QuestionOutside temperature over a day can be modeled as a sinusoidal function. Suppose you know the temperature is 65 degrees at midnight and the high and low temperature during the day are 80 and 50 degrees, respectively. Assuming $t$ is the nulnber of hours since midnight, find an equation for the temperature, $D$ , in terms of $t$ . $D(t)=$ $\square$

Studdy Solution

STEP 1

1. The temperature varies sinusoidally over time.
2. The temperature is 65 degrees at midnight.
3. The high temperature is 80 degrees, and the low temperature is 50 degrees.
4. The period of the sinusoidal function is 24 hours.

STEP 2

1. Determine the amplitude and vertical shift of the sinusoidal function.
2. Determine the phase shift and period of the function.
3. Construct the sinusoidal function equation.

STEP 3

The amplitude $A$ of the sinusoidal function is half the difference between the high and low temperatures:
$A = \frac{80 - 50}{2} = 15$
The vertical shift $C$ is the average of the high and low temperatures:
$C = \frac{80 + 50}{2} = 65$

STEP 4

The period $P$ of the sinusoidal function is 24 hours, which corresponds to a full cycle of the sine function. Therefore, the angular frequency $B$ is given by:
$B = \frac{2\pi}{P} = \frac{2\pi}{24} = \frac{\pi}{12}$
Since the temperature is 65 degrees at midnight, which is the average temperature, there is no horizontal phase shift needed for the sine function starting at its midline.

STEP 5

The general form of the sinusoidal function is:
$D(t) = A \sin(Bt + \phi) + C$
Since there is no phase shift ( $\phi = 0$ ), the equation simplifies to:
$D(t) = 15 \sin\left(\frac{\pi}{12}t\right) + 65$
The equation for the temperature $D$ in terms of $t$ is:
$D(t) = 15 \sin\left(\frac{\pi}{12}t\right) + 65$

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