Solved on Dec 10, 2023

Find a sinusoidal function for the outside temperature over a day given it is 80 degrees at midnight, with a high of 92 and low of 68 degrees. Let $t$ be hours since midnight, find $D(t)$ .

STEP 1

Assumptions
1. The temperature varies sinusoidally over a day.
2. The temperature is 80 degrees at midnight ( $t = 0$ hours).
3. The high temperature during the day is 92 degrees.
4. The low temperature during the day is 68 degrees.
5. The period of the sinusoidal function is 24 hours since the pattern repeats every day.

STEP 2

The general form of a sinusoidal function is given by:
$D(t) = A \sin(B(t - C)) + D$
where: - $A$ is the amplitude of the function. - $B$ determines the period of the function. - $C$ is the horizontal shift (phase shift). - $D$ is the vertical shift (midline).

STEP 3

Calculate the amplitude $A$ of the function. The amplitude is half the distance between the high and low temperatures.
$A = \frac{High\, temperature - Low\, temperature}{2}$

STEP 4

Plug in the values for the high and low temperatures to calculate the amplitude.
$A = \frac{92 - 68}{2}$

STEP 5

Calculate the amplitude $A$ .
$A = \frac{24}{2} = 12$

STEP 6

Determine the vertical shift $D$ . The vertical shift is the average of the high and low temperatures.
$D = \frac{High\, temperature + Low\, temperature}{2}$

STEP 7

Plug in the values for the high and low temperatures to calculate the vertical shift.
$D = \frac{92 + 68}{2}$

STEP 8

Calculate the vertical shift $D$ .
$D = \frac{160}{2} = 80$

STEP 9

Determine the value of $B$ which is related to the period $T$ of the function. The period $T$ is the time it takes for one complete cycle, which is 24 hours for a daily temperature cycle.
$T = \frac{2\pi}{B}$

STEP 10

Solve for $B$ using the period $T = 24$ hours.
$B = \frac{2\pi}{T}$

STEP 11

Plug in the value for the period to calculate $B$ .
$B = \frac{2\pi}{24}$

STEP 12

Calculate the value of $B$ .
$B = \frac{\pi}{12}$

STEP 13

Determine the horizontal shift $C$ . Since the temperature is 80 degrees at midnight and this is also the vertical shift, the sinusoidal function must start at the midline at $t = 0$ . Therefore, we need to use cosine instead of sine to ensure the function starts at the midline.
$D(t) = A \cos(B(t - C)) + D$

STEP 14

Since the function starts at the midline at $t = 0$ , we do not need a horizontal shift, so $C = 0$ .

STEP 15

Now, we have all the parameters to write the equation for the temperature $D(t)$ in terms of $t$ .
$D(t) = A \cos(Bt) + D$

STEP 16

Plug in the values for $A$ , $B$ , and $D$ to write the final equation.
$D(t) = 12 \cos\left(\frac{\pi}{12}t\right) + 80$
The equation for the temperature $D$ in terms of the number of hours since midnight $t$ is:
$D(t) = 12 \cos\left(\frac{\pi}{12}t\right) + 80$

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Find a sinusoidal function for the outside temperature over a day given it is 80 degrees at midnight, with a high of 92 and low of 68 degrees. Let $t$ be hours since midnight, find $D(t)$ .

STEP 1

STEP 2

The general form of a sinusoidal function is given by:
$D(t) = A \sin(B(t - C)) + D$
where: - $A$ is the amplitude of the function. - $B$ determines the period of the function. - $C$ is the horizontal shift (phase shift). - $D$ is the vertical shift (midline).

STEP 3

Calculate the amplitude $A$ of the function. The amplitude is half the distance between the high and low temperatures.
$A = \frac{High\, temperature - Low\, temperature}{2}$

STEP 4

Plug in the values for the high and low temperatures to calculate the amplitude.
$A = \frac{92 - 68}{2}$

STEP 5

Calculate the amplitude $A$ .
$A = \frac{24}{2} = 12$

STEP 6

Determine the vertical shift $D$ . The vertical shift is the average of the high and low temperatures.
$D = \frac{High\, temperature + Low\, temperature}{2}$

STEP 7

Plug in the values for the high and low temperatures to calculate the vertical shift.
$D = \frac{92 + 68}{2}$

STEP 8

Calculate the vertical shift $D$ .
$D = \frac{160}{2} = 80$

STEP 9

Determine the value of $B$ which is related to the period $T$ of the function. The period $T$ is the time it takes for one complete cycle, which is 24 hours for a daily temperature cycle.
$T = \frac{2\pi}{B}$

STEP 10

Solve for $B$ using the period $T = 24$ hours.
$B = \frac{2\pi}{T}$

STEP 11

Plug in the value for the period to calculate $B$ .
$B = \frac{2\pi}{24}$

STEP 12

Calculate the value of $B$ .
$B = \frac{\pi}{12}$

STEP 13

STEP 14

Since the function starts at the midline at $t = 0$ , we do not need a horizontal shift, so $C = 0$ .

