In the month of March, the temperature at the South Pole varies over the day in a periodic way that can be modeled approximately by a trigonometric function. The highest temperature is about $-50^\circ\text{C}$ , and it is reached around $2\text{p.m.}$ The lowest temperature is about $-54^\circ\text{C}$ and it is reached half a day apart from the highest temperature, at $2\text{a.m.}$ Find the formula of the trigonometric function that models the temperature $T$ in the South Pole in March $t$ hours after midnight. Define the function using radians. $\newline$ $T(t)=\square$ $\newline$ What is the temperature at $5\text{p.m.}$ ? Round your answer, if necessary, to two decimal places. $\newline$ $^\circ\text{C}$

View step-by-step help

Home

Math Problems

Precalculus

Inverses of trigonometric functions

Full solution

Q. In the month of March, the temperature at the South Pole varies over the day in a periodic way that can be modeled approximately by a trigonometric function. The highest temperature is about

-50^\circ\text{C}

, and it is reached around

2\text{p.m.}

The lowest temperature is about

-54^\circ\text{C}

and it is reached half a day apart from the highest temperature, at

2\text{a.m.}

Find the formula of the trigonometric function that models the temperature

T

in the South Pole in March

t

hours after midnight. Define the function using radians.

\newline

T(t)=\square

\newline

What is the temperature at

5\text{p.m.}

? Round your answer, if necessary, to two decimal places.

\newline

^\circ\text{C}

Trigonometric Function Definition: The trigonometric function that models temperature variations is typically a cosine function because it starts at a maximum value at $t=0$ . We will use the cosine function in the form $T(t) = A \cdot \cos(B(t - C)) + D$ , where: $\newline$ - $A$ is the amplitude of the function, $\newline$ - $B$ is the frequency, $\newline$ - $C$ is the horizontal shift, $\newline$ - $D$ is the vertical shift.
Find Amplitude: First, we find the amplitude $A$ , which is half the difference between the maximum and minimum temperatures. The maximum temperature is $-50$ °C, and the minimum is $-54$ °C. So, $A = (-50 - (-54)) / 2 = (54 - 50) / 2 = 4 / 2 = 2$ .
Find Vertical Shift: Next, we find the vertical shift $D$ , which is the average of the maximum and minimum temperatures. $D = (-50 + (-54)) / 2 = (-50 - 54) / 2 = -104 / 2 = -52$ .
Calculate Frequency: The period of the temperature function is $24$ hours (one day), but we need to express it in radians since the function will be defined in radians. There are $2\pi$ radians in a full cycle ( $24$ hours), so the frequency $B$ is $2\pi / 24 = \pi / 12$ .
Calculate Horizontal Shift: The horizontal shift $C$ corresponds to the time when the maximum temperature occurs. Since the maximum temperature is at $2$ p.m. ( $14$ hours after midnight), we need to convert this to radians. $C = 14 \times (\pi / 12) = \frac{7\pi}{6}$ .
Write Function: Now we can write the function as $T(t) = 2 \cdot \cos\left(\frac{\pi}{12} \cdot (t - \frac{7\pi}{6})\right) - 52$ .
Substitute Time: To find the temperature at $5$ p.m., we need to substitute $t = 17$ (since $5$ p.m. is $17$ hours after midnight) into the function $T(t)$ . So, $T(17) = 2 \cdot \cos\left(\frac{\pi}{12} \cdot (17 - \frac{7\pi}{6})\right) - 52$ .
Calculate Argument: We calculate the argument of the cosine function: $\frac{\pi}{12} \times (17 - \frac{7\pi}{6}) = \frac{\pi}{12} \times (17 - \frac{42}{6}) = \frac{\pi}{12} \times (17 - 7) = \frac{\pi}{12} \times 10 = \frac{5\pi}{6}$ .
Substitute into Function: Now we substitute $5\pi/6$ into the cosine function: $T(17) = 2 \cdot \cos(5\pi/6) - 52$ . The cosine of $5\pi/6$ is $-\sqrt{3}/2$ .
Calculate Temperature: We calculate the temperature: $T(17) = 2 \times \left(-\frac{\sqrt{3}}{2}\right) - 52 = -\sqrt{3} - 52$ . Since $\sqrt{3}$ is approximately $1.732$ , we have $T(17) \approx -1.732 - 52 \approx -53.732$ .