Question

In the month of June, the temperature in Johannesburg, South Africa, varies over the day in a periodic way that can be modeled approximately by a trigonometric function. The lowest temperature is usually around 3^\circ C3 ? C3, degree, C, and the highest temperature is around 18^\circ C18 ? C18, degree, C. The temperature is typically halfway between the daily high and daily low at both 10\text{ a.M.}10 a.M.10, space, a, point, m, point and 10\text{ p.M.}10 p.M.10, space, p, point, m, point, and the highest temperatures are in the afternoon. Find the formula of the trigonometric function that models the temperature TTT in Johannesburg ttt hours after midnight. Define the function using radians

Answer 1

Answer:

T = - 7.5 Cos π/12( t - 4 ) + 10.5

Step-by-step explanation:

Given that the

Maximum temp. = 18 degree Celsius

Minimum temp. = 3 degree Celsius

The half way between 10 am and 10 pm is 4 am

The sine and cosine functions can be used to model fluctuations  in temperature data through out the year. An equation that can be used to model these data is of the form:

T = A cos B(t - C) + D, where A,B,C,D, are constants, T is the  temperature in °C and t is the hour (1–24)

A = amplitude = (Tmax - Tmin)/2

A = (3 - 18)/2 = - 15/2 = -7.5 ( note : after midnight)

B = 2π/24 = π/12

C = units translated to the right

C = 4

D = ymin + amplitude = units translated up

D = 7.5 + 3 = 10.5

The formula of the trigonometric function that models the temperature T in Johannesburg t hours after midnight we be

T = - 7.5 Cos π/12( t - 4 ) + 10.5