)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?
3 answers
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.
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.
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.
good answers -- I guess.
It would be nice if they were actually legible.
It would be nice if they were actually legible.
2 answers