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 | 1610 | 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)$P(t)= (1002.29) (1.09976^x)+(-2.26115)$

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(c\) What will the population be after 100 years under this model?
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(c\)$P(t)=(1002.29)(1.09976^{100})+(-2.26115)= 13,514,042.82077471$ people.

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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 | 115.4 | 127 | 139.5 | 153.3 |169 |
The interpertation of $P'(5)$ is the central difference from the values based off the calculations we got in part a. The units would be population of the settlement/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)$f"(3)=\frac{f'\left(4\right)-f'\left(2\right)}{4-2}$ = $12.05$
This number would be the increasing rate of 12.05 people 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)For my calculations, I got $k=0.0953719$. I got thay by using 115.4=k(1210). Solving for k I got =0.095371. ​

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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)$D(128)=0.025(128)^2+(-0.5(128))+10=355.6mg$

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(c\) What is the interpretation of the value $D'(128)$.
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(c\)The interpertation would give you the instanneous rate for a dosage for a person weighing 128lbs.

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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)$D'(128)=5.9mg/lb$ This was found by using the limit definiton of the derivitave.

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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)$L(x)=f(130)+f'(130)(x-130)$
$L(x)=367.5+6(x-130)$
This was found by using the equation above, $f(130)$ was found using the graph, and $f'(130)$ was given, plugging that into the equation is how I arrived at my answer.

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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)$L(128)=367.5+6(128-130)$=$367.5-12)=359.9$
Using the tangent line equation from above (since 128 is close to 130), we were able to solve and got 359.5. This is a good estimate since 128 is close to 130 and is a close amount to the one is Desmos which was 355.6.

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