Math 181 Miniproject 2: Population and Dosage.md
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Math 181 Miniproject 2: Population and Dosage
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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--1.6 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\. A settlement starts out with a population of 1000. Each year the population increases by $10\%$. Let $P(t)$ be the function that gives the population in the settlement after $t$ years.
(a) Find the missing values in the table below.
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(a)
| $t$ | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
|--------|------|---|---|---|---|---|---|---|
| $P(t)$ | 1000 | 1100 | 1210 | 1331 | 1464.1| 1610.51 | 1771.561 |1948.7171 |

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(b) Find a formula for $P(t)$. You can reason it out directly or you can have Desmos find it for you by creating the table of values above (using $x_1$ and $y_1$ as the column labels) and noting that the exponential growth of the data should be modeled using an exponential model of the form
\\[
y_1\sim a\cdot b^{x_1}+c
\\]
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(b) By inputting the data into desmos with the following exponential model:
\\[
y_1\sim a\cdot b^{x_1}+c
\\]
The following when typed into desmos, would give you values for a, b, and c. Which looks like this.
\\[
P(t)=1000\cdot1.1^{100}+\left(1.2866\cdot10^{-12}\right)
\\]

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(c\) What will the population be after 100 years under this model?
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(c\)The population after 100 years using the previous model is
\\[P(t)=1000\cdot1.1^{100}+\left(1.2866\cdot10^{-12}\right)
\\]
This would get you $P(t)=13,780,612.3398$

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(d) Use a central difference to estimate the values of $P'(t)$ in the table below. What is the interpretation of the value $P'(5)$?
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(d)
| $t$ | 1 | 2 | 3 | 4 | 5 | 6 |
|--- |---|---|---|---|---|---|
| $P'(t)$ | 100| 110 | 121 | 133.1 | 146.41|161.051 |

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(e) Use a central difference to estimate the values of $P''(3)$. What is the interpretation of this value?
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(e) Using the central difference for $P''(3)$ which is equal to
$P''\left(3\right)=\frac{133.1-110}{2\ }=\frac{23.1}{2}=11.55$ people per year per year.
This means that the population slope is positive and increases by approximately 11.55 people per year per year

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(f) **Cool Fact:** There is a constant $k$ such that $P'(t)=k\cdot P(t)$. In other words, $P$ and $P'$ are multiples of each other.

What is the value of $k$? (You could try creating a slider and playing with the graphs or you can try an algebraic approach.)
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(f) The value of k varies depending on what values you plug into your equation.
$k=\frac{P'\left(t\right)}{P\left(t\right)}$

$k=\frac{P'\left(2\right)}{P\left(2\right)}$

$k=\frac{110}{1210}= 0.091$

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2\. The dosage recommendations for a certain drug are based on weight.
| Weight (lbs)| 20 | 40 | 60 | 80 | 100 | 120 | 140 | 160 | 180 |
|--- |--- |--- |--- |--- |--- |--- |--- |--- |--- |
| Dosage (mg) | 10 | 30 | 70 | 130 | 210 | 310 | 430 | 570 | 730 |
(a) Find a function D(x) that approximates the dosage when you input the weight of the individual. (Make a table in Desmos using $x_1$ and $y_1$ as the column labels and you will see that the points seem to form a parabola. Use Desmos to find a model of the form
\\[
y_1\sim ax_1^2+bx_1+c
\\]
and define $D(x)=ax^2+bx+c$.)
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(a) When inputting the table data and given model you get the values for a, b, and c which is \\[
y_1\sim 0.025x_1^2-0.5x_1+10
\\]

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(b) Find the proper dosage for a 128 lb individual.
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(b) Using the given equation and pluggin in 128 for x like so
$D(x)=0.025\left(128\right)^2-0.5\left(128\right)+10$
=$409.6-54$
=$355.6$ mg/lb

So the proper dosage for a 128 lb individual would be $355.6$ mg/lb.

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(c\) What is the interpretation of the value $D'(128)$.
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(c\) D'(128) just means the slope or increase for the dosage in mg of the drug for a 128lb person.

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(d) Estimate the value of $D'(128)$ using viable techniques from our calculus class. Be sure to explain how you came up with your estimate.
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(d) To determine $D'(128)$ you would need the formula from getting the derivative. The input 128 as x.
$D'(128)=0.05(128)-0.5=5.9 mg/lb$

This describes that the dosage would increase by 5.9 milligram per pound.

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(e) Given the value $D'(130)=6$, find an equation of the tangent line to the curve $y=D(x)$ at the point where $x=130$ lbs.
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(e)The equation for the tangent line would be $L(x)$. You would get these values when looking at the charts, using central difference, and plugging in $x=130 lbs$
$L(x)=D(x)+D'(x)(x-a)$

$L(x)=367.5+6(x-130)$

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(f) Find the point on the tangent line in the previous part that has $x$-coordinate $x=128$. Does the output value on the tangent line for $x=128$ lbs give a good estimate for the dosage for a 128 lb individual?
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(f) Using the equation from the previous step and inputting 128 as your x. It is conclusive that the dosage for a 128 lb individual is accurate with examination of the values on the chart.
$L(x)=367.5+6(x-130)$

$L(x)=367.5+6(128-130)=355.5 mg/lb$

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