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# Designing functions: Find a trigonometric function f that ISBN: 9780321570567 2

## Solution for problem 27RE Chapter 1

Calculus: Early Transcendentals | 1st Edition

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Problem 27RE

Designing functions: Find a trigonometric function ?f? that satisfies each set of properties. Answers are not unique. a. It has a period of 6 hr with a minimum value of ?2 at ?t? = 0 hr and a maximum value of 2 at ?t? = 3 hr. b. It has a period of 24 hr with a maximum value of 20 at ?t? = 6 hr and a minimum value of 10 at ?t? = 18 hr.

Step-by-Step Solution:

Step-by-step solution Step 1 a) We need to find a trigonometric function f that has a period of 6 hr with a minimum value of -2 at t = 0 hr and a maximum value of 2 at t = 3 hr. Step 2 To solve the problem, we first assume that the function f is a sine function. Next, we use the general sine function f(t) = Asin B(tC) +D  where A : amplitude is |A|, B : period is |B| C : phase shift is C, D : vertical shift is D. Step 3 Solving for A: Since it is given that the function has a minimum value of -2 at t = 0 hr and a maximum value of 2 at t = 3 hr, we can now get a value for A. Amplitude = maxmin 2 2(2) 4 Amplitude = 2 = 2 Amplitude = A = 2 Step 4 Solving for D: The vertical shift D is defined as: D = min_value+amplitude Hence, we got: D =2+2 D = 0 Step 5 Solving for B: Since it is given that the function has a period of 6 hr, we can now get a value for B. Period = |B| 2 6 = |B| B = 2 6 B = 3 Step 6 Solving for C: Since it is given that the function has a minimum value of -2 at t = 0 hr, we use the values we got for A, B, and D to solve for C . At t = 0 : 2 = 2sin (t3C) +0 2 = 2sin (03) C 2 = 2sin ( 3 ) C = 3 2 Step 7 Substituting all the values we got into , we get: 3 f(t) = 2sin (3 ) 2 or f(t) = 2sin ( ) 3 2 Step 8 Next, we check if f(t) = 2sin (t ) satisfies the given information by graphing. 3 2 The function f(t) = 2sin (t3 ) i2deed satisfies the given information.

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##### ISBN: 9780321570567

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