Over a 24 hour period the temperature in a town can be modeled by one period of a sinusoidal function. The temperature measures 70*F in the morning, rises to a high of 80*F, falls to a low of 60*F, and then rises to 70*F by the next morning. What is the equation for the sine function f(x), where x represents time on hours since the beginning of the 24 hour period that models the situation​

Start with an initial temperature of T0 = 70° at t = 0 hours, and add to that a sine function that does the following, in order:

Rises by 10°, reaching 80° after 6 hours

Drops by 10º, reaching 70° after 12 hours

Drops by 10°, reaching 60° after 18 hours

Rises by 10°, reaching  70° after 24 hours

Note that the amplitude of the sine function is A = 10°.

T = period of the sine function

T = 24 hours

f(x) = T0 + A sin[(2π/T)x]

where

T0 = 70°

A = 10°

T = 24 hours

x = elapsed time in hours

f(x) = 70º + (10°)sin[(π/12)x]

with x in hours.

Test this solution and see that it fits the data.  (Make sure your calculator is in the radian mode!)

10sin(πx/12)+70

Step-by-step explanation:

I just took a quiz and this was the correct answer.

it could be 1,3 and more others to create a cylinder.

step-by-step explanation:

part a

answer: a reasonable domain is where n is a real number. so n can be between 0 and 10. both endpoints are included.

work shown:

n is the number of days after the experiment starts. the smallest n can be is n = 0 which means that 0 days have gone by, and we're at the start. to find out how large n should be, then replace f(n) with 16.13 and solve for n. use logarithms to isolate the exponent.

f(n) = 12*(1.03)^n

16.13 = 12*(1.03)^n

16.13/12 = (1.03)^n

1.34416666666667 = (1.03)^n

(1.03)^n = 1.34416666666667

log[ (1.03)^n ] = log[ 1.34416666666667 ]

n*log[ 1.03 ] = log[ 1.34416666666667 ]

n = log[ 1.34416666666667 ]/log[ 1.03 ]

n = 10.0062999823929

this rounds to n = 10 which is fairly close but not 100% perfect. so this is the largest n can be.

note: if you plug n = 10 into f(n), you'll get roughly 16.127 which rounds to 16.13 (this comes up again in part c)

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part b

answer: the y intercept is 12. it represents the starting height of the plant in cm.

work shown:

plug n = 0 into the f(n) function. simplify

f(n) = 12*(1.03)^n

f(0) = 12*(1.03)^0

f(0) = 12*(1)

f(0) = 12

on day n = 0, aka the starting point, the height f(n) is 12 cm

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part c

answer: the average rate of change is approximately 0.43061036458991 (round however you need to). this represents the average growth rate from day n = 3 to day n = 10. so the plant grew roughly 0.43 cm per day during this timespan, assuming you round to 2 decimal places.

work shown:

compute f(3)

f(n) = 12*(1.03)^n

f(3) = 12*(1.03)^3

f(3) = 13.112724 < we'll use this later

compute f(10)

f(n) = 12*(1.03)^n

f(10) = 12*(1.03)^10

f(10) = 16.1269965521294 < we'll use this later

now use the formula below with a = 3 and b = 10

aroc = average rate of change

aroc = [ f(b) - f(a) ]/[ b - a ]

aroc = [ f(10) - f(3) ]/[ 10 - 3 ]

aroc = (16.1269965521294 - 13.112724)/(10 - 3)

aroc = 3.0142725521294/7

aroc = 0.43061036458991

round this however you need to

note: the plant grew approximately 3.01 cm over 7 days, so roughly 0.43 cm per day is the average growth rate (if you were to round to 2 decimal places).