Solutions to Exercises, Section 5 - UW-Madison …park/Fall2014/precalculus/5.2sol.pdfInstructor’s...

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Instructor’s Solutions Manual, Section 5.2 Exercise 1 Solutions to Exercises, Section 5.2 In Exercises 1–8, convert each angle to radians. 1. 15 solution Start with the equation 360 = 2π radians. Divide both sides by 360 to obtain 1 = π 180 radians. Now multiply both sides by 15, obtaining 15 = 15π 180 radians = π 12 radians.

Transcript of Solutions to Exercises, Section 5 - UW-Madison …park/Fall2014/precalculus/5.2sol.pdfInstructor’s...

Page 1: Solutions to Exercises, Section 5 - UW-Madison …park/Fall2014/precalculus/5.2sol.pdfInstructor’s Solutions Manual, Section 5.2 Exercise 1 Solutions to Exercises, Section 5.2 In

Instructor’s Solutions Manual, Section 5.2 Exercise 1

Solutions to Exercises, Section 5.2

In Exercises 1–8, convert each angle to radians.

1. 15◦

solution Start with the equation

360◦ = 2π radians.

Divide both sides by 360 to obtain

1◦ = π180

radians.

Now multiply both sides by 15, obtaining

15◦ = 15π180

radians = π12

radians.

Page 2: Solutions to Exercises, Section 5 - UW-Madison …park/Fall2014/precalculus/5.2sol.pdfInstructor’s Solutions Manual, Section 5.2 Exercise 1 Solutions to Exercises, Section 5.2 In

Instructor’s Solutions Manual, Section 5.2 Exercise 2

2. 40◦

solution Start with the equation

360◦ = 2π radians.

Divide both sides by 360 to obtain

1◦ = π180

radians.

Now multiply both sides by 40, obtaining

40◦ = 40π180

radians = 2π9

radians.

Page 3: Solutions to Exercises, Section 5 - UW-Madison …park/Fall2014/precalculus/5.2sol.pdfInstructor’s Solutions Manual, Section 5.2 Exercise 1 Solutions to Exercises, Section 5.2 In

Instructor’s Solutions Manual, Section 5.2 Exercise 3

3. −45◦

solution Start with the equation

360◦ = 2π radians.

Divide both sides by 360 to obtain

1◦ = π180

radians.

Now multiply both sides by −45, obtaining

−45◦ = −45π180

radians = −π4

radians.

Page 4: Solutions to Exercises, Section 5 - UW-Madison …park/Fall2014/precalculus/5.2sol.pdfInstructor’s Solutions Manual, Section 5.2 Exercise 1 Solutions to Exercises, Section 5.2 In

Instructor’s Solutions Manual, Section 5.2 Exercise 4

4. −60◦

solution Start with the equation

360◦ = 2π radians.

Divide both sides by 360 to obtain

1◦ = π180

radians.

Now multiply both sides by −60, obtaining

−60◦ = −60π180

radians = −π3

radians.

Page 5: Solutions to Exercises, Section 5 - UW-Madison …park/Fall2014/precalculus/5.2sol.pdfInstructor’s Solutions Manual, Section 5.2 Exercise 1 Solutions to Exercises, Section 5.2 In

Instructor’s Solutions Manual, Section 5.2 Exercise 5

5. 270◦

solution Start with the equation

360◦ = 2π radians.

Divide both sides by 360 to obtain

1◦ = π180

radians.

Now multiply both sides by 270, obtaining

270◦ = 270π180

radians = 3π2

radians.

Page 6: Solutions to Exercises, Section 5 - UW-Madison …park/Fall2014/precalculus/5.2sol.pdfInstructor’s Solutions Manual, Section 5.2 Exercise 1 Solutions to Exercises, Section 5.2 In

Instructor’s Solutions Manual, Section 5.2 Exercise 6

6. 240◦

solution Start with the equation

360◦ = 2π radians.

Divide both sides by 360 to obtain

1◦ = π180

radians.

Now multiply both sides by 240, obtaining

240◦ = 240π180

radians = 4π3

radians.

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Instructor’s Solutions Manual, Section 5.2 Exercise 7

7. 1080◦

solution Start with the equation

360◦ = 2π radians.

Divide both sides by 360 to obtain

1◦ = π180

radians.

Now multiply both sides by 1080, obtaining

1080◦ = 1080π180

radians = 6π radians.

