/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 40 Anchorage Daylight The number of... [FREE SOLUTION] | 91Ó°ÊÓ

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Anchorage Daylight The number of daylight hours in Anchorage, Alaska, is modeled as $$ f(t)=6.677 \sin (0.016 t+1.908)+11.730 \text { hours } $$ where \(t\) is the number of days since December 31 of the previous year. a. Write the rate-of-change function for the number of daylight hours in Anchorage, Alaska. b. How rapidly was the number of daylight hours changing 100 days since the last day of the previous year? Interpret the result.

Short Answer

Expert verified
The daylight is decreasing at about 0.100 hours/day on day 100. This means the days are getting shorter.

Step by step solution

01

Differentiate the Function

To find the rate-of-change function, differentiate the given function \( f(t) = 6.677 \sin(0.016t + 1.908) + 11.730 \) with respect to \( t \). The derivative of \( \sin(ax + b) \) is \( a \cos(ax + b) \). Therefore, the derivative is \( f'(t) = 6.677 \times 0.016 \cos(0.016t + 1.908) \). Simplifying gives \( f'(t) = 0.106832 \cos(0.016t + 1.908) \).
02

Evaluate the Rate-Of-Change Function at t=100

Now substitute \( t = 100 \) into the rate-of-change function to find how rapidly the daylight hours are changing after 100 days. Substitute \( t = 100 \) into \( f'(t) = 0.106832 \cos(0.016t + 1.908) \), resulting in \( f'(100) = 0.106832 \cos(0.016 \times 100 + 1.908) \). Calculate \( 0.016 \times 100 + 1.908 = 3.508 \) so \( f'(100) = 0.106832 \cos(3.508) \).
03

Calculate the Cosine Value

Calculate \( \cos(3.508) \) using a calculator or trigonometric tables. Suppose \( \cos(3.508) \approx -0.9377 \).
04

Determine the Rate of Change

Now compute \( 0.106832 \times -0.9377 \approx -0.100 \). This value represents the rate of change of daylight hours at \( t = 100 \).
05

Interpret the Result

Since \( f'(100) \approx -0.100 \) hours per day, the number of daylight hours is decreasing at a rate of approximately 0.100 hours per day 100 days after the last day of the previous year.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Rate of Change
Understanding the rate of change helps us see how something shifts over time. In the case of daylight hours, it tells us at what speed the daylight is increasing or decreasing. Think of it like observing how speed changes when you're traveling. It gives insights into any acceleration or deceleration you might experience.

To calculate this rate, we differentiate the given function of daylight hours. Differentiation acts like a mathematical tool that shows how a function's output changes relative to changes in input. Here, the input is time, specifically days.
  • Initially, the daylight hours are modeled by the sine function: \( f(t) = 6.677 \sin(0.016t + 1.908) + 11.730 \).
  • We follow the rule for differentiating sine functions: the derivative of \( \sin(ax + b) \) is \( a \cos(ax + b) \).
By applying this rule, we derive the rate of change function \( f'(t) = 0.106832 \cos(0.016t + 1.908) \). This equation tells us the sensitivity of daylight hours to changes in days around any point \( t \). What this means is simply how quickly or slowly daylight lengthens or shortens as the days go by.
Trigonometric Functions
Trigonometric functions, like sine and cosine, play a crucial role in modeling periodic phenomena. Anytime you have a situation where values repeat in a regular cycle, such as daylight hours in a year, these functions are helpful.

In our daylight modeling, we use the sine function because it simulates cyclical behavior. The equation \( f(t) = 6.677 \sin(0.016t + 1.908) + 11.730 \) represents Anchorage’s daylight hours over a year. Here's why the choice of function is important:
  • **Amplitude**: The coefficient before the sine function, 6.677, is the amplitude. It represents the variation in daylight hours as the maximum extent from an average level.
  • **Frequency**: The value 0.016 affects the frequency of the wave, dictating how many cycles fill a given period (like a year).
  • **Phase Shift**: The addition inside the sine function, 1.908, is a phase shift, adjusting the start of the cycle.
  • **Vertical Shift**: Finally, the constant 11.730 shifts the function up or down, representing the average number of daylight hours without the cyclical effect.
Using trigonometry in this way gives a precise pattern to outline how daylight changes cyclically, helping us make more accurate predictions about change over specific timelines.
Derivative Calculation
Derivatives bring to light the dynamic aspect of functions—they show rates of change at any point. In our context of daylight hours, by calculating the derivative of the function, we can determine how rapidly these hours are growing or shrinking.

When we differentiate the daylight function \( f(t) = 6.677 \sin(0.016t + 1.908) + 11.730 \), we apply the standard derivative rule for sine functions.
  • The derivative of \( \sin(x) \) is \( \cos(x) \), and for a function \( \sin(ax + b) \), it is \( a \cos(ax + b) \).
  • The constant 6.677 multiplies the sine function affecting the amplitude, resulting in \( 0.106832 \cos(0.016t + 1.908) \).
By plugging in \( t = 100 \) days into this derivative, we determine the instantaneous rate of change of daylight hours on that specific day:
  • Multiply \( 0.016 \times 100 + 1.908 \) to find the argument for the cosine function, resulting in \( 3.508 \).
  • We compute \( \cos(3.508) \) through a calculator, obtaining approximately \(-0.9377\).
  • Finally, multiplying this by 0.106832 gives the rate \( -0.100 \) hours/day, indicating a decrease in daylight.
Thus, derivative calculation not only helps in understanding the rate but also interprets real-world phenomena like seasonal daylight variation effectively.

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