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)
$$y=\left(.047619\right)x^{2}+\left(.0119048\right)x+\left(-.0714286\right)$$

$$Where:$$
$$a=.047619$$
$$b=.0119048$$
$$c=-.0714286$$

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(b) Estimate the stopping distance for a car that is traveling 43 mi/h.
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(b)
Using the same Quadrative formula as in step a. We will be substituting $x$ with $43$ and compute:
$$y=\left(a\right)43^{2}+\left(b\right)43+\left(c\right)$$
The Output value when the input ($x$) is 43 will equal $88.488$:
$$f(43)=88.488$$

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(c\) Estimate the stopping distance for a car that is traveling 100 mi/h.
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(c\)
Using the same quadrative formula as in step a. We will be substituting $x$ with $100$ and compute:
$$y=\left(a\right)100^{2}+\left(b\right)100+\left(c\right)$$
The Output value when $x$ will equal $477.309$:

$$f(100)=477.309$$

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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)
Step 1:
$$\frac{f(b)-
f(a)}{b-a}$$

Step 2:
a. Subsitute $f(a)$ for the quadratic equation provided above (in part (a) of question 1)while $x$ equals 40.
b. Subsitute $f(b)$ for the quadratic equation provided above while $x$ equals 50.
c.b-a in the denominater is: $40-50$
$$\frac{[\left(a\right)\left(50\right)^{2}+\left(b\right)\left(50\right)+\left(c\right)]-
[\left(a\right)\left(40\right)^{2}+\left(b\right)\left(40\right)+\left(c\right)]}{50-40}$$

Step 3:
Once you have substituted all the variables for the values of a, b, and c [into the Average Velocity formula], your values should apear as:
$$\frac{
120-76.5}{10}$$

Step 4:
The Average distance of stop when a car is traveling between intervals 40 and 50 is an average of : $$4.35 mi/ft$$

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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)
Step 1.
a. Solve for $f(44)$ by using the quadratic formula provided for us above. $$y=\left(.047619\right)x^{2}+\left(.0119048\right)x+\left(-.0714286\right)$$
b.Solve for $f(44)$ by substitute $x$ for 44:
$$y=\left(.047619\right)(44)^{2}+\left(.0119048\right)(44)+\left(-.0714286\right)$$ which will equal $$92.6431$$
Step 2.
a. Solve for $f(46)$ by using the quadratic formula provided for us above. $$y=\left(.047619\right)x^{2}+\left(.0119048\right)x+\left(-.0714286\right)$$
b.Solve for $f(46)$ by substitute $x$ for 46:
$$y=\left(.047619\right)(46)^{2}+\left(.0119048\right)(46)+\left(-.0714286\right)$$ which will equal $$101.2383$$

Step 3.
a.) Subsitute $f(a)$ for the quadratic equation provided above while $x$ equals 44.
b.) Subsitute $f(b)$ for the quadratic equation provided above while $x$ equals 46.
c.) b-a in the denominater is: $46-44$
$$\frac{[\left(a\right)\left(46\right)^{2}+\left(b\right)\left(46\right)+\left(c\right)]-
[\left(a\right)\left(44\right)^{2}+\left(b\right)\left(44\right)+\left(c\right)]}{46-44}$$

d.) Once you have substituted all the variables into the Average Velocity formula, your values should apear as:
$$\frac{101.2383-92.643}{2}$$

f.) The value for the Average Velocity of the stopping distance of a car traveling between the intervals of 44 and 46 will be: $$4.29 {ft}/{mph}$$
This value of the average velocity from more exact intervals $[44,46]$ give a more precise value of $f'(45)$ which change the value by .6 from the original $4.35$.

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(f) Find the exact value of $f'(45)$ using the limit definition of derivative.
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f.) The value for the Average Velocity between interval 44 and 46 will be: $$4.29$$
This value of the average velocity from more exact intervals $[44,46]$ give a more precise value of $f'(45)$ which change the value by .6 from the original $4.35$.

For the equation with numbers:
$\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}$
$\lim_{h \to 0}\frac{[.047619(45+h)^2+.0119048(45+h)+(-.0714286)]-[.047619(45)^2-.0119048(45)+(-.0714286)]}{h}$
Solving the equation above resulted in:
$$4.297 {ft}/{mph}$$

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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)
The values of the f(x) inputs are found below
$$g(1)=1$$
$$g(2)=5$$
$$g(4)=30$$
$$g(5)=55$$

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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)
The Graph below represents the number of squares (of any size) there are when we zoom out from 1x1 squares to the 50x50 square grid.

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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\)
Once we applied the formula above unto desmos
\\[
y_1\sim ax_1^3+bx_1^2+cx_1+d
\\], and inputed the values for y(1) and x(1) we were given the approximate values for $a$,$b$,$c$, and $d$. Where:
$$a=.333333$$
$$b=.5$$
$$c=.166667$$
$$d=0$$
To have the quadratic formula with the values given to be:
$$f\left(x\right)=\left(.333333\right)x^{3}+\left(.5\right)x^{2}+\left(.166667\right)x$$

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(d) How many squares are in a $50\times50$ grid?
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(d)
Computing the equation when the value of x equals 50 our answer should be

approximatly $42925$ squares.

Our equation when we have a $50x50$ grid should be:
$$f\left(50\right)=\left(.333333\right)50^{3}+\left(.5\right)50^{2}+\left(.166667\right)50$$

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(e) How many squares are in a $2000\times2000$ grid?
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(e)
Similar to part (d) of question number 2, our equation should be:

$$f\left(2000\right)=\left(.333333\right)2000^{3}+\left(.5\right)2000^{2}+\left(.166667\right)2000$$

There are Approx. $2.6686643333×109$ squares in this 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)
The formula to find the central difference is the same as the Average Velocity formula.
$$ AV_[{3,5}_]\frac{f\left(b\right)-f\left(a\right)}{b-a}$$

Once we have the formula f(b) should be replaced by 55 and f(a) for f(3).

$$ AV_[{3,5}_]\frac{f\left(5\right)-f\left(3\right)}{5-3}$$

Because we already know the value for the output of $f(3)$ and $f(5)$ (we already solved this in part a of question 2), we can just plug in the outputs instead of $f(x)$.
$$ AV_[{3,5}_]\frac{55-14}{5-3}$$
Which gives an approximate of:
$$g'(4)=20.5 Squares/Length~of~Grid$$

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