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 | 1,100 |1,210 | 1,331 | 1,464 |1,610 | 1,771 | 1,948 |

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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)
The formula for $P(t)= 1002.29\times 1.09976^x -2.26115$

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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 grow to 14 million people following this model:
$P(t)=1002.29\times 1.09976^x -2.26115= 14million $\frac{people}{year}$.

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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 |
|--- |---|---|---|---|---|---|
| $P'(t)$ |95.3 | 115.2 | 126.7 | 139.4 | 153.3 | 168.6 |

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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)
The central difference to estimate the values of $P"(3)$ is going $12.1t$
to be with values $P'(2) and P'(4)$. Which looks like $\frac{s\left(4\right)-s\left(2\right)}{4-2}=\frac{24.2}{2}=12.1\frac{people}{\frac{year}{year}}$. This means that 12.1 is the rate of which the population increases per 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)
$P'(t)+k\cdot P(t)$
$P'(1)=k\cdot P(1)$
$95.5= k\cdot P(1100)$
$\frac{95.5}{1100}$$=k$
$k= 0.085$

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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)

The propar dosage is 355.6mg for a 128lb individual.
$f\left(128\right)=0.025\left(128\right)^2-0.5\left(128\right)+10$
$f(128)=355.6$mg for 128 lb individual.

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(c\) What is the interpretation of the value $D'(128)$.
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(c\)
The interpretation of the value D'(128) is $5.9 mg/lbs$, the weight per unit of dosage.

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

$\lim_{h \to 0}\frac{f(128+h)-f(128)}{h}$ = $\lim_{h \to 0}\frac{f(0.025(128+h)^2-0.5(128+h)+10)-f(0.025(128)^2-0.5(128)−10)}{h}$
$5.75\frac{mg}{lb}$

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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) The equation for the tangent line to the curve $y=D(x)$ is $6x-y=412.5$
$y\left(130\right)=D\left(130\right)$
$=0.025\left(130\right)^2-0.5\left(130\right)+10$
$\left(y_{130}\right)=m\left(x-x_0\right)$
$y-367.5=\ 6\left(x-130\right)$
$y-367.5=6x-780$
$=6x-y=412.5$
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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)A good estimate for a person who weighs 128 lb is about 355.5mg for dosage.

$6x-y=412.5\ at\ x=128\ lb$
$y=6x-412.5$
$y=\ 6\left(128\right)-412.5$
$y\approx 355.5mg$ for a person who weighs $128 lb$.

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