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title: 'Math 181 Miniproject 2: Population and Dosage'

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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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:::info
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.
:::

(a)
| $t$    | 0    | 1   | 2   | 3 | 4 | 5 | 6 | 7 |
|--------|------|-----|-----|---|---|---|---|---|
| $P(t)$ | 1000 | 1100| 1210|1331|1464|1610 	|   1771|1948

*0.10 was multiplied by the population for each year then added to the total population to give the population for the following year. Decimals were rounded down to give a whole number of people per year. 



:::info
(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
\\]
:::

(b) Using Desmos to plot the given points and then approximate the function, P(t) was estimated to be:

$P(t)=1002.29\cdot(1.09976)^t-2.26115$




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(c\) What will the population be after 100 years under this model?
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(c\)
$P\left(100\right)=1002.29\cdot(1.09976)^{100}-2.26115=1.351x10^{7}$

After 100 years, the population with have $1.351x10^{7}$ people. 



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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 |
|---      |---|---|---|---|---|---|
| $P'(t)$ | 105 |115.5   |127   |139.5   |153.5   |169   |

$\frac{1210-1000}{2}=105$

$\frac{1331-1100}{2}=115.5$

$\frac{1464-1210}{2}=127$

$\frac{1610-1331}{2}=139.5$

$\frac{1771-1464}{2}=153.5$

$\frac{1948-1610}{2}=169$

The interpretation of $P'(5)$ is that the population is changing at approximately a rate of 153.5 people/year at the 5 year mark. 


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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)
$\frac{139.5-115.5}{2}=12\frac{people/year}{year}$

The interpretation of $P''(3)$ is that the rate of population change per year is changing at a rate of 12 $\frac{people/year}{year}$ (the rate of population growth is "speeding up" at this rate.).

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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)
$k$ could be solved for algebraically given that $P'(t)$ was said to be a multiple of $P(t)$ by $k$.

Given: $P'(t)=k(P(t))$

Year 1: $1100\cdot k=105$

$k=\frac{105}{1100}=0.0954$



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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 $D(x)=ax^2+bx+c$.)
:::

(a) After using Desmos to approximate a function for the points provided, D(x) was approximately defined to be:

$D\left(x\right)=0.025x^2-0.5x+10$



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(b) Find the proper dosage for a 128 lb individual.
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(b)
$D\left(128\right)=0.025(128)^2-0.5(128)+10=355.6$

For a 128 Ib person, they would receive a dosage of 355.6 mg.



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(c\) What is the interpretation of the value $D'(128)$. 
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(c\) The interpretation of $D'(128)$ is that this is the rate at which the drug dosage in mg is changing per Ib of body weight, $\frac{mg}{Ib}$. 

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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)
$D'(x)=\lim_{h \to 0}\frac{D(x+h)-D(x)}{h}$

$=\lim_{h \to 0}\frac{(0.025(x+h)^2-0.5(x+h)+10)-(0.025(x)^2-0.5(x)+10)}{h}$

$=\lim_{h \to 0}\frac{(0.025x^2+0.05xh-0.025h^2-0.5x-0.5h+10-0.025x^2+0.5x-10}{h}$

$=\lim_{h \to 0}\frac{h(0.05x+0.025h-0.5)}{h}$

$D'(x)=0.05x-0.5$

$D'(128)=0.05(128)-0.5=5.9 \frac{mg}{Ib}$

Taking the limit as h (the change in interval) approaches 0 of D(x), allows you to find the slope at a specific point. 


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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 of the tangent line at x=130 Ibs in point-slope form would be:
$y=367.5+6(x-130)$


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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)
$y=367.5+6((128)-130)=355.5 mg$

This is a mostly accurate estimation of the medication (in mg) required for someone who is 128 Ibs. Since the tangent line approximation was used for a point relatively close to 128 Ibs (130 was used) the approximation resulted in a mass of medication that was close to the mass calculated using the function D(x) approximated by Desmos (Part A):

$D(128)=0.025(128)^2-0.5(128)+10=355.6 mg$



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