Calculus concepts and contexts 4th edition stewart solutions manual

Calculus Concepts and Contexts 4th Edition Stewart Solutions Manual Visit to download the full and correct content document: https://testbankdeal.com/dow nload/calculus-concepts-and-contexts-4th-edition-stewart-solutions-manual/

Applications of Integration 6.1

6

More About Areas

Suggested Time and Emphasis 1 class

Essential material (areas enclosed by parametric curves is optional, but recommended)

Points to Stress 1. Integration used to determine the area between graphs of functions.

2. Integration used to determine the area enclosed within a parametric curve, or several such curves.

Text Discussion • Why is it sometimes easier to integrate with respect to y than with respect to x to find the area between two curves?

ANSWER In order to integrate with respect to x, function formulas have to be solved for y . In cases like x = y3 + y this is very hard to do. b β • How do we go from a y dx to α g (t) f (t) dt when computing the area under the parametric curve x = f (t), y = g (t), α ≤ t ≤ β ?

• In Example 6, we have x = r (θ − sin θ), dx = r (1 − cos θ) dθ. Why have we used the derivative with respect to θ, rather than the derivative with respect to r?

Materials for Lecture b • Note that if f ≥ g the area between f and g is a [f (x) − g (x)] dx regardless of the signs of f and g . Check this statement with the examples f (x) = x, g (x) = −2 cos x, 0 ≤ x ≤ π2 , or f (x) = −x, g (x) = 12 (x − 2)2 − 2, 0 ≤ x ≤ 2.

• Compute the area enclosed by the curves y = x2 , y = x3 , x = −1, and x = 2 as two separate integrals:

1 2 2 3 3 2 −1 x − x dx + 1 x − x dx. Point out that we could also write this as a single integral, namely 2 2 3 −1 x − x dx.

ANSWER 25 12

377

CHAPTER 6 APPLICATIONS OF INTEGRATION

• Give an example, such as the one below, where changing from y = f (x) and y = g (x) to x = h (y) and x = k (y) makes integration easier. y

1

y=(x+1)! ?%

x=-y@+3y 0

_1

1

x

2

1.261 2 ANSWER The area is approximately 0 −y + 3y − y5 − 1 dy ≈ 2.3077.

• Discuss how the area of a region such as the one below can be found using integration. y 3

f(x)

2 1 1

0

1.5

_1

ANSWER 0 [f (x) − (−2x + 2)] dx − 1.5 2 0 [f (x) − (−2x + 2)] dx.

2

3 x

3

1.5 [f (x) − (2x − 4)] dx,

or

use

symmetry

to

get

• Develop the formula for the area enclosed by a parametric curve.

• Foreshadow polar coordinates by finding the area of the ellipse 14 x2 + y2 = 1 using the parametrization x = 2 cos θ, y = sin θ.

Workshop/Discussion • Numerically estimate the points where f (x) = sin x and g (x) = x3 − 2x − 1 intersect, and then compute the area between the curves.

ANSWER The intersection points are x ≈ −1.413, x ≈ −0.350, and x ≈ 1.767. The area is approximately

1.767 −0.350 3 x − 2x − 1 − (sin x) dx + −0.350 (sin x) − x3 − 2x − 1 dx ≈ 4.418. −1.413 y 2

g

1

_2

f 0

_1 _1 _2

378

1

2

x

SECTION 6.1 MORE ABOUT AREAS

• Compute the area between f (x) = − sin x and g (x) = −2 sin x from 0 to π . A picture of this area is shown below. Have the students estimate the answer before computing it.

ANSWER Surprisingly, the area is exactly 2.

• Compute the area between the parabola 2x = y 2 − 4 and the line y = −3x + 2. Note that these curves intersect at the point (0, 2); the other point of intersection can then be computed. Point out that integrating with respect to y makes this problem easier.

2 686 ANSWER −8/3 − 13 (y − 2) − 12 y 2 − 4 dy = 81 . Integrating with respect to x gives √

√ 0 √ 14/9 2x + 4 − − 2x + 4 dx + 0 (−3x + 2) − − 2x + 4 dx = 686 81 . −2 • Compute the area between the curves f (x) = ex − 1 and g (x) = 2 ln (x + 1) for − 12 ≤ x ≤ 12 . Point out what happens at x = 0. y

g

1

f

1 __ 2 1 _ 2

x

_1

_2

−1/2 ANSWER 2 + ln 27 16 − (1 + e) e • The graph of x (t) = t2 − 5, y (t) = sin (t), −20 ≤ t ≤ 20 is a chain-like series of loops. Demonstrate

the method of computing areas of parametric curves by finding the area enclosed by the leftmost loop (between t = −π and t = π ).

Group Work 1: Practice with Areas The emphasis of this exercise should be on how to set up an integral to compute area, not on the computation itself. You may want the students to compute a few, such as in Problems 1, 3 and 6, but they shouldn’t get so bogged down in the calculations that they miss the point of the activity.

The fifth problem requires the use of technology to find the points of intersection. After students have completed the sixth problem, show them how to simplify the computation using symmetry.

ANSWERS (for graphs see next page) 3 1. 0 (8 − 2x ) dx = 24 − ln72 1 √ √ 2 3. 0 1 − x2 − (1 − x) dx = 14 π − 16

2 2. 0 3x dx = ln83

2 4. 0 3 1 − 14 y 2 − − 14 y 2 + 1 dy = − 43 + 32 π 379

CHAPTER 6 APPLICATIONS OF INTEGRATION

5.

1.714 cos x − 12 x − 1 dx ≈ 1.969 0

6.

0 1 −2

2x + 1

2 − x2 − 4 dx + 0 − 12 x + 1 − x2 − 4 dx

0 = 2 −2 12 x + 1 − x2 − 4 dx = 38 3

1. y

2.

8

1

8

6

6

4

4

2

2

0

3. y

y

1

2

3

0

x

4. y

5.

2

1

2

x

0

1 x

6.

y

y 1

1

_2

_1

0

1

2 x

_1

1

0

1

x

_2 _3

0

1

2

3 x

_4

_1

Group Work 2: Clarifying Areas WARNING This one is tricky.

