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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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:::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)**

| $t$    | 0    | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
|--------|------|---|---|---|---|---|---|---|
| $P(t)$ | **1000** | **1100** | **1210**  | **1331**  |  **1404.1**	|  **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
\\]
:::

**(B)**

P(t) = 1000 x $1.1^1$ + (-1.6399 x $10^{-12}$)


![](https://i.imgur.com/H94uIKz.png) ![](https://i.imgur.com/pj94J2x.png)

$f(x) = 1000 \times 1.1^1 + (-1.6399 \times 10^{-12})$ 






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

P(t) = 1000 x $1.1^{100}$ + (-1.6399 x $10^{-12}$) **= 13,780,612 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)**
**Central Difference = 153.7305 people/ additional year**

P'(5) = AV [4, 6] = $\frac{f(6)-f(4)}{6-4}$ = $\frac{1771.561- 1464.1}{2}$ = $\frac{307.461}{2}$

P'(5) **= 153.7305**

| $t$     | 1 | 2 | 3 | 4 | 5 | 6 |
|---      |---|---|---|---|---|---|
| $P'(t)$ |**55**   | **115.5**  | **127.05**  | **139.755**  |  **153.7305** |  **169.10355** |




:::info
(e) Use a central difference to estimate the values of $P''(3)$. What is the interpretation of this value?
:::


**(E)**

P(t) values = people / $year^2$ 

P'(1) = AV[0, 2] = $\frac{f(2) - f(0)}{2 - 0}$ = $\frac{1210 - 1100}{2}$ = $\frac{110}{2}$ 
**= 55**

P'(2) = AV[3, 1 ] = $\frac{f(3) - f(1)}{3-1}$ = $\frac{1331 - 1100}{2}$ = $\frac{231}{2}$ 
**= 115.5**

P'(3) = AV[2, 4] = $\frac{f(4) - f(2)}{4 - 2}$ = $\frac{1464.1 - 1210}{2}$ = $\frac{254.1}{2}$
**= 127.05**

P'(4) = AV[3, 5] = $\frac{f(5) - f(3)}{5 - 3}$ = $\frac{1610.51 - 1331}{2}$ = $\frac{279.51}{2}$  
**= 139.755**

P'(6) = AV[5, 7] = $\frac{f(7) - f(5)}{7 - 5}$ = $\frac{1948.7171 - 1610.51}{2}$ = $\frac{338.2071}{2}$ 
**= 169.10355**


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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.)
:::

**(F)**

$P^"$(3) = AV [2, 4] = $\frac{f'(4)-f'(2)}{4-2}$ = $\frac{139.755 - 115.5}{2}$ = $\frac{24.255}{2}$


$P^"$(3) = 12.1275 people / $year^2$

*The value of $P^"$(3) delineates a yearly increase in population by 12.1275 people.*


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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)**

$D(x) = 0.025x^2_1 - 0.5x_1 + 10$ 

![](https://i.imgur.com/EL5eRxK.png) ![](https://i.imgur.com/TXeonSR.png)




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

![](https://i.imgur.com/0CzbLnb.png)


**The proper dosage for an individual weighing 128 lbs would be 355.6 mg.**






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(c\) What is the interpretation of the value $D'(128)$. 
:::


**(c\)**

![](https://i.imgur.com/S72WNX4.png)



:::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.
:::


**(d)**

D'(128) AV[127, 129] = $\frac{f(129) - f(127)}{129 - 127}$ = $\frac{361.525 -349.725}{2}$ = $\frac{11.8}{2}$ **= 5.9 mg/lb**

![](https://i.imgur.com/uiYX5AI.png)



Via Desmos, I calculated proper dosage proportions for individuals who  weighed 126 and 129 lbs. Incorporating that into the central difference formula, I was able to determine that dosage amounts will increase by 5.9 mg/lb upon exceeding the weight of 128 lbs. 



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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)**

![](https://i.imgur.com/iUif8hu.png)


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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)**

On a tangent line in which x = 128 lbs, the output value (y point) will be 355.6 mg (the proper dosage amount), in correspondence with the previous estimation. 

![](https://i.imgur.com/YTnsjQw.png)




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To submit this assignment click on the Publish button ![Publish button icon](https://i.imgur.com/Qk7vi9V.png). Then copy the url of the final document and submit it in Canvas.

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