)utside temperature over a day can be modeled as a sinusoidal function. Suppose you know the temperature varies between 28 and 62 degrees during the day and the average daily temperature first occurs at 8 AM. How many hours after midnight, to two decimal places, does the temperature first reach 41 degrees?
The average temperature is $(28+62)/2=45$ degrees, so the temperature varies by $45-28=17$ degrees around the average temperature. The temperature first reaches 41 degrees when it is $45-(41-45)=49$ degrees, which is half a period after the average daily temperature. The period of the temperature is $12$ hours, so the temperature first reaches 41 degrees $\frac{1}{2}\cdot12=\boxed{6.00}$ hours after midnight.
good answers -- I guess.
It would be nice if they were actually legible.
To find out how many hours after midnight the temperature first reaches 41 degrees, we need to model the given information using a sinusoidal function and then solve for the time.
Step 1: Determine the amplitude and midline of the sinusoidal function.
Since the temperature varies between 28 and 62 degrees, the amplitude of the function is (62 - 28) / 2 = 17. The midline is the average of the highest and lowest temperature, which is (62 + 28) / 2 = 45.
Step 2: Find the period of the sinusoidal function.
The period is the length of one full cycle of the sinusoid. Since the temperature cycle repeats daily, the period is 24 hours.
Step 3: Determine the phase shift of the sinusoidal function.
The average daily temperature first occurs at 8 AM, which is 8 hours after midnight. Therefore, the sinusoidal function is shifted to the right by 8 hours.
Step 4: Form the equation of the sinusoidal function.
The general form of a sinusoidal function is given by: f(x) = A sin(B(x - C)) + D
A is the amplitude, B is the frequency (which affects the period), C is the phase shift, and D is the midline.
In this case, the equation becomes: f(x) = 17 sin(2π/24(x - 8)) + 45
Step 5: Solve for x when the temperature is 41 degrees.
Set the equation equal to 41 and solve for x:
17 sin(2π/24(x - 8)) + 45 = 41
17 sin(2π/24(x - 8)) = -4
Step 6: Solve for x using inverse sine.
Divide both sides by 17: sin(2π/24(x - 8)) = -4/17
Now, take the inverse sine of both sides to isolate the expression:
2π/24(x - 8) = arcsin(-4/17)
x - 8 = (24 / 2π) * arcsin(-4/17)
x - 8 = (12/π) * arcsin(-4/17)
x = (12/π) * arcsin(-4/17) + 8
Step 7: Calculate the value of x.
Using a calculator, plug in the value of arcsin(-4/17) and solve for x to obtain the final answer, rounded to two decimal places.
We can write the temperature as a function of time $t$ in hours since midnight as $T(t) = 28 + 34 \sin \left( \frac{t\pi}{12} \right)$, where the period of the function is $24$ hours.
The average temperature is the average of the minimum and maximum values, which is $\frac{28 + 62}{2} = 45$. Since the average temperature first occurs at 8 AM, the average temperature occurs at $t = \frac{8 \pi}{12} = \frac{2 \pi}{3}$ hours. Therefore, the temperature first reaches 41 degrees when
[T \left( \frac{2 \pi}{3} \right) = 41.]Solving, we find
[41 = 28 + 34 \sin \left( \frac{2 \pi}{3} \right) = 28 + 34 \left( \frac{\sqrt{3}}{2} \right) = \frac{92}{2} + \frac{87}{2} = \frac{179}{2} = 89.5.]This does not make sense, so we must have made a mistake. We realized that the average temperature is not 45, but rather the average of the minimum and maximum temperatures at a given time. Thus, the average temperature at 8 AM is $\frac{T_{\min} + T_{\max}}{2} = \frac{28 + 62}{2} = 45$. This means that the temperature first reaches 41 degrees when
[T \left( \frac{2 \pi}{3} \right) = 41 = 45 + 34 \sin \left( \frac{2 \pi}{3} \right) = 45 - 34 \left( \frac{\sqrt{3}}{2} \right) = \frac{79}{2} - \frac{87}{2} = \frac{-8}{2} = -4.]Since this does not make sense, we must have made another mistake. We realized that the temperature first reaches 41 degrees at the time when the average temperature is 41. Thus, the temperature first reaches 41 degrees when
[T \left( \frac{t \pi}{12} \right) = 41 = 28 + 34 \sin \left( \frac{t \pi}{12} \right) = 28 + 34 \cos \left( \frac{\pi}{2} - \frac{t \pi}{12} \right) = 28 + 34 \sin \left( \frac{\pi}{6} - \frac{t \pi}{12} \right).]Therefore,
[0 = 34 \sin \left( \frac{\pi}{6} - \frac{t \pi}{12} \right) - 13.]The solutions to this equation are $\frac{t \pi}{12} = \frac{\pi}{6} + k \pi$ for some integer $k$, so $t = \frac{1}{2} + k$. Since $t$ is the number of hours after midnight, the only solution that is positive is $t = \frac{5}{2}$, so the temperature first reaches 41 degrees $\boxed{\frac{5}{2}}$ hours after midnight.