Temperature Precalc

Question

Hi I have been having problems with making the equation.

Outside temperature over a day can be modeled as a sinusoidal function. Suppose you know the high temperature of 69 degrees occurs at 4 PM and the average temperature for the day is 60 degrees. Find the temperature, to the nearest degree, at 5 AM.

Answer 1

let t be time in hours from midnight of some day

and F the temperature

F(t)=9cos((2π/24)(t-16))+60 ,,,,60 is midline,,,9 is amplitude

,,,,period is 24, freq=2π/period,,,,cosine needs to be 1 at +16 hours from midnight

F as a function of t derived from given data points

t=5, F=9cos(-11π/12)+60

which is the same value as t=29 (5AM 24 hours later)

t=29, F=9cos(13π/12)+60

Answer 2

Hi again, Everett.

Consider the hints within the problem.

We're told that the function is sinusoidal.

If the high temperature of 69°, and the average temperature is 60°,

then the midline is 60 and the amplitude is 9.

Since the problem begins with a high temperature, perhaps it makes more sense to use the Cosine function, instead of Sine.

There are 24 hours in a day, so our period is 24 and each hour is 1/24th of a cycle.

5 AM is 13 hours after a known maximum or 13π/12 into the cycle.

We can now enter 9 Cos (13π/12) + 60 into a calculator to get ≈ 51.307°.

Since the problem asked to the nearest degree, our submitted answer is 51°.

I hope this helps.