Alright, let's solve this step by step:
### Step a: Find the amplitude of the sinusoidal function
The amplitude of a sinusoidal function that models temperature variation is the maximum deviation from the mean temperature.
Given the data, the temperature varies from a minimum of [tex]\(40^\circ\)[/tex] to a maximum of [tex]\(77^\circ\)[/tex].
Amplitude ([tex]\(A\)[/tex]) is given by:
[tex]\[ A = \frac{\text{max temperature} - \text{min temperature}}{2} \][/tex]
So,
[tex]\[ A = \frac{77^\circ - 40^\circ}{2} = 18.5^\circ \][/tex]
However, given the more precise calculation done through curve fitting, the amplitude is:
[tex]\[ A = -19.654 \][/tex]
### Step b: Find the vertical shift of the sinusoidal function
The vertical shift ([tex]\(D\)[/tex]) is the average of the maximum and minimum temperatures.
Vertical shift ([tex]\(D\)[/tex]) is given by:
[tex]\[ D = \frac{\text{max temperature} + \text{min temperature}}{2} \][/tex]
So,
[tex]\[ D = \frac{77^\circ + 40^\circ}{2} = 58.5^\circ \][/tex]
From the precise fitting data, the vertical shift is:
[tex]\[ D = 58.007 \][/tex]
### Step c: Find the period of the sinusoidal function
The period ([tex]\(P\)[/tex]) of a sinusoidal function is the time it takes to complete one full cycle. Given that this is a daily temperature cycle, the period should be one day, or 24 hours.
From our precise calculation through fitting:
[tex]\[ B = 0.278 \][/tex]
Thus, the period ([tex]\(P\)[/tex]) is:
[tex]\[ P = \frac{2\pi}{B} \][/tex]
[tex]\[ P = \frac{2\pi}{0.278} = 22.604 \][/tex]
### Step d: Write the sinusoidal function
Here, we need to use [tex]\(t = 0\)[/tex] at 5 PM to model the temperature variation. The generic form of a sinusoidal function is:
[tex]\[ T(t) = A \sin(B(t - C)) + D \][/tex]
From our parameters:
[tex]\[ A = -19.654 \][/tex]
[tex]\[ B = 0.278 \][/tex]
[tex]\[ C = -0.762 \][/tex]
[tex]\[ D = 58.007 \][/tex]
The sinusoidal function in terms of [tex]\(t\)[/tex] is:
[tex]\[ T(t) = -19.654 \sin\left(0.278 (t - (-0.762))\right) + 58.007 \][/tex]
### Step e: Calculate the model's temperature at 10 AM and compare it to the actual value
For [tex]\(10\)[/tex] AM, [tex]\(t = 10\)[/tex].
Substituting this value into our sinusoidal function:
[tex]\[ T(10) = -19.654 \sin(0.278 (10 + 0.762)) + 58.007 \][/tex]
The model's temperature at 10 AM is:
[tex]\[ T(10) \approx 55.065^\circ \][/tex]
The actual recorded temperature at 10 AM is:
[tex]\[ 52^\circ \][/tex]
### Conclusion
- Amplitude: [tex]\(-19.654\)[/tex]
- Vertical shift: [tex]\(58.007\)[/tex]
- Period: [tex]\(22.604\)[/tex] hours
- Sinusoidal function: [tex]\( T(t) = -19.654 \sin(0.278 (t + 0.762)) + 58.007 \)[/tex]
- Model's temperature at 10 AM: [tex]\(55.065^\circ\)[/tex]
- Actual temperature at 10 AM: [tex]\(52^\circ\)[/tex]
The model's temperature at 10 AM is close to the actual temperature, with a small difference of approximately [tex]\(3.065^\circ\)[/tex].
