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|1771.56|1948.72|

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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)
$P\left(t\right)=1000\cdot\left(1.1^{x}\right)+\left(1.25\cdot10^{-12}\right)$

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(c\) What will the population be after 100 years under this model?
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(c\)
$P\left(t\right)=1000\cdot\left(1.1^{10}\right)+\left(1.25\cdot10^{-12}\right)=2593.74$

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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|127.05|139.76|153.73|169.10|

$P'(1)=\frac{\left(1000\cdot\left(1.1^{2}\right)+\left(1.25\cdot10^{-12}\right)\right)-\left(1000\cdot\left(1.1^{0}\right)+\left(1.25\cdot10^{-12}\right)\right)}{2}=105$
$P'(2)=\frac{\left(1000\cdot\left(1.1^{3}\right)+\left(1.25\cdot10^{-12}\right)\right)-\left(1000\cdot\left(1.1^{1}\right)+\left(1.25\cdot10^{-12}\right)\right)}{2}=115.5$
$P'(3)=\frac{\left(1000\cdot\left(1.1^{4}\right)+\left(1.25\cdot10^{-12}\right)\right)-\left(1000\cdot\left(1.1^{2}\right)+\left(1.25\cdot10^{-12}\right)\right)}{2}=127.05$
$P'(4)=\frac{\left(1000\cdot\left(1.1^{5}\right)+\left(1.25\cdot10^{-12}\right)\right)-\left(1000\cdot\left(1.1^{3}\right)+\left(1.25\cdot10^{-12}\right)\right)}{2}=139.76$
$P'(5)=\frac{\left(1000\cdot\left(1.1^{6}\right)+\left(1.25\cdot10^{-12}\right)\right)-\left(1000\cdot\left(1.1^{4}\right)+\left(1.25\cdot10^{-12}\right)\right)}{2}=153.73$
$P'(6)=\frac{\left(1000\cdot\left(1.1^{7}\right)+\left(1.25\cdot10^{-12}\right)\right)-\left(1000\cdot\left(1.1^{5}\right)+\left(1.25\cdot10^{-12}\right)\right)}{2}=169.10$

The population is changing by 153.73 people per year at 5 years time.

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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)
$P''(3)=\frac{P'\left(4\right)-P'\left(2\right)}{h}$
$=\frac{139.76-115.5}{\left(2\right)}$
$=12.13$ people/year per year
The derivative is changing by 12.13 people per year at 3 years time.

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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)
$105=k(1100)$
$k=\frac{105}{1100}$
$=0.09545$

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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)
$D(x)=0.025x^{2}-0.5x+10$

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

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(c\) What is the interpretation of the value $D'(128)$.
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(c\)
$D'(128)$ is the recommended drug dosage in mg based on a person weighing $128 lb$.

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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)
$D'(128)=\frac{D\left(120\right)-D\left(128\right)}{120-128}$
$=\frac{310-\left(0.025\left(128^{2}\right)-0.5\left(128\right)+10\right)}{120-128}$
$=5.7 mg/lb$
The backward difference was used to estimate the value of $D'(128)$ because 120 and 128 are close to each other compared to 128 and 140. The difference from 128 to 120 and 140 are not the same so the central difference was not used.

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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)
$0.025\left(130^{2}\right)-0.5\left(130\right)+10=367.5$

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

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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)
$D(128)=367.5+6\left(128-130\right)=355.5 mg$
Yes, the output value on the tangent line for $x=128 lbs$ does give a good estimate for the dosage for a $128 lbs$ indiviudal.

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