Outside temperature over a day can be modelled as a sinusoidal function. Suppose you know the high temperature for the day is 83 degrees and the low temperature of 67 degrees occurs at 5 AM. Assuming t is the number of hours since midnight, find an equation for the temperature, D, in terms of t.
P =period = 24 hours
D = 67 + a sin [ ( t-5 )/24]
83 - 67 = 16
D = 67 + 16 sin [ ( t-5 )/24]
Now when is it hottest ?
left 2 pi out
D = 67 + 16 sin [ 2 pi ( t-5 )/24]
whoops again amplitude is half range
D = 75 + 8 sin [ 2 pi ( t-5 )/24]
To find an equation for the temperature, D, in terms of time, t, we can use a sinusoidal function that models the temperature changes throughout the day. The general form of a sinusoidal function is:
D(t) = A * sin(B(t - C)) + D₀
Where:
- A is the amplitude of the function (half the difference between the maximum and minimum values of the temperature)
- B determines the period of the function (the amount of time it takes to complete one cycle, in this case, 24 hours)
- C is the horizontal shift of the function, representing the time at which the maximum temperature occurs
- D₀ represents the vertical shift of the function, representing the average temperature over the day
In this case, we know that the high temperature is 83 degrees, the low temperature is 67 degrees at 5 AM (which is 5 hours after midnight), and we want to find the equation in terms of t (the number of hours since midnight).
Step 1: Find the amplitude (A)
The amplitude is half the difference between the maximum and minimum values of the temperature:
A = (83 - 67) / 2 = 8
Step 2: Find the period (B)
The period represents the time it takes for one complete cycle, which is 24 hours in this case:
B = 2π / 24 = π / 12
Step 3: Find the horizontal shift (C)
The maximum temperature occurs at 12 noon, which is 12 hours after midnight:
C = 12
Step 4: Find the vertical shift (D₀)
Since the average temperature over the day is not given, we can assume it to be the midpoint between the high and low temperatures:
D₀ = (83 + 67) / 2 = 75
Now we can substitute the values into the general form of the sinusoidal function:
D(t) = 8 * sin((π / 12)(t - 12)) + 75
Therefore, the equation for the temperature, D, in terms of t is D(t) = 8 * sin((π / 12)(t - 12)) + 75.
let's pick a sine function
the range is 9-70 = 20
so a = 10
we know that the period is 24 hrs
so 2π/k=24
24k = 2π
k = π/12
so far we have:
D = 10 sin(π/12)(t) + 80
giving us a range from 70 to 90
But obviously the temp after midnight would decrease, whereas our function has it increasing to 90 when t = 6 (6:00 am)
We could do a phase shift, or more simply, just flip the function to
D = -10sin(π/12)t + 80
check some values
t = 0 , D = -10sin0 + 80 = 80 , ok
t=6 , (6:00 am) D = -10 sin (π/2) + 80 = 70 , ok
t = 12 (noon), D = -10 sin π + 80 = 80 , ok
t = 18 , (6:00 pm) , D = -10sin 3π/2 + 80 = 90
Hope this helps