Math Problem Statement

Outside temperature over a day can be modeled as a sinusoidal function. Suppose you know the temperature is 75 degrees at midnight and the high and low temperatures during the day are 95 and 55 degrees, respectively. Assuming t is the number of hours since midnight, find an equation for the temperature, D, in terms of t.

Solution

This problem involves modeling temperature over a day as a sinusoidal function. Let's break it down step by step to construct the equation for temperature DD as a function of time tt (in hours since midnight):


Step 1: Identify Key Information

  1. Midline temperature: The temperature at midnight is 7575^\circ, and the temperature oscillates around this value. This is the average of the high and low temperatures: Midline=High+Low2=95+552=75.\text{Midline} = \frac{\text{High} + \text{Low}}{2} = \frac{95 + 55}{2} = 75. So, 7575 is the vertical shift of the sinusoidal function.

  2. Amplitude: The amplitude is half the difference between the high and low temperatures: Amplitude=HighLow2=95552=20.\text{Amplitude} = \frac{\text{High} - \text{Low}}{2} = \frac{95 - 55}{2} = 20. This represents how far the temperature oscillates above and below the midline.

  3. Period: A full cycle of temperature variation occurs over 24 hours. The general period formula for a sinusoidal function is: Period=2πB,\text{Period} = \frac{2\pi}{B}, where BB is the frequency. Setting the period to 2424: 24=2πB    B=2π24=π12.24 = \frac{2\pi}{B} \implies B = \frac{2\pi}{24} = \frac{\pi}{12}.

  4. Phase shift: At t=0t = 0 (midnight), the temperature is 7575^\circ, which is the midline value. This indicates a cosine function, as cos(0)=1\cos(0) = 1. No horizontal shift is needed.


Step 2: Write the Equation

The general form of a sinusoidal function is: D(t)=Acos(Bt+C)+D.D(t) = A \cos(Bt + C) + D.

  • A=20A = 20 (amplitude),
  • B=π12B = \frac{\pi}{12} (frequency),
  • C=0C = 0 (no phase shift),
  • D=75D = 75 (midline).

Thus, the equation becomes: D(t)=20cos(π12t)+75.D(t) = 20 \cos\left(\frac{\pi}{12} t\right) + 75.


Final Answer

The temperature DD as a function of time tt is: D(t)=20cos(π12t)+75.\boxed{D(t) = 20 \cos\left(\frac{\pi}{12} t\right) + 75.}


Would you like a detailed explanation of the graph or any specific part of this derivation?

Here are 5 follow-up questions to consider:

  1. How would the equation change if the temperature were modeled using a sine function instead of cosine?
  2. Can you determine the temperature at t=6t = 6 hours (6:00 AM) using this equation?
  3. How would the equation adjust if the high and low temperatures changed?
  4. What does the frequency π12\frac{\pi}{12} represent in practical terms?
  5. Can you verify that this function correctly models the temperature at both the high and low points?

Tip: The cosine function is often used when a cycle starts at a maximum or midline value, while sine is used for cycles starting at a minimum.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Trigonometry
Periodic Functions
Sinusoidal Modeling

Formulas

Midline = (High + Low) / 2
Amplitude = (High - Low) / 2
Period = 2π / B
Sinusoidal function: D(t) = A cos(Bt + C) + D

Theorems

Properties of the Cosine Function
Periodicity of Trigonometric Functions

Suitable Grade Level

Grades 9-12