Tempuratures using Trig Function

Question

Outside temperature over a day can be modeled as a sinusoidal function. Suppose you know the temperature varies between 38 and 52 degrees during the day and the average daily temperature first occurs at 10 AM. How many hours after midnight, to two decimal places, does the temperature first reach 43 degrees?

 

Hours = ?

Answer 1

Hi Rebecca:

The sinusoidal function is represented as: y = A sin (ω(x - α)) + C

where: y = temperature in degrees; x = hours after midnight; A = amplitude; ω = 2π/period; α = Y-axis phase shift; and C = the midline.

In this problem, midline C = (38 + 52) / 2 = 45

Amplitude A = 52 - 45 = 7

Period = 24 hours

ω = 2π/24 = π/12

α = 10 (hours after midnight)

Therefore: y = A sin (ω(x - α)) + C

= 7 sin (π/12 (x - 10)) + 45

Solving this equation for y = 43, gives: 43 = 7 sin (π/12 (x - 10)) + 45

-2 = 7 sin (π/12 (x - 10))

-2/7 = sin (π/12 (x - 10))

(π/12 (x - 10)) = arcsin (-2/7)

And I leave solving for x to you, as an exercise.

The sinusoidal wave can be represented by the equation:y=A∗sin[ω(x−α)]+Cy=A∗sin[ω(x−α)]+C

where, A is the amplitude; ω=2π/periodω=2π/period; α=α= phase shift on the Y-axis; and C = midline.

With the information given in this problem,

Midline (C) is the average calculated as: (72+38)/2=55(72+38)/2=55;

Amplitude (A) is 72-55= 17;

Period = 24 hours;

ω=2π/24ω=2π/24;

α=10α=10;

Substituting in the equation,

y=17∗sin[2π/24(x−10)]+55y=17∗sin[2π/24(x−10)]+55

Solving this equation for y=51y=51 gives the value of x as 9.09.

Thus, the temperature first reaches 51 degrees about 9.09 hours after midnight.