Math 181 Miniproject 5: Hours of Daylight.md
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tags: MATH 181
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Math 181 Miniproject 5: Hours of Daylight
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**Overview:** This miniproject will apply what you've learned about derivatives so far, especially the Chain Rule, to analyze the change the hours of daylight.

**Prerequisites:** The computational methods of Sections 2.1--2.5 of *Active Calculus*, especially Section 2.5 (The Chain Rule).

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:::info
The number of hours of daylight in Las Vegas on the $x$-th day of the year ($x=1$ for Jan 1) is given by the function together with a best fit curve from Desmos.}[^first]
[^first]: The model comes from some data at http://www.timeanddate.com/sun/usa/las-vegas?

\\[
D(x)=12.1-2.4\cos \left(\frac{2\pi \left(x+10\right)}{365}\right).
\\]
(1) Plot a graph of the function $D(x)$. Be sure to follow the guidelines for formatting graphs from the specifications page for miniprojects.
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(1)

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(2) According to this model how many hours of daylight will there be on July 19 (day 200)?
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(2) According to the model, there will be approximately 9.7 hours of daylight on July 19.
Work: D(200)=12.1-2.4cos((2pi(210))/365)
=12.1-2.4 cos(3.614983327)
=12.1-2.4(0.998010274)
=12.1-2.395224658
=9.704775342

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(3) Go to http://www.timeanddate.com/sun/usa/las-vegas? and look up the actual number of hours of daylight for July 19 of this year. By how many minutes is the model's prediction off of the actual number of minutes of daylight?
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(3)
The website states that there will be 14 hours and 16 minutes of daylight on July 19th. That is 856 minutes. Using the model, it predicted that there would only be 582 minutes of daylight which is 274 minutes off.
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(4) Compute $D'(x)$. Show all work.
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(4)
D'(x)=-sin((365-2pi)/365)
Here's how I got that:
I had to use the chain rule because there is a division problem inside of a multiplication problem inside a subtraction problem. Because constants become 0 when they are differentiated, we can drop off the first chunk of the problem and have cos[(2pi(x+10)/365)]
=-sin[(d/dx2pi(x+10)+2pid/dx(x+10))]/365
Then, we note that 2pi is the same as 6.28 and it is just a number and a constant. It becomes zero.
-sin(0*(x+10)+(2pi*1) all over 365.
Then, I used to quotient rule.
d'(x)=-sin[(d/dx 2pi *365-(2pi*d/dx 365)] over 365^2.

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(5) Find the rate at which the number of hours of daylight are changing on July 19. Give your answer in minutes/day and interpret the results.
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(5) Derivative means slope so we can utilize the derivative formula we just calculated. In my D'x forumla I dont know how to keep the x in the formula so I got this D'(x) formula online: =
$[-24cos((2pi(x+10)/365)-121]/365$
When I plugged in 200 (july 19th is the 200th day of the year) I got 1.2192. When we add units to that statement, we say that the hours of daylight are changing at a rate of 1.2 min/day.

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(6) Note that near the center of the year the day will reach its maximum length when the slope of $D(x)$ is zero. Find the day of the year that will be longest by setting $D'(x)=0$ and solving.
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(6 If I set the equation D'(x) to 0 and follow the steps of getting x by itself , $x=cos((2pi(x+10))/365)=-5.042$
The longest day of the year will be June 20th.

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(7) Write an explanation of how you could find the day of the year when the number of hours of daylight is increasing most rapidly.
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(7) You could find the day of the year when the number of daylight hours is increasing most rapidly by looking on a graph of the daylight hours and finding where the slope is the steepest. The steeper the slope, the more drastic the increase in daylight hours is.

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