Graphs of the Sine and Cosine Function

Question

Outside temperature over a day can be modeled as a sinusoidal function. Suppose you know the temperature is 58 degrees at midnight and the low and high temperature during the day are 51 and 65 degrees, respectively. Assuming t is the number of hours since midnight, find an equation for the temperature, D, in terms of t.

D(t) =

Answer 1

Let t be the number of hours since midnight.

As t runs from 0 to 24, we go through an entire day.

In order to go through an entire sine wave, the argument of the sine function must run from 0 to 2π.

Consider the expression (t/24)2π, our candidate for the argument we will give to sine.

As t runs from 0 to 24, the fraction t/24 would run from 0 to 1. (That's because 0/24 is 0 and 24/24 is 1.)

Therefore as t runs from 0 to 24, the value of the entire expression (t/24)2π would run from 0 to 2π.

So this expression (t/24)2π is a good candidate for the argument of sine.

We can write this expression (t/24)2π more simply as this to 2πt/24 or even πt/12.

As x runs from 0 to 2π, the values of a normal sine function start at 0 (at x=0) and go up and down between a low of -1 and a high of +1, a distance of 2.

As t runs from 0 to 24, the temperature starts at 58 (at midnight) and goes up and down between a low of 51 degrees and a high of 65 degrees, a distance of 14 degrees.

14 is seven times as great as 2.

So let's investigate what 7sin(x) would do. As x runs from 0 to 2π, the value of 7sin(x) would start at 0 (at x=0) and go up and down between a low of -7 and a high of +7.

Let's add 58 to 7sin(x) and see what happens.

As x runs from 0 to 2π, 7sin(x) + 58 would start at 58 (at x=0) and go up and down between a low of 51 degrees and a high of 65 degrees. That's exactly the range of temperature that we want.

Now let's put in the argument πt/12 we discovered earlier.

And as t runs from 0 to 24, the expression πt/12 would go from 0 to 2π.

Therefore, as t runs from 0 to 24, the expression sin(πt/12) would start at 58 (when t=0, i.e., at midnight) and go up and down between a low of 51 and a high of 65 degrees.

That does everything the problem asks for, so the answer is

D(t) = sin(πt/12)

Something extra. D(t) = sin(πt/12), like a normal sine wave, begins (at midnight) by going up. But the temperature probably continues to drop during the hours after midnight, so we might want a sine wave that begins (at midnight) by going down.

So a more realistic choice would be

D(t) = -sin(πt/12), which begins (at midnight) by going down. It goes down because it is a mirror image of the function D(t) = sin(πt/12), reversed from top to bottom.