Sinusoidal Functions

Question

The temperature of London fluctuates by approximately 20*c throughout the course of a year. The average temperature of London is approximately 12*c. The highest temperature occurs halfway through the year in June (m=6).

Determine T(m) the function that models the temperature of London in a year.

Answer 1

Hi Shayan,

The standard equation for a sine wave or cosine wave is the following:

y = A*sin(Bx - C) + D

y = A*cos(Bx - C) + D

There are four components to a sine or cosine wave: amplitude, period, phase shift, vertical shift.

The amplitude measures how far up the wave goes and how far down the wave goes. The problem says the fluctuation in temperature 20 degrees, meaning it goes 10 degrees up and 10 degrees down. This is an amplitude of 10. In the standard equation above, A represents amplitude. Therefore, A = 10.

The period measures how long it takes for the wave to complete a full cycle. The timeframe for this problem is 12 months. Thus, the period is 12. In the standard equation above, the period is determined by the equation: Period = 2π/B. Since the period is 12, B = π/6

The phase shift is the same as the horizontal shift. We're going to try to avoid this. Let's say for the sake of simplicity that the phase shift is 0. The formula for phase shift using the equation above is C/B. Since the phase shift is 0, C = 0.

The vertical shift for this problem is the midline, or the average temperature. In this case, 12 degrees. The vertical shift in the equation above is denoted by D. So, D = 12.

So we've got a couple options.

y = 10sin(π/6*x) + 12

or

y = 10cos(π/6*x) + 12

Your problem states that at the halfway point, the temperature is at its max. Both of the equations above fail to reflect that.

If we plug in x = 6 for the sin equation, sin will go to zero, leaving y = 12.

If we plug in x = 6 for the cos equation, cos will go to -1, leaving y = 2.

We can correct this by either phase shifting sin (difficult and not recommended), or by inverting cosine. Since cos bottoms out when x = 6, flipping the equation over the x-axis by making the amplitude negative will make the cos max out instead.

y = -10cos(π/6*x) + 12

If we plug x = 6 into this equation, we get y = 22, which is our max result.

Hope this helps.