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)
Taking the base formula of $y_{1}\sim ax_{1}^{2}+bx_{1}+c$ and graphing. The shape attemps to match the data set which is a parabola. Points a,b, and c are determined to be:

$$a=0.047619$$$$b=0.0119048$$$$c=−0.0714286$$

By plugging this into the formula we get:
$y=\left(0.047619\right)x_{1}^{2}+\left(0.0119048\right)x_{1}-0.0714286$
where $x_{1}$ is the speed the car is going in mi/hr and the outcome of the problem y= is breaking distance in ft.

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(b) Estimate the stopping distance for a car that is traveling 43 mi/h.
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(b)
$y=\left(0.047619\right)\left(43\right)^{2}+\left(0.0119048\right)\left(43\right)-0.0714286$

$y=88.4880088 ft$

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(c\) Estimate the stopping distance for a car that is traveling 100 mi/h.
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(c.)
$y=\left(0.047619\right)\left(100\right)^{2}+\left(0.0119048\right)\left(100\right)-0.0714286$

$y=477.3090514 ft$

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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)
as in this problem $$y=\left(0.047619\right)\left(40\right)^{2}+\left(0.0119048\right)\left(40\right)-0.0714286$$
$$76.5951634$$
$$y=\left(0.047619\right)\left(50\right)^{2}+\left(0.0119048\right)\left(50\right)-0.0714286$$
$$96.8927624$$
then we take the difference of the two answers which both represent the number of ft needed to stop at their respective speed in mph.
$$119.5713114-76.5951634=42.976148$$
this answer represents the difference in ft extra that is needed when increasing from 40 mph to 50 mph. To calculate the amount of ft needed per each 1 mile extra from 40 mph to 50 mph, we must divide this answer by 10 because there is a 10 mph difference between 40 mph to 50 mph. 42.976148 ft per extra needed to break from 40 to 50 mph /10 mph or $$42.976148/10=4.2976148$$ we get 4.2976148 ft needed to break per 1 mile over 40 mph to 50 mph. This means we would multiple this answer by 5 to go from 40 mph to 45 mph the 5 is the difference between the two numbers. We now get $$4.2976148\cdot5=21.488074$$
This answer represents the increase in ft needed to break going from 40mph to 45 mph so we must add this number to the breaking distance given at 40 mph. $$76.5951634+21.488074=98.0832374$$

98.0832374 represents the estimated breaking distance in ft needed to stop when a car is moving 45 mph. Additionally if solve using central difference we must re-write the formula.

Once the formula is re-written as $f\left(x\right)=\left(0.047619\right)\left(x\right)^{2}+\left(0.0119048\right)\left(x\right)-0.0714286$
the x in f(x) can be used to plug in the x coordinate in this case we have an interval [40,50] and the question is asking for an estimate utilizing central difference. Solving the equation with a substitution of the number 50 and then the number 40 and then taking the final answer subtracting point a from point b (in this case 50 is point b and 40 is point a) this equals 10.

$$\frac{\left(f\left(50\right)-f\left(40\right)\right)}{50-40}$$
$$\frac{\left(120-76.5)\right)}{10}$$
$$=4.35 ft/mph$$

The estimation of f'(45) or the car at 45 miles/hr given the interval is =4.35 ft stopping distance extra per 1 extra mile this is the rate at the 45mph speed, although this is not a precise answer it is a good estimation, but could still be better if a closer set of points are chosen. so by using this formula we get the answer of 4.35 and then we must account for miles increased and all the rest of the variables.
$$(4.35*5)+76.5=98.25$$
This 98.25 is estimated the amount of ft needed to stop at 45 mph in a car and it is similar to the first estimation.

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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)
In this next section we are asked to use an interval with closer points to the desired f'(45). The points are [44,46]. When placed in desmos an even closer estimation is calculated.
$$\frac{\left(f\left(46\right)-f\left(44\right)\right)}{46-44}$$
$$\frac{\left(101.238005)-92.642775)\right)}{2}$$
$$=4.2976148 ft/mph$$
The estimation of f'(45) or the car at 45 miles/hr given the interval is =4.2976148 ft extra per 1 mph extra at 45 mph, although this is not a precise answer it is a good estimation, but could still be better if closer points are chosen.

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

$lim_{h→0\left(\left(\left(0.047619\right)\left(x+h\right)^{2}+\left(0.0119048\right)\left(x+h\right)-0.0714286\right)-\left(\left(0.047619\right)\left(x\right)^{2}+\left(0.0119048\right)\left(x\right)-0.0714286\right)\right)}/h$

Once all is simplified and then substituted at the end with h=0, derivative at f'(45) will equal 4.2976148 extra ft per every 1 mph extra

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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 because it is one singular square.
g(2)= 5 because a 2x2 grid is comprised up of four 1x1 squares and one 2x2 square.
g(3)= is given
g(4)= 30 because a 4x4 grid consists of sixteen 1x1 squares, nine 2x2 squares, four 3x3 squares and one 4x4 square, totaling to 30.
g(5)= 55 because there is one square that is 5x5, four squares that are 4x4, nine squares that are 3x3, sixteen 2x2 squares and twenty five 1x1 squares

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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\)
$y_{1}\sim ax_{1}^{3}+bx_{1}^{2}+cx_{1}+d$

$a=0.333333$
$b=0.5$
$c=0.166667$
$d=0$

Now plugging values back into the equation we get:

$$y_{1}\sim\left(0.333333\right)x_{1}^{3}+\left(0.5\right)x_{1}^{2}+\left(0.166667\right)x_{1}+(0)$$
or
$y=\left(n(n+1)(2n+1)\right)/6$


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(d) How many squares are in a $50\times50$ grid?
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(d)
$y=\left(n(n+1)(2n+1)\right)/6$
$y=\left((50)((50)+1))(2(50)+1)\right)/6$
$y=42925$

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(e) How many squares are in a $2000\times2000$ grid?
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(e)
$y=\left(n(n+1)(2n+1)\right)/6$
$y=\left((2000)((2000)+1))(2(2000)+1)\right)/6$
$y=2.668667×10^9$

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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 x=3 then y=14 and when x=5 then y=55
so then we can solve taking the difference of point b from point ain the form of slope. Slope is rise over run or y value/x value. So then:
$(y_2 -y_1)/(x_2 -x_1)$
$(55-14)/(5-3)$
$41/2$
$=20.5$
This answer is interpreted as such because they are the point before and after g'(4) which basically means the average rate of change. This means block counts are increasing in tandem with increases in grid size at a predictable rate. The value calculated for g'(4) suggests that when we increase from a grid size of 3x3 to a grid size of 4x4 then the amount of squares within the grid increase by 20.5 or over 1 along the x-axis (an increase in grid size by +1) and up 20.5 squares along the y-axis (the number of total countable squares increased).

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