Precalc 2 Homework

Question

Outside temperature over a day can be modeled as a sinusoidal function. Suppose you know the high temperature of 67 degrees occurs at 6 PM and the average temperature for the day is 55 degrees. Find the temperature, to the nearest degree, at 4 AM.

Answer 1

General equation for sine:

y = asin[k(x - b)] + h

Where:

Amplitude = |a| (always positive)

Period = 2π/k

Phase Shift = b (left if b < 0, right if b > 0)

Vertical Shift = h (down if h < 0, up if h > 0)

Note:

The sine graph starts at midline value at x = b

then goes up if a > 0, or goes down if a < 0

From the information given:

Max. Value = 67 @ 6 PM (18:00 hours)

Avg. Value (Midline) = 55 = h = vertical shift

Amplitude = |67 – 55| = 12 degrees

Period = 24 hrs.

Assuming a > 0, sine value starts at midline ¼ of the period prior to the maximum value

Sine wave starts at: 18 - 0.25(24) = 12 (Noon)

Horizontal shift: b = 12

Period = 2πpi/k = 24

k =2π/24

k= π/12

Equation is:

y = asin[k(x - b)] + h

f(x) = 12sin((π/12)(x – 12)) + 55

f(x) = 12sin(π(x – 12)/12) + 55

Find temperature at 4 AM (4:00)

f(4) = 12sin(π(-8)/12) + 55

f(4) = 12sin(-2π/3) + 55

f(4) = 12(-√3/2) + 55

f(4) = 55 - 6√3

f(4) = 44.61 ºF