Question

1.Outside temperature over a day can be modelled as a sinusoidal function. Suppose you know the high temperature for the day is 62 degrees and the low temperature of 48 degrees occurs at 4 AM. Assuming t is the number of hours since midnight, find an equation for the temperature, D, in terms of t.

2. Outside temperature over a day can be modeled as a sinusoidal function. Suppose you know the high temperature of 75 degrees occurs at 3 PM and the average temperature for the day is 55 degrees. Find the temperature, to the nearest degree, at 6 AM.

Answer 1

High temperature = 62°

Low temperature = 48 ° at 4 AM

Range (62-48)/2 = 14/2= 7

Therefore amplitude A = 7

Period = 24 hrs

(2π/B ) = 24

B = (π/12)

Midline (62 + 48)/2 = 55

D (4)= 48 ......the interval can be written as (4,48)

A day is 24 hours (periodic function )

Then then next will occur at 24 +4 = 28 hrs

ie, D(28) = 48 ........the interval can be written as (28,48)

Since period is 24 hours .The maximum and minimum must occur at 12 hours apart .

4 + 12 = 16

D(16) = 62 .... maximum temperature

Therefore the interval can be written as (16, 62)

Now we have to plot the curve with the points (4,48), (28,48),(16,62)

Here we can have an option of using either sine or cosine function . We can consider it as a basic cosine function. We can write the function as

Y = Acos (B (x - D))+ C

Here the amplitude is shifted downwards so A = - 7

D is the horizontal shift = 4

C is the vertical shift = midline in the graph = 55

So we have

D(t) = -7 cos (π/12(x-4))+ 55

2.

We have high temperature is 75 ° @ 3 PM

Average temperature = 55 day. This makes midline 55 and amplitude 6.

So let's let's choose a simple function

F(x) = 6 sin (x) + 55

The time at which we have to find the temperature at 6 AM is 9 hours from 3PM , which is the maximum temperature. 9 hours is 3/8 of the day . So we want to find 3/8 of a cycle before 3 PM.

Since sine function with no horizontal shift, the time is maximum at 3 pm, corresponds to 5/8 of the way of the cycle .

5/8 of a cycle = 4π / 5

We want 3/ 8 of a cycle, or 6π/8 from π/2 ; that put at - π/4

Therefore the temperature corresponding to 6 AM

D(6) = 6 sin (- π/4) + 55 = 54.91°

The temperature at 6 AM is 54.91°