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 | 1330 | 1463 | 1609 | 1771 | 1948 |

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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)
$y=984.672\ \cdot\ 1.10111^x+15.6151$

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(c\) What will the population be after 100 years under this model?
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(c\)
$y=984.672\ \cdot\ 1.10111^{100}+15.6151$
$y=15,009,379$ population
Under this model, after 100 years, the population will be at 15,009,379.

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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 | 126.5 | 139.5 | 154 | 169.5 |
$P'\left(1\right)=\frac{\left(P\left(2\right)-P\left(0\right)\right)}{2-0}=\frac{\left(1210-1000\right)}{2}=105$ population per year
$P'\left(2\right)=\frac{\left(P\left(3\right)-P\left(1\right)\right)}{3-1}=\frac{\left(1330-1100\right)}{2}=115$ population per year
$P'\left(3\right)=\frac{\left(P\left(4\right)-P\left(2\right)\right)}{4-2}=\frac{\left(1463-1210\right)}{2}=126.5$ population per year
$P'\left(4\right)=\frac{\left(P\left(5\right)-P\left(3\right)\right)}{5-3}=\frac{\left(1609-1330\right)}{2}=139.5$ population per year
$P'\left(5\right)=\frac{\left(P\left(6\right)-P\left(4\right)\right)}{6-4}=\frac{\left(1771-1463\right)}{2}=154$ population per year
$P'\left(6\right)=\frac{\left(P\left(7\right)-P\left(5\right)\right)}{7-5}=\frac{\left(1948-1609\right)}{2}=169.5$ population per year
At 5 years, the population is increasing by 154 population per year.

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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)
$P"\left(3\right)=\frac{\left(P'\left(4\right)-P\left(2\right)\right)}{4-2}=\frac{\left(139.5-115\right)}{2}=12.25$ population/year per year
After 3 years, the population is increasing at a rate of 12.25 population a 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)
$P'\left(t\right)\ =\ k\ \cdot\ P\left(t\right)$
$k=\frac{P'\left(t\right)}{P\left(t\right)}$
$k=\frac{P'\left(1\right)}{P\left(1\right)}=\frac{105}{1100}=0.095$
$k=\frac{P'\left(2\right)}{P\left(2\right)}=\frac{115}{1210}=0.095$
$k=\frac{P'\left(3\right)}{P\left(3\right)}=\frac{126.5}{1330}=0.095$
$k=\frac{P'\left(4\right)}{P\left(4\right)}=\frac{139.5}{1463}=0.095$
$k=\frac{P'\left(5\right)}{P\left(5\right)}=\frac{154}{1609}=0.096$
$k=\frac{P'\left(6\right)}{P\left(6\right)}=\frac{169.5}{1771}=0.096$
By taking the average, the value of k is at 0.095.

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

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

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(c\) What is the interpretation of the value $D'(128)$.
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(c\)
As the weight of an individual increases, the dosage also increases per pound of weight.

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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 estimate D'(128) we could use the limit definition.
$D'(128)=\lim_{h \to 0}\frac{f(128+h)-f(128)}{h}$
$=\lim_{h \to 0}\frac{[0.025(128+h)^2-0.5(128+h)+10]-[0.025(128)^2-0.5(128)+10]}{h}$
$=\lim_{h \to 0}\frac{[0.025(16384 + 256h + h^2)-(64-0.5h)+10]-[0.025(16384)-64+10]}{h}$
$=\lim_{h \to 0}\frac{[409.6 + 6.4h + 0.025 h^2-54-0.5h]-[409.6-54]}{h}$
$=\lim_{h \to 0}\frac{5.9h + 0.025h^2}{h}$
$=\lim_{h \to 0}\frac{h(5.9 + 0.025h)}{h}$
$=\lim_{h \to 0}5.9 + 0.025h$
$=\lim_{h \to 0}5.9 + 0.025(0)$
$=5.925 mg/lbs$
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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)
Use the linear localization formula $L(x)=D(a)+D'(a)(x-a)$
$L(x)=D(130)+D'(130)(x-130)$
$L(x)=367.5+6(x-130)$

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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)
$L(128)=367.5+6(128-130)$
$L(128)=367.5+6(-2)$
$L(128)=367.5-12$
$L(128)=355.5$

By solving the value tangent line for x=128 lbs, it gives a point at 355.5 mg. The proper dosage that was solved was at 355.6 mg. Comparing the two values, it provides a good estimate for the dosage of a 128 lb individual.

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