Outside temperature over a day can be modeled as a sinusoidal function. Suppose you know the high temperature of 52 degrees occurs at 6 PM and the average temperature for the day is 45 degrees. Find the temperature, to the nearest degree, at 8 AM.
24 hour period
t =0 is midnight
t = 8 is 8 am
t = 18 = 6 pm
temp = y
y = 45 + (52 - 45) cos ( (2 pi /24)(t-18) )
y = 45 + 7 cos ( pi t/12 - 3/2 pi)
y = 45 + 7 cos (pi *8/12 - 18 pi/12) = 45 + 7 cos(-10 pi/12)
= 45 + 7 cos (- 5 pi/6) = 45 + 7 cos (-150 deg)
= 45 + 7 (-0.866) = 45 - 6.06 = 38.9
the period is 24 hours
since cosine has a maximum at t=0, shift that to 6 pm.
y = 45 + 7cos(π/12 (t-18))
now find y(8)
Well, let's put on our mathematical clown nose and get started, shall we?
Since we know that the function models a sinusoidal wave, it will have the form:
T(t) = A * sin(B * t + C) + D
Where:
A is the amplitude of the wave
B is the period of the wave
C is the phase shift of the wave
D is the vertical shift of the wave
We're given that the high temperature occurs at 6 PM, which is 12 hours after 8 AM, so the period of the wave is 24 hours.
We're also given that the average temperature for the day is 45 degrees, so D = 45.
The high temperature is 52 degrees, so we can infer that the amplitude is (52 - 45) / 2 = 3.5.
Finally, we need to determine the phase shift, C. Since the high temperature occurs at 6 PM (which is halfway through the day), the phase shift, C, is 12.
Putting it all together, we have:
T(t) = 3.5 * sin(2π/24 * t + 12) + 45
Now, we just need to plug in t = 8 to find the temperature at 8 AM.
T(8) = 3.5 * sin(2π/24 * 8 + 12) + 45
After some calculations, we find that T(8) is approximately 43 degrees.
So, my dear human friend, the temperature at 8 AM is approximately 43 degrees. Time to bring out the hot coffee and warm blankets!
To find the temperature at 8 AM, we can use a sinusoidal function of the form:
T(t) = A * sin(B * (t - C)) + D
where:
T(t) represents the temperature at time t
A represents the amplitude (half the difference between the maximum and minimum temperatures)
B represents the period (the time it takes for the temperature function to complete one full cycle)
C represents the phase shift (the horizontal displacement of the function)
D represents the vertical displacement (the average temperature)
Given that the high temperature of 52 degrees occurs at 6 PM, which is 12 hours after midnight (t = 0), the phase shift C is equal to -12.
We're also given that the average temperature for the day is 45 degrees, so the vertical displacement D is equal to 45.
To solve for the amplitude A and period B, we need more information. We'll assume the temperature follows a 24-hour cycle, so the period B is equal to 2π/24.
Now we can calculate the amplitude A:
A = (52 - 45)/2 = 7/2 = 3.5
Finally, we can substitute these values into the temperature function to find the temperature at 8 AM (t = 8):
T(8) = 3.5 * sin((2π/24) * (8 - (-12))) + 45
Simplifying:
T(8) = 3.5 * sin((2π/24) * 20) + 45
T(8) = 3.5 * sin(10π/12) + 45
T(8) = 3.5 * sin(5π/6) + 45
Using a calculator, we find that sin(5π/6) is approximately 0.866.
T(8) = 3.5 * 0.866 + 45
T(8) ≈ 3.03 + 45
T(8) ≈ 48.03
Therefore, the temperature at 8 AM would be approximately 48 degrees.
To find the temperature at 8 AM, we need to determine the equation of the sinusoidal function that models the temperature over a day. We are given that the high temperature of 52 degrees occurs at 6 PM, which is 12 hours after 6 AM.
Let's start by assuming the equation for the sinusoidal function is of the form:
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 determines the period of the function (in this case, it will be 2π since there are 24 hours in a day),
C represents the horizontal shift (time at which the sine curve is at its maximum or minimum),
D is the average temperature.
Since we know the average temperature is 45 degrees, we have D = 45.
Next, we need to find the amplitude, A. The high temperature is given as 52 degrees, and the low temperature is not provided. However, since the function is sinusoidal, we know that the difference between the high and low temperatures is twice the amplitude. Therefore, the amplitude is (52 - low temperature)/2.
Now, the time at which the high temperature occurs (6 PM) is 12 hours after 6 AM. Since the period of the function is 24 hours, we can conclude that the horizontal shift, C, is 12.
Finally, we can write the equation for the temperature using the given information:
T(t) = (52 - low temperature)/2 * sin(2π(t - 12)) + 45.
To find the temperature at 8 AM, we substitute t = 8 into the equation above:
T(8) ≈ (52 - low temperature)/2 * sin(2π(8 - 12)) + 45.
Simplifying further:
T(8) ≈ (52 - low temperature)/2 * sin(2π(-4)) + 45.
Since sin(2π(-4)) = sin(-8π) = sin(0) = 0, the equation becomes:
T(8) ≈ (52 - low temperature)/2 * 0 + 45.
Therefore, the temperature at 8 AM is approximately equal to the average temperature of 45 degrees.
Please note that this is an approximation, as we do not have the exact value of the low temperature for the day.