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 labeled $x_1$ 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 quadratic 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)**

Using a quadratic model for the function f(x) in order to produce corresponding output values when are associated with their corrosponding speed values will calculate the distance a vehicle travels once the brakes have been applied.

**a = 0.04762
b = 0.011905
c = -0.071429**

The quadratic formula used to calculate respective output values to input values is:
**f(x):(0.04762)$x^2$+(0.011905)x-0.071429**

*when graphed, this creates a parabola:*

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(B) Estimate the stopping distance for a car that is traveling 43 mi/h.
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**(B)**

f(x)=(0.04762)x$^2$ + (0.011905)x - 0.071429

*(replace with 43mi/hr)*

f(x)=(0.04762)(43)$^2$ + (0.011905)(43) - 0.071429

**=88.489 ft**

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( C ) Estimate the stopping distance for a car that is traveling 100 mi/h.
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**( C )**

f(x)=(0.04762)x$^2$ + (0.011905)x - 0.071429

*(replace with 100mi/hr)*

f(x)=(0.04762)(100)$^2$ + (0.011905)(100) - 0.071429

**=477.319 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)**

**Interval [40, 50]
h = 5**

Using formula: f(x) = $\frac{f(x+h)-f(x-h)}{2h}$

--> *plug in 45 for x, and 5 for h*

f'(45) = $\frac{f(45+5)-f(45-5)}{10}$

f'(45) = $\frac{f(50)-f(40)}{10}$

f'(45) = $\frac{120-76.5}{10}$

**f'(45) $\approx$ 4.35 ft/mi/hr**

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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)**

**Interval [44, 46]**

**When x = 44**
**f(44)** = (0.04762)(44)$^2$ + (0.011905)(44) - 0.071429
**f(44) = 92.6447 ft**

**When x = 46**
**f(46)** = (0.04762)(46)$^2$ + (0.011905)(46) - 0.071429
**f(46) = 101.2401 ft**

*--> using central line differance formula:*

f(x) = $\frac{f(x+h)-f(x-h)}{2h}$

f'(45) =$\frac{f(45+1)-f(45-1)}{2(1)}$

f'(45) =$\frac{f(46)-f(44)}{2}$

f'(45) = $\frac{101.2401 - 92.6447}{2}$

**f'(45) $\approx$ 4.29771 ft/mi/hr**

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(F) Find the exact value of $f'(45)$ using the limit definition of derivative.
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**(F)**

f(n) = (0.04762)(2n) + (0.011905)(1) - 0

f(n) = (0.04762)n + (0.011905)

**When n = 45**

f'(45) = (0.04762)(45) + (0.011905)

**f'(45) = 4.297705 ft/mi/hr**

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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(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 9x9, 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)**

*input and output values of g(x)*

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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\)**


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(d) How many squares are in a $50\times50$ grid?
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**(d)**

$\sum$n$^2$ = $\frac{n(n+1)(2n+1)}{6}$

--> *plug in 50 for n*

$\frac{50(51)(101)}{6}$

**= 257550 squares**

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(e) How many squares are in a $2000\times2000$ grid?
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**(e)**

$\sum$n$^2$ = $\frac{n(n+1)(2n+1)}{6}$

--> *plug in 2000 for n*

$\frac{2000(2001)(4001)}{6}$

**=266866700 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)**

g'(4) = AV[3, 5] = $\frac{g(5) - g(3)}{(5 - 3)}$ = $\frac{53.75 - 13.41}{2}$ = $\frac{40.34}{2}$
**g'(4) = 20.17**

The value of g'(4) represents the the increased quantity of squares (20.17) upon exceeding the dimensions 4 x 4.

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