Page 8: Solutions to Exercises, Section 5 - UW-Madison …park/Fall2014/precalculus/5.2sol.pdfInstructor’s Solutions Manual, Section 5.2 Exercise 1 Solutions to Exercises, Section 5.2 In

Instructor’s Solutions Manual, Section 5.2 Exercise 8

8. 1440◦

solution Start with the equation

360◦ = 2π radians.

Divide both sides by 360 to obtain

1◦ = π180

radians.

Now multiply both sides by 1440, obtaining

1440◦ = 1440π180

radians = 8π radians.

Page 9: Solutions to Exercises, Section 5 - UW-Madison …park/Fall2014/precalculus/5.2sol.pdfInstructor’s Solutions Manual, Section 5.2 Exercise 1 Solutions to Exercises, Section 5.2 In

Instructor’s Solutions Manual, Section 5.2 Exercise 9

In Exercises 9–16, convert each angle to degrees.

9. 4π radians

solution Start with the equation

2π radians = 360◦.

Multiply both sides by 2, obtaining

4π radians = 2 · 360◦ = 720◦.

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Instructor’s Solutions Manual, Section 5.2 Exercise 10

10. 6π radians

solution Start with the equation

2π radians = 360◦.

Multiply both sides by 3, obtaining

6π radians = 3 · 360◦ = 1080◦.

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Instructor’s Solutions Manual, Section 5.2 Exercise 11

11. π9 radians

solution Start with the equation

2π radians = 360◦.

Divide both sides by 2 to obtain

π radians = 180◦.

Now divide both sides by 9, obtaining

π9

radians = 1809

◦= 20◦.

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Instructor’s Solutions Manual, Section 5.2 Exercise 12

12. π10 radians

solution Start with the equation

2π radians = 360◦.

Divide both sides by 2 to obtain

π radians = 180◦.

Now divide both sides by 10, obtaining

π10

radians = 18010

◦= 18◦.

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Instructor’s Solutions Manual, Section 5.2 Exercise 13

13. 3 radians

solution Start with the equation

2π radians = 360◦.

Divide both sides by 2π to obtain

1 radian = 180π

◦.

Now multiply both sides by 3, obtaining

3 radians = 3 · 180π

◦= 540

π

◦.

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Instructor’s Solutions Manual, Section 5.2 Exercise 14

14. 5 radians

solution Start with the equation

2π radians = 360◦.

Divide both sides by 2π to obtain

1 radian = 180π

◦.

Now multiply both sides by 5, obtaining

5 radians = 5 · 180π

◦= 900

π

◦.

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Instructor’s Solutions Manual, Section 5.2 Exercise 15

15. −2π3 radians

solution Start with the equation

2π radians = 360◦.

Divide both sides by 2 to obtain

π radians = 180◦.

Now multiply both sides by −23 , obtaining

−2π3

radians = −23· 180◦ = −120◦.

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Instructor’s Solutions Manual, Section 5.2 Exercise 16

16. −3π4 radians

solution Start with the equation

2π radians = 360◦.

Divide both sides by 2 to obtain

π radians = 180◦.

Now multiply both sides by −34 , obtaining

−3π4

radians = −34· 180◦ = −135◦.

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Instructor’s Solutions Manual, Section 5.2 Exercise 17

17. Suppose an ant walks counterclockwise on the unit circle from thepoint (0,1) to the endpoint of the radius that forms an angle of 5π

4radians with the positive horizontal axis. How far has the ant walked?

solution The radius whose endpoint equals (0,1) makes an angle ofπ2 radians with the positive horizontal axis. This radius corresponds tothe smaller angle shown below.

Because 5π4 = π + π

4 , the radius that forms an angle of 5π4 radians with

the positive horizontal axis lies π4 radians beyond the negative

horizontal axis (half-way between the negative horizontal axis and thenegative vertical axis). Thus the ant ends its walk at the endpoint of theradius corresponding to the larger angle shown below:

1

The ant walks along the thickened circular arc shown above. Thiscircular arc corresponds to an angle of 5π

4 − π2 radians, which equals 3π

4radians. Thus the distance walked by the ant is 3π

4 .

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Instructor’s Solutions Manual, Section 5.2 Exercise 18

18. Suppose an ant walks counterclockwise on the unit circle from thepoint (−1,0) to the endpoint of the radius that forms an angle of 6radians with the positive horizontal axis. How far has the ant walked?

solution The radius whose endpoint equals (−1,0) makes an angleof π radians with the positive horizontal axis. This radius lies on thenegative horizontal axis.

