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, degrees, C, and the highest temperature is around 18^\circ C18 ∘ C18, degrees, C. The temperature is typically halfway between the daily high and daily low at both 10\text{ a.M.}10 a.M.10, start text, space, a, point, m, point, end text and 10\text{ p.M.}10 p.M.10, start text, space, p, point, m, point, end text, and the highest temperatures are in the afternoon.

Answer 1

Answer:

The answer is "$\bold{T = - 7.5 \cos \frac{\pi}{12}( t - 4 )+ 10.5}$"

Step-by-step explanation:

Given value:

Temp-maximum=$18^{\circ}$  

Temp. minimum = $3^{\circ}$

It is halfway between 10 am and 10 pm to 4 am.  

The sinus and cosine roles could be used throughout the year to predict fluctuations in climate models. Its type of formula that can be used to model such information is:  

$T = A \cos B(t-C) + D,$ where parameters are A, B , C, D, T is the ° C temperature and t is the time (1-24)

$A = amplitude = \frac{(T_{max} - T_{min})}{2}$

                       $= \frac{(3 - 18)}{2}\\\\= - \frac{15}{2}\\\\ = -7.5$

$B = \frac{2 \pi}{24}$

   $= \frac{\pi}{12}$

$C = \text{ units translated to the right}= 4$

$D = y_{min} + amplitude = units \ translated \ up$

D = 7.5 + 3 = 10.5

Its trigonometric function equation that model temperature T hours after midnight in Johannesburg t.

$T = - 7.5 \cos \frac{\pi}{12}( t - 4 )+ 10.5$