Let's analyze the problem step-by-step.

You are given the temperature function:

$T(t) = 60 + 9 \sin \left(\frac{\pi t}{12}\right)$

Where $t$ is the number of hours after 9 AM.

1. Find the temperature at 9 AM (i.e., when $t = 0$):

Substitute $t = 0$ into the function:

$T(0) = 60 + 9 \sin \left(\frac{\pi \cdot 0}{12}\right)$ $T(0) = 60 + 9 \sin(0)$ $T(0) = 60 + 9 \times 0$ $T(0) = 60$

So, the temperature at 9 AM is 60°F.

2. Find the temperature at 3 PM (i.e., when $t = 6$):

Substitute $t = 6$ into the function:

$T(6) = 60 + 9 \sin \left(\frac{\pi \cdot 6}{12}\right)$ $T(6) = 60 + 9 \sin \left(\frac{\pi}{2}\right)$ $T(6) = 60 + 9 \times 1$ $T(6) = 60 + 9$ $T(6) = 69$

So, the temperature at 3 PM is 69°F.

3. Find the average temperature from 9 AM to 9 PM:

The average temperature over an interval from $t = 0$ (9 AM) to $t = 12$ (9 PM) is given by the formula:

$\text{Average Temperature} = \frac{1}{12} \int_0^{12} T(t) \, dt$

First, we express the temperature function:

$T(t) = 60 + 9 \sin \left(\frac{\pi t}{12}\right)$

Now, we compute the integral:

$\int_0^{12} T(t) \, dt = \int_0^{12} \left(60 + 9 \sin \left(\frac{\pi t}{12}\right)\right) \, dt$

This breaks into two integrals:

$\int_0^{12} 60 \, dt + \int_0^{12} 9 \sin \left(\frac{\pi t}{12}\right) \, dt$

The first integral is straightforward:

$\int_0^{12} 60 \, dt = 60t \Big|_0^{12} = 60 \times 12 = 720$

For the second integral, we make a substitution to handle the sine term:

Let $u = \frac{\pi t}{12}$, so that $du = \frac{\pi}{12} dt$, and the limits of integration change accordingly.

The integral becomes:

$\int_0^{12} 9 \sin \left(\frac{\pi t}{12}\right) \, dt = 9 \times \frac{12}{\pi} \int_0^\pi \sin(u) \, du$

We know:

$\int_0^\pi \sin(u) \, du = 2$

So, the integral is:

$9 \times \frac{12}{\pi} \times 2 = \frac{216}{\pi}$

Now, putting everything together:

$\int_0^{12} T(t) \, dt = 720 + \frac{216}{\pi}$

The average temperature is:

$\frac{1}{12} \left( 720 + \frac{216}{\pi} \right)$

Simplifying:

$\text{Average Temperature} = 60 + \frac{18}{\pi}$

Approximating $\pi$ as 3.1416:

$\text{Average Temperature} \approx 60 + \frac{18}{3.1416} \approx 60 + 5.73 = 65.73 \text{°F}$

Thus, the average temperature from 9 AM to 9 PM is approximately 65.73°F.

Would you like further details or clarifications on any part of the solution? Here are 5 related questions that could expand on this:

1. How do you interpret the sine function in temperature models like this one?
2. What are other methods for calculating the average temperature over an interval?
3. How would you interpret the significance of the 60 constant in the temperature equation?
4. What would the temperature be at 12 PM (noon) in this model?
5. How does changing the amplitude of the sine function affect the temperature variation?

Tip: When dealing with periodic functions like sine, the period often corresponds to a complete cycle. In this case, the temperature oscillates every 12 hours, representing a daily temperature cycle.