Because 2π ≈ 6.28, the radius that forms an angle of 6 radians with thepositive horizontal axis lies slightly below the positive horizontal axis,as shown below:

1

The ant walks along the thickened circular arc shown above. Thiscircular arc corresponds to an angle of 6−π radians. Thus the distancewalked by the ant is 6−π .

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Instructor’s Solutions Manual, Section 5.2 Exercise 19

19. Find the lengths of both circular arcs of the unit circle connectingthe point (1,0) and the endpoint of the radius that makes an angle of 3radians with the positive horizontal axis.

solution Because 3 is a bit less than π , the radius that makes anangle of 3 radians with the positive horizontal axis lies a bit above thenegative horizontal axis, as shown below. The thickened circular arccorresponds to an angle of 3 radians and thus has length 3. The entireunit circle has length 2π . Thus the length of the other circular arc is2π − 3, which is approximately 3.28.

1

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Instructor’s Solutions Manual, Section 5.2 Exercise 20

20. Find the lengths of both circular arcs of the unit circle connectingthe point (1,0) and the endpoint of the radius that makes an angle of 4radians with the positive horizontal axis.

solution The radius that makes an angle of 4 radians with thepositive horizontal axis is shown below. The thickened circular arccorresponds to an angle of 4 radians and thus has length 4. The entireunit circle has length 2π . Thus the length of the other circular arc is2π − 4, which is approximately 2.28.

1

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Instructor’s Solutions Manual, Section 5.2 Exercise 21

21. Find the lengths of both circular arcs of the unit circle connectingthe point

(√22 ,−

√2

2

)and the point whose radius makes an angle of 1

radian with the positive horizontal axis.

solution The radius of the unit circle whose endpoint equals(√22 ,−

√2

2

)makes an angle of −π4 radians with the positive horizontal

axis, as shown with the clockwise arrow below. The radius that makesan angle of 1 radian with the positive horizontal axis is shown with acounterclockwise arrow.

1

Thus the thickened circular arc above corresponds to an angle of 1+ π4

and thus has length 1+ π4 , which is approximately 1.79. The entire unit

circle has length 2π . Thus the length of the other circular arc below is2π − (1+ π

4 ), which equals 7π4 − 1, which is approximately 4.50.

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Instructor’s Solutions Manual, Section 5.2 Exercise 22

22. Find the lengths of both circular arcs of the unit circle connectingthe point

(−√22 ,

√2

2

)and the point whose radius makes an angle of 2

radians with the positive horizontal axis.

solution The radius of the unit circle whose endpoint equals(−√22 ,

√2

2

)makes an angle of 3π

4 radians with the positive horizontalaxis, as shown with the longer arrow below. The radius that makes anangle of 2 radian with the positive horizontal axis is shown with theshorter arrow below.

1

Thus the thickened circular arc above corresponds to an angle of 3π4 − 2

and thus has length 3π4 − 2, which is approximately 0.356. The entire

unit circle has length 2π . Thus the length of the other circular arcbelow is 2π − (3π

4 − 2), which equals 5π4 + 2, which is approximately

5.927.

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Instructor’s Solutions Manual, Section 5.2 Exercise 23

23. For a 16-inch pizza, find the area of a slice with angle 34 radians.

solution Pizzas are measured by their diameters; thus this pizza hasa radius of 8 inches. Thus the area of the slice is 1

2 · 34 · 82, which equals

24 square inches.

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Instructor’s Solutions Manual, Section 5.2 Exercise 24

24. For a 14-inch pizza, find the area of a slice with angle 45 radians.

solution Pizzas are measured by their diameters; thus this pizza hasa radius of 7 inches. Thus the area of the slice is 1

2 · 45 · 72, which equals

985 square inches.

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Instructor’s Solutions Manual, Section 5.2 Exercise 25

25. Suppose a slice of a 12-inch pizza has an area of 20 square inches.What is the angle of this slice?

solution This pizza has a radius of 6 inches. Let θ denote the angleof this slice, measured in radians. Then

20 = 12θ · 62.

Solving this equation for θ, we get θ = 109 radians.

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Instructor’s Solutions Manual, Section 5.2 Exercise 26

26. Suppose a slice of a 10-inch pizza has an area of 15 square inches.What is the angle of this slice?

solution This pizza has a radius of 5 inches. Let θ denote the angleof this slice, measured in radians. Then

15 = 12θ · 52.

Solving this equation for θ, we get θ = 65 radians.

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Instructor’s Solutions Manual, Section 5.2 Exercise 27

27. Suppose a slice of pizza with an angle of 56 radians has an area of 21

square inches. What is the diameter of this pizza?

solution Let r denote the radius of this pizza. Thus

21 = 12 · 5

6r2.