This exercise, Problem 12 in the Focus on Problem Solving section after this chapter, is for classes that have covered parametric curves, and the areas enclosed by parametric regions. It develops the equations of an involute of a circle. String, tape, and paper towel rolls should be made available to the students for Problem 2.

Note that although θ is the parameter, the curve is not given in polar form because the cow is tethered to a point on the circle of radius r rather than to the origin.

The picture in the answer to Problem 2 below may be given as a hint for Problem 4. A1 + A2 can be computed by integrating y dx from θ = π to θ = π2 and subtracting off the integral from θ = 0 to θ = π2 .

380

SECTION 6.1 MORE ABOUT AREAS

ANSWERS 1. πr 2.

3. The coordinates of T are (r cos θ, r sin θ). Since T P was unwound from arc T A, T P has length rθ, the length of the sector. Also, P T Q = P T R − QT R = π2 − θ, so P has coordinates

x = r cos θ + |T P | cos P T Q = r cos θ + rθ cos π2 − θ = r (cos θ + θ sin θ) and y = r sin θ + |T P | sin P T Q =

r sin θ + rθ sin π2 − θ = r (sin θ − θ cos θ). π/2 π/2 0 π 3 r2 4. A3 = 14 π (πr)2 = . A1 + A2 = π y dx − 0 y dx = π y dx. Using the results from Problem 3 4

0

π π3 + gives A1 + A2 = r2 −θ cos2 θ − 12 θ2 − 1 sin θ cos θ − 16 θ3 + 12 θ π = r2 . Therefore, the 2 6 total area is 56 π 3 r2 .

Homework Problems CORE EXERCISES 2, 6, 15, 21, 23, 27, 29, 33, 40 SAMPLE ASSIGNMENT 2, 3, 6, 9, 15, 17, 21, 22, 23, 26, 27, 29, 30, 33, 35, 38, 40, 43 EXERCISE 2 3 6 9 15 17 21 22 23

D

A × × × × × × × × ×

N × × × × × × × × ×

G × × × × × × × × ×

EXERCISE 26 27 29 30 33 35 38 40 43

381

D × × ×

A

N

× × × × × × × ×

× × × × ×

G × ×

GROUP WORK 1, SECTION 6.1 Practice with Areas For each of the following problems, first sketch the relevant area, then write out the definite integral that will give its exact value.

1. The area bounded by y = 2x , y = 8, and the y-axis.

2. The area bounded by y = 3x , x = 2, the x-axis, and the y -axis.

3. The area in the first quadrant between x2 + y2 = 1 and x1/2 + y1/2 = 1.

4. The area in the first quadrant bounded by the curves

x2 y2 y2 + = 1 and x = − + 1.

9 4 4

5. The area between the curves y = cos x and y = 12 x − 1, bounded on the left by the y-axis.

6. The area bounded by the curves y = x2 − 4 and y =

382

1

if x ≤ 0 2x + 1 − 12 x + 1 if x > 0

GROUP WORK 2, SECTION 6.1 Clarifying Areas Clara the Calculus Cow has been tied to a silo with radius r by a rope just long enough to reach a point diametrically opposite to the point where she is tied, as depicted in the diagram below:

If she goes to the left side of the silo, she can stand far away from the silo, while at the right side, she can only graze right next to the silo. We wish to compute the total area of the region upon which she can graze.

1. How far from the silo can sweet Clara stand when she is to its left?

2. Draw a picture of the shape of the region that she can reach. It may help to actually tie a string to a cardboard tube, and model the situation.

383

Clarifying Areas

3. Assume that a very long rope is wound around the circular silo, and then unwound while being held taut.

The curve traced by the end of the rope is called the involute of the circle. If the silo has radius r and center O, as shown in the figure below, and if the parameter θ is chosen as in the figure, show that the parametric equations of the involute are y = r (sin θ − θ cos θ)

x = r (cos θ + θ sin θ) y

T r ¬

P x

O

4. What is the area of the grazing region available to Clara the Calculus Cow?

384

6.2

Volumes

Suggested Time and Emphasis 1–1 12 classes

Essential material

Points to Stress 1. The general volume formula V =

b

a A (x) dx.

2. The application of the general formula to various types of solids, such as solids with given cross-sections and solids of revolution.

Text Discussion

2 2

• In Example 4, why is it true that A (x) = πx2 − π x2 instead of A (x) = π x − x2 ?

ANSWER The cross-sections are shaped like concentric circles. Their area is the area of the outer disk, πR2 , minus the area of the inner disk, πr2 .

• Examples 4 and 5 use a method sometimes called “the method of washers”. What is the reason for the choice of this term?

ANSWER The cross sections resemble washers.

Materials for Lecture • This is a good time to revisit the ideas given in “Principles of Problem Solving” after Chapter 1.

Demonstrate with several examples how these steps can be used to find a formula for the cross-sectional area A (x) of a solid.

• Define a solid beam with cross-sections as in the figure below at left: x x x

x

x

Find the cross-sectional area A (x), 0 ≤ x ≤ 5, and then compute the volume.

5 ANSWER A (x) = 6x2 − πx2 ⇒ V = 0 (6 − π) x2 dx = 250 − 125 3 π • Describe the general approach for rotating y = f (x) about the line y = k over the interval [a, b], and similarly for x = g (y) rotated around the line x = m.

• Many party stores have wedding or Halloween decorations that are good for demonstrating rotated regions.

They are sold as flat cross-sections, but unfold accordion-style when rotated around a central axis.

Common shapes are Halloween pumpkins or wedding bells.

385

CHAPTER 6 APPLICATIONS OF INTEGRATION

• Demonstrate how to find the volume of the solid obtained by rotating the indicated area about the line y = 2. Use the method of

slicing, but emphasize the value of deriving an expression for A (x), as opposed to just plugging numbers into a formula. Then set up the corresponding integrals for the solid obtained by rotating the same area around x = 0.

ANSWER

1

0 π

2

22 − (e−x )

1 1/e dx = 12 π 7 + e−2 , 0 π (1)2 dx + 1/e π (− ln y)2 dy = −4e−1 π + 2π

Workshop/Discussion

• Rotate the region below about the x-axis, and compute the volume of the resulting solid. Then do the same for the line x = −1 and the line y = −1.

