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. $\newline$ The lowest temperature is usually around $3^{\circ} \mathrm{C}$ , and the highest temperature is around $18^{\circ} \mathrm{C}$ . The temperature is typically halfway between the daily high and daily low at both $10$ a.m. and $10$ p.m., and the highest temperatures are in the afternoon. $\newline$ Find the formula of the trigonometric function that models the temperature $T$ in Johannesburg $t$ hours after midnight. Define the function using radians. $\newline$ $T(t)=\square$

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Q. 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.

\newline

The lowest temperature is usually around

3^{\circ} \mathrm{C}

, and the highest temperature is around

18^{\circ} \mathrm{C}

. The temperature is typically halfway between the daily high and daily low at both

10

a.m. and

10

p.m., and the highest temperatures are in the afternoon.

\newline

Find the formula of the trigonometric function that models the temperature

T

in Johannesburg

t

hours after midnight. Define the function using radians.

\newline

T(t)=\square

Calculate Amplitude: The amplitude of the temperature variation is half the difference between the highest and lowest temperatures. $\newline$ Amplitude = $(18 - 3) / 2 = 7.5$ °C.
Calculate Average Temperature: The average temperature is the midpoint between the highest and lowest temperatures. $\newline$ Average temperature = $(18 + 3) / 2 = 10.5$ °C.
Choose Cosine Function: Since the temperature is periodic and the highest temperatures are in the afternoon, we'll use a cosine function that starts at its maximum at $t=0$ .
Convert Period to Radians: The period of the function is the length of the day, which is $24$ hours. But we need to convert this into radians since the function uses radians. $\newline$ Period in radians = $2\pi$ radians = $24$ hours. $\newline$ So, $1$ hour = $\frac{2\pi}{24}$ radians.
Write Function with Parameters: Now we can write the function using the amplitude, average temperature, and period in radians. $\newline$ $T(t) = 7.5 \times \cos\left(\frac{2\pi}{24}t\right) + 10.5$ .
Adjust Phase Shift: However, we need to adjust the phase shift to match the given information that the temperature is halfway between the high and low at $10$ a.m. and $10$ p.m. Since the cosine function starts at its maximum, we need to shift it to the right by $10$ hours to match the $10$ a.m. halfway point. $\newline$ Phase shift in hours = $10$ hours. $\newline$ Phase shift in radians = $10 \times (2\pi/24)$ radians.
Final Function with Phase Shift: The final function with the phase shift is: $T(t) = 7.5 \times \cos\left(\frac{2\pi}{24}(t - 10)\right) + 10.5$ .