x24dx,x>2
  2. 11e2xdx11e2xdx

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.

An image of chains hanging between posts that all take the shape of a catenary.
Figure 2.83 Chains between these posts take the shape of a catenary. (credit: modification of work by OKFoundryCompany, Flickr)

Hyperbolic functions can be used to model catenaries. Specifically, functions of the form y=acosh(x/a)y=acosh(x/a) are catenaries. Figure 2.84 shows the graph of y=2cosh(x/2).y=2cosh(x/2).

This figure is a graph. It is of the function f(x)=2cosh(x/2). The curve decreases in the second quadrant to the y-axis. It intersects the y-axis at y=2. Then the curve becomes increasing.
Figure 2.84 A hyperbolic cosine function forms the shape of a catenary.

Example 2.51

Using a Catenary to Find the Length of a Cable

Assume a hanging cable has the shape 10cosh(x/10)10cosh(x/10) for −15x15,−15x15, where xx is measured in feet. Determine the length of the cable (in feet).

Solution

Recall from Section 6.46.4 that the formula for arc length is

Arc Length=ab1+[f(x)]2dx.Arc Length=ab1+[f(x)]2dx.

We have f(x)=10cosh(x/10),f(x)=10cosh(x/10), so f(x)=sinh(x/10).f(x)=sinh(x/10). Then

Arc Length=ab1+[f(x)]2dx=−15151+sinh2(x10)dx.Arc Length=ab1+[f(x)]2dx=−15151+sinh2(x10)dx.

Now recall that 1+sinh2x=cosh2x,1+sinh2x=cosh2x, so we have

Arc Length=−15151+sinh2(x10)dx=−1515cosh(x10)dx=10sinh(x10)|−1515=10[sinh(32)sinh(32)]=20sinh(32)42.586ft.Arc Length=−15151+sinh2(x10)dx=−1515cosh(x10)dx=10sinh(x10)|−1515=10[sinh(32)sinh(32)]=20sinh(32)42.586ft.
Checkpoint 2.51

Assume a hanging cable has the shape 15cosh(x/15)15cosh(x/15) for −20x20.−20x20. Determine the length of the cable (in feet).

Section 2.9 Exercises

377.

[T] Find expressions for coshx+sinhxcoshx+sinhx and coshxsinhx.coshxsinhx. Use a calculator to graph these functions and ensure your expression is correct.

378.

From the definitions of cosh(x)cosh(x) and sinh(x),sinh(x), find their antiderivatives.

379.

Show that cosh(x)cosh(x) and sinh(x)sinh(x) satisfy y=y.y=y.

380.

Use the quotient rule to verify that tanh(x)=sech2(x).tanh(x)=sech2(x).

381.

Derive cosh2(x)+sinh2(x)=cosh(2x)cosh2(x)+sinh2(x)=cosh(2x) from the definition.

382.

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

383.

Prove sinh(x+y)=sinh(x)cosh(y)+cosh(x)sinh(y)sinh(x+y)=sinh(x)cosh(y)+cosh(x)sinh(y) by changing the expression to exponentials.

384.

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.

385.

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

386.

[T] sinh(x2)sinh(x2)

387.

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

388.

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

389.

[T] cosh2(x)+sinh2(x)cosh2(x)+sinh2(x)

390.

[T] cosh2(x)sinh2(x)cosh2(x)sinh2(x)

391.

[T] tanh(x2+1)tanh(x2+1)

392.

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

393.

[T] sinh6(x)sinh6(x)

394.

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

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

395.

cosh(2x+1)cosh(2x+1)

396.

tanh(3x+2)tanh(3x+2)

397.

xcosh(x2)xcosh(x2)

398.

3x3tanh(x4)3x3tanh(x4)

399.

cosh2(x)sinh(x)cosh2(x)sinh(x)

400.

tanh2(x)sech2(x)tanh2(x)sech2(x)

401.

sinh(x)1+cosh(x)sinh(x)1+cosh(x)

402.

coth(x)coth(x)

403.

cosh(x)+sinh(x)cosh(x)+sinh(x)

404.

(cosh(x)+sinh(x))n(cosh(x)+sinh(x))n

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

405.

tanh−1(4x)tanh−1(4x)

406.

sinh−1(x2)sinh−1(x2)

407.

sinh−1(cosh(x))sinh−1(cosh(x))

408.

cosh−1(x3)cosh−1(x3)

409.

tanh−1(cos(x))tanh−1(cos(x))

410.

esinh−1(x)esinh−1(x)

411.

ln(tanh−1(x))ln(tanh−1(x))

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

412.

dx4x2dx4x2

413.

dxa2x2dxa2x2

414.

dxx2+1dxx2+1

415.

xdxx2+1xdxx2+1

416.

dxx1x2dxx1x2

417.

exe2x1exe2x1

418.

2xx412xx41

For the following exercises, use the fact that a falling body with friction equal to velocity squared obeys the equation dv/dt=gv2.dv/dt=gv2.

419.

Show that v(t)=gtanh(gt)v(t)=gtanh(gt) satisfies this equation.

420.

Derive the previous expression for v(t)v(t) by integrating dvgv2=dt.dvgv2=dt.

421.

[T] Estimate how far a body has fallen in 1212 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=dy/dxS=dy/dx that satisfies dS/dx=c1+S2.dS/dx=c1+S2. The constant cc is the ratio of cable density to tension.

422.

Show that S=sinh(cx)S=sinh(cx) satisfies this equation.

423.

Integrate dy/dx=sinh(cx)dy/dx=sinh(cx) to find the cable height y(x)y(x) if y(0)=1/c.y(0)=1/c.

424.

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

For the following exercises, solve each problem.

425.

[T] A chain hangs from two posts 22 m apart to form a catenary described by the equation y=2cosh(x/2)1.y=2cosh(x/2)1. Find the slope of the catenary at the left fence post.

426.

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

427.

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

428.

A telephone line is a catenary described by y=acosh(x/a).y=acosh(x/a). Find the ratio of the area under the catenary to its arc length. Does this confirm your answer for the previous question?

429.

Prove the formula for the derivative of y=sinh−1(x)y=sinh−1(x) by differentiating x=sinh(y).x=sinh(y). (Hint: Use hyperbolic trigonometric identities.)

430.

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.)

431.

Prove the formula for the derivative of y=sech−1(x)y=sech−1(x) by differentiating x=sech(y).x=sech(y). (Hint: Use hyperbolic trigonometric identities.)

432.

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

433.

Prove the expression for sinh−1(x).sinh−1(x). Multiply x=sinh(y)=(1/2)(eyey)x=sinh(y)=(1/2)(eyey) by 2ey2ey and solve for y.y. Does your expression match the textbook?

434.

Prove the expression for cosh−1(x).cosh−1(x). Multiply x=cosh(y)=(1/2)(eyey)x=cosh(y)=(1/2)(eyey) by 2ey2ey and solve for y.y. Does your expression match the textbook?

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