The number of hours of daylight at any point on Earth fluctuates throughout the year. In the northern​ hemisphere, the shortest day is on the winter solstice and the longest day is on the summer solstice. At 40 degrees north​ latitude, the length of a day is approximated by Upper D left parenthesis t right parenthesis equals 12 minus 3 cosine left bracket StartFraction 2 pi left parenthesis t plus 10 right parenthesis Over 365 EndFraction right bracket
​, where D is measured in hours and 0less than or equalstless than or equals365 is measured in​ days, with tequals0 corresponding to January 1. Complete parts​ (a) through​ (e) below.
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Part 3
c. Find the rate at which the daylight function changes on October 28. Convert your answer to units of​ min/day and explain what this result means.
The rate at which the daylight function changes on October 28 is 
  
enter your response here ​min/day.
​(Round to the nearest hundredth as​ needed.)

We are given the function that describes the number of hours of daylight at 40 degrees north latitude:

\[
D(t) = 12 - 3 \cos\left(\frac{2\pi (t + 10)}{365}\right)
\]
where \( t \) represents the day of the year with \( t = 0 \) corresponding to January 1.

We want to find the rate at which the daylight function changes on October 28, which corresponds to \( t = 300 \) (since October 28 is the 301st day of the year, and \( t \) is based on \( t = 0 \) for January 1).

### Step 1: Differentiate the function

The rate at which daylight changes is the derivative of the function \( D(t) \) with respect to \( t \). So, we need to find \( D'(t) \):

\[
D'(t) = \frac{d}{dt} \left(12 - 3 \cos\left(\frac{2\pi (t + 10)}{365}\right)\right)
\]

We know the derivative of a cosine function is:

\[
\frac{d}{dt} \left( \cos(u) \right) = -\sin(u) \cdot \frac{du}{dt}
\]

So, applying this to our function, we get:

\[
D'(t) = 3 \sin\left(\frac{2\pi (t + 10)}{365}\right) \cdot \frac{2\pi}{365}
\]

Thus, the derivative becomes:

\[
D'(t) = \frac{6\pi}{365} \sin\left(\frac{2\pi (t + 10)}{365}\right)
\]

### Step 2: Evaluate \( D'(t) \) at \( t = 300 \)

Now, we substitute \( t = 300 \) into the derivative:

\[
D'(300) = \frac{6\pi}{365} \sin\left(\frac{2\pi (300 + 10)}{365}\right)
= \frac{6\pi}{365} \sin\left(\frac{2\pi (310)}{365}\right)
\]

### Step 3: Convert the rate of change to minutes per day

Since \( D'(t) \) is currently in hours per day, we need to multiply by 60 to convert the result to minutes per day.

Let me compute this value.The rate at which the daylight function changes on October 28 is approximately \(-2.51\) minutes per day. 

This means that on October 28, the length of daylight is decreasing by about 2.51 minutes per day as the days approach the winter solstice.

Would you like more details on this calculation or have any other questions?

Here are 5 related questions for further exploration:
1. How would the daylight rate change on the summer solstice?
2. Can you calculate the rate of change in daylight on March 21?
3. What would the total daylight hours be on October 28?
4. How does the rate of daylight change vary across the year?
5. Can you find when the rate of change in daylight is the fastest?

**Tip:** The rate of daylight change is usually the highest around the equinoxes (March 21 and September 23).

The number of hours of daylight at any point on Earth fluctuates throughout the year. In the northern hemisphere, the shortest day is on the winter solstice and the longest day is on the summer solstice. At 40 degrees north latitude, the length of a day is approximated by D(t) = 12 - 3cos[(2π(t + 10) / 365)], where D is measured in hours and 0 ≤ t ≤ 365 is measured in days, with t = 0 corresponding to January 1. Complete parts (a) through (e) below.

Part 3c. Find the rate at which the daylight function changes on October 28. Convert your answer to units of min/day and explain what this result means.

Learn how to find the rate of change in the length of daylight at 40 degrees north latitude on October 28. This problem involves trigonometric functions, differentiation, and converting units to min/day. Suitable for students in Grades 11-12 studying calculus.

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