Math Problem Statement

Outside temperature over a day can be modeled as a sinusoidal function. Suppose you know the high temperature for the day is 81 degrees and the low temperature of 59 degrees occurs at 5 AM. Assuming t is the number of hours since midnight, find an equation for the temperature, D(t), in terms of t.

Solution

The problem you provided states:

  • The temperature over a day can be modeled as a sinusoidal function.
  • The high temperature is 81°F.
  • The low temperature is 59°F, and this low occurs at 5 AM.
  • tt represents the number of hours since midnight.

We are tasked with finding an equation for the temperature, D(t)D(t), in terms of tt.

Step-by-Step Solution:

1. General Sinusoidal Function:

A sinusoidal function modeling temperature can be written as: D(t)=Asin(B(tC))+D0D(t) = A \sin(B(t - C)) + D_0 where:

  • AA is the amplitude,
  • BB affects the period (since temperature repeats every 24 hours, the period is 24),
  • CC is the horizontal shift (the time at which the minimum or maximum occurs),
  • D0D_0 is the vertical shift (midline of the sinusoid).

2. Finding Amplitude AA:

The amplitude AA is half the difference between the maximum and minimum temperatures: A=81592=11A = \frac{81 - 59}{2} = 11

3. Finding the Vertical Shift D0D_0:

The vertical shift D0D_0 is the average of the maximum and minimum temperatures: D0=81+592=70D_0 = \frac{81 + 59}{2} = 70

4. Determining the Period:

Since temperature repeats every 24 hours, the period TT is 24 hours. The formula relating BB to the period is: B=2πT=2π24=π12B = \frac{2\pi}{T} = \frac{2\pi}{24} = \frac{\pi}{12}

5. Horizontal Shift CC:

The low temperature occurs at 5 AM, which corresponds to a minimum point on the sinusoidal curve. A sine function normally has its minimum at 3π2\frac{3\pi}{2}, but since the low temperature occurs at t=5t = 5, the function is shifted horizontally by 5 hours. Therefore, C=5C = 5.

6. Complete Equation:

Putting everything together, the equation for D(t)D(t) becomes: D(t)=11sin(π12(t5))+70D(t) = 11 \sin\left(\frac{\pi}{12}(t - 5)\right) + 70

Final Equation:

D(t)=11sin(π12(t5))+70D(t) = 11 \sin\left(\frac{\pi}{12}(t - 5)\right) + 70

This is the sinusoidal function modeling the temperature in terms of tt, the number of hours since midnight.

Would you like more details on any of the steps, or have any further questions?


Here are 5 related questions to deepen your understanding:

  1. How would the equation change if the low temperature occurred at 3 AM instead of 5 AM?
  2. What would happen to the sinusoidal model if the high temperature was lower, say 75°F?
  3. How would you modify the equation if the period of the temperature cycle was 12 hours instead of 24?
  4. What is the temperature at 12 PM (noon) according to this model?
  5. How would the model change if the temperature variation was smaller, say between 70°F and 60°F?

Tip: When working with sinusoidal functions, always identify the maximum, minimum, period, and phase shift carefully, as these will affect the shape and position of the graph.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Sinusoidal Functions
Amplitude
Period
Phase Shift
Vertical Shift

Formulas

General sinusoidal function: D(t) = A sin(B(t - C)) + D_0
Amplitude: A = (Max - Min) / 2
Period: B = 2π / T
Vertical shift: D_0 = (Max + Min) / 2

Theorems

Trigonometric modeling of periodic phenomena

Suitable Grade Level

Grades 10-12