2 2 1 dx = ANSWER 0 π x1/4 − π x4 2 4 2 1 1/4 +1 −π y +1 dy = 79 45 π 0 π y

5 9 π,

1 0

• Describe a water tower modeled by rotating the function ⎧ 2 if 0 < x ≤ 1 ⎪ ⎨2−x 1 if 1 < x ≤ 5 f (x) = ⎪ ⎩ 4 − (x − 7)2 + 1 if 5 ≤ x ≤ 9

2

2 π x1/4 + 1 − π x4 + 1 dx

79 45 π ,

=

y 4 3 2 1

about the x-axis, and set up the integrals necessary to compute its 0 volume.

2 4

2 2 1 5 9 ANSWER V = 0 π 2 − x2 dx + 1 π (1)2 dx + 5 π 4 − (x − 7)2 + 1 dx 323

= 15 π + 4π 2 ≈ 107.127 386

6

8

x

SECTION 6.2 VOLUMES

• Write the formula (without computing the integral) for the volume obtained by rotating the region below about the x-axis.

y=x

y=(x-1)#+1

ANSWER

"

" 2 2 2 3 3 2 2 π (x − 1) + 1 − πx dx + 1 πx − π (x − 1) + 1 dx (= π ) 0

1

• Demonstrate how to find the volume of the solid whose base is the region lying between

ANSWER y

the parabolas x = y 2 and x = 3 − 2y 2 and

1

whose cross-sections taken perpendicular to the x-axis are squares.

0

z

1

2

3 x

y

_1 x=y@

1 0

√ 2 (2 x) dx +

The Base 3 1

2

3−x 2

2 dx = 6

x

x=3-2y@ x=0

x=1

x=3

• Set up an integral for the volume of the solid described in the following problem: A wedge is cut out of a log of radius 1 foot by cutting two slices to the center, each making an angle of 45◦ to the ground. What is the volume of the resulting wedge?

45º

45º

(Cross-sections of the wedge taken perpendicular to the axis of the log are isosceles right triangles. If we model a cross-section of the log with the circle x2 + y 2 = 1, then the altitude of these triangles at a √ horizontal distance x from the axis of the log is 1 − x2 .)

387

CHAPTER 6 APPLICATIONS OF INTEGRATION

• Revisit the problem of computing the volume formed by rotating y = e−x , 0 ≤ x ≤ 1 about the yaxis, using cylindrical shells. Compare this computation to the slicing method (which would require two integrals in this case).

2

• Compute the volume generated by rotating y = ex , 0 ≤ x ≤ 1 about the y -axis. Discuss the problems with trying to compute the volume generated by rotating this curve about the x-axis.

Group Work 1: The Revolution Will Not Be Televised This series of drill problems gives the students practice with regions rotated about different lines. If a group finishes early, have them either approximate or solve the integrals.

ANSWERS 2 2 dx = 512 1. 0 π (4x)2 − π x3 21 π ≈ 76.595

2 2 dx = 680 2. 0 π (4x + 1)2 − π x3 + 1 21 π ≈ 101.728

2 2 3. 0 π 9 − x3 − π (9 − 4x)2 dx = 1000 21 π ≈ 149.600 y 2 8 √ 2 dy = 128 4. 0 π 3 y − π 4 15 π ≈ 26.808

8

√ 2 2 dy = 112 5. 0 π 2 − y4 − π 2 − 3 y 15 π ≈ 23.457 2 y 2 8 √ dy = 248 6. 0 π 3 y + 1 − π 4 + 1 15 π ≈ 51.941

Group Work 2: Edible Volumes The students should be alerted to the fact that the volume integrals in Problems 2(b) and 3 involve the use of

a2 − y2 dy . Time can be saved by using trigonometric substitutions in computing integrals of the form technology to do the definite integrals. This exercise can also be reinforced by assigning Exercise 35.

ANSWERS 1. Solids of revolution can be used to describe the doughnut, bagel, bowl of soup, watermelon, carafe, certain sausages, and the cupcake, if it is frosted symmetrically.

2. The problem is that there is no proper hole, just a depression. (a) The Californian bagel has a bigger hole.

2

2 3/4 9 3 9 3 2 − π − 16 − y2 + 4 + 1 π dy ≈ 19.431 (b) Standard: 2 0 16 − y + 4 + 1

2

2 3/4 9 3 9 3 2 2 π dy ≈ 30.534 Californian: 2 0 − π − 16 − y + 4 + 2 16 − y + 4 + 2 3., 4. Answers will vary.

5. A good strategy is to set up a volume as 2

2 $ 3/4 # 9 3 9 2 2 − π − 16 − y2 + 34 + K π dy 16 − y + 4 + K 0

placing this nasty-looking integral on a calculator with a numerical integration feature. Then, using a key such as ENTRY , vary the values of K until your answer is close to your opponents’.

388

SECTION 6.2 VOLUMES

Group Work 3: Astroid Fragments Set up this activity by sketching x1/2 + y 1/2 = 1 on the board. Close by describing the relevant regions.

ANSWERS 2 1 π 1. 2π 0 x 1 − x1/2 dx = 15 2 4 1 1 π 2. 2π 0 y 1 − y 1/2 dy = π 0 1 − x1/2 dx = 15 3. The volumes are the same because the shapes are identical.

Group Work 4: Geometric Volumes This group work gives students practice with finding expressions for cross-sectional areas of familiar shapes.

ANSWERS 4 2 1. 0 14 x dx = 43

2.

4 1 1 2 dx = 83 0 2 2x

3.

3

0 (2.5)

2 − 23 x dx = 7.5

4.

3 1 x 2 dx = 12 π 0 2π 3

Homework Problems CORE EXERCISES 1, 3, 9, 16, 20, 25, 28, 31, 39, 45, 49 SAMPLE ASSIGNMENT 1, 2, 3, 5, 9, 14, 16, 20, 21, 25, 26, 28, 31, 33, 36, 39, 42, 45, 47, 49, 51 EXERCISE 1 2 3 5 9 14 16 20 21 25 26 28 31 33 36 39 42 45 47 49 51

D

× ×

389

A × × × × × × × × ×

N × × × × × × ×

× × × × × × × × × ×

× ×

G × × × × ×

×

× × ×

GROUP WORK 1, SECTION 6.2 The Revolution Will Not Be Televised Consider the following region: y 8 6 4

y=4x y=x#

2 0

1

2

x

For each problem below, set up an integral to compute the volume of the solid obtained by rotating this region about the given line.

