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 | 1771.56 | 1948.71 |

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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) Using the given formula for exponential growth, I was able to create the formual:

$P\left(t\right)=\ 1000.06\left(1.1^x\right)-0.088024$

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(c\) What will the population be after 100 years under this model?
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(c\) $P\left(100\right)=\ 1000.06\left(1.1^{100}\right)-0.088024$

After 100 years, the population will grow $13781439.0885$ 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) P'(5) can be interpreted by saying that at the five year mark, the population is growing at a rate of about 154 people per year.

| $t$ | 1 | 2 | 3 | 4 | 5 | 6 |
|--- |---|---|---|---|---|---|
| $P'(t)$ | 210 | 115.5 | 127.05 | 139.775 | 153.73 | 169.1 |

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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) $p''\left(3\right)\ =\frac{\ p'\left(4\right)-p'\left(2\right)}{4-2}$

$p''\left(3\right)\ =\frac{\ 139775-115.5}{2}$

$p''\left(3\right)\ =121.375$

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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)

To find the constant k, we would need to divide both sides P(a) so that k equals:

$k=\frac{P'\left(t\right)}{p\left(t\right)}$

Then pluggin in values from our table for P'(t) and P(t), I found that the constant k equals:

$k=0.0954$

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

This quaderatic fuction is telling us that we can imput weight in pounds for x and get an ouput of dosage in in mg.

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(b) Find the proper dosage for a 128 lb individual.
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(b) For a person who weighs 120 punds, they will need a dosage of 355.6 mg.

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(c\) What is the interpretation of the value $D'(128)$.
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(c\) We can interpret D'(128) by saying that at the 128lb mark, the rate that the dosage is incresing is (------) 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)
I decided to use the limit definition of derivative to find D'(128).

$D'\left(x\right)=\frac{\left[0.025\left(x+h\right)^2-0.5\left(x+h\right)+10\right]-\left[0.025x^2-0.5x+10\right]}{h}$

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

$D'\left(x\right)=\frac{0.05xh+0.025h^2-0.5h}{h}$

$D'(x)=0.05x-0.5$

$D'(128)=5.9$

We can say that at the 128lb mark, the dosage needed is increasing by 5.9 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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$L(x)= f(x)+f'(x)(x-a)$

$=f(130)+f'(130)(x-a)$

$=367.5+6(x-130)$

$=6x-412.5$
(e)

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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)

Using the previous equation of a tangent line:
$6x-412.5$

We can find the tangent line for x=128.

$6(128)-412.5$

$=355.5$ mg/lb.

This is a good estimate for the dosage at x=128 because the values x=130 and x=128 are close enough that the tangent line doesn't change much. to check, I put in the value x=128 back into the function for d(x), and found that d(x)= 355.6. these values are off by 0.1. I don't know how much a difference this makes in the medical field, but I think that's close enough (I hope).

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