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)

:::info
(2) According to this model how many hours of daylight will there be on July 19 (day 200)?
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(2)
$$14.236 hours $$

:::info
(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 actual number of hours of daylight for July 19th is 14.17 hours. The model estimated 14.236 hours. The model's prediction was off by 3.96 minutes. $$14.236-14.17=0.066*60=3.96hours$$

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(4) Compute $D'(x)$. Show all work.
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(4) \\[
D(x)=12.1-2.4\cos \left(\frac{2\pi \left(x+10\right)}{365}\right)
\\]
$$D'\left(x\right)=0+2.4\sin\left(\frac{2\pi\left(x\right)}{365}\right)+\left(\frac{20\pi}{365}\right)$$
$$D'\left(x\right)=2.4\sin\left(\frac{2\pi\left(x+10\right)}{365}\right)\left(\frac{2\pi}{365}\right)$$

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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)
$$D'(200)=-0.019x(60)$$
$$=-1.13 min/day$$
On july the hours of daylight is decreasing at a rate of 1.13 minutes per 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)

$$D'\left(x\right)=2.4\sin\left(\frac{2\pi\left(x+10\right)}{365}\right)\left(\frac{2\pi}{365}\right)$$
$$0=\sin\left(\frac{2\pi\left(x+10\right)}{365}\right)\left(\frac{4.8\pi}{365}\right)$$
$$\left(\frac{365}{4.8\pi}\right)0=\sin\left(\frac{2\pi\left(x+10\right)}{365}\right)\left(\frac{4.8\pi}{365}\right)\left(\frac{365}{4.8\pi}\right)$$
$$\left(\arcsin\right)0=\arcsin\cdot\sin\left(\frac{2\pi\left(x+10\right)}{365}\right)$$
The middle of the year on the unit circle is $\pi$ so:
$$\pi=\left(\frac{2\pi}{365}\right)\left(x+10\right)$$
$$\pi\cdot\frac{365}{2\pi}=\left(\frac{2\pi}{365}\right)\left(x+10\right)\left(\frac{365}{2\pi}\right)$$
$$\frac{365}{2}-10=x$$
$$x=172.5$$
So day 172.5 is going to be the longest day, which is June 21.

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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)
When the number of hours of daylight is increasing rapidly, one can find the day of the year ($D(x)$), by finding $D'(x)$ and comparing it to $0$.When $D'(x)>0$, then $D(x)$ is increasing. So when $D(x)$ is increasing then the slope is positive and one can look at a point from which the slope is at its greatest.

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