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)
$y=0.047619x^2+0.0119048x-0.0714286$

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(b) Estimate the stopping distance for a car that is traveling 43 mi/h.
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(b)
y=0.047619$(43 mi/hr)^2$+0.0119048$(43 mi/hr)$-0.0714286
y= 88.4880088 ft.

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(c\) Estimate the stopping distance for a car that is traveling 100 mi/h.
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(c\)
y=0.047619$(100 mi/hr)^2$+0.0119048$(100 mi/hr)$-0.0714286
y= 477.3090514 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)$f'(45)$= $\frac{f(50)-f(40)}{50-40}$ = $\frac{(120)-(76.5)}{10}$ = 4.35 $\frac{ft}{\frac{mi}{hr}}$
The interpretation of this value is a positive value since the interval $[40,50]$ to estimate 45 is positive and increasing on the graph, so I would imagine the value of $f'(45)$ is also positive.

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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)
$f'(45)$= $\frac{f(46)-f(44)}{46-44}$ = $\frac{(101.2379962)-(92.6427666)}{2}$
=4.2976148 $\frac{ft}{\frac{mi}{hr}}$

The value of $f'(45)$ found with the interval $[44,46]$ ended up being $f'(45)$=4.2976148, which is the same value as previous estimate in part d

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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. $\lim_{h \to 0}\frac{f(45+h)-f(45)}{h}$

$\lim_{h \to 0}\frac{[(0.047619(45+h)^2+0.0119048(45+h)-0.0714286]-[0.047619^2(45)+0.0119048(45)-0.07144286])}{h}$
$\lim_{h \to 0}\frac{[(4.2857h+0.047619h^2+0.0119048h])}{h}$
$\lim_{h \to 0}\frac{[h(4.2857+0.047619h+0.0119048])}{h}$
= $4.297659$$\frac{ft}{\frac{mi}{hr}}$
After using the limit definition formula the final answer I got was $4.297659$, This answer is only slightly higher than the previous estimated values, over all very very close to the exact number. In conclusion $f'(45)$= 4.297659

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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)
$g(1)$= 1
$g(2)$= 1+4=5
$g(3)$= 1+4+9=14
$g(4)$= 1+4+9+16= 30
$g(5)$= 1+4+9+16+25= 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)
<iframe src="https://www.desmos.com/calculator/quylnkp5dd?embed" width="500px" height="500px" style="border: 1px solid #ccc" frameborder=0></iframe>

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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\)
$g(x)= 0.33x^3+0.5x^2+0.167x+0$

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(d) How many squares are in a $50\times50$ grid?
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(d)
$g(50)= 0.333333(50)^3+0.5(50)^2+0.166667(50)+0$
$g(50)= 42924.95835$
$g(50)~42925$ (Rounded to a full square)
In a $50\times50$ grid there will be 42925 squares

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(e) How many squares are in a $2000\times2000$ grid?
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(e)
$g(2000)= 0.33(2000)^3+0.5(2000)^2+0.167(2000)+0$
$g(2000)= 2.6686643333\times10^9$
Without scientific notation the answer would be 2,668,664,333.3 If we were to round to a full square the answer would now be $2,668,664,333.0$ which is how many squares are in a grid of $2000\times2000$

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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
$f'(4)$= $\frac{f(5)-f(3)}{5-3}$ = $\frac{(55)-(14)}{2}$ $= 20.5$$\frac{squares}{grid}$
Since this value is positive, it can be interpreted that $f(4)$ is a positive value that is increasing.

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