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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
===

**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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:::info
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)
Table 2.1
| $t$    | 0    | 1    | 2   | 3   | 4   | 5   | 6   | 7  |
| ------ |----|----|----|---| --- | --- | --- | --- |
| $P(t)$ | 1000|1100|1210|1331|1464.1|1610.51| 1771.561|1948.7171 |





:::info
(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
\\]
:::

(b)
The Function after I approximated the function unto the points in Step a is:
$$f\left(x\right)=999.767\cdot1.10006^{x}+.182344$$



:::info
(c\) What will the population be after 100 years under this model?
:::


(c\)
To solve for the input $100$, we will substitute $x$ for $100$ so the equation looks like this:
$$f\left(x\right)=999.767\cdot1.10006^{100}+.182344$$
Once one calculates the equation, one can see that at $100$ years after the settlement, the population is approximately: $$13,852,754$$



:::info
(d) Use a central difference to estimate the values of $P'(t)$ in the table below. What is the interpretation of the value $P'(5)$?
:::
(d)
In order to solve for the table below, we will first consider the values given above in Step a.
Step 1:
$$\frac{f(b)-
f(a)}{b-a}$$ Will be are Formula.

Step 2.
Once We have our formula, we can use the input and output values from Table 2.1 to solve for the instantaneous rate of change for each $P'(t)$.

Step 3.
Example: Step 1:
$$\frac{f(2)-
f(0)}{2-0}$$
$$=\frac{1210-
1000}{2-0}$$
$$=\frac{210}{2}$$
$$105$$ is the instantaneous rate of change for each $P'(1)$.

| $t$     | 1 | 2 | 3 | 4 | 5 | 6 |
|---      |---|---|---|---|---|---|
| $P'(t)$ |$105$|$115$|$127$|$140$|$154$|169|

For $p'(5)$ the Derivative (the slope)at point $(5,p'(5))$ is 154 people per year after settlement.



:::info
(e) Use a central difference to estimate the values of $P''(3)$. What is the interpretation of this value?
:::
(e)
$p''(3)={139.755-115.5}/{4-2}$
$p''(3)={24.255}/{2}$
$p''(3)=12.1275{people}/{year^2}$
After the 3 year mark after settlement, the rate of change increases at approximatly 12 people per year.
:::info 
(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.)
:::

(f)
Variable $k$ is $10%$, If we use say $p(1)$ which is $1100$, $p'(1)$ is 110, the output being $10%$ of the function input.



:::success
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)
The Function which best describes the chart is: 
$$D\left(x\right)=\left(.025\right)x^{2}+\left(-.5\right)x+10$$




:::success
(b) Find the proper dosage for a 128 lb individual.
:::

(b)
The Proper Dosage for an individual who weighs 128lb should be:
$$D\left(x\right)=\left(.025\right)(128)^{2}+\left(-.5\right)(128)+10$$
Which Equals approximetly $$350mg$$




:::success
(c\) What is the interpretation of the value $D'(128)$. 
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(c\)
$D'(128)$ is the Instantaneous rate of change at exactly 128lbs. The rate at which the dosage increases at 128lbs is $5.9 mg/lb$

:::success
(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)
$D'(x)={P(b)-P(a)}/{b-a}$
$D'(130)={P(140)-P(120)}/{140-120}$
$D'(130)={430-310}/{20}$
$D'(130)={120}/{20}$
$D'(130)=6{mg}/{lb}$


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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. 
:::

(e)
$L(x)=d(a)+d'(a)(x-a)$
$L(x)=d(130)+d'(130)(x-130)$
$L(x)=[.025(130)^2-.5(130)+10]+6(x-130)$
$L(x)=367.5+6(x-130)$
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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(128)=367.5+6(128-130)$
$L(128)=355.5mg$

The Output value is approximetly $355.5mg$ for an individual weighing $128lbs$. This is about one hundreths more accurate than part (e) of question 2.

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