EXAMPLE 7 In-Class Example 7 See Example 7. Change the numbers in a)–c) to the following: a) 30 sec b) 2 min c) 3,087,000 samples of sound Answer: a) 1,323,000 b) 5,292,000 c) 70 sec A compact disk is read at 44.1 kHz (kilohertz). This means that a CD player scans 44,100 samples of sound per second on a CD to produce the sound that we hear. The function S(t) 44,100t tells us how many samples of sound, S(t), are read after t seconds. (www.mediatechnics.com) a) How many samples of sound are read in 20 sec? b) How many samples of sound are read in 1.5 min? c) How long would it take the CD player to scan 1,764,000 samples of sound? d) What is the smallest value t could equal? Solution a) To determine how much sound is read in 20 sec, let t 20 and fi nd S(20). S(t) 44,100t S(20) 44,100(20) Substitute 20 for t. S(20) 882,000 Multiply. The CD player has read 882,000 samples of sound. b) To determine how much sound is read in 1.5 min, do we let t 1.5 and fi nd S(1.5)? No. Recall that t is in seconds. Change 1.5 min to seconds before substituting for t. We must use the correct units in the function. 1.5 min 90 sec Let t 90 and fi nd S(90). S(t) 44,100t S(90) 44,100(90) S(90) 3,969,000 It has read 3,969,000 samples of sound. c) Since we are asked to determine how long it would take a CD player to scan 1,764,000 samples of sound, we will be solving for t. What do we substitute for S(t)? We substitute 1,764,000 for S(t) and fi nd t. That is, fi nd t when S (t) 1,764,000. S(t) 44,100t 1,764,000 44,100t Substitute 1,764,000 for S(t). 40 t Divide by 44,100. It will take 40 sec for the CD player to scan 1,764,000 samples of sound. d) Since t represents the number of seconds a CD has been playing, the smallest value that makes sense for t is 0. S(t), read as “S of t,” means “Sound as a function of time.” www.mhhe.com/messersmith SECTION 10.5 Introduction to Functions 647
messersmith_power_introductory_algebra_1e_ch4_7_10
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