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)
$f(x)=.05x^{2}$

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(b) Estimate the stopping distance for a car that is traveling 43 mi/h.
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(b)
The actual number I got was 92.45 ft. So estimating the stopping distance at 43 mi/h would be about 92 feet.

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(c\) Estimate the stopping distance for a car that is traveling 100 mi/h.
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(c)
500 ft would be the estimated stopping distance for a car traveling 100mi/h.

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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)
The central difference with the interval [40,50] is 4.35.
$f'(x)=\frac{1}{10}x$
$f'(45)=4.5$
Using the table's values and the estimated function I made based on the table, I found that the estimated value of $f'(45)$ is about 4.4. I noticed that when comparing my estimated function to the values on desmos, that my funciton has a slightly smaller width. Based on that I rounded up to be in between both the estimated function and the values in desmos.
The interpretation of this value means that the value of speed in mi/h at 45 feet is increasing at a rate of 4.4 ft per mi/h. $f'(x)$ is the rate of change or average of how many feet it takes to stop after going a certain speed in ft per mi/h after going x mi/h.

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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)
The value I got from finding the central difference on the interval [44,46] to estimate the value of $f'(45)$ is 4.5 ft per mi/h. This was the same number I got in the previous question by finding the derivative of the estimated function and then inputing 45 as x. I still rounded down from 4.5 to 4.4 because my estimated function is a little off from the function that we were given values from.

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

After using the limit definition of derivative, the value I got for $f'(45)$ is 4.5 ft per mi/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)
$g(1)= 0.4$
$g(2)=4$
$g(4)=33$
$g(5)=64$

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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)
https://www.desmos.com/calculator/hez8je6afc

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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\)
$g(x)=.515x^{3}-.25$

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(d) How many squares are in a $50\times50$ grid?
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(d)
There are 64375 squares in a 50x50 grid.

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(e) How many squares are in a $2000\times2000$ grid?
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(e)
There are 4,120,000,000 squares in a 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)
When using the central difference on the interval [5,3] I got 25. I also used the limit definition of a derivative ($f'(x)=\frac{309}{200}x^{2}$) and got 24.72 which rounds to 25. The interpretation of this value is that the number of squares inputed into the slope of a line connecting the points with the x values being 3 and 5, is 25. The rate of change of the tangent line with the x values being 3 and 5 is 25.

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