STEP 15

Now, we have all the parameters to write the equation for the temperature $D(t)$ in terms of $t$ .
$D(t) = A \cos(Bt) + D$

STEP 16

Plug in the values for $A$ , $B$ , and $D$ to write the final equation.
$D(t) = 12 \cos\left(\frac{\pi}{12}t\right) + 80$
The equation for the temperature $D$ in terms of the number of hours since midnight $t$ is:
$D(t) = 12 \cos\left(\frac{\pi}{12}t\right) + 80$

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Find a sinusoidal function for the outside temperature over a day given it is 80 degrees at midnight, with a high of 92 and low of 68 degrees. Let ttt be hours since midnight, find D(t)D(t)D(t).

STEP 1

STEP 2

STEP 3

Calculate the amplitude AAA of the function. The amplitude is half the distance between the high and low temperatures.A=High temperature−Low temperature2A = \frac{High\, temperature - Low\, temperature}{2}A=2Hightemperature−Lowtemperature​

STEP 4

Plug in the values for the high and low temperatures to calculate the amplitude.A=92−682A = \frac{92 - 68}{2}A=292−68​

STEP 5

Calculate the amplitude AAA.A=242=12A = \frac{24}{2} = 12A=224​=12

STEP 6

Determine the vertical shift DDD. The vertical shift is the average of the high and low temperatures.D=High temperature+Low temperature2D = \frac{High\, temperature + Low\, temperature}{2}D=2Hightemperature+Lowtemperature​

STEP 7

Plug in the values for the high and low temperatures to calculate the vertical shift.D=92+682D = \frac{92 + 68}{2}D=292+68​

STEP 8

Calculate the vertical shift DDD.D=1602=80D = \frac{160}{2} = 80D=2160​=80

STEP 9

Determine the value of BBB which is related to the period TTT of the function. The period TTT is the time it takes for one complete cycle, which is 24 hours for a daily temperature cycle.T=2πBT = \frac{2\pi}{B}T=B2π​

STEP 10

Solve for BBB using the period T=24T = 24T=24 hours.B=2πTB = \frac{2\pi}{T}B=T2π​

STEP 11

Plug in the value for the period to calculate BBB.B=2π24B = \frac{2\pi}{24}B=242π​

STEP 12

Calculate the value of BBB.B=π12B = \frac{\pi}{12}B=12π​

STEP 13

STEP 14

Since the function starts at the midline at t=0t = 0t=0, we do not need a horizontal shift, so C=0C = 0C=0.

STEP 15

Now, we have all the parameters to write the equation for the temperature D(t)D(t)D(t) in terms of ttt.D(t)=Acos⁡(Bt)+DD(t) = A \cos(Bt) + DD(t)=Acos(Bt)+D

STEP 16

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Find a sinusoidal function for the outside temperature over a day given it is 80 degrees at midnight, with a high of 92 and low of 68 degrees. Let $t$ be hours since midnight, find $D(t)$ .

Calculate the amplitude $A$ of the function. The amplitude is half the distance between the high and low temperatures.
$A = \frac{High\, temperature - Low\, temperature}{2}$

Plug in the values for the high and low temperatures to calculate the amplitude.
$A = \frac{92 - 68}{2}$

Calculate the amplitude $A$ .
$A = \frac{24}{2} = 12$

Determine the vertical shift $D$ . The vertical shift is the average of the high and low temperatures.
$D = \frac{High\, temperature + Low\, temperature}{2}$

Plug in the values for the high and low temperatures to calculate the vertical shift.
$D = \frac{92 + 68}{2}$

Calculate the vertical shift $D$ .
$D = \frac{160}{2} = 80$

Determine the value of $B$ which is related to the period $T$ of the function. The period $T$ is the time it takes for one complete cycle, which is 24 hours for a daily temperature cycle.
$T = \frac{2\pi}{B}$

Solve for $B$ using the period $T = 24$ hours.
$B = \frac{2\pi}{T}$

Plug in the value for the period to calculate $B$ .
$B = \frac{2\pi}{24}$

Calculate the value of $B$ .
$B = \frac{\pi}{12}$

Since the function starts at the midline at $t = 0$ , we do not need a horizontal shift, so $C = 0$ .

Now, we have all the parameters to write the equation for the temperature $D(t)$ in terms of $t$ .
$D(t) = A \cos(Bt) + D$