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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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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	| 1610.51 	|  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)
 $p(t)= 999.319\cdot1.10005^{x}+0.692754$


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(c\) What will the population be after 100 years under this model?
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(c\)
$p(t)= 999.319\cdot1.10005^{100}+0.692754$
$=999.319\cdot3.890871512\cdot10^{41}+0.692754$
$=13833965.9768$
After 100 years the population will be at 13833965.9768.



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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 |  139.755 | 153.7305  |  169.10355 |
The value of 

$p'(5)$ is just the slope or rate of change of of $p(t)$ at t=5. Which after 5 years the number of people increased at the rate 153 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)
$P''(3)= \frac{p\left(4\right)-p\left(2\right)}{4-2}$
$= \frac{139.755-115.5}{2}$
$=\frac{24.255}{2}$
$=12.1275 \frac{people}{year^{2}}$
After 3 years, the rate at which the population is increasing is increasing at a rate of 12.1275 $\frac{people}{year^{2}}$.


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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)
I think the constant would be .1 or 10% we use 10% to get our table answers. In question one part A the population increased by 10%. For part D we did the exact same for part A. 



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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)

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



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(b) Find the proper dosage for a 128 lb individual.
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(b)
$D(128)=0.025\left(128\right)^{2}+\left(-0.5\left(128\right)\right)+10$
$D(128)=0.025(16384)-64+10$
$D(128)=409.6-64+10$
$=355.6 mg$




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(c\) What is the interpretation of the value $D'(128)$. 
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(c\)
D'(128) means, the value of the rate of change of the dosage at the exact point of x=128 in 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)
$D'(128)=\frac{d\left(140\right)-d\left(140\right)}{140-120}$
$=\frac{430-310}{20}$
$=\frac{120}{20}$
$=6 \frac{mg}{lb}$


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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)
$L(x)=d(a)+d'(a)(x-a)$
$L(130)=d(130)+d'(30)(x-130)$
$=(0.025\left(130\right)^{2}+\left(-0.5\left(130\right)\right)+10)+6(x-130)$
$=422.5-65+10+6(x-130)$
$=367.5+6(x-130)$

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


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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)

$L(x)=367.5+6(x-130)$
$L(x)=367.5+6(128-130)$
$=367.5+6(-2)$
$=367.5-12$
$=355.5 mg$
The estimated value of dosage for a 128 lb person is 355.6 mg. 


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