1. The x-axis

2. y = −1

3. y = 9

390

The Revolution Will Not Be Televised

4. The y -axis

5. x = 2

6. x = −1

391

GROUP WORK 2, SECTION 6.2 Edible Volumes (With thanks to Luis Valero-Elizando) Solids of revolution can be used to model many things, from tiny bolts to giant pillars. Today we will use them to model food items.

1. Decide as a group which of the following items can be described using solids of revolution. a doughnut a pretzel a curly fry

a cup of cocoa a bagel a bowl of soup

a turkey a watermelon a carafe of orange juice

a square cake a sausage a frosted cupcake

2. Now we’re going to make bagels, but instead of kneading dough for countless hours, we’ll simply use solids of revolution. As every cook knows, you need to rotate a circle in order to make a bagel, but the subtleties of rotating circles often elude novice cooks. The diagram illustrates the most common mistake when making bagels. Describe what the problem is and how to correct it.

The “standard” bagel has always been the one obtained by rotating the circle of radius 34 inch centered

at 34 , 0 around the line x = −1. Recently, a Californian chef created a bagel with the same circle but rotated around the line x = −2. (a) Physically, what’s the difference between the two bagels?

(b) What are the volumes of the standard and Californian bagels?

392

Edible Volumes

Cooking Contest!

3. Create your own bagel by rotating the circle of radius 34 centered at preference. Find the volume of your creation.

3

4, 0

around the vertical line of your

4. Swap answers with another team. The first one to uncover the other team’s recipe wins.

5. What is the “winning strategy” in this game?

393

GROUP WORK 3, SECTION 6.2 Astroid Fragments Consider the astroid x1/2 + y 1/2 = 1. y 1

x!?@+y!?@=1

0

1

x

1. Using cylindrical shells, compute the volume generated by rotating the portion of the curve in the first quadrant about the y-axis.

2. Compute the volume generated by rotating the portion of the curve in the first quadrant about the x-axis.

3. Compare your answers to Problems 1 and 2. Can you explain your conclusions geometrically?

394

GROUP WORK 4, SECTION 6.2 Geometric Volumes Find the volume of the solids below whose cross-sections perpendicular to the x-axis are 1. Squares y 1

z

4 x

2. Isosceles right triangles y 2

z 4

x

3. Rectangles y 2

z

3

2.5

x

4. Half-circles y 1 z 3 x 395

Discovery Project: Rotating on a Slant This project tests the students’ understanding of the mathematics behind the volume and surface area formulas.

As an extended project, it would make a good challenge for an advanced group of students. Problems 1 and 2 could be assigned as an in-class exercise, while the later problems will require a good deal of time and effort.

Be prepared to give assistance where needed.

396

6.3

Volumes by Cylindrical Shells

Suggested Time and Emphasis 1 class

Optional material

Points to Stress 1. The method of using cylindrical shells to compute volumes of revolution.

Text Discussion • Why does the factor 2π appear with cylindrical shells, while π appears with washers?

ANSWER 2π comes from the circumference of a circle (2πr). π comes from the area of a circle (πr2 ).

Materials for Lecture • Outline the shell method and use it to compute some standard volumes, such as the volume generated by rotating y = cos x, 0 ≤ x ≤ π2 about the y-axis.

• A good (albeit pungent) solid that can be used to demonstrate shells is an onion with the top and the bottom cut off.

• Emphasize that the method of shells now allows us two options for every rotated region—we now get to choose whether we want to integrate with respect to x or with respect to y .

Workshop/Discussion • Revisit the problem of computing the volume formed by rotating y = e−x , 0 ≤ x ≤ 1 about the yaxis, using cylindrical shells. Compare this computation to the slicing method (which would require two integrals in this case).

• Have the students construct some solids for which computing the volume using shells is the easier method, and other solids for which computing the volume using washers is the easier method.

Group Work: Astroid Fragments Set up this activity by sketching x1/2 + y 1/2 = 1 on the board. Close by describing the relevant regions.

ANSWERS 2 1 π 1. 2π 0 x 1 − x1/2 dx = 15 2 4 1 1 π 2. 2π 0 y 1 − y 1/2 dy = π 0 1 − x1/2 dx = 15

3. The volumes are the same because the shapes are identical.

397

CHAPTER 6 APPLICATIONS OF INTEGRATION

Homework Problems CORE EXERCISES 1, 2, 5, 15, 20, 23, 31, 36 SAMPLE ASSIGNMENT 1, 2, 3, 5, 7, 9, 13, 15, 19, 20, 22, 23, 31, 32, 35, 36 EXERCISE 1 2 3 5 7 9 13 15 19 20 22 23 31 32 35 36

D × ×

×

398

A × × × × × × × × × × × × × × ×

N

× × × × × × × × ×

G × × × × × × ×

GROUP WORK, SECTION 6.3 Astroid Fragments Consider the astroid x1/2 + y 1/2 = 1. y 1

x!?@+y!?@=1

0

1

x

1. Using cylindrical shells, compute the volume generated by rotating the portion of the curve in the first quadrant about the y-axis.

2. Compute the volume generated by rotating the portion of the curve in the first quadrant about the x-axis.

3. Compare your answers to Problems 1 and 2. Can you explain your conclusions geometrically?

399

6.4

Arc Length

Suggested Time and Emphasis 1 2 –1 class

Essential material

Points to Stress 1. The arc length of a parametric curve, and the formulation of an integral to compute that length.

2. The connection between the two formulas for arc length:

b β 2 2 2 L= (dx/dt) + (dy/dt) dt = 1 + (dy/dx) dx α

a

3. The uses of the arc length function.

Text Discussion • So far, we have mainly used integrals to compute areas (for two-dimensional quantities) or volumes (for three-dimensional quantities). Why do integrals come into play when computing the length of a curve?

ANSWER We are adding up infinitely many infinitesimal quantities, which constitutes an integration.

• What are the parametric equations used to derive Equation 2?

Materials for Lecture • Derive the equations for arc length. End with a discussion about why these integrals are usually difficult to compute exactly.

