https://file.aiquickdraw.com/mathbot/file/1727798855913y6m6vofo.jpg

Let's analyze the problem based on the image you've provided. 

The temperature in Middletown Park is modeled by the equation:

\[
r(t) = 3.44 \cos\left( \frac{\pi t}{12} \right) + 0.044
\]

Where:
- \( r(t) \) is the rate of temperature change at time \( t \) hours after 6:00 AM last Sunday.
- The temperature at 6:00 AM last Sunday is 50.2°F.

### 1. First Question: Rate of Temperature Increase at 10:00 AM

To find the rate at which the temperature is increasing at 10:00 AM, we need to calculate the derivative of \( r(t) \), which will give the rate of change of temperature. We will then evaluate this derivative at \( t = 4 \) hours (since 10:00 AM is 4 hours after 6:00 AM).

The derivative of \( r(t) \) is:

\[
r'(t) = -3.44 \cdot \frac{\pi}{12} \sin\left( \frac{\pi t}{12} \right)
\]

Now, evaluate \( r'(4) \):

\[
r'(4) = -3.44 \cdot \frac{\pi}{12} \sin\left( \frac{\pi \cdot 4}{12} \right)
= -3.44 \cdot \frac{\pi}{12} \sin\left( \frac{\pi}{3} \right)
\]

Since \( \sin\left( \frac{\pi}{3} \right) = \frac{\sqrt{3}}{2} \), we get:

\[
r'(4) = -3.44 \cdot \frac{\pi}{12} \cdot \frac{\sqrt{3}}{2}
\]

Now let’s calculate this:

\[
r'(4) \approx -3.44 \cdot 0.2618 \cdot 0.866
\]

\[
r'(4) \approx -0.777 \text{ degrees per hour}
\]

So, the temperature was decreasing by approximately \( -0.78 \) degrees per hour at 10:00 AM.

### 2. Second Question: Temperature Increase from 6:00 AM to 1:00 PM

The total increase in temperature is found by calculating the change in temperature between 6:00 AM (when \( t = 0 \)) and 1:00 PM (when \( t = 7 \)).

We compute \( r(7) - r(0) \).

At \( t = 0 \):

\[
r(0) = 3.44 \cos(0) + 0.044 = 3.44 \cdot 1 + 0.044 = 3.484
\]

At \( t = 7 \):

\[
r(7) = 3.44 \cos\left( \frac{7\pi}{12} \right) + 0.044
\]

Now, let’s calculate the cosine term:

\[
\cos\left( \frac{7\pi}{12} \right) \approx -0.2588
\]

So:

\[
r(7) = 3.44 \cdot (-0.2588) + 0.044 = -0.890 \, + \, 0.044 = -0.846
\]

Now, the total temperature increase from 6:00 AM to 1:00 PM is:

\[
r(7) - r(0) = -0.846 - 3.484 = -4.33 \text{ degrees}
\]

The temperature actually decreased by approximately \( 4.33 \) degrees over this time.

### 3. Third Question: Temperature at 1:00 PM Last Sunday

We already computed \( r(7) = -0.846 \), so the rate at 1:00 PM is:

\[
\text{Temperature at 1:00 PM} = 50.2 - 0.846 = 49.35^\circ F
\]

### 4. Fourth Question: Temperature at 4:00 PM Last Friday

Friday is 5 days (or 120 hours) before last Sunday. Therefore, 4:00 PM on Friday is:

\[
t = 6 + 120 + 10 = 136 \text{ hours}
\]

We calculate \( r(136) \):

\[
r(136) = 3.44 \cos\left( \frac{136\pi}{12} \right) + 0.044
\]

Simplifying the angle:

\[
\frac{136\pi}{12} = \frac{34\pi}{3} = 11\pi + \frac{\pi}{3}
\]

Thus:

\[
r(136) = 3.44 \cos\left( \frac{\pi}{3} \right) + 0.044 = 3.44 \cdot 0.5 + 0.044 = 1.72 + 0.044 = 1.764
\]

The temperature at 4:00 PM on Friday would be:

\[
\text{Temperature at 4:00 PM on Friday} = 50.2 + 1.764 = 51.96^\circ F
\]

---

Let me know if you'd like more details or further clarifications.

### Follow-up questions:
1. What is the rate of temperature change at 12:00 PM last Sunday?
2. How would the equation look if we changed the amplitude of the cosine term?
3. What is the maximum temperature change possible according to the given equation?
4. Can we predict when the temperature will be the same as it was at 6:00 AM?
5. How would the temperature change if the cosine term's period was altered?

**Tip:** For sinusoidal functions, the amplitude determines how much fluctuation occurs from the mean value.

The temperature in Middletown Park at 6:00 AM last Sunday was 50.2 degrees Fahrenheit. The temperature was changing at a rate given by r(t) = 3.44 cos((πt/12)) + 0.044, where t is in hours after 6:00 AM last Sunday. Calculate the following: 
1. Rate of temperature increase at 10:00 AM. 
2. Temperature increase from 6:00 AM to 1:00 PM. 
3. Temperature at 1:00 PM last Sunday. 
4. Temperature at 4:00 PM last Friday (5 days earlier).

Learn how to model temperature changes using a cosine function in Middletown Park. This problem demonstrates the application of trigonometric functions to find the rate of temperature change at specific times, calculate total temperature increases, and predict future temperatures. Suitable for high school students in Grades 10-12.

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