Solving this equation for r , we get r =√

2525 ≈ 7.1. Thus the diameter of

the pizza is approximately 14.2 inches.

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Instructor’s Solutions Manual, Section 5.2 Exercise 28

28. Suppose a slice of pizza with an angle of 1.1 radians has an area of25 square inches. What is the diameter of this pizza?

solution Let r denote the radius of this pizza. Thus

25 = 12 × 1.1r 2.

Solving this equation for r , we get r =√

501.1 ≈ 6.74. Thus the diameter

of the pizza is approximately 13.48 inches.

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Instructor’s Solutions Manual, Section 5.2 Exercise 29

For each of the angles in Exercises 29–34, find the endpoint of the radiusof the unit circle that makes the given angle with the positive horizontalaxis.

29. 5π6 radians

solution For this exercise it may be easier to convert to degrees.Thus we translate 5π

6 radians to 150◦.

The radius making a 150◦ angle with the positive horizontal axis isshown below. The angle from this radius to the negative horizontal axisequals 180◦ − 150◦, which equals 30◦ as shown in the figure below.Drop a perpendicular line segment from the endpoint of the radius tothe horizontal axis, forming a right triangle as shown below. Wealready know that one angle of this right triangle is 30◦; thus the otherangle must be 60◦, as labeled below.

The side of the right triangle opposite the 30◦ angle has length 12 ; the

side of the right triangle opposite the 60◦ angle has length√

32 . Looking

at the figure below, we see that the first coordinate of the endpoint ofthe radius is the negative of the length of the side opposite the 60◦

angle, and the second coordinate of the endpoint of the radius is thelength of the side opposite the 30◦ angle. Thus the endpoint of theradius is

(−√32 ,

12

).

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Instructor’s Solutions Manual, Section 5.2 Exercise 29

150�

30�

60�

1

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Instructor’s Solutions Manual, Section 5.2 Exercise 30

30. 7π6 radians

solution For this exercise it may be easier to convert to degrees.Thus we translate 7π

6 radians to 210◦.

The radius making a 210◦ angle with the positive horizontal axis isshown below. The angle from the negative horizontal axis to this radiusequals 210◦ − 180◦, which equals 30◦ as shown in the figure below.Draw a perpendicular line segment from the endpoint of the radius tothe horizontal axis, forming a right triangle as shown below. Wealready know that one angle of this right triangle is 30◦; thus the otherangle must be 60◦, as labeled below.

The side of the right triangle opposite the 30◦ angle has length 12 ; the

side of the right triangle opposite the 60◦ angle has length√

32 . Looking

at the figure below, we see that the first coordinate of the endpoint ofthe radius is the negative of the length of the side opposite the 60◦

angle, and the second coordinate of the endpoint of the radius is thenegative of the length of the side opposite the 30◦ angle. Thus theendpoint of the radius is

(−√32 ,−1

2

).

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Instructor’s Solutions Manual, Section 5.2 Exercise 30

210�

30�

60�

1

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Instructor’s Solutions Manual, Section 5.2 Exercise 31

31. −π4 radians

solution For this exercise it may be easier to convert to degrees.Thus we translate −π4 radians to −45◦.

The radius making a −45◦ angle with the positive horizontal axis isshown below. Draw a perpendicular line segment from the endpoint ofthe radius to the horizontal axis, forming a right triangle as shownbelow. We already know that one angle of this right triangle is 45◦; thusthe other angle must also be 45◦, as labeled below.

The hypotenuse of this right triangle is a radius of the unit circle andthus has length 1. The other two sides each have length

√2

2 . Looking atthe figure below, we see that the first coordinate of the endpoint of theradius is

√2

2 and the second coordinate of the endpoint of the radius is

−√

22 . Thus the endpoint of the radius is

(√22 ,−

√2

2

).

45�

45� 1

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Instructor’s Solutions Manual, Section 5.2 Exercise 32

32. −3π4 radians

solution For this exercise it may be easier to convert to degrees.Thus we translate −3π

4 radians to −135◦.

The radius making a −135◦ angle with the positive horizontal axis isshown below. Draw a perpendicular line segment from the endpoint ofthe radius to the horizontal axis, forming a right triangle as shownbelow. Because 180− 135 = 45, one angle of this triangle equals 45◦;thus the other angle must also be 45◦, as labeled below.