• Show that the area under the curve at right is exactly 2, and comment on the simplicity of the area integral. Then set up the integral for the arc length. This integral cannot be computed directly, but numerical methods give π 1 + (cos x)2 dx ≈ 3.8202. Comment on the fact that the arc 0

length integral is much more complicated than the area integral.

400

SECTION 6.4 ARC LENGTH

• Estimate the arc length of the part of the implicitly defined function x2/3 + y2/3 = 1 that lies in the first quadrant, then show that the length is 32 . y 1

0

3/2 ANSWER y = 1 − x2/3

1 x

1/2 ⇒ y = − x−2/3 − 1

⇒ L=

1√ x−2/3 dx = 32 0

• Compute the length of the curve x = 2 cos2 θ, y = 2 cos θ sin θ, 0 ≤ θ ≤ π. Check the result by graphing the curve and showing that it is a circle of radius 2.

Workshop/Discussion • Use the arc length formula to compute the length along the straight line y = 3x from x = 0 to x = 2.

Then compute the length using the distance formula and show that you get the same number. √ √ 2√ ANSWER 0 1 + 9dx = 2 10, (2 − 0)2 + (6 − 0)2 = 2 10 • Compute the total length of the curve x3/5 + y3/5 = 1.

• Present a straightforward parametric arc length problem such as x = t2 + 1, y = t2 − 1, 0 ≤ t ≤ 2. √ (The computations work out nicely for this one, giving L = 4 2.)

• Graph the portion of the spiral x = t sin t, y = t cos t from t = 0 to t = 2π . Set up, and if time permits, evaluate an integral to find the length of the curve. The integral requires the use of the trigonometric substitution tan θ = t.

Group Work 1: The Bizarre Coastline Closure is important to this exercise. Make sure the students get some idea of the shape of the curve for large k. Technically, the island of Ytinifni is not well–defined, but there are many shapes in advanced mathematics that have finite area with infinite circumference [for example, an island whose top border is y = sin (1/x)]. In the real world, coastline figures are notoriously inaccurate, because the more accurately people try to measure a coast, the more crinkles they find, and the larger the result gets.

ANSWERS

2π √ 1. L = 4 + 2π + 0 1 + cos2 x dx ≈ 17.924, A = 4π 2π √ 2. L = 4 + 2π + 0 1 + 4 cos2 2x dx ≈ 20.824, A = 4π 2π √ 3. L = 4 + 2π + 0 1 + 9 cos2 3x dx ≈ 24.258, A = 4π 2π √ 4. L = 4 + 2π + 0 1 + 100 cos2 10x dx ≈ 51.120, A = 4π

5. The areas remain constant, while the coastline lengths increase.

6. L = ∞, A = 4π 401

2π 0

sin (kx) dx = 0 for any integer k.

CHAPTER 6 APPLICATIONS OF INTEGRATION

Group Work 2: Cable Guy This exercise requires the students to make several assumptions. For example, when computing the length of cable needed for the sample segment shown on the left, they will have to decide which of the three figures on the right most accurately represents the cross-sectional side view of the lake bottom: 10

10

10

20

20

20

10

20

Have them make an assumption and provide justification for it. Also remind students that the path of steepest descent always runs perpendicular to the contour lines.

Notice that this exercise involves many computations and estimates. The students must divide up the work in order to finish in a reasonable amount of time. Make sure to have several rulers on hand.

After the exercise is completed, poll the groups to see whose path used the least amount of cable, and whose used the most.

ANSWERS (will vary) Lake Wobegone

1.

Bill Turner’s House Cable Box 12

10

11 20

10 9 8

30 40 50

7 6 4 3

60

5

50 40 30 20 10

2 1

Cable Source

The path above was chosen so that the line segments are approximately orthogonal to the contour lines.

2.

1 4 inch on the map represents 100 ft. In estimating the amount of

10

cable needed, we will assume that a cross-sectional view of the path between two contour lines looks like the figure at right.

Segment Horizontal Vertical Length

1 30 10 31.6

2 25 10 26.9

3 25 10 26.9

4 25 10 26.9

5 50 10 51.0

6 125 10 125.4

7 200 10 200.2

8 75 10 75.7

9 75 10 75.7

10 90 10 90.6

The total length is about 966.8 ft, so we should probably order at least 1000 ft to be safe.

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20

11 150 10 150.3

12 85 10 85.6

SECTION 6.4 ARC LENGTH

Group Work 3: Find the Function

1 1 + [f (x)]2 = x + for f (x) and then 4x

integrating. The fact that the graph of f passes through the point 1, 12 is key, as it allows the students to

This is a tricky problem that reduces to solving the equation

determine the constant of integration.

x 1 1 ⇔ f (x) = x − 1 + [f (x)]2 = x + ANSWER 1 1 + [f (t)]2 dt = 12 x2 + 14 ln x ⇔ 4x 4x

⇔ f (x) = 12 x2 − 14 ln x + C . Because the graph of f passes through the point 1, 12 , C = 0, so f (x) = 1 2 1 2 x − 4 ln x.

Special Project: How to Define π This project allows the students to deduce a historic geometric result using techniques from this section.

They start by defining π as the circumference of a semi-circle of radius 1, and then show that the integral representing this quantity is the same as the integral representing the area of a circle of radius 1.

Note that the simplification in Problem 3 requires an algebraic trick. The students may need the hint that 2

x −1+1 x2 √ √ dx = dx. They will also need to be able to integrate by parts.

2 1−x 1 − x2 ANSWERS

1 1 2 dx d √ 2 √ 1.

1+ 1−x dx = 2 dx −1 −1 1 − x 2. The term on the left is the length of a semiscircle of radius 1, and the term on the right is an expression of the area of a circle of radius 1. √ √ x dx 1 − x2 dx. Let u = 1 − x2 , dv = dx ⇒ du = − √ , v = x. Then I = 3. I = 1 − x2

√ √ √ x2 dx dx = x 1 − x2 + √ − 1 − x2 dx. x 1 − x2 + √ 1 − x2 1 − x2 1 1 √ 1 √ dx 2 2 √ + and the result follows.

