Precalc 2 Homework

Question

Outside temperature over a day can be modeled as a sinusoidal function. Suppose you know the temperature varies between 44 and 56 degrees during the day and the average daily temperature first occurs at 8 AM. How many hours after midnight, to two decimal places, does the temperature first reach 48 degrees?

Answer 1

it helps to check the math calculations with reality

realistically, if the average was at 8am with 50 degrees,

then 48 degrees occurred a little before 8am, such as maybe 6 or 7 am

temperatures increase in the morning, with minimum 44 degrees about 2 am

maximum at 2 pm

average occurs at the midpoint between max and min, which is 8am and 8 pm

it may help to use "military time" using 2 for 2am but 14 for 2pm

8 for 8 am but 20 for 8pm

add 12 to the pm numbers

48 is between 44 and 50, but closer to 50

44 is at 2am, 50 at 8am

44 is between 5pm and 8am

after doing the sine wave calculations, you should get somewhere in that range, with time x, 5<x < 8am

44=min

56=max

A=amplitude= (56-44)/2 =6= half the range

B=2pi/period = 2pi/24= pi/12

24 hour temperature cycle

C= phase shift= 8

D= midline =average =(56+44)/2=50

f(x)=AsinB(x-C)+D

plug in the numbers

6sin[(pi/12)(x-8)+50

at 8am f(x) = 50

sinB(x-C) = 0

x-C = 0

C=x = 8

f(x)= 6sin[pi/12)(x-8)]+50

48 = 6sin[pi/12(x-8]+50

(48-50)/6 = sin[pi/12(x-8)]

(pi/12)(x-8) = sin^-1(-1/3)

x-8 = -(12/pi)sin^-1(1/3)

x = 8 - (12/pi)sin^-1(1/3)

x = 8 - (12/180)sin^-1(1/3)

= 8 - (1/15)19.471 = 8-1.298

=6.702 = 6 hours and .702(60) minuts

= 6:42 am when x = 48 degrees

you could have worked this problem

using

f(x) =AcosB(x-C)+D

it doesn't matter whether you use a cosine or a sine wave

you get the same answer, but C will have a different value

A,B, and D will be the same

the difference between sinx and cosx is just a phase shift

Answer 2