Math problems please help

Question

During a polar vortex last winter, the temperatures followed a regular pattern each day for a week: the maximum temperature was –4° F each day at 5:00 pm and the minimum temperature each day was – 24° F each day at 5:00 am. Make a cosine function that models the temperature as a function of time throughout the week.

 

1.Define variables. State exactly what real world quantities are represented by each variable.
x =
y =

 

 

2.Find the four features of the periodic function.

 

 

3.Write a cosine equation that matches the four features.

 

 

4.The function you wrote for question 3 models the temperature for the week. Use it to predict the
temperature at noon.

 

Answer 1

In order to make a cosine function for this problem, we first have to define our variables. Let x be the number of hours after midnight on the first day of the week, and let y be the temperature in degrees Fahrenheit at that time. So, x = 0 is midnight, x = 5 is 5:00 am, x = 12 is noon and x = 17 is 5:00 pm.

Now we need the four features of the periodic function, which are the amplitude, the midline, the period and the horizontal shift. The amplitude is half the distance between the maximum and the minimum so it becomes:

A = (−4 − (−24)) / 2 = 20 / 2 = 10

The midline is the average of the maximum and the minimum since it sits right in the middle of the two so it becomes:

D = (−4 + (−24)) / 2 = −28 / 2 = −14

The period is 24 hours since the pattern repeats itself every day. If you notice, the maximum at 5:00 pm and the minimum at 5:00 am are 12 hours apart, which is exactly half of one period, so that is a good way to check that 24 is correct. We use the period to find B so it becomes:

B = 2π / 24 = π / 12

Lastly, the horizontal shift. Remember, a regular cosine graph starts at its maximum, so we want our graph to start when the temperature is at its maximum, which is 5:00 pm or x = 17. So, the horizontal shift is 17 to the right.

Now we can put all four features into y = A cos(B(x − C)) + D so it becomes:

y = 10 cos(π/12 (x − 17)) − 14

We can check it by plugging in x = 17, which gives 10 cos(0) − 14 = 10 − 14 = −4, and x = 5, which gives 10 cos(−π) − 14 = −10 − 14 = −24. So both the maximum and the minimum match what the problem told us.

To predict the temperature at noon, we plug in x = 12 so it becomes:

y = 10 cos(π/12 (12 − 17)) − 14 = 10 cos(−5π/12) − 14

cos(−5π/12) is the same as cos(75°), which is about 0.2588, so it becomes:

y ≈ 10(0.2588) − 14 ≈ 2.588 − 14 ≈ −11.4

This makes sense since noon comes after the coldest time at 5:00 am and before the warmest time at 5:00 pm, and it is closer to 5:00 pm, so the temperature should be a little above the midline of −14.

So, our final answer is y = 10 cos(π/12 (x − 17)) − 14, and the temperature at noon is about −11.4°F.