Math 181 Miniproject 1: Modeling and Calculus.md
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Math 181 Miniproject 1: Modeling and Calculus
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**Overview:** In this miniproject you will use technological tools to turn data and into models of real-world quantitative phenomena, then apply the principles of the derivative to them to extract information about how the quantitative relationship changes.

**Prerequisites:** Sections 1.1--1.5 in *Active Calculus*, specifically the concept of the derivative and how to construct estimates of the derivative using forward, backward and central differences. Also basic knowledge of how to use Desmos.

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1\. The table below gives the distance that a car will travel after applying the brakes at a given speed.
| Speed (in mi/h) | Distance to stop (in ft) |
|----------------- |-------------------------- |
| 10 | 5 |
| 20 | 19 |
| 30 | 43 |
| 40 | 76.5 |
| 50 | 120 |
| 60 | 172 |
| 70 | 234 |

(a) Find a function $f(x)$ that outputs stopping distance when you input speed. This will just be an approximation. To obtain this function we will first make a table in Desmos. The columns should be labled $x_1$ and $y_1$. Note that the points are plotted nicely when you enter them into the table. Click on the wrench to change the scale of the graph to fit the data better. Since the graph has the shape of a parabola we hope to find a quadratric formula for $f(x)$. In a new cell in Desmos type
\\[
y_1\sim ax_1^2+bx_1+c
\\]
and let it come up with the best possible quadratic model. Use the suggested values of $a$, $b$, and $c$ to make a formula for $f(x)$.
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(a)
Work and answers go here. **Replace this paragraph of text with your solution.** You can use the editor buttons above to import images. You type fancy math input between dollar signs like $x^2$. You can get the code for the fancy math input by copying and pasting from a cell in Desmos.

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(b) Estimate the stopping distance for a car that is traveling 43 mi/h.
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(b) Since f(x) outputs the stopping distance at speed x, I just plugged in the 43 into f(x) to determine the stopping distance. Once I did that I determined that the stopping distance at 43 mi/h would be 88.4 ft.

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(c\) Estimate the stopping distance for a car that is traveling 100 mi/h.
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(c) Since f(x) outputs the stopping distance at speed x, I again just plugged in the value given which was 100 to determine the stopping distance. Once I put in that value for x I determined that the stopping distance at 100 mi/h would be 477.3 ft.

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(d) Use the interval $[40,50]$ and a central difference to estimate the value of $f'(45)$. What is the interpretation of this value?
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(d) For this question we would use the central difference formula, which is:$f'\left(x_{n}\right)\sim\frac{f\left(x_{n+1}\right)-f\left(x_{n-1}\right)}{2Δx}$
With this we will have to figure out our variables so we can solve.
$x_{n\ }=45$ $x_{n+1}=50$ $x_{n-1}=40$
With this we can figure out what $Δx$ is, it's just the change in our x value so the interval given:
$Δx = 50-45 =5$
And from here we have the variables needed to complete the equation:
**$f'\left(45\right)\sim\frac{f\left(50\right)-f\left(40\right)}{2\left(5\right)}$**
So just solving numerator first using our desmos graph so $f(50)=120$ and $f(40)=76.5$ and again we just put these values in our formula to get our final answer:
$\frac{120-76.5}{2\left(5\right)}= 4.35$ ft.

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(e) Use your function $f(x)$ on the interval $[44,46]$ and a central difference to estimate the value of $f'(45)$. How did this value compare to your estimate in the previous part?
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(e) For this question we will also use the central difference formula we used in part d, so we can start again by finding our variables for the equation:
$x_{n\ }=45$ $x_{n+1}=46$ $x_{n-1}=44$
Next we need to find $Δx$ so it would be: $Δx = 46-45 = 1$
With this we now have the variables needed to use the formula, so we just plug in our values:
$f'\left(45\right)\sim\frac{f\left(46\right)-f\left(44\right)}{2\left(1\right)}$
And we can simplify the $f(46)$ and $f(44)$ using the graph again to solve:
$\frac{101.24-92.64}{2} = 4.3$ ft.
We can see that the answer from part d and this question are extremely close

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(f) Find the exact value of $f'(45)$ using the limit definition of derivative.
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(f)Here's a sample of how to write a limit using LaTeX code. $\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}$

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2\. Suppose that we want to know the number of squares inside a $50\times50$ grid. It doesn't seem practical to try to count them all. Notice that the squares come in many sizes.

(a) Let $g(x)$ be the function that gives the number of squares in an $x\times x$ grid. Then $g(3)=14$ because there are $9+4+1=14$ squares in a $3\times 3$ grid as pictured below.
Find $g(1)$, $g(2)$, $g(4)$, and $g(5)$.
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(a) In order to find $g(1)$ we need to count up all the squares on the grid. However, since $g(1)$ is just a single square $g(1)=1$. To find $g(2)$ it's essentially the same premise but with a 2x2 grid. We can see that there are 5 squares in that so, $g(2)=5$. We already found that $g(3)=14$. Finally, $g(4)=30$ and $g(5)=55.

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(b) Enter the input and output values of $g(x)$ into a table in Desmos. Then adjust the window to display the plotted data. Include an image of the plot of the data (which be exported from Desmos using the share button ). Be sure to label your axes appropriately using the settings under the wrench icon .
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(b) 

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(c\) Use a cubic function to approximate the data by entering
\\[
y_1\sim ax_1^3+bx_1^2+cx_1+d
\\]
into a new cell of Desmos (assuming the columns are labeled $x_1$ and $y_1$). Find an exact formula for $g(x)$.
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(c\) The exact formula which we used Desmos to figure out is:
$f\left(x\right)=0x^{3}+3.64286x^{2}+\left(-8.35714\right)x+5.6$
Desmos gave us the variables:
$a=0$, $b=3.64286$, $c=-8.35714$, and $d=5.6$.

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(d) How many squares are in a $50\times50$ grid?
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(d) In order to determine how many squares are in a 50x50 grid we can simply put the 50 in the place of x in our formula we got from Desmos. What we get by using that formula is 8.695. It isn't a whole number but we can still use it and estimate that there are 8,695 squares in the 50x50 grid.

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(e) How many squares are in a $2000\times2000$ grid?
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(e) To determine how many squares are in a 2000x2000 grid, we would do the same thing that we did in part d. We would input 2000 in place of x in our formula from part c. So the total we get by using that equation is $14,554,731.32$. To round that and get our final answer we can say there are about $14,554,731$ squares in the 2000x2000 grid.

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(f) Use a central difference on an appropriate interval to estimate $g'(4)$. What is the interpretation of this value?
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(f) So for this question we are going to use the formula from part one d and e. We are trying to find $g'(4)$ so 4 is our x and h is 1. The only thing that really differs from the formula we used in part one and the one we're using here is that the function name is $g'$ not $f'$. We are going to use the other form of the central difference formula:
$g'\left(x\right)\sim\frac{g\left(x+h\right)-f\left(x-h\right)}{2h}$
We then plug these variables into the formula and solve:
$g'\left(4\right)\sim\frac{g\left(4+1\right)-g\left(4-1\right)}{2}\ =\frac{g\left(5\right)-g\left(3\right)\ }{2}=\frac{\left(55-13\right)}{2}=21$.

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