4. Using the answer to Problem 3, 2 −1 1 − x dx = x 1 − x 2 −1 −1 1 − x

Homework Problems CORE EXERCISES 2, 5, 11, 15, 22, 27, 28 SAMPLE ASSIGNMENT 1, 2, 4, 5, 7, 9, 11, 14, 15, 16, 22, 27, 28, 29, 30 EXERCISE 1 2 4 5 7 9 11 14

D

A × × × × × × × ×

N × × × × × ×

G

EXERCISE 15 16 22 27 28 29 30

× 403

D

A × × × × × × ×

N × × × × × ×

G × × ×

GROUP WORK 1, SECTION 6.4 The Bizarre Coastline In this exercise we are going to consider the areas and coastlines of some bizarre islands, and end by analyzing a type of shape that has intrigued mathematicians for many years. Put on your shorts and sunglasses; we are heading to the Pacific Ocean.

1. The first island on our trip is the island of Eno. The best way to model this island is to shade the region between the curves x = 0, x = 2π , y = 0, and y = sin x + 2.

The length of the coastline of this island is merely the sum of four arc lengths. The southern border has length 2π , and the east and west borders each have length 2. You can compute the north border’s length by using the arc length formula to obtain a numerical estimate accurate to three decimal places.

Coastline Length of Eno: Area of Eno: 2. Let us reluctantly leave the island of Eno, and wander over to the neighboring island of Owt. The south, east, and west borders of Owt look the same as Eno’s do, but the northern border is given by y = sin 2x+2.

In order to be able to deduct the cost of our trip, we need to do some research here as well. Please compute the coastline length of Owt and the area of Owt. Coastline Length of Owt: Area of Owt:

404

The Bizarre Coastline

3. As you probably expect, the island of Eerht is similar to the islands of Eno and Owt, except that the northern border is given by y = sin 3x + 2. Please sketch Eerht, and compute its coastline and area.

Coastline Length of Eerht: Area of Eerht: 4. Ah, if only we had the time to fully explore the islands of Ruof, Evif, and Xis... But there are a lot of islands in this chain, and we don’t have all the time in the world. Let’s skip ahead a bit, and look at the island of Net. Graph the northernmost border of Net (y = sin 10x + 2) on your calculator, and estimate the length of the coastline of Net before computing.

Coastline Length of Net: Area of Net: 5. You have probably noticed that there has been a pattern in the areas of this chain of islands, and the use of this pattern will save us some in-person visits. Explain what you have observed, and show that your conjecture holds for any one of the islands. What happens to the coastlines as we go farther along the chain of islands?

405

The Bizarre Coastline

6. For our final stop, let’s go to the mysterious and oft misunderstood island of Ytinifni. Try your best to sketch this mysterious island:

Notice that it is very hard to draw. One of the reasons people don’t like to travel here is that if you are in the northern part of the region, it is impossible to tell if you are on land or in the water! Many point out the impossibility of such an island even existing; but the inhabitants of the island merely smile enigmatically at any such objections.

Now, using geometric reasoning, complete the following: Coastline Length of Ytinifni: Area of Ytinifni: 7. Is this result counterintuitive? Try to explain it, and ask your teacher for help if you cannot do so.

406

GROUP WORK 2, SECTION 6.4 Cable Guy Billionaire businessman Bill Turner (the president of MacroHard) lives on a small island in the middle of Lake Wobegone. What follows is a topographical map of the region.

Lake Wobegone

Bill Turner’s House Cable Box 10

20 30 40 50

60 50 40 30 20 10

Cable Source

1. As the Cable Guy, your job is to wire up Mr. Turner’s home for cable television. There is a cable box on the shore, as indicated. The cable will be unwound on the bottom of the lake, from the shore to the island.

Draw a path that will minimize the amount of cable necessary to use. Don’t worry about being 100% sure, make a guess and go with it.

2. Given that 14 inch on this map represents 100 feet, what is the amount of cable needed to connect Mr. Turner’s house using your path?

407

GROUP WORK 3, SECTION 6.4 Find the Function

Find a function f (x) whose arc length L (x) from 1, 12 to (x, f (x)), x > 1, is 12 x2 + 14 ln x.

408

SPECIAL PROJECT FOR SECTION 6.4 How to Define π (Idea due to H. Wu, The Joy of Lecturing, pages 5–6) In this age of calculators, people sometimes get confused about how the number π is defined. The constant π is not the number 22 7 , nor 3.1415926, nor even 3.14159265358979323846264338327950288419716939937510582097

All of those are just approximations.

One can find many expressions that equal π. For example,

2 × 2 × 4 × 4 × 6 × 6 × 8 × ··· π = 2 1 × 3 × 3 × 5 × 5 × 7 × 7 × ···

1 1 1 1 1 π = 4 1− + − + − + ··· 3 5 7 9 11

6 6 6 6 π = + 2 + 2 + 2 + ··· 2 1 2 3 4

12 12 × 32 12 × 32 × 52 π = 3 1+ + + + ··· 4 × 6 4 × 6 × 8 × 10 4 × 6 × 8 × 10 × 12 × 14 Usually, however, people define π as the ratio of the circumference of a circle to its diameter. Then we say that the circumference of a circle is 2πr, and its area is πr2 . But it is easy to forget how amazing an idea this originally was: That the circumference, the area, and the radius of a general circle are so tightly related. In this exercise, we go back to basics. We define π as the length of a semicircle of radius 1 (that is, one-half the circumference) and then prove that it is equal to the area of a circle of radius 1.

1. Use the arc length formula to show that the length of the semicircle of radius 1, x2 + y 2 = 1, y ≥ 0, can 1 dx √ . be written as 2 −1 1 − x

2. Show that the statement “The length of a semicircle of radius 1 equals the area of a circle of radius 1” can be written as 1 1 dx √ =2 1 − x2 dx 2 −1 1 − x −1

409

How to Define π

√ 3. Using integration by parts with u = 1 − x2 , show that

dx 2 2 1 − x dx = x 1 − x + √ − 1 − x2 dx 1 − x2

4. Show that the formula of Problem 2 is true.

410

Discovery Project: Arc Length Contest This interesting project requires students to discover formulas for two positive functions with area 1 over [0, 1] satisfying f (0) = f (1) = 1. The basic idea is to find the graph with minimal arc length. Since one example is given with L < 3, this is a challenging project.

