Can someone please help me? I don't know how to solve for time under 23 units. When I try I think I get the answer for above 23 units and when I try to subtract from the whole period my answer is wrong. The correct answer is 5.1337 hours.

Fidel is measuring the light intensity of a distant star. He has found that the intensity varies between a minimum of 19 units and a maximum of 75 units. At 7:00 AM today, he found the intensity was 19. The intensity then increased, reaching an intensity of 47 at 11:00 AM today.
Assume that the light intensity is a sinusoidal function of time. For what length of time today will the intensity be below 23 units?

To determine the length of time the intensity is below 23 units, we need to find the time period when the function representing the intensity is below 23. Let's break down the problem step-by-step to solve it:

Step 1: Identify the variables and given values:
- Minimum intensity: 19 units
- Maximum intensity: 75 units
- Initial intensity at 7:00 AM: 19 units
- Intensity at 11:00 AM: 47 units
- Intensity threshold (below which we want to measure time): 23 units

Step 2: Determine the amplitude and midline:
The amplitude (A) is half the difference between the maximum and minimum intensities. In this case, it is (75 - 19) / 2 = 28 units.
The midline (B) is the average of the maximum and minimum intensities. In this case, it is (75 + 19) / 2 = 47 units.

Step 3: Determine the angular frequency (ω):
The angular frequency (ω) can be calculated using the formula ω = 2π / T, where T is the time period of one complete cycle. In this case, T is 24 hours because the intensity repeats itself every 24 hours from the given information.

ω = 2π / 24 = π / 12 radians per hour.

Step 4: Determine the phase shift (C):
The phase shift (C) refers to a horizontal shift or delay in the function. Since the minimum intensity occurs at 7:00 AM, the phase shift can be found by calculating the time difference between the minimum intensity and the starting time (7:00 AM).

Phase shift (C) = (7:00 AM - 12:00 AM) = -5 hours.

Step 5: Write the equation representing the light intensity as a function of time:
The equation can be written as:
Int(t) = A * sin(ω(t - C)) + B

Int(t) = 28 * sin(π/12 * (t - (-5))) + 47
Int(t) = 28 * sin(π/12 * (t + 5)) + 47

(Note: The function is in the form A * sin(B(t - C)) + D, where A is the amplitude, B is the angular frequency, C is the phase shift, and D is the midline.)

Step 6: Solve for the time when the intensity is below 23 units:
Setting the equation equal to 23 units and solving for t:

23 = 28 * sin(π/12 * (t + 5)) + 47

Rearranging the equation gives us:

28 * sin(π/12 * (t + 5)) = -24

Divide both sides by 28:

sin(π/12 * (t + 5)) = -24/28
sin(π/12 * (t + 5)) = -6/7

Taking the inverse sine (arcsin) of both sides gives us:

π/12 * (t + 5) = arcsin(-6/7)

Now solve for (t + 5):

(t + 5) = arcsin(-6/7) * 12/π

Subtract 5 from both sides:

t = (arcsin(-6/7) * 12/π) - 5

Using a calculator, the approximate value of t is 5.1337 hours.

Therefore, the length of time today that the intensity is below 23 units is approximately 5.1337 hours.

To find the time period during which the light intensity is below 23 units, we need to analyze the given information and make use of the concept of sinusoidal functions.

The given information tells us that the light intensity varies between a minimum of 19 units and a maximum of 75 units. We are also given specific intensity values at two different times: 19 units at 7:00 AM and 47 units at 11:00 AM.

To solve for the time period during which the intensity is below 23 units, we can start by finding the equation of the sinusoidal function that represents the light intensity variation.

Step 1: Find the amplitude
The amplitude of a sinusoidal function is half the difference between the maximum and minimum values. In this case, the amplitude is (75 - 19) / 2 = 28 units.

Step 2: Find the period
The period of a sinusoidal function represents the length of one complete cycle. To find the period, we can note that the intensity reaches its maximum value of 75 units at some time after 11:00 AM. Let's assume it happens at 1:00 PM. Therefore, the length of time from 11:00 AM to 1:00 PM is half the period.

However, we need to convert these times into decimals based on a 24-hour clock. Let's assume that 12:00 PM is represented as 12.00 and add 12 to the hours to convert them to decimals. So, 11:00 AM becomes 11.00 and 1:00 PM becomes 13.00.

Now, the length of time from 11.00 AM to 1:00 PM is 13.00 - 11.00 = 2.00 hours, which represents half of the period. Therefore, the period is 2.00 * 2 = 4.00 hours.

Step 3: Determine the phase shift
The phase shift represents the amount of time the sinusoidal function is shifted horizontally. In this case, since the minimum intensity occurs at 7:00 AM and we are given the value at that time, there is no phase shift.

Step 4: Determine the equation
The equation of a sinusoidal function is given by:
Intensity = amplitude * sin((2π / period) * (time - phase shift)) + midpoint.

Since the midpoint is halfway between the minimum and maximum values, it is given by (maximum + minimum) / 2 = (75 + 19) / 2 = 47 units.

Putting it all together, the equation becomes:
Intensity = 28 * sin((2π / 4) * (time - 0)) + 47.

To find the length of time during which the intensity is below 23 units, we can set up the following inequality:
23 > 28 * sin((2π / 4) * (time - 0)) + 47.

Now we need to solve this inequality for time.

Unfortunately, there is no direct algebraic solution for this type of transcendental equation. However, we can use numerical methods or graphing techniques to solve it.

One numerical method is to use trial and error or approximation techniques to find values of time where the inequality holds true.

An alternative approach is to graph the sinusoidal function and visually identify the regions where the intensity is below 23 units.

Therefore, to find the specific time period during which the intensity is below 23 units, numerical methods or graphing techniques need to be used.