Outside temperature over a day can be modeled as a sinusoidal function. Suppose you know the high temperature of 85 degrees occurs at 6 PM and the average temperature for the day is 65 degrees. Find the temperature, to the nearest degree, at 9 AM.
(max + min)/2 = avg
(85+min)/2 = 65
min = 45
amplitude = 20
period is 24 hours,
24 = 2π/k, k = π/12
start with t = 0 representing midnight, and the temp would logically
start to decrease from there.
temp = -20sin (π/12 t) + 65 would be a good start
For this equation the max happens when t = 18 ,(6 pm)
which is what we want
so at 9:00 am, t = 9
temp = -20sin(π/12(9)) + 65
= 50.9°
Well, well, well, let's put on our mathematical clown shoes and solve this problem!
Since we're dealing with a sinusoidal function, we can express it as:
T(t) = A*sin(B(t - C)) + D
Where:
T(t) is the temperature at time t
A is the amplitude (half the difference between the high and low temperatures)
B is the frequency (2π divided by the period)
C is the horizontal shift (time when the temperature reaches its maximum value)
D is the vertical shift (the average temperature for the day)
So, in our scenario:
A = (85 - 65)/2 = 10
B = 2π/24 (since we have a 24-hour day)
C = 6 PM
D = 65
Now, let's calculate the temperature at 9 AM, which is 15 hours after the temperature reaches its maximum at 6 PM:
T(9 AM) = 10*sin((2π/24)(9 - 6)) + 65
Calculating this clown-like equation, we get:
T(9 AM) ≈ 10*sin(π/8) + 65 ≈ 10*0.383 + 65 ≈ 3.83 + 65 ≈ 68.83
So, according to my calculations, the temperature at 9 AM would be approximately 69 degrees. Keep in mind that this is an estimation, and your actual weather clown might have other plans for the day!
To find the temperature at 9 AM, we need to consider the period of the sinusoidal function and the time difference between 6 PM and 9 AM.
The period of a sinusoidal function is the length of one complete cycle. Since we are dealing with a day's temperature, the period will be 24 hours.
The time difference between 6 PM and 9 AM is 15 hours. We need to find the temperature at 9 AM, which is 15 hours after the high temperature at 6 PM.
Now, let's set up a sinusoidal function to model the temperature:
T(t) = A * sin(2π/P * (t - t₀) + φ) + C
Where:
- T(t) represents the temperature at time t
- A is the amplitude of the function (half the difference between the maximum and minimum)
- P is the period of the function
- t₀ is the time at which the maximum temperature occurs
- φ is the phase shift of the function
- C is the average temperature for the day
From the information provided, we know that the high temperature occurs at 6 PM (18:00) with a temperature of 85 degrees. So we have t₀ = 18 and A = (85 - 65) / 2 = 10.
To find the temperature at 9 AM (time t = 9), we need to find φ and plug it into the function with the given values.
15 = 2π/P * (9 - 18) + φ
We can rearrange the equation to solve for φ:
15 - 2π/P * (-9) = φ
Now, we need to find the value of P.
P represents the period, which is 24 hours. So P = 24.
Let's plug these values into the equation to find φ:
15 - 2π/24 * (-9) = φ
15 - (3π/4) ≈ 4.536
Now, we can substitute these values into the function to find the temperature at 9 AM (t = 9):
T(9) = 10 * sin(2π/24 * (9 - 18) + 4.536) + 65
Calculating this expression:
T(9) ≈ 10 * sin(-π/2 + 4.536) + 65
T(9) ≈ 10 * sin(2.536) + 65
T(9) ≈ 10 * 0.604 + 65
T(9) ≈ 6.04 + 65
T(9) ≈ 71.04
Therefore, the temperature at 9 AM will be approximately 71 degrees Fahrenheit.
To solve this problem, we can use the concept of a sinusoidal function to model the temperature over the course of a day. A sinusoidal function can be represented in the form:
T(t) = A sin(B(t - C)) + D
Where:
- T(t) represents the temperature at time t
- A represents the amplitude of the function, which is half of the difference between the highest and lowest temperatures
- B represents the period of the function, which is the length of one complete cycle (in this case, 24 hours)
- C represents the horizontal shift of the function, which is the time at which the cycle begins (in this case, 6 PM)
- D represents the vertical shift of the function, which is the average temperature for the day (in this case, 65 degrees)
From the given information, we know that the high temperature of 85 degrees occurs at 6 PM, which is 12 hours after midnight. Therefore, we can set t = 12 in the equation and solve for A. We can substitute the other known values into the equation to find the values of B, C, and D.
T(12) = A sin(B(12 - C)) + D
85 = A sin(B(12 - C)) + 65
Subtracting 65 from both sides of the equation gives:
20 = A sin(B(12 - C))
Next, we can use the fact that the period of the function is 24 hours, meaning the time difference between two consecutive high temperatures is 24 hours. From 6 PM to 9 AM the next day, there are 15 hours. Therefore, we can set t = 15 and solve for A.
T(15) = A sin(B(15 - C)) + D
T(15) = A sin(B(3 - C)) + D
Since the temperature at 9 AM is not given, we need to solve for A. Rearranging the equation, we get:
A = (T(15) - D)/sin(B(3 - C))
The temperature at 9 AM would be:
T(9) = A sin(B(9 - C)) + D
To get an approximate value, substitute the values of A, B, C, and D into the equation. Keep in mind that A, B, C, and D are all positive values in this case. You can make an educated guess or use trial and error with different values to get closest to the given high temperature and average temperature.
Using this approach, you can find the temperature at 9 AM to the nearest degree.