Type:
Solution Manual
Resource:
Precalculus A Prelude to Calculus
Edition:
2nd Edition
Author(s):
Sheldon Axler
Instructor’s Solutions Manual, Appendix A
Exercise 1
Solutions to Exercises, Appendix A 1 Find the radius of a circle that has circumference 12 inches. solution Let r denote the radius of this circle in inches. Thus 2π r = 12, 6 which implies that r = π inches.
Instructor’s Solutions Manual, Appendix A
Exercise 2
2 Find the radius of a circle that has circumference 20 feet. solution Let r denote the radius of this circle in feet. Thus 2π r = 20, which implies that r = 10 π feet.
Instructor’s Solutions Manual, Appendix A
Exercise 3
3 Find the radius of a circle that has circumference 8 more than its diameter. solution Let r denote the radius of this circle. Thus the circle has circumference 2π r and has diameter 2r . Because the circumference is 8 more than diameter, we have 2π r = 2r + 8. Thus (2π − 2)r = 8, which implies that r = π 4−1 .
Instructor’s Solutions Manual, Appendix A
Exercise 4
4 Find the radius of a circle that has circumference 12 more than its diameter. solution Let r denote the radius of this circle. Thus the circle has circumference 2π r and has diameter 2r . Because the circumference is 12 more than diameter, we have 2π r = 2r + 12. Thus (2π − 2)r = 12, which implies that r = π 6−1 .
Instructor’s Solutions Manual, Appendix A
Exercise 5
5 Suppose you want to design a 400-meter track consisting of two halfcircles connected by parallel line segments. Suppose also that you want the total length of the curved part of the track to equal half the total length of the straight part of the track. What dimensions should the track have? solution Let t equal the total length of the straight part of the track in meters. We want the curved part of the track to have total length 2t . Thus we want t t + 2 = 400. Solving this equation for t, we get t = 800 3 meters. Thus each of the two
400 straight pieces must be 400 3 meters long. Hence we take c = 3 ≈ 133.33 meters in the figure in the below.
d
c
We also want the total length of the two half-circles to be 400 3 meters. 400 Thus we want π d = 400 3 . Hence we take d = 3π ≈ 42.44 meters in the figure above.
Instructor’s Solutions Manual, Appendix A
Exercise 6
6 Suppose you want to design a 200-meter indoor track consisting of two half-circles connected by parallel line segments. Suppose also that you want the total length of the curved part of the track to equal three-fourths the total length of the straight part of the track. What dimensions should the track have? solution Let t equal the total length of the straight part of the track in meters. We want the curved part of the track to have total length 3t 4 . Thus we want 3t t + 4 = 200. Solving this equation for t, we get t = 800 7 meters. Thus each of the two 400 straight pieces must be 7 meters long. Hence we take c = 400 7 ≈ 57.14 meters in the figure in the below. c d
3 We also want the total length of the two half-circles to be 800 7 × 4 meters, 600 600 which equals 7 meters. Thus we want π d = 7 . Hence we take d = 600 7π ≈ 27.28 meters in the figure above.