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)
We will be using a $cos$ function considering there is no daylight in the start of the year.
$D\left(x\right)=12.1-2.4\cos\left(\frac{2\pi\left(x+10\right)}{365}\right)$

Once we solve for the formula given above, we can input 200, and solve.

$D\left(200\right)=12.1-2.4\cos\left(\frac{2\pi\left(200+10\right)}{365}\right)$

$$D\left(200\right)=14.23606618$$

:::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 website states that in July 19, 2020 there will be approximetly 14.16 hours of daylight. Considering our calculations we were only about .076066 of a minutes off.

:::info
(4) Compute $D'(x)$. Show all work.
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(4)
Now to solve for any input value, we will substitute 200 for $x$.
$D\left(x\right)=12.1-2.4\cos\left(\frac{2\pi\left(x+10\right)}{365}\right)$

In this step we will be using the chain rule first.

$D'\left(x\right)=2.4\sin\left(\frac{2\pi\left(x+10\right)}{365}\right)\cdot\frac{d}{dx}\left[\frac{2\pi\left(x\cdot10\right)}{365}\right]$

In the second part, we see that we will need to use the quotient rule for the inner function.

$D'\left(x\right)=2.4\sin\left(\frac{2\pi\left(x+10\right)}{365}\right)\cdot\frac{\left(\frac{d}{dx}\left[2\pi x+20\pi\right]\cdot365-\left(2\pi x+20\pi\right)\cdot\frac{d}{dx}\left[365\right]\right)}{\left(365\right)^{2}}$

$D'\left(x\right)=2.4\sin\left(\frac{2\pi\left(x+10\right)}{365}\right)\cdot\frac{\left(2\pi\cdot365-\left(2\pi x+20\pi\right)\cdot0\right)}{\left(365\right)^{2}}$

Leaving us with this formula:

$$D'\left(x\right)=2.4\sin\left(\frac{2\pi\left(x+10\right)}{365}\right)\cdot\left(\frac{730\pi}{365^{2}}\right)$$

:::info
(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)
$D'\left(x\right)=2.4\sin\left(\frac{2\pi\left(x+10\right)}{365}\right)\cdot\left(\frac{730\pi}{365^{2}}\right)$

$D'\left(200\right)=2.4\sin\left(\frac{2\pi\left(200+10\right)}{365}\right)\cdot\left(\frac{730\pi}{365^{2}}\right)$

$D'\left(200\right)=-0.01883537\frac{hr}{day}$

$=-\frac{0.01883537hr}{1\ day}\cdot\ \frac{60\min}{1\ hr}=-\frac{1.1301222\min}{1\ day}$

According to our calculations, after July 19th of year 2020, daylight during the day will decrease at a rate of:
$$-1.1301222 ~ mins/day$$

:::info
(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)
To find where there is a derivative of zero in the year, we will just equal $D'(x)$ to $0$.

$D'\left(x\right)=2.4\sin\left(\frac{2\pi\left(x+10\right)}{365}\right)\cdot\left(\frac{730\pi}{365^{2}}\right)$

$0=2.4\sin\left(\frac{2\pi\left(x+10\right)}{365}\right)\cdot\left(\frac{730\pi}{365^{2}}\right)$

$0=2.4\sin\left(\frac{2\pi x+20\pi}{365}\right)$

$0=\sin\left(\frac{2\pi x+20\pi}{365}\right)$

Once we solve for sin, to solve for the exact radian(in the year this would be the date and time), we will use sin of inverse.

$\sin^{-1}\left(0\right)=\frac{2\pi x+20\pi}{365}$

After we are done with that step, we will simply solve for x as done below:

$365\sin^{-1}\left(0\right)=2\pi x+20\pi$

$365\sin^{-1}\left(0\right)-20=2\pi x$

$\frac{\left(365\sin^{-1}\left(0\right)-20\right)}{2\pi}=x$

$x=-10$
Taking 365/2=182.10
$$182.10 - 10 = 172.10$$
This coordinates to the point on the graph.
According to the website the longest day of the year is June 20, 2020 with 14.37:11 hours of daylight. June 20 is day 172.5 out of 365 on a leap year.

According to the website, because year 2020 is a leap year, the day with longest daylight is June 20th with about 14.37 hours; The x-axis in the graph will be 172.5 when there will be the most ddaylight.

:::info
(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)

Using the graph, we can look for a point in where we can pressume the slope is moving more rapidly, Estimating an exact x value. Another method is using the Average velocity formula between 2 points in the graph where we notice the slope is much more steeper than in other places.

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