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The problem asks for the instantaneous rate of change of the temperature \( H(t) \), which is modeled by the function

\[
H(t) = 55 - 9 \cos\left(\frac{2\pi}{365}(t + 10)\right),
\]

where \( t \) represents the number of days since January 1. We are tasked with finding the instantaneous rate of change at \( t = 90 \), which means we need to compute the derivative \( H'(t) \) and evaluate it at \( t = 90 \).

### Step 1: Compute the derivative \( H'(t) \)
We begin by differentiating \( H(t) \) with respect to \( t \). The derivative of \( H(t) = 55 - 9 \cos\left(\frac{2\pi}{365}(t + 10)\right) \) can be found using the chain rule. 

\[
H'(t) = \frac{d}{dt} \left[ 55 - 9 \cos\left( \frac{2\pi}{365}(t + 10) \right) \right].
\]

The derivative of the constant 55 is 0, and the derivative of \( -9 \cos\left(\frac{2\pi}{365}(t + 10)\right) \) is:

\[
H'(t) = -9 \cdot \left(- \sin\left(\frac{2\pi}{365}(t + 10)\right)\right) \cdot \frac{2\pi}{365}.
\]

Thus, we get:

\[
H'(t) = 9 \cdot \frac{2\pi}{365} \sin\left(\frac{2\pi}{365}(t + 10)\right).
\]

### Step 2: Evaluate \( H'(t) \) at \( t = 90 \)
Now we substitute \( t = 90 \) into the derivative expression:

\[
H'(90) = 9 \cdot \frac{2\pi}{365} \sin\left(\frac{2\pi}{365}(90 + 10)\right).
\]

This simplifies to:

\[
H'(90) = 9 \cdot \frac{2\pi}{365} \sin\left(\frac{2\pi}{365} \cdot 100\right).
\]

### Step 3: Approximate the value
We now compute the value of \( H'(90) \):

1. \( \frac{2\pi}{365} \cdot 100 \approx 1.72 \) radians.
2. \( \sin(1.72) \approx 0.988 \).
3. Therefore, \( H'(90) \approx 9 \cdot \frac{2\pi}{365} \cdot 0.988 \).

Finally, approximate:

\[
H'(90) \approx 9 \cdot 0.0172 \cdot 0.988 \approx 0.153 \, ^\circ F/\text{day}.
\]

Thus, the instantaneous rate of change of the temperature at \( t = 90 \) days is approximately **0.153°F/day**, which corresponds to option (B).

---

Would you like further details or have any questions?

### 5 related questions:
1. How do we determine the instantaneous rate of change from a function?
2. Why is the chain rule necessary in this problem?
3. What role does the sine function play in modeling temperature change?
4. How would the rate of change vary at different times of the year (different \( t \))?
5. How does the amplitude of the cosine function affect the temperature's rate of change?

### Tip:
Remember that the derivative of a trigonometric function often involves the chain rule, especially when the argument is a function of \( t \).

The temperature, in degrees Fahrenheit (°F), of water in a pond is modeled by the function H given by H(t) = 55 - 9cos( (2π/365)(t + 10) ), where t is the number of days since January 1 (t = 0). What is the instantaneous rate of change of the temperature of the water at time t = 90 days?

Learn how to find the instantaneous rate of change of a temperature function H(t) using differentiation. This problem involves key calculus concepts like the chain rule and derivatives of trigonometric functions. Suitable for students in Grades 11-12.

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