a. Explanation Given: Each week crew digs 0.95 of the distance of its previous week and the first week the crew constructed 100m of tunnel. Result used: The sum of the geometric series is ∑ k = 0 n a r k = a ⋅ 1 − r n 1 − r if | r | < 1 . Calculation: Let T n be the amount of additional tunnel dug during week n . From the given, each week crew digs 0.95 of the distance of its previous week. That, is, T n = ( 0.95 ) T n − 1 . In the first week, T 0 = 100 . Compute the total distance dug in N weeks. T n = ( 0.95 ) T n − 1 T n = ( 0.95 ) n T 0 T n = 100 ( 0.95 ) n [ ∵ T 0 = 100 ] Thus, the total distance dug in N weeks is S N = 100 ∑ k = 1 N − 1 ( 0.95 ) k . S N = 100 ∑ k = 1 N − 1 ( 0.95 ) k = 100 ( 1 − ( 0 b. To determine To find: The longest tunnel the crew can build at this rate.

Student Solutions Manual, Single Variable for Calculus: Early Transcendentals

2nd Edition

ISBN: 9780321954329

Author: William L. Briggs, Lyle Cochran, Bernard Gillett

Publisher: PEARSON

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Chapter 8, Problem 65RE

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To find: The distance of crew dig in 10 weeks, 20 weeks and N weeks.

To determine

To find: The longest tunnel the crew can build at this rate.

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Chapter 8, Problem 64RE

Chapter 8, Problem 66RE

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Student Solutions Manual, Single Variable for Calculus: Early Transcendentals

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The distance of crew dig in 10 weeks, 20 weeks and N weeks.

Solution Summary: The author calculates the distance of crew dig in 10 weeks, 20 weeks and N weeks.