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6.5

Average Value of a Function

Suggested Time and Emphasis 1 2 class

Optional material

Points to Stress 1. The definition of average value.

2. A geometric interpretation of average value and the Mean Value Theorem for Integrals.

Text Discussion area • The text defines average height (for positive functions) as = average height. It also says that width b 1 fave = f (x) dx. How are these two definitions related? b−a a ANSWER Answers will vary, given the open-endedness of the question. One possible correct answer b would be that the width is b − a, and the area is a f (x) dx.

• Estimate the average value of the function below: y 5 4 3 2 1 0

1

2

3

4

5

6

x

ANSWER The average value is 3. Anything between 2.5 and 3.5 is a good estimate.

Materials for Lecture • Draw an analogy between the average of a collection of discrete quantities, and the definition of the average value of a function. Ask the students questions like, “What corresponds to adding up the quantities? What corresponds to dividing by the number of quantities?” b x1 + x2 + x3 + · · · + xn 1 1 corresponds to f (x) dx. n is analogous to , ANSWER The sum n b−a a b−a b and x1 + x2 + x3 + · · · + xn is analogous to a f (x) dx.

412

SECTION 6.5 AVERAGE VALUE OF A FUNCTION

• If desired, make the previous analogy more mathematically rigorous. For a continuous function, choose n sample points. Then

n

n n

b−a 1 1 1 ∗ ∗ ∗ f (xi ) = f (xi ) = f (xi ) ∆xi n i=1 b−a n i=1 b − a i=1 b 1 As the number of sample points n goes to infinity, we get f (x) dx. b−a a • Show the Mean Value Theorem for Integrals by an alternate proof such as the following: Let f be a continuous function on [a, b]. By the Extreme Value Theorem, there exist m and M with m ≤ f (x) ≤ M for all x ∈ [a, b]. Then b b 1 m (b − a) ≤ f (x) dx ≤ M (b − a) ⇒ m ≤ f (x) dx ≤ M b−a a a b 1 Finally, by the Intermediate Value Theorem, there exists c ∈ [a, b] with f (c) = f (x) dx. b−a a • After a straightforward average value computation with a function such as y = x2 , show the students that it is possible to find a value of b such that f (x) = x2 + b has an average value of 8 on [0, 2]. Show geometrically that for any continuous function f and any real number a, it is possible to find a value of b such that g (x) = f (x) + b has an average value of a on [0, 2].

ANSWER b = 20 3 . Assume the average value of f on [0, 2] is A. Then the average value of g (x) on [0, 2] is A + b as seen below. y

y

6

6

5

5

4

4

3

3

f

2

2 b 1

1 0

1

2

3

4

g

5

0

6 x

1

2

3

4

5

6 x

Workshop/Discussion • Compute the average values of sin x, cos x, and cos2 x on [0, π]. Compare the average values of sin x, 2 sin x, and sin 2x on [0, 2π]. Foreshadow standard deviation.

ANSWER On [0, π]: π2 , 0, 12 . On [0, 2π]: 0, 0, 12 .

5 • Suppose f is a continuous function and 0 f (x) dx = 15. Ask the students what we can determine about f (x) from the Mean Value Theorem for Integrals.

ANSWER We know that there is a c in (0, 5) with f (c) = 3, and that’s all.

413

CHAPTER 6 APPLICATIONS OF INTEGRATION

• Suppose f is a continuous function and that the average value of f on [−3, 3] is 0. Ask the students why 3 we can then conclude that −3 f (x) dx = 0. How do we know that there is some c in [−3, 3] such that f (c) = 0? Does this mean that f (0) = 0?

3 ANSWER −3 f (x) dx = 0 by definition of the average value. The c exists by the Intermediate Value

Theorem. f (0) may or may not be 0.

• Find values of a and b so that f (x) = x2 + ax + b has an average value of 2 on [0, 1] and an average value of 1 on [0, 2].

ANSWER a = −4, b = 11 3

Group Work 1: Average Value: Theory and Practice Problem 1(d) is impossible. Let the students discover this for themselves, and then have them explain why the problem cannot be done. When all of the students have reached this stage, conduct a class discussion in which each group explains its reasoning. If a group finishes the second part early, ask them to compute the time interval during which the temperature could possibly be 80◦ , given that the average temperature is 84◦ .

ANSWERS (will vary) 1 x 1. (a) f (x) = 50

(b) f (x) = 23 x

1 98 (c) f (x) = 50 x + 100

(d) This is impossible, by the Mean Value Theorem for Integrals, and the condition that f (x) ≤ 1.

2. (a) The temperature graph may look like this: y

(c) If the temperature is 80◦ at 6:00 P. M . , the average temperature cannot exceed 82 23 ◦ .

80

If the temperature is 80◦ at 4:00 P. M . , the average

60

temperature can range between 66 23 ◦ and 85 13 ◦ .

40

The temperature cannot be 80◦ any time after 5:00 P. M . if the average temperature is to be 84◦ .

20 0

(b) The average temperature is between 60◦ and 96◦ .

2

4

6

8 10 12 x

Group Work 2: Choosing a Bank When this exercise is completed, make sure the students understand that the average value of sin

2πt 365

for

0 ≤ t ≤ 365 is 0. The students may be interested to know that both of the systems of granting interest are used in commercial banks. The first method is often used on interest-bearing checking accounts, and the separate “interest fund” is forfeited in its entirety if the checking balance falls below a certain amount.

414

SECTION 6.5 AVERAGE VALUE OF A FUNCTION

ANSWERS

1 15.21 1 1. Consolidated: I = P rt = (100) 15.21 I (t) dt 24 ≈ 26 cents 0 1 Bailey: I = (100) (0.06) 24 = 25 cents 1 30.42 1 I (t) dt 12 2. Consolidated: I = P rt = (100) 30.42 ≈ 52 cents 0 1 Bailey: I = (100) (0.06) 12 = 50 cents 1 182.5 1 I (t) dt 3. Consolidated: I = P rt = (100) 182.5 2 ≈ $3.31 0 1 Bailey: I = (100) (0.06) 2 = $3.00 4. The average rate is 6% at Consolidated. During the first six months, Consolidated will do better than Bailey. During the second six months, Bailey will catch up. Consolidated is a better deal, unless the money is to be saved for a whole number of years, in which case they are equal.

