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)
After creating a talbe and using the given the template $y_1\sim\ ax_1^2+bx_1+c$ I was give A= 0.047619, B= 0.0119048, and C=-0.714286.

With this, I was able to create a quadratic formula for f(x)
$f\left(x\right)=0.047619\left(x\right)^2+0.0119048\left(x\right)-0.714286$

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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 formula f(x), we can find the stopping distance of a car moving at 43mi/h by finding f(43)

$f(43)= 0.047619(43)^2+0.0119048(43)-0.714286$
$=87.845$

A car moving at 43 miles per hour will need about 88 feet of stoping distance.

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(c\) Estimate the stopping distance for a car that is traveling 100 mi/h.
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Using the formula f(x), we can find the stopping distance of a car moving at 100mi/h by finding f(100).

$f(100)= 0.047619(100)^2+0.0119048(100)-0.714286$
$=476.666$

A car moving at 100 miles per hour will need a stoping distance of about 477 feet.

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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)
We can estimate the value of f'(45) unsing the central differnece formula:
$f'(x)= (f(b)-f(a))/(b-a)$

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

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

$f'(45)=4.35$

At the 45 mile per hour mark, rate at which the car is sotping is 4.35 feet/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)
First finding f(44) and f(46) I found:
$f(44)=91.999$
$f(46)=100.595$
Next, using the central differnece formula again, I estimate f'(45):

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

$f'\left(45\right)=\frac{100.595-91.999}{2}$

$f'(45)=4.30$

I found that this estimate and the estimate made in the previous section are both very similar with only a difference of .05.

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(f) Find the exact value of $f'(45)$ using the limit definition of derivative.
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$f'\left(x\right)=limh\to 0 =\frac{\left(0.047619\left(x+h\right)^2+0.0119048\left(x+h\right)-0.714286\right)-\left(0.047619\left(x\right)^2+0.0119048\left(x\right)-0.714286\right)}{h}$

$f'\left(x\right)=limh\to 0=\frac{0.095238xh+0.047619h^2+0.0119048h}{h}$

$f'\left(x\right)=\ 0.095238x+0.0119048$

$f'\left(45\right)=0.095238\left(45\right)+0.0119048$

$f'(45)= 4.2976148$

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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)1 = 1
g(2) = 1+4 = 5
g(3) = 1+4+9 = 14
g(4) = 1+4+9+16 = 30
g(5) = 1+4+9+25 = 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\) With the given cubic function, I was able to find that
$a= 0.333333$
$b= 0.5$
$c= 0.166667$
$d= -3.5644\left(10\right)^{-15}$

The exact formual for $g\left(x\right)=\ 0.3333\left(x\right)^3+0.5\left(x\right)^2+0.166667\left(x\right)+\left(-3.5644\left(10\right)^{-15}\right)$

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(d) How many squares are in a $50\times50$ grid?
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(d) Now with a formula for g(x), we can find the amount of square in a 50x50 grid by finding g(50).

$g\left(50\right)=\ 0.3333\left(50\right)^3+0.5\left(50\right)^2+0.166667\left(50\right)+\left(-3.5644\left(10\right)^{-15}\right)$

$g(50)=42920$ squares

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

Using the formula for g(x) we can find g(2000)

$g\left(2000\right)=\ 0.3333\left(2000\right)^3+0.5\left(2000\right)^2+0.166667\left(2000\right)+\left(-3.5644\left(10\right)^{-15}\right)$

$g(2000)= 2.6684003333(10)^9$ 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)

$g'\left(4\right)=\ \frac{f\left(5\right)-f\left(3\right)}{5-3}$

$g'\left(4\right)=\ \frac{55-14}{2}$

$g'\left(4\right)=20.5$

At the exactly the 4x4 mark, the rate at which the sqaures are increasing is 20.5 squares per grid length.

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