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)

$y=1000(1.1^x)+1.2757*EE-12$

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(c\) What will the population be after 100 years under this model?
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(c\)
After 100 years the population will be 13,780,612

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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)
$p'(5)=153.7305$ means the population after 5 years has an institanious growth rate of 153.7 people/year.
| $t$ | 1 | 2 | 3 | 4 | 5 | 6 |
|--- |---|---|---|---|---|---|
| $P'(t)$ |105|115.5|127.05|139.755|153.7305|169.10355 |

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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)
$P"(3)=12.1275$ is the instataneous rate after 3 years at which the population rate is growing 12 people 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)
$k=P'(t)/P(t)$
$k=.09545$

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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)

$y=.025x^2-.5x+10$

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(b) Find the proper dosage for a 128 lb individual.
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(b)
The proper dosage of a 128lb patient would be 355.6mg

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(c\) What is the interpretation of the value $D'(128)$.
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(c\)
$D'(128)$ is the interpretation that at 128 lbs the rate of the dosage is an amount dosage per pound, or mg/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)
to find $D'(128)$ I used the limit function $D'(x)=lim h-0 (D(x+h)-D(x)/h)$ And plugged it into our formula for $D(x)=.025x^2-.5x+10$ to get $D'(x)=.05x-.475$ Then I used x=128lbs to find $D'(128)=5.925 mg/lb$

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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 where x=130 would be $L(x)=D(130)+D'(130)(x-130)$

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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)

$L(x)=D(130)+D'(130)(128-130)$
$=367.5+6.025(-2)$
$=367.5-12.05$
$=355.45$ mg
Which is a good estimate, because if we look at part B we will see when x=128 on our function D(x) y=355.6 which is .15 away from our estimation on our tangent line.

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