Homework Problems CORE EXERCISES 2, 5, 8, 14, 15 SAMPLE ASSIGNMENT 1, 2, 5, 6, 7, 8, 14, 15, 16 EXERCISE 1 2 5 6 7 8 14 15 16

D

A × × × × × × × ×

415

N × × × × × ×

G

× × ×

GROUP WORK 1, SECTION 6.5 Average Value: Theory and Practice 1. For this part of the exercise, we consider continuous functions y = f (x) defined on [0, 1] such that f (x) is always increasing, positive, and f (x) ≤ 1. Functions such as these can be used to model, for example, the fraction of students who scored at or below a certain percentage score on a test.

1 (a) Construct such a function f with the average value of f on [0, 1] equal to 100 .

(b) Construct another such function with the average value on [0, 1] equal to 13 .

99 (c) Construct another such function with the average value on [0, 1] equal to 100 .

(d) Construct one more such function with the average value on [0, 1] equal to 101 100 .

2. The temperature on a July day starts at 60◦ at 8 A . M . , and rises to 96◦ at 8 P. M . (a) Why can’t you say with certainty that the average temperature between 8 A . M . and 8 P. M . was 78◦ ?

(b) What can you say about the average temperature during this 12-hour period?

(c) Suppose you also know that the average temperature during this period was 84◦ . Is it possible that the temperature was 80◦ at 6 P. M . ? At 4 P. M . ?

416

GROUP WORK 2, SECTION 6.5 Choosing a Bank Suppose that the variable interest rate I at Consolidated Savings and Loan, Inc. can be modeled by 2π I (t) = 0.06 + 0.01 sin 365 t for 0 ≤ t ≤ 365, where t is measured in days. This interest is not compounded, but put in a separate interest fund.

Another bank, the Bailey Building and Loan Company, offers a plain old simple rate of 6%.

1. How much interest have you earned (on $100, say) at Consolidated when t = 365 24 ≈ 15.21 days? How much have you earned at Bailey after 15.21 days?

2. How much interest have you earned at Consolidated when t = 365 12 ≈ 30.42 days? How much have you earned at Bailey after this time?

3. Answer Question 2 if t = 365 2 ≈ 182.5 days. Can you explain the answer?

4. What is the average interest rate at Consolidated? Which bank offers the better deal, and why?

417

Applied Project: Where to Sit at the Movies It is hard to imagine an undergraduate student who has not had an argument with his or her friends concerning the best seat in the movie theater. “Up front” connoisseurs, “Middle center” partisans and “Balcony” fetishists have nearly come to blows over this decision. In this exercise, the students use the mathematics they’ve learned to settle the question for a model movie theater.

Problems 3 and 4 can be done with a scientific calculator if a CAS is not available. (The derivative will be very messy, but still within the students’ abilities.)

If a more in-depth experience is desired, students could be assigned to perform measurements at a local theater, and thus determine the actual “best seat in the house”.

418

6.6

Applications to Physics and Engineering

NOTE It is recommended that at least one of Sections 6.5 and 6.6 is covered.

Suggested Time and Emphasis 1 2 –1 class

Optional material

Points to Stress 1. Three applications (work, hydrostatic pressure, and centers of mass) are discussed in the text. It is recommended that the instructor stress one, and perhaps touch on the other two.

2. Centroids and the symmetry principle, if moments and centers of mass are covered.

Text Discussion • All three applications in the text start by a summation of discrete or sampled data, and eventually replace the summation by an integral. We have seen this technique earlier in the course. Name two previous occasions where we’ve used the technique of changing the sum of sample values into an integral of a continuous function.

ANSWER Answers include finding the area under a curve, finding average values of a function, and finding volumes.

Materials for Lecture, Workshop/Discussion Coverage will be determined by which applications are to be emphasized. We recommend an in-depth treatment of at most one of the three applications, perhaps touching on the other two. Throughout, emphasize problem-solving skills such as drawing a picture and setting up the problem carefully.

• Work: Foreshadow Exercise 21 by looking at it from a slightly different perspective: Newton’s Law of k Gravitation says that the force F needed to counteract Earth’s gravity is F (x) = 2 , where x is the x distance from the center of the Earth. Using this information, compute how much work is required to lift a 500-pound payload 1000 miles, 10,000 miles, 100,000 miles, and then “infinitely” far away from the Earth.

[Use the fact that the object weighs 500 pounds on the Earth’s surface (x = 4000) to obtain k = 8 × 109 , final height F (x) dx, converting mile-lbs to foot-lbs.] and then compute W = 4000 ANSWERS 1000 mi: 400,000 mi-lb = 2,112,000,000 ft-lb; 10,000 mi: 1,428,571 mi-lb = 7,542,854,880 ft-lb; 100,000 mi: 1,923,077 mi-lb = 10,153,846,560 ft-lb; ∞ 8,000,000,000 dx = 2,000,000 mi-lb = 10,560,000,000 ft-lb “infinitely” far away: x2 4000 419

CHAPTER 6 APPLICATIONS OF INTEGRATION

• Work: Find the work required to pump all of the water out of the top of the completely filled tank pictured below, given that the density of water is 62.6 lbs per cubic foot.

10»

3»

ANSWER

13 3

3 62.6π (9) dx + 0 62.6π 9 − (3 − x)2 dx ≈ 21,240 ft-lb

• Hydrostatic Pressure and Force: Work through Exercise 25. This problem requires some geometry to calculate the forces on the side.

• Center of Mass: Model the state of Minnesota as shown below, and compute the center of mass of the state. (Many other states can be modeled similarly.) This problem requires many computations; it may be advisable to break the problem into parts, and have different students work on different parts simultaneously.

ANSWER x ≈ 2.160, y ≈ 3.861

• Center of Mass: Find the center of mass (x, y) of an equilateral triangular plate of constant density and

√ side length r with vertices at (0, 0), (r, 0), and 12 r, 23 r .

ANSWER x = 12 r, y =

√ 3 6 r

• Center of Mass: Bring in an oddly-shaped flat piece of wood and experimentally find its center of mass.

420