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) By inputting the data into desmos with the following quadratic model, $y_1\sim ax_1^2+bx_1+c$. The suggested values for a, b, and c is $f(x)=0.047619x^2+0.0119048x-0.0714286$

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(b) Estimate the stopping distance for a car that is traveling 43 mi/h.
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(b) The stopping distance for a car traveling at 43 mi/h is $88.488 ft$. This can be found by plugging in 43 as your x in the f(x) equation.
$f(43)=0.047619(43)^2+0.0119048(43)-0.0714286$

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(c\) Estimate the stopping distance for a car that is traveling 100 mi/h.
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(c)The stopping distance for a car traveling at 100 mi/h is $477.309 ft$. Similar to part b) you would just plug in 100 for your x in the f(x) equation. $f(100)=0.047619(100)^2+0.0119048(100)-0.0714286$

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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) Use the information from the graph for intervals [40,50] and the central difference to estimate the value of f'(45).
$f'\left(45\right)=\frac{120-76.5}{10}$

which would make the value of $f'(45)= 4.35$ ft per mile per hour

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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)To find the exact value of $f'(45)$ using the interval [44,46] and the central difference you would get $f'(45)=4.3$
First by using the formula from a) and replacing x with the interval numbers 44 and 46. $f(x)=0.047619(x)^2+0.0119048(x)-0.0714286$

which makes $f(44)=92.64$ and $f(46)=101.24$

Plug that into the central difference.
$f'\left(45\right)=\frac{101.24-92.64}{2}$

This makes $f'(45)$ which is equivalent to our previous answer.
$f'(45)=4.3$ ft per mile per hour

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(f) Find the exact value of $f'(45)$ using the limit definition of derivative.
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(f) To find the limit derivative of $f'(45)$ you would use
$\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}$

which would look like this.

$\lim_{h \to 0}\frac{f(45+h)-f(45)}{h}=4.30$ ft per mile per hour

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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)
$g(1)=1$
$g(2)=5$
$g(3)=14$
$g(4)=30$
$g(5)=55$

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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\)
$g(x)=0.333333x^3 +0.5x^2 +0.166667x +0$

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(d) How many squares are in a $50\times50$ grid?
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(d) By using our equation from part c) and inputting 50 in for x the answer to how many squares are in a $50 x 50$ grid can be found.
$g(50)= 0.333333(50)^3 +0.5(50)^2 +0.166667(50)+0= 42,924.96$ squares.

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(e) How many squares are in a $2000\times2000$ grid?
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(e) Similar to our previous question, you would use the equation from part c) and plug in 2000 for x to find how many squares are in a $2000 x 2000$ grid.
$g(2000)=0.333333(2000)^3 +0.5(2000)^2 +0.16667(2000)+0= 2,668,664,333$ squares.

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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)
When using the central difference formula.
$g'\left(4\right)=\frac{g\left(5\right)-g\left(3\right)}{5-3}$
You would then input the numbers from the graph for g(5) and g(3).
$g'\left(4\right)=\frac{55-14}{2}$
Which would make $g'(4)= 20.5$

This means that the derivative/slope of g(4) is 20.5, from that point.

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