The hypotenuse of this right triangle is a radius of the unit circle andthus has length 1. The other two sides each have length

√2

2 . Looking atthe figure below, we see that the first coordinate of the endpoint of theradius is −

√2

2 and the second coordinate of the endpoint of the radius

is also −√

22 . Thus the endpoint of the radius is

(−√22 ,−

√2

2

).

45�

45� �135�

1

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Instructor’s Solutions Manual, Section 5.2 Exercise 33

33. 5π2 radians

solution Note that 5π2 = 2π + π

2 . Thus the radius making an angle of5π2 radians with the positive horizontal axis is obtained by starting at

the horizontal axis, making one complete counterclockwise rotation(which is 2π radians), and then continuing for another π

2 radians. Theresulting radius is shown below. Its endpoint is (0,1).

1

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Instructor’s Solutions Manual, Section 5.2 Exercise 34

34. 11π2 radians

solution Note that 11π2 = 4π + 3π

2 . Thus the radius making an angleof 11π

2 radians with the positive horizontal axis is obtained by startingat the horizontal axis, making two complete counterclockwise rotations(which is 4π radians), and then continuing for another 3π

2 radians. Theresulting radius is shown below. Its endpoint is (0,−1).

1

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Instructor’s Solutions Manual, Section 5.2 Problem 35

Solutions to Problems, Section 5.2

For each of the angles listed in Problems 35–42, sketch the unit circleand the radius that makes the indicated angle with the positivehorizontal axis. Be sure to include an arrow to show the direction inwhich the angle is measured from the positive horizontal axis.

35. 5π18 radians

solution

5Π�18

1

An angle of 5π18 radians, which equals 50◦.

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Instructor’s Solutions Manual, Section 5.2 Problem 36

36. 12 radian

solution

1�2

1

An angle of 12 radian, which approximately equals 28.6◦.

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Instructor’s Solutions Manual, Section 5.2 Problem 37

37. 2 radians

solution

2

1

An angle of 2 radians, which approximately equals 114.6◦.

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Instructor’s Solutions Manual, Section 5.2 Problem 38

38. 5 radians

solution

5

1

An angle of 5 radians, which approximately equals 286.5◦.

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Instructor’s Solutions Manual, Section 5.2 Problem 39

39. 11π5 radians

solution

11Π�5

1

An angle of 11π5 radians, which equals 396◦.

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Instructor’s Solutions Manual, Section 5.2 Problem 40

40. − π12 radians

solution

�Π�12 1

An angle of − π12 radians, which equals −15◦.

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Instructor’s Solutions Manual, Section 5.2 Problem 41

41. −1 radian

solution

�1

1

An angle of −1 radian, which approximately equals −57.3◦.

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Instructor’s Solutions Manual, Section 5.2 Problem 42

42. −8π9 radians

solution

�8Π�9

1

An angle of −8π9 radians, which equals −160◦.

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Instructor’s Solutions Manual, Section 5.2 Problem 43

43. Find the formula for the length of a circular arc corresponding to anangle of θ radians on a circle of radius r .

solution Suppose 0 < θ ≤ 2π and consider a circular arc on a unitcircle of radius r corresponding to an angle of θ radians, as shown inthe thickened part of the unit circle below:

Θ

r

The length of this circular arc equals the fraction of the entire circletaken up by this circular arc times the circumference of the entirecircle. In other words, the length of this circular arc equals θ

2π times2πr , which equals θr .

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Instructor’s Solutions Manual, Section 5.2 Problem 44

44. Most dictionaries define acute angles and obtuse angles in terms ofdegrees. Restate these definitions in terms of radians.

solution An angle between 0 radians and π2 radians is called an acute

angle.

An angle between π2 radians and π radians is called an obtuse angle.

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Instructor’s Solutions Manual, Section 5.2 Problem 45

45. Find a formula (in terms of θ) for the area of the region bounded by thethickened radii and the thickened circular arc shown below.

Θ

1

Here 0 < θ < 2π and the radius shownabove makes an angle of θ radians with thepositive horizontal axis.

solution The area of this region equals the fraction of the area insidethe circle taken up by this region times the area inside the entire circle.In other words, the area of this region equals θ

2π times π , which equalsθ2 .

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Instructor’s Solutions Manual, Section 5.2 Problem 46

46. Suppose the region bounded by the thickened radii and circular arcshown above is removed. Find a formula (in terms of θ) for theperimeter of the remaining region inside the unit circle.

solution The length of the thickened arc above is θ, and thus thelength of the remaining part of the circle equals 2π − θ. Each of thetwo radii has length 1. Thus the perimeter of the region in question is